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**Figure 1.** **A**—As the magnetic field rotates inside the generator housing, a three-phase electrical output develops**. B**—The rotating field inside the stationary armature ac generator consists of coils formed and connected to slip rings. The fields are powerful electromagnets. **C**—Each phase has the same voltage peak and frequency. The peak positive polarity occurs 120 electrical degrees apart.

Look again at **Figure 1, part A**. The sine wave that develops from three-phase electrical systems is unique. The three windings of the armature are connected as pairs. They are called phase A, phase B, and phase C. As the magnetic field rotates inside the generator armature windings, three different sine waves develop. The three waves are 120 electrical degrees apart. See **Figure 1**, **part C**.

It is important to note that each of the three phases develops the same frequency and same peak voltage. The difference between the three phases is that each phase develops positive and negative voltage peaks at different times, making each phase unique.

The coils of the three-phase generator can be connected in series or parallel. When connected in series it is called a delta connection. When connected in parallel, it is called a wye or star connection. The reasons for their names can be seen in **Figure 2**.

**Figure 2.** Generator windings are connected as either delta or wye.

You may have heard the terms 120/240 volt and 120/208 volt and wondered which is correct. Both terms are correct. The voltage present is determined by the type of generator connection used. A wye connection provides 120/208 volts, and delta connection provides 240 volts with no common, or neutral, connection.

Delta is used mainly for heavy industrial loads, while the wye connection is used to support both power and lighting circuits. While some equipment will operate on either voltage, it is best to connect a piece of machinery to the voltage level specified in its operator’s manual.

When generators are connected in parallel, **three electrical characteristics** of each generator must match. These characteristics are the generator’s **frequency, voltage, and instantaneous polarity**. If all three items do not match, and the switch is closed to connect the two generators, an electrical fire and explosion are likely to occur.

If the voltages of two parallel generators do not match, they can be made to match by adjusting one of the field pole rheostats. A slightly trickier part in paralleling generators is matching the correct frequencies of the two generators. The frequency or hertz of each generator can be compared using a **frequency meter**. If the frequencies do not match, we need to speed up or slow down one of the generators until they match.

**Figure 3** shows sine waves of matching and non-matching instantaneous polarity. Each generator is producing a 120-volt output at 60 Hz. However, in the waves shown on the top, the third requirement, instantaneous polarity, does not match. The two sine waves in the figure are slightly off. Notice how their peaks do not align on the time axis.

The instantaneous polarity of the ac generators refers to that part of the sine wave that is rep- resented at the output terminals of each generator at that instant of time. These outputs could both be positive, or one positive and one negative. Ideally, if both sine wave patterns match, the generators have identical instantaneous polarity. These generators can be connected together electrically.

**Figure 3**. The two generators outputs shown on the top have the same voltage peaks and the same frequency, but their instantaneous polarities do not match. Notice how their peaks come at different times. The two outputs on the bottom match.

A simple way to tell if two three-phase generator outputs match is called the “two dark–one light method.”

Two lights are connected across the phase A and B out- puts of both generators in such a way that they match in polarity, voltage, and frequency. Both lamps should not light since there would be no difference in potential across the lamps. Another lamp is connected across two out of phase phases of the generator. This lamp should light, indicating that the two phases are out of phase with each other. See **Figure 4**.

**Figure 4**. The two dark–one light method of paralleling generators. The two lamps connected across in phase lines from the two generators remain dark. The third lamp is connected across two out of phase lines. This lamp will glow. These two generators can be connected in parallel safely.

Generators generally fail due to one of three things:

- Excessive brush wear.
- Excessive bearing wear.
- Electrical overload.

**Excessive brush wear**

As the brushes wear down, more arcing is produced between the brushes and commutator segments or slip rings. Excessive wear will damage the surface of either the commutator segments or the slip rings. A bluish burn pattern will be evident.

Excessive brush wear can be determined by physical inspection. The measurement for minimum brush length for a generator can be found in the owner’s manual. The brushes should be replaced if excessive wear is found. Brushes should be dressed when being replaced.

Dressing brushes refers to shaping the end of the brush to match the surface of the commutator or slip ring. A fine grade of sandpaper or emery cloth can be used for this purpose.

The sandpaper is placed between the brush and commutator with the rough side facing the brush. Rotating the commutator causes the end of the brush to wear down to match the shape of the surface it is riding against. See Figure 5.

**Figure 5.** The brushes of a dc motor or generator need to be shaped to match the contour of the commutator surface.

**Excessive bearing wear**

Worn bearings will cause a roaring sound. Excessive vibration can also be present. Another sign of excessive bearing wear is a shiny spot appearing on the armature. This spot is caused by the armature rubbing against a pole piece as the bearing wears out.

Bearings should be lubricated on a regular schedule as indicated in the owner’s manual. Some bearings are permanently sealed and require no lubrication. If brushes and bearings are properly maintained, an electrical generator can easily provide more than 20 years of service.

**Electrical overload**

Another cause of failure is electrical overload. Electrical overload causes the electrical insulation to break down, resulting in a short circuit or ground condition. Most times the damage can be detected by the naked eye.

A dark discoloration is a sign of excessive overload on a conductor. Test instruments are needed to determine if the armature or field windings are good or bad. Usually a voltage equal to the generator voltage is used to check the windings for grounds. Most multimeters only use a 9-volt battery and cannot detect a ground created when 120 or 240 volts are present during generator operation.

Heat from an excessive overload can also cause the generator pole pieces to lose their residual magnetism. This is very damaging to the **self-excited generator**. If the residual magnetism is lost, the generator will fail to build up sufficient voltage. Residual magnetism can be restored by connecting a separate dc power supply, such as a battery, to the field windings.

**LESSON IN SAFETY**

Before connecting the separate dc power supply to the field windings, the armature leads must be disconnected to prevent the generator from rotating. When voltage is applied to a generator, it will revolve as a motor.

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]]>Further, while the power produced by a single-phase system has a pulsating nature a three-phase system can deliver a steady, constant supply of power. For example, later in this article it will be shown that a three-phase generator producing three **balanced voltages**—that is, voltages of equal amplitude and frequency displaced in phase by 120°—has the property of delivering constant instantaneous power.

The change to three-phase AC power systems from the early DC system proposed by Edison was due to a number of reasons: the efficiency resulting from transforming voltages up and down to minimize transmission losses over long distances, the ability to deliver constant power, a more efficient use of conductors, and the ability to provide starting torque for industrial motors.

Consider a three-phase source connected in a **wye (star) configuration,** as shown in **Figure 1**. Each of the three voltages is 120° out of phase with the others, such that:

**Figure 1 **Balanced three-phase AC circuit

\[\begin{matrix}\begin{align}& \overset{\text{ }\!\!\tilde{\ }\!\!\text{ }}{\mathop{{{\text{V}}_{\text{an}}}}}\,=\overset{\tilde{\ }}{\mathop{{{V}_{an}}}}\,\angle {{0}^{^{o}}} \\& \overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{bn}}}}}\,=\overset{\tilde{\ }}{\mathop{{{V}_{bn}}}}\,\angle -({{120}^{^{o}}}) \\& \overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{cn}}}}}\,=\overset{\tilde{\ }}{\mathop{{{V}_{cn}}}}\,\angle (-{{240}^{^{o}}})=\overset{\sim }{\mathop{{{V}_{cn}}}}\,\angle {{120}^{o}} \\\end{align} & {} & (1) \\\end{matrix}\]

If the three-phase source is balanced, then:

\[\overset{\text{ }\!\!\tilde{\ }\!\!\text{ }}{\mathop{{{\text{V}}_{\text{an}}}}}\,\text{+}\overset{\text{ }\!\!\tilde{\ }\!\!\text{ }}{\mathop{{{\text{V}}_{\text{bn}}}}}\,\text{+}\overset{\text{ }\!\!\tilde{\ }\!\!\text{ }}{\mathop{{{\text{V}}_{\text{cn}}}}}\,\text{=0}\begin{matrix}{} & Balanced\text{ }phase\text{ }voltages\begin{matrix}{} & {} & (2) \\\end{matrix} \\\end{matrix}\]

The result is the so-called **positive abc sequence,** as shown in **Figure 2.** In the wye (star) configuration, the three phase voltages share a common *neutral node,* denoted by *n.*

**Figure 2 **Positive, or abc, sequence for balanced three-phase voltages

It is also possible to define **line voltages** as the potential differences between lines aa′ and bb′, lines aa′ and cc′, and lines bb′ and cc’. Each **line voltage** is related to the phase voltages by:

\[\begin{matrix}\overset{\text{ }\!\!\tilde{\ }\!\!\text{ }}{\mathop{{{\text{V}}_{\text{ab}}}}}\,\text{=}\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{an}}}}}\,\text{-}\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{bn}}}}}\,\text{=}\sqrt{3}\overset{\tilde{\ }}{\mathop{V}}\,\angle {{30}^{o}} & {} & {} \\\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{bc}}}}}\,\text{=}\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{bn}}}}}\,\text{-}\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{cn}}}}}\,=\sqrt{3}\overset{\tilde{\ }}{\mathop{V}}\,\angle (-{{90}^{o}}) & {} & (4) \\\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{ca}}}}}\,\text{=}\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{cn}}}}}\,\text{-}\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{an}}}}}\,=\sqrt{3}\overset{\tilde{\ }}{\mathop{V}}\,\angle {{150}^{o}} & {} & {} \\\end{matrix}\]

It is instructive to note that the circuit of **Figure 1** can be redrawn as shown in **Figure 3**, where it is clear that the three branches are in parallel.

**Figure 3 **Balanced three-phase AC circuit (redrawn)

When **Z**_{a} = **Z**_{b} = **Z**_{c} = **Z**, the wye (star) load configuration is also balanced. When both the source and load networks are balanced, KCL requires that the current **Ĩ**_{n} in the neutral line n − n′ be identically zero.

\[\overset{\tilde{\ }}{\mathop{{{\text{I}}_{\text{n}}}}}\,\text{=}\overset{\tilde{\ }}{\mathop{{{\text{I}}_{\text{a}}}}}\,\text{+}\overset{\tilde{\ }}{\mathop{{{\text{I}}_{\text{b}}}}}\,\text{+}\overset{\tilde{\ }}{\mathop{{{\text{I}}_{\text{c}}}}}\,\text{=}\frac{\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{an}}}}}\,\text{+}\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{bn}}}}}\,\text{+}\overset{\tilde{\ }}{\mathop{{{\text{V}}_{\text{cn}}}}}\,}{\text{ }\!\!Z\!\!\text{ }}\text{=0}\begin{matrix}{} & \text{(5)} \\\end{matrix}\]

Another important characteristic of a balanced three-phase power system is illustrated by a simplified version of **Figure 3**, where the balanced load impedances are replaced by three equal resistors R. Since 𝜃_{R} = 0, the instantaneous power p(t) delivered to each resistor is given by equation [with 𝜃_{V} = 𝜃_{I} and with the freely chosen reference (𝜃_{V})_{a} = 0)] to be:

\[\begin{matrix}{{p}_{a}}(t)=\frac{\overset{\tilde{\ }}{\mathop{{{V}^{2}}}}\,}{R}(1+\cos 2\omega t) & {} & {} \\{{p}_{b}}(t)=\frac{\overset{\tilde{\ }}{\mathop{{{V}^{2}}}}\,}{R}[(1+\cos (2\omega t-{{120}^{o}})] & {} & (6) \\{{p}_{c}}(t)=\frac{\overset{\tilde{\ }}{\mathop{{{V}^{2}}}}\,}{R}[(1+\cos (2\omega t-{{120}^{o}})] & {} & {} \\\end{matrix}\]

The total instantaneous power p(t) delivered to the total load is the sum:

\[\begin{matrix}p(t)={{p}_{a}}(t)+{{p}_{b}}(t)+{{p}_{c}}(t) & {} & {} \\=\frac{\overset{\tilde{\ }}{\mathop{{{V}^{2}}}}\,}{R}[(3+\cos 2\omega t+\cos (2\omega t-{{120}^{o}})+\cos (2\omega t+{{120}^{o}})] & {} & (7) \\=\frac{\overset{\tilde{\ }}{\mathop{3{{V}^{2}}}}\,}{R}=\text{constant!} & {} & {} \\\end{matrix}\]

It is worthwhile to verify that the sum of the three cosine terms is identically zero. (Hint: Consider the phasor sum of e^{j(2ωt)}, e^{j(2ωt–π/3)} and e^{j(2ωt+π/3)}.)

Thus, with the simplified balanced resistive load, the total power delivered to the load by the balanced three-phase source is constant. This is an extremely important result, for a very practical reason: Delivering power in a steady fashion (as opposed to the pulsating nature of single-phase power) reduces “wear and tear” on the source and load.

It is also possible to connect three AC sources in a **delta** (**Δ**) **configuration,** as shown in **Figure 4** although it is rarely used in practice.

**Figure 4 **Delta configuration

These results for purely resistive loads can be generalized for any arbitrary balanced complex load. Consider again in **Figure 5**, where now the balanced load consists of three complex impedances:

\[{{Z }_{a}}={{Z }_{b}}={{Z }_{c}}={{Z }_{y}}=\left| {{Z }_{y}} \right|\angle \theta \begin{matrix}{} & {} & (8) \\\end{matrix}\]

Because of the common neutral line n − n′, each impedance sees the corresponding phase voltage across itself. Therefore, since $\overset{\tilde{\ }}{\mathop{{{V}_{an}}}}\,=\overset{\tilde{\ }}{\mathop{{{V}_{bn}}}}\,=\overset{\tilde{\ }}{\mathop{{{V}_{cn}}}}\,$, it is also true that **Ĩ**_{a} = **Ĩ**_{b} = **Ĩ**_{c} and the phase angles of the currents will differ by ±120°. Consequently, it is possible to compute the power for each phase from the phase voltage and the associated line current. Denote the complex power for each phase by **S**, where:

\[\begin{matrix}\begin{align}& S=P+jQ \\& \overset{\tilde{\ }}{\mathop{I}}\,\cos \theta +j\overset{\tilde{\ }}{\mathop{V}}\,\overset{\tilde{\ }}{\mathop{I}}\,\sin \theta \\\end{align} & {} & (9) \\\end{matrix}\]

The total real power delivered to the balanced wye (star) load is 3P, and the total reactive power is 3Q. The total complex power **S**_{T} is

\[\begin{matrix}\begin{align}& {{S}_{T}}={{P}_{T}}+j{{Q}_{T}}=3P+j3Q \\& =\sqrt{{{(3P)}^{2}}+{{(3Q)}^{2}}}\angle \theta \\\end{align} & {} & (10) \\\end{matrix}\]

The apparent power |**S**_{T}**|** is:

\[\begin{matrix}\begin{align}& \begin{matrix}\left| {{S}_{T}} \right| & =3\sqrt{{{(\overset{\tilde{\ }}{\mathop{V}}\,\overset{\tilde{\ }}{\mathop{I}}\,)}^{2}}{{\cos }^{2}}\theta +{{(\overset{\tilde{\ }}{\mathop{V}}\,\overset{\tilde{\ }}{\mathop{I}}\,)}^{2}}{{\sin }^{2}}\theta } \\\end{matrix} \\& =3\overset{\tilde{\ }}{\mathop{V}}\,\overset{\tilde{\ }}{\mathop{I}}\, \\\end{align} & {} & (11) \\\end{matrix}\]

Such that:

\[\begin{matrix}\begin{align}& {{P}_{T}}=\left| {{S}_{T}} \right|\cos \theta \\& {{Q}_{T}}=\left| {{S}_{T}} \right|\sin \theta \\\end{align} & {} & (12) \\\end{matrix}\]

It is also possible to assemble a balanced load in a delta configuration. A wye (star) generator and a delta load are shown in **Figure 5**.

**Figure 5 **Balanced wye (star) generators with balanced delta load

Note immediately that each impedance **Z**_{Δ} sees a corresponding line voltage, rather than a phase voltage. For example, the voltage across **Z**_{c′a′} is $\overset{\tilde{\ }}{\mathop{{{V}_{ca}}}}\,$. Thus, the three load currents are:

\[\begin{align}& {{\overset{\sim }{\mathop{I }}\,}_{ab}}=\frac{{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{ab}}}}{{{\text{ }\!\!Z\!\!\text{ }}_{\text{ }\!\!\Delta\!\!\text{ }}}}=\frac{\sqrt{3}\overset{\tilde{\ }}{\mathop{V}}\,\angle (\pi /6)}{\left| {{Z }_{\Delta }} \right|\angle \theta } \\& {{\overset{\sim }{\mathop{I }}\,}_{bc}}=\frac{{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{bc}}}}{{{\text{ }\!\!Z\!\!\text{ }}_{\text{ }\!\!\Delta\!\!\text{ }}}}=\frac{\sqrt{3}\overset{\tilde{\ }}{\mathop{V}}\,\angle (-\pi /2)}{\left| {{Z }_{\Delta }} \right|\angle \theta }\begin{matrix}{} & {} & (13) \\\end{matrix} \\& {{\overset{\sim }{\mathop{I }}\,}_{ca}}=\frac{{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{ca}}}}{{{\text{ }\!\!Z\!\!\text{ }}_{\text{ }\!\!\Delta\!\!\text{ }}}}=\frac{\sqrt{3}\overset{\tilde{\ }}{\mathop{V}}\,\angle (5\pi /6)}{\left| {{Z }_{\Delta }} \right|\angle \theta } \\\end{align}\]

The relationship between a delta load and a wye (star) load can be illustrated by determining the delta load **Z**_{Δ} that would draw the same amount of current as a wye (star) load **Z*** _{y}* assuming a given source voltage. Consider the circuits shown in

\[{{({{\overset{\tilde{\ }}{\mathop{I }}\,}_{a}})}_{y}}=\frac{{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{an}}}}{\text{ }\!\!Z\!\!\text{ }}=\frac{\overset{\tilde{\ }}{\mathop{V}}\,}{\left| {{Z }_{y}} \right|}\angle (-\theta )\begin{matrix}{} & {} & (14) \\\end{matrix}\]

The current drawn by a delta load is:

\[\begin{matrix}\begin{align}& {{\text{(}{{\overset{\text{ }\!\!\tilde{\ }\!\!\text{ }}{\mathop{\text{ }\!\!I\!\!\text{ }}}\,}_{\text{a}}}\text{)}}_{\text{ }\!\!\Delta\!\!\text{ }}}\text{=}{{\overset{\text{ }\!\!\tilde{\ }\!\!\text{ }}{\mathop{\text{ }\!\!I\!\!\text{ }}}\,}_{\text{ab}}}\text{-}{{\overset{\text{ }\!\!\tilde{\ }\!\!\text{ }}{\mathop{\text{ }\!\!I\!\!\text{ }}}\,}_{\text{ca}}} \\& \text{=}\frac{{{\overset{\text{ }\!\!\tilde{\ }\!\!\text{ }}{\mathop{\text{V}}}\,}_{\text{ab}}}}{{{\text{ }\!\!Z\!\!\text{ }}_{\text{ }\!\!\Delta\!\!\text{ }}}}\text{-}\frac{{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{ca}}}}{{{\text{ }\!\!Z\!\!\text{ }}_{\text{ }\!\!\Delta\!\!\text{ }}}} \\& =\frac{\text{1}}{{{\text{ }\!\!Z\!\!\text{ }}_{\text{ }\!\!\Delta\!\!\text{ }}}}\text{(}{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{an}}}\text{-}{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{bn}}}\text{-}{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{cn}}}\text{+}{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{an}}}\text{)} \\& \text{=}\frac{\text{1}}{{{\text{ }\!\!Z\!\!\text{ }}_{\text{ }\!\!\Delta\!\!\text{ }}}}\text{(2}{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{an}}}\text{-}{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{bn}}}\text{-}{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{cn}}}\text{)} \\& =\frac{{{\overset{\tilde{\ }}{\mathop{\text{3V}}}\,}_{\text{an}}}}{{{\text{ }\!\!Z\!\!\text{ }}_{\text{ }\!\!\Delta\!\!\text{ }}}}\text{=}\frac{\text{3}{{\overset{\tilde{\ }}{\mathop{\text{V}}}\,}_{\text{an}}}}{\left| {{\text{ }\!\!Z\!\!\text{ }}_{\text{ }\!\!\Delta\!\!\text{ }}} \right|}\angle \text{(- }\!\!\theta\!\!\text{ )} \\\end{align} & {} & \text{(15)} \\\end{matrix}\]

The two currents (**Ĩ**_{a})_{Δ} and (**Ĩ**_{a})_{y} are equal if:

\[{{Z }_{\Delta }}=3{{Z }_{y}}\begin{matrix}{} & {} & (16) \\\end{matrix}\]

This result also implies that a delta load will draw three times as much current and absorb three times as much power as a wye (star) load with the same branch impedance.

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**Figure 1** Time and frequency domain representations of an AC circuit. The phase angle of the load is θ_{Z} = θ_{V} − θ_{I}.

The most general expressions for the voltage and current delivered to an arbitrary load are as follows:

\[\begin{matrix}\begin{align}& v(t)=V\cos (\omega t+{{\theta }_{V}}) \\& i(t)=I\cos (\omega t+{{\theta }_{I}}) \\\end{align} & {} & (1) \\\end{matrix}\]

Where *V* and *I* are the peak amplitudes of the sinusoidal voltage and current, respectively, and θ* _{V}* and θ

**Figure 2** Current and voltage waveforms with unit amplitude and a phase shift of 60°.

Notice that the current leads the voltage; or equivalently, the voltage lags the current. Keep in mind that all phase angles are relative to some reference, which is usually chosen to be the phase angle of a source. The reference phase angle is freely chosen and therefore usually set to zero for simplicity. Also keep in mind that a phase angle represents a time delay of one sinusoid relative to its reference sinusoid.

The instantaneous power dissipated by any element is the product of its instantaneous voltage and current.

\[\begin{matrix}p(t)=v(t)i(t)=VI\cos (\omega t+{{\theta }_{V}})\cos (\omega t+{{\theta }_{I}}) & {} & (2) \\\end{matrix}\]

This expression is further simplified with the aid of the trigonometric identity:

\[\begin{matrix}2\cos (x)cos(y)=\cos (x+y)+\cos (x-y) & {} & (3) \\\end{matrix}\]

Let *x=ωt + *θ* _{V}* and

\[\begin{matrix}\begin{align}& p(t)=\frac{VI}{2}[\cos (2\omega t+{{\theta }_{V}}+{{\theta }_{I}})+\cos ({{\theta }_{V}}-{{\theta }_{I}})] \\& =\frac{VI}{2}[\cos (2\omega t+{{\theta }_{V}}+{{\theta }_{I}})+\cos ({{\theta }_{Z}})] \\\end{align} & {} & \left( 4 \right) \\\end{matrix}\]

**Equation 4** illustrates that the total instantaneous power dissipated by an element is equal to the sum of a constant $\frac{1}{2}$VI cos (θ_{Z}) and a sinusoidal $\frac{1}{2}$ VI cos (2ωt + θ_{V} + θ_{I}), which oscillates at twice the frequency of the source. Since the time average of a sinusoid is zero over one period or over a sufficiently long interval, the constant $\frac{1}{2}$VI cos (θ_{Z}) is the time averaged power dissipated by a complex load Z, where θ_{Z} is the phase angle of that load.

**Figure 3** shows the instantaneous and average power corresponding to the voltage and current signals of **Figure 2**.

**Figure 3** Instantaneous and average power corresponding to the signals

These observations can be confirmed mathematically by noting that the time average of the instantaneous power is defined by:

\[\begin{matrix}{{P}_{avg}}\equiv \frac{1}{T}\int\limits_{{{t}_{0}}}^{{{t}_{0}}+T}{P(t)dt} & {} & (5) \\\end{matrix}\]

Where *T* is one period of *p*(*t*). Use **equation 4** to substitute for *p*(*t*) and yield:

\[\begin{matrix}\begin{align}& {{P}_{avg}}=\frac{1}{T}\int\limits_{{{t}_{0}}}^{{{t}_{0}}+T}{\frac{VI}{2}\left[ \cos (2\omega t+{{\theta }_{v}}+{{\theta }_{t}})+\cos ({{\theta }_{Z}}) \right]dt} \\& =\frac{VI}{2T}\int\limits_{{{t}_{0}}}^{{{t}_{0}}+T}{\left[ \cos (2\omega t+{{\theta }_{v}}+{{\theta }_{t}})+\cos ({{\theta }_{Z}}) \right]dt} \\\end{align} & {} & (6) \\\end{matrix}\]

The integral of the first part cos (2ωt + θ_{V} + θ_{I}) is zero while the integral of the second part (a constant) is T cos (θ_{Z}). Thus, the time averaged power P_{avg} is:

\[{{P}_{avg}}=\frac{VI}{2}\cos ({{\theta }_{Z}})=\frac{1}{2}\frac{{{V}^{2}}}{\left| Z \right|}\cos ({{\theta }_{Z}})=\frac{1}{2}{{I}^{2}}\left| Z \right|\cos ({{\theta }_{Z}})\begin{matrix}{} & (7) \\\end{matrix}\]

where

\[\left| Z \right|=\frac{\left| V \right|}{\left| I \right|}=\frac{V}{I}\begin{matrix}{} & and\begin{matrix}{} & {{\theta }_{Z}}={{\theta }_{V}}-{{\theta }_{I}}\begin{matrix}{} & (8) \\\end{matrix} \\\end{matrix} \\\end{matrix}\]

**Effective or Root Mean Square Value**

In North America, AC power systems operate at a fixed frequency of 60 cycles per second, or hertz (Hz), which corresponds to an angular (radian) frequency ω given by:

\[\begin{matrix}\omega =2\pi .60=377rad/s & AC\text{ }power\text{ }frequency & (9) \\\end{matrix}\]

It is customary in AC power analysis to employ the *effective* or *root*–*mean*–*square* (rms) amplitude rather than the peak amplitude for AC voltages and currents. In the case of a sinusoidal waveform, the effective voltage $\overset{\tilde{\ }}{\mathop{V}}\,\equiv {{V}_{rms}}$ is related to the peak voltage *V* by:

\[\begin{matrix}\overset{\sim }{\mathop{V}}\,={{V}_{rms}}=\frac{V}{\sqrt{2}} & {} & (10) \\\end{matrix}\]

Likewise, the effective current $\overset{\tilde{\ }}{\mathop{I}}\,\equiv {{I}_{rms}}$ is related to the peak current *I* by:

\[\begin{matrix}\overset{\sim }{\mathop{I}}\,={{I}_{rms}}=\frac{I}{\sqrt{2}} & {} & (11) \\\end{matrix}\]

The rms, or effective, value of an AC source is the DC value that produces the same average power to be dissipated by a common resistor.

The average power can be expressed in terms of effective voltage and current by plugging

$\begin{matrix} V=\sqrt{2}\overset{\tilde{\ }}{\mathop{V}}\, & and & I=\sqrt{2}\overset{\tilde{\ }}{\mathop{I}}\, \\\end{matrix}$

into **equation 7** to find:

\[\begin{matrix}{{P}_{avg}}=\overset{\sim }{\mathop{V}}\,\overset{\sim }{\mathop{I}}\,\cos ({{\theta }_{Z}})=\frac{\overset{\tilde{\ }}{\mathop{{{V}^{2}}}}\,}{\left| Z \right|}\cos ({{\theta }_{Z}})=\overset{\sim }{\mathop{{{I}^{2}}}}\,\left| Z \right|\cos ({{\theta }_{Z}}) & {} & (12) \\\end{matrix}\]

Voltage and current phasors are also represented with effective amplitudes by the notation:

\[\overset{\tilde{\ }}{\mathop{\text{V}}}\,=\overset{\tilde{\ }}{\mathop{V}}\,{{e}^{j{{\theta }_{V}}}}=\overset{\tilde{\ }}{\mathop{V}}\,\angle {{\theta }_{V}}\begin{matrix}{} & (13) \\\end{matrix}\]

AND

\[\overset{\tilde{\ }}{\mathop{\text{I}}}\,=\overset{\tilde{\ }}{\mathop{I}}\,{{e}^{j{{\theta }_{V}}}}=\overset{\tilde{\ }}{\mathop{I}}\,\angle {{\theta }_{I}}\begin{matrix}{} & (14) \\\end{matrix}\]

It is critical to pay close attention to the mathematical notation, namely that complex quantities, such as **V**, **I**, and **Z** are boldface. On the other hand, scalar quantities, such as *V, I*,$\overset{\tilde{\ }}{\mathop{V}}\,$ and $\overset{\tilde{\ }}{\mathop{I}}\,$are italic.

**Impedance Triangle**

**Figure 4** illustrates the concept of the impedance triangle, which is an important graphical representation of impedance as a vector in the complex plane.

**Figure 4** Impedance triangle

Basic trigonometry yields:

\[R=\left| \text{ }\!\!Z\!\!\text{ } \right|\cos \theta \begin{matrix}{} & (15) \\\end{matrix}\]

\[X=\left| \text{ }\!\!Z\!\!\text{ } \right|\sin \theta \begin{matrix}{} & (16) \\\end{matrix}\]

where *R* is the *resistance* and *X* is the *reactance.* Notice that both *R* and *P*_{avg} are proportional to cos (*𝜃 _{Z}*), which suggests that a triangle similar to (i.e., the same shape as) the impedance triangle could be constructed with

**Power Factor**

The phase angle *𝜃 _{Z}* of the load impedance plays a very important role in AC power circuits. From

\[\begin{matrix}{{\theta }_{Z}}=0 & \to & pf=1 & \begin{matrix}\text{Resistive Load} & (17) \\\end{matrix} \\\end{matrix}\]

For purely inductive or capacitive loads:

\[\begin{matrix}{{\theta }_{Z}}=+\pi /2 & \to & pf=0 & \begin{matrix}Inductive\text{ }Load & (18) \\\end{matrix} \\\end{matrix}\]\[\begin{matrix}{{\theta }_{Z}}=-\pi /2 & \to & pf=0 & \begin{matrix}Capacitive\text{ }Load & (19) \\\end{matrix} \\\end{matrix}\]

For loads with non-zero resistive (real) and reactive (imaginary) parts:

\[\begin{matrix}0<\left| {{\theta }_{Z}} \right|<\pi /2 & \to & \begin{matrix}0<pf<1 & \begin{matrix}Complex\text{ }Load & (20) \\\end{matrix} \\\end{matrix} \\\end{matrix}\]

Using the definition pf = cos(*𝜃 _{Z}*) the average power can be expressed as:

\[{{P}_{avg}}=\overset{\tilde{\ }}{\mathop{V}}\,\overset{\tilde{\ }}{\mathop{I}}\,pf\begin{matrix}{} & {} & (21) \\\end{matrix}\]

Thus, average power dissipated by a resistor is:

\[{{({{P}_{avg}})}_{R}}=\overset{\tilde{\ }}{\mathop{{{V}_{R}}}}\,\overset{\tilde{\ }}{\mathop{{{I}_{R}}}}\,p{{f}_{R}}=\overset{\tilde{\ }}{\mathop{{{V}_{R}}}}\,\overset{\tilde{\ }}{\mathop{{{I}_{R}}}}\,\begin{matrix}{} & (22) \\\end{matrix}\]

Because pf* _{R}* = 1. By contrast, the average power dissipated by a capacitor or inductor is:

\[{{({{P}_{avg}})}_{X}}=\overset{\tilde{\ }}{\mathop{{{V}_{X}}}}\,\overset{\tilde{\ }}{\mathop{{{I}_{X}}}}\,p{{f}_{X}}=0\begin{matrix}{} & (23) \\\end{matrix}\]

Because pf_{X} = 0, where the subscript X indicates a reactive element (i.e., either a capacitor or inductor). It is important to note that although capacitors and inductors are lossless (i.e., they store and release energy but do not dissipate energy), they do influence power dissipation in a circuit by affecting the voltage across and the current through resistors in the circuit.

When θz is positive, the load is inductive and the power factor is said to be lagging; when θz is negative, the load is capacitive and the power factor is said to be leading.

It is important to keep in mind that pf = cos(θ_{Z}) = cos(−θ_{Z}) because the cosine is an even function. Thus, while it may be important to know whether a load is inductive or capacitive, the value of the power factor only indicates the extent to which a load is inductive or capacitive.

To know whether a load is inductive or capacitive, one must know whether the power factor is leading or lagging.

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]]>The post Resistance and Impedance in an AC Circuit appeared first on Electrical A2Z.

]]>**$\text{V=IZ}$**

where the phasor quantity **Z** is known as impedance. For a resistor, inductor, and capacitor, the impedances are, respectively:

$\begin{matrix}{{Z}_{R}}=R & {{Z}_{L}}=j\omega L & {{Z}_{C}}=\frac{1}{j\omega C} \\\end{matrix}=\frac{-j}{\omega C}$

Combinations of resistors, inductors, and capacitance can be represented by a single equivalent impedance of the form:

\[\begin{matrix}Z\left( j\omega \right)=R\left( j\omega \right)+jX\left( j\omega \right) & {} & \text{units of }\!\!\Omega\!\!\text{ (ohms)} \\\end{matrix}\]

Where *R* (jω) and *X* (jω) are known as the “resistance” and “reactance” portions, respectively, of the equivalent impedance **Z**. Both terms are, in general, functions of frequency ω.

The admittance is defined as the inverse of impedance.

$\begin{matrix}\text{Y=}\frac{\text{1}}{\text{Z}} & {} & \text{units of S }\left( Siemens \right) \\\end{matrix}$

Consequently, all the DC circuit relations and techniques introduced in Chapter 3 can be extended to AC circuits. Thus, it is not necessary to learn new techniques and formulas to solve AC circuits; it is only necessary to learn to use the same techniques and formulas with phasors.

**Generalized Ohm’s Law**

The impedance concept reflects the fact that capacitors and inductors act as frequency- dependent resistors. **Figure 1** depicts a generic AC circuit with a sinusoidal voltage source **V**_{S }phasor and an impedance load **Z**, which is also a phasor and represents the effect of a generic network of resistors, capacitors, and inductors.

**Figure 1** The impedance concept

The resulting current **I **is a phasor determined by:

$\begin{matrix}\operatorname{V}=IZ & Generalized\text{ }Ohms\text{ }Law~ & \left( 1 \right) \\\end{matrix}$

A specific expression for the impedance **Z** is found for each specific network of resistors, capacitors, and inductors attached to the source. To determine **Z** it is first necessary to determine the impedance of resistors, capacitors, and inductors using:

$\begin{matrix}\text{Z=}\frac{\text{V}}{\text{I}} & Definition\text{ }of\text{ }impedance & \left( 2 \right) \\\end{matrix}$

Once the impedance of each resistor, capacitor, and inductor in a network is known, they can be combined in **series and parallel** (using the usual rules for resistors) to form an equivalent impedance “seen” by the source.

The *i-v* relationship for a resistor is, of course, **Ohm’s law**, which in the case of sinusoidal sources is written as (see **Figure 2):**

**Figure 2** For a resistor, V_{R}(t)=i_{R}(t)R

$\begin{matrix}{{v}_{R}}(t)={{i}_{R}}\left( t \right)R & {} & \left( 3 \right) \\\end{matrix}$

or, in phasor form,

\[{{\text{V}}_{\text{R}}}{{e}^{j\omega t}}={{\text{I}}_{\text{R}}}{{e}^{j\omega t}}R\]

Where ${{V}_{R}}={{V}_{\operatorname{R}}}{{e}^{j\theta t}}$ and ${{I}_{R}}={{I}_{\operatorname{R}}}{{e}^{j\theta t}}$ are phasors.

Both sides of the above equation can be divided by e^{jωt} to yield:

$\begin{matrix}{{\text{V}}_{\text{R}}}\text{=}{{\text{I}}_{\text{R}}}R & {} & \left( 4 \right) \\\end{matrix}$

The impedance of a resistor is then determined from the definition of impedance:

$\begin{matrix}{{\text{Z}}_{\text{R}}}=\frac{{{\text{V}}_{\text{R}}}}{{{\text{I}}_{\text{R}}}}=R & {} & \left( 5 \right) \\\end{matrix}$

Thus:

**Z _{R} = R Impedance of a resistor**

The impedance of a resistor is a real number; that is, it has a magnitude *R* and a zero phase, as shown in **Figure 2**. The phase of the impedance is equal to the phase difference between the voltage across an element and the current through the same element.

In the case of a resistor, the voltage is completely in phase with the current, which means that there is no time delay or time shift between the voltage waveform and the current waveform in the time domain.

**Figure 2** Phasor diagram of the impedance of a resistor. Remember that Z=V/L

It is important to keep in mind that the phasor voltages and currents in AC circuits are functions of frequency, **V** =** V **(jω) and **I **= **I** (jω). This fact is crucial for determining the impedance of capacitors and inductors, as shown below.

The *i-v* relationship for an inductor is (see **Figure 3):**

**Figure 3** For an inductor

\[\begin{matrix}{{v}_{L}}\left( t \right)=L\frac{d{{i}_{L}}\left( t \right)}{dt} & {} & \left( 6 \right) \\\end{matrix}\]

At this point, it is important to proceed carefully. The time-domain expression for the current through the inductor is:

\[\begin{matrix}{{i}_{L}}\left( t \right)={{I}_{L}}\cos \left( \omega t+\theta \right) & {} & \left( 7 \right) \\\end{matrix}\]

Such that

\[\begin{align}& \frac{d}{dt}{{i}_{L}}\left( t \right)=-{{I}_{L}}\omega \sin \left( \omega t+\theta \right) \\& ={{I}_{L}}\omega \cos \left( \omega t+\theta +\pi /2 \right) \\& =\operatorname{Re}({{I}_{L}}\omega {{e}^{j\pi /2}}{{e}^{j\omega t+\theta }}) \\& =\operatorname{Re}\left[ {{I}_{L}}\left( j\omega \right){{e}^{j\omega t+\theta }} \right] \\\end{align}\]

Notice that the net effect of the time derivative is to produce an extra *( j ω)* term along with the complex exponential expression of i_{L}(t). That is:

Time Domain |
Frequency Domain |

${}^{d}/{}_{dt}$ | $j\omega $ |

Therefore, the phasor equivalent of the *i-v* relationship for an inductor is:

$\begin{matrix}{{\text{V}}_{\text{L}}}=L\left( j\omega \right){{I}_{L}} & {} & \left( 8 \right) \\\end{matrix}$

The impedance of an inductor is then determined from the definition of impedance:

$\begin{matrix}{{\text{Z}}_{\text{L}}}=\frac{{{\text{V}}_{\text{L}}}}{{{\text{I}}_{\text{L}}}}=j\omega L & {} & \left( 9 \right) \\\end{matrix}$

Thus:

$\begin{matrix}{{Z}_{L}}=j\omega L=\omega L\angle \frac{\pi }{2} & ~Impedance\text{ }of\text{ }an\text{ }inductor~ & \left( 10 \right) \\\end{matrix}$

The impedance of an inductor is a positive, purely imaginary number; that is, it has a magnitude of *ωL* and a phase of *π/*2 radians or 90◦, as shown in **Figure 4***.* As before, the phase of the impedance is equal to the phase difference between the voltage across an element and the current through the same element.

In the case of an inductor, the voltage leads the current by π/2 radians, which means that a feature (e.g., a zero crossing point) of the voltage waveform occurs *T /*4 seconds earlier than the same feature of the current waveform. *T *is the common period.

Note that the inductor behaves as a complex frequency-dependent resistor and that its magnitude *ωL* is proportional to the angular frequency ω. Thus, an inductor will “impede” current flow in proportion to the frequency of the source signal. At low frequencies, an inductor acts like a short-circuit; at high frequencies, it acts like an open-circuit.

**Figure 4** Phasor diagram of the impedance of an inductor. Remember that Z=V/L

The principle of duality suggests that the procedure to derive the impedance of a capacitor should be a mirror image of the procedure shown above for an inductor. The i-v relationship for a capacitor is (see **Figure 5**):

**Figure 5** For a capacitor

\[\begin{matrix}{{i}_{C}}\left( t \right)=C\frac{d{{v}_{C}}\left( t \right)}{dt} & {} & \left( 11 \right) \\\end{matrix}\]

The time-domain expression for the voltage across the capacitor is:

\[\begin{matrix}{{v}_{C}}\left( t \right)={{V}_{C}}\cos \left( \omega t+\theta \right) & {} & \left( 12 \right) \\\end{matrix}\]

Such that

\[\begin{align}& \frac{d}{dt}{{v}_{C}}\left( t \right)=-{{V}_{C}}\omega \sin \left( \omega t+\theta \right) \\& ={{V}_{C}}\omega \cos \left( \omega t+\theta +\pi /2 \right) \\& =\operatorname{Re}({{V}_{C}}\omega {{e}^{j\pi /2}}{{e}^{j\omega t+\theta }}) \\& =\operatorname{Re}\left[ {{V}_{C}}\left( j\omega \right){{e}^{j\omega t+\theta }} \right] \\\end{align}\]

Notice that the *net* effect of the time derivative is to produce an extra *( j ω*) term along with the complex exponential expression of v_{C}(*t*). Therefore, the phasor equivalent of the *i-v* relationship for a capacitor is:

$\begin{matrix}{{\text{I}}_{\text{C}}}=C\left( j\omega \right){{\text{V}}_{\text{C}}} & {} & \left( 13 \right) \\\end{matrix}$

The impedance of an inductor is then determined from the definition of impedance:

\[\begin{matrix}{{\text{Z}}_{\text{C}}}=\frac{{{\text{V}}_{\text{C}}}}{{{\text{I}}_{\text{C}}}}=\frac{1}{j\omega C}=\frac{-j}{\omega C} & {} & \left( 14 \right) \\\end{matrix}\]

Thus:

\[\begin{matrix}{{Z}_{C}}=\frac{1}{j\omega C}=\frac{-j}{\omega C}=\frac{1}{\omega C}\angle -\frac{\pi }{2} & {} & \left( 15 \right) \\\end{matrix}\]

The impedance of a capacitor is a negative, purely imaginary number; that is, it has a magnitude of 1/*ωC* and a phase of −*π/*2 radians or −90^{o}, as shown in **Figure 6.** As before, the phase of the impedance is equal to the phase difference between the voltage across an element and the current through the same element. In the case of a capacitor, the voltage lags the current by *π/*2 radians, which means that a feature (e.g., a zero-crossing point) of the voltage waveform occurs *T/4* seconds later than the same feature of the current waveform. *T* is the common period of each waveform.

**Figure 6** Phasor diagram of the impedance of a capacitor. Remember that Z=V/L

Note that the capacitor also behaves as a complex frequency-dependent resistor, except that its magnitude 1/*ωC* is inversely proportional to the angular frequency *ω*.

Thus, a capacitor will “impede” current flow in inverse proportion to the frequency of the source. At low frequencies, a capacitor acts like an open-circuit; at high frequencies, it acts like a short-circuit.

The impedance concept is very useful in solving AC circuit analysis problems. It allows network theorems developed for DC circuits to be applied to AC circuits. The only difference is that complex arithmetic, rather than scalar arithmetic, must be employed to find the equivalent impedance.

**Figure 7** depicts Z_{R}(*jω*), Z_{L}(*jω*), and Z_{C}(*jω*) in the complex plane. It is important to emphasize that although the impedance of resistors is purely real and the impedance of capacitors and inductors is purely imaginary, the equivalent impedance seen by a source in an arbitrary circuit can be complex.

**Figure 7** The impedance of R, L and C are shown in the complex plane. Impedances in the upper right quadrant are inductive while those in the lower right quadrant are capacitive.

$\begin{matrix}\text{Z}\left( j\omega \right)=R+X\left( j\omega \right) & {} & \left( 16 \right) \\\end{matrix}$

Here, *R* is resistance and *X* is reactance. The unit of *R*, *X*, and *Z* is the ohm.

It was suggested that the solution of certain circuit analysis problems was handled more easily in terms of conductance’s than resistances. This is true, for example, when one is using node analysis, or in circuits with many parallel elements, since conductance in parallel add as resistors in series do. In AC circuit analysis, an analogous quantity may be defined—the reciprocal of complex impedance. Just as conductance *G* was defined as the inverse of resistance, admittance **Y** is defined as the inverse of impedance:

$\begin{matrix}Y=\frac{1}{Z} & \text{units of S (Siemens)} & \left( 17 \right) \\\end{matrix}$

Whenever the impedance **Z** is purely real, the admittance **Y** is identical to the conductance **G**. In general, however, **Y** is complex.

$\begin{matrix}\text{Y}=G+jB & {} & (18) \\\end{matrix}$

where *G* is the AC conductance and *B* is the susceptance, which is analogous to reactance. Clearly, *G* and *B* are related to *R* and *X*; however, the relationship is not a simple inverse. If **Z** = *R* + *j*X , then the admittance is:

$\begin{matrix}Y=\frac{1}{Z}=\frac{1}{R+jX} & {} & \left( 19 \right) \\\end{matrix}$

Multiply the numerator and denominator by the complex conjugate **Z ̄** = *R* − *j*X:

$\begin{matrix}Y=\frac{\overline{Z}}{\overline{Z}Z}=\frac{R-jX}{{{R}^{2}}+{{X}^{2}}} & {} & \left( 20 \right) \\\end{matrix}$

and conclude that

$\begin{align}& \begin{matrix}G=\frac{R}{{{R}^{2}}+{{X}^{2}}} & {} & \left( 21 \right) \\\end{matrix} \\& B=\frac{-X}{{{R}^{2}}+{{X}^{2}}} \\\end{align}$

Notice in particular that *G *is not the reciprocal of R in the general case!

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]]>The post Solid State Motor Starters appeared first on Electrical A2Z.

]]>Solid-state motor starters include motor overload protection and are controlled by the same switches (pushbuttons, pressure switches, etc.) as electromechanical starters. See **Figure 1.**

Solid-state motor starters consist of solid-state components such as SCRs or triacs that allow current flow when they are conducting and stop current flow when they are not conducting.

**Figure 1.** Solid-state motor starters eliminate electromechanical components by using solid-state components to turn a motor on and off.

Solid-state motor starters have terminals for connecting the incoming supply power (L1/R, L2/S, L3/T) and terminals for connecting a motor (T1/U, T2/V, T3/W). Solid-state motor starters also include a terminal strip for connecting external inputs (pushbuttons, proximity switches, etc.).

Solid-state motor starters also include a dual inline package (DIP) switchboard for programming starter functions (starting mode/time, stopping mode, etc.), and potentiometers for adjusting motor full-load current (in amps) and trip class. Solid-state motor starters may also include LEDs to provide a visual indication of circuit conditions. See **Figure 2.**

**Figure 2**. Solid-state motor starters have a control terminal strip, input and output power terminals, dials for current and trip class adjustment, and programming DIP switches.

**Wiring Power and Control Circuits**

The power circuit of a solid-state motor starter is wired by bringing power from the fuses/circuit breaker into the starter. The incoming power must be at the same voltage level for which the motor is rated or wired. See **Figure 3.**

The control circuit is wired to the control terminal strip located on the starter. The control circuit voltage is less than the power circuit voltage (typically 12 VDC, 12 VAC, 24 VDC, or 24 VAC).

The control terminal strip includes a connection for external control voltage (when required), connections for external control switches (pushbuttons, temperature switches, etc.), and connections for output contacts (alarms, indicating lamps, etc.) that can be used for controlling external loads.

**Setting Overload Protection**

Motors must be protected from overcurrents and overloads. Fuses and circuit breakers (normally located in the motor disconnect) are used to protect a motor from overcurrents (short circuits and high operating currents).

Overloads located in the motor starter protect the motor from overload current caused when the load on the motor is greater than the design torque rating of the motor. Overloads can be thermal overloads (heaters) or solid-state overloads.

**Figure 3.** The power circuit of a solid-state motor starter is wired by bringing power from the fuses/circuit breaker into the starter. The control circuit is wired to the control terminal strip located on the starter.

Solid-state overloads use a current **transformer** (CT) to monitor each power line. Solid-state overloads are set by selecting a current limit based on full-load current ratings listed on the motor nameplate and the trip class setting (class 10, 15, 20, or 30). The current limit is set by adjusting the current adjustment dial located on the starter.

The trip class setting is the length of time it takes for an overload relay to trip and remove power from the motor.

The lower the trip class setting, the faster the trip time of the solid-state overload. The higher the trip class setting, the slower the trip time of the solid-state overload. The trip class setting is based on the motor application (a type of load placed on the motor). The trip class setting may be adjusted using a trip class setting dial located near the current adjustment dial or by using DIP switches. **See Figure 4.**

**Cold trip** is the trip point from the time the motor starts until the first time the overloads trip (motor operating below nameplate rated current). **Hot trip** is the trip point after the overloads have tripped and have been reset (motor operating near or over nameplate rated current).

**Tech Fact**

Magnetic motor starter overloads usually have a class 20 trip rating. Solid-state starters or drives have a class 10 trip rating. A slower trip time is usually programmed by selecting a class number or, if a number is unavailable, a “fast/minimum” or “slow/ maximum” rating.

A solid-state reduced-voltage starter ramps up the motor voltage as the motor accelerates, instead of applying full voltage instantaneously like across-the-line starters do. Solid-state starters reduce inrush current compared to the high inrush current across-the-line starters produce.

Solid-state starters also minimize starting torque, which can damage some loads connected to the motor and smooth motor acceleration. **See Figure 5.**

**Figure 4.** The trip class setting of solid-state overloads is based on the motor application (a type of load placed on the motor.

**Figure 5.** A solid-state starter ramps up the voltage reduces inrush current, minimizes starting torque, and smooths acceleration.

Solid-state starting provides a smooth, stepless acceleration in applications that require it, such as starting conveyors, compressors, pumps, and a wide range of other industrial applications because of its unique switching capability.

**Electronic Control Circuitry**

A solid-state controller determines to what degree the SCRs should be triggered on to control the voltage, current, and torque applied to a motor.

A solid-state controller also includes current-limiting fuses and current transformers for protection of the unit. The current-limiting fuses are used to protect the SCRs from excess current. The current transformers are used to feed information back to the controller. Heat sinks and thermostat switches are also used to protect the SCRs from high temperatures.

The controller also provides the sequential logic necessary for interfacing other control functions of the starter, such as line loss detection during acceleration. The controller is turned off if any voltage is lost or too low on any one line. This may happen if one line opens or a fuse blow.

A typical solid-state starting circuit consists of both start and run contactors connected in the circuit. The start contactor contacts C_{1} are in series with the SCRs and the run contactor contacts C_{2} are in parallel with the SCRs. **See Figure 6.**

**Figure 6.** An SCR circuit with reverse-parallel wiring of SCRs provides maximum control of an AC load.

The start contacts C_{1} close and the acceleration of the motor is controlled by triggering on the SCRs when the starter is energized. The SCRs control the motor until it approaches full speed, at which time the run contacts C_{2} close, connecting the motor directly across the power line. At this point, the SCRs are turned off, and the motor runs with full power applied to the motor terminals.

Electromechanical and solid-state motor starters can be used to start a motor. When an electromechanical motor starter is used, the motor is connected to the full supply voltage.

When a motor is connected to full supply voltage, the motor has the highest possible current draw, the highest possible torque applied to the load, and the shortest acceleration time. This operating condition may be acceptable for some loads. **However**, many loads cannot be started with high starting torque because they control light loads (small parts, etc.) or delicate loads (paper rolls, etc.).

High starting current can also damage the power distribution system and trip breakers or blow fuses. Solid- state motor starters can be programmed for different starting modes to help reduce problems caused by full-voltage starting.

Motor starting modes include soft start, soft start with start boost, and current-limit start. See Figure 7.

**Figure 7.** Solid-state motor starters can be programmed for different starting modes to help reduce problems caused by full-voltage starting.

Soft start is the most common solid-state starting method. When a starter is set for soft start, the motor is gradually accelerated over a programmable time period, normally 0 sec to 30 sec.

Common start time periods include 2 sec, 4 sec, 6 sec, 8 sec, or 16 sec. The starting torque is adjustable to a percent of the motor’s locked rotor torque. Common starting torque settings include 15%, 25%, 50%, or 60%. A soft start helps cushion the stress applied to loads connected to the motor.

Soft starting is achieved by increasing the motor voltage in accordance with the setting of the ramp-up control. A **potentiometer** is used to set the ramp-up time (normally 1 sec to 20 sec). Soft stopping is achieved by decreasing the motor voltage in accordance with the setting of the ramp-down control.

A **second** **potentiometer** is used to set the ramp-down time (normally 1 sec to 20 sec). A **third** **potentiometer** is used to adjust the starting level of motor voltage to a value at which the motor starts to rotate immediately when soft starting is applied. See **Figure 8.**

**Figure 8.** A soft starter is a device that provides a gradual voltage increase (ramp up) during motor starting and a gradual voltage decrease (ramp down) during motor stopping.

Like any solid-state switch, a soft starter produces heat that must be dissipated for proper operation. The heat dissipation requires large heat sinks when high-current loads (motors) are controlled. For this reason, a contractor is often added in parallel with a soft starter.

The soft starter is used to control the motor when the motor is starting or stopping. The contactor is used to short out the soft starter when the motor is running. This allows for soft starting and soft stopping without the need for large heat sinks during motor running. The soft starter includes an output signal that is used to control the time when the contactor is ON or OFF. See **Figure 9.**

**Figure 9.** A contactor is used with a soft starter to control the voltage to the motor when the motor is running.

When a solid-state motor starter is set for soft start with start boost, the motor is given a current pulse during starting to provide additional starting torque for loads that are difficult to start.

The boost time is usually adjustable from 0 sec to 2 sec. The start boost is normally applied when there is a problem starting a motor using only a soft start.

A solid-state motor starter must be programmed for proper operation before any power is applied to the starter. A solid-state motor starter is programmed by setting each DIP switch to a predetermined position based on motor and application requirements.

The number of DIP switch parameters can range from a few parameters (4 to 6) to numerous parameters. The higher the number of parameters that are available, the greater the number of applications for which the starter can be used. See **Figure 10.**

**Figure 10.** A solid-state motor starter is programmed by setting each DIP switch to a predetermined position based on motor and application requirements.

DIP switch parameters include motor starting mode (start time, soft start, start boost, etc.) and the operation of auxiliary contacts (when they are open or closed).

Each DIP switch setting must be understood and checked before any power is applied to the starter because some settings can be critical to protecting workers, the motor, and the system.

**For example,** the overload reset function can be placed in a manual or automatic mode. In the manual mode, the reset button on the starter must be pressed before the motor can be restarted manually (by external pushbuttons, etc.).

**However**, in the automatic reset mode, the starter automatically restarts the motor after a short time period if the external control switch (pressure switch, etc.) is still closed. This can cause a safety hazard if the person working on or around the system does not know the motor may automatically restart. **For this reason**, it is important to always refer to the manufacturer literature regarding the setting and meaning of each DIP switch position.

Electromechanical and solid-state motor starters can be used to stop a motor when power is removed. When an electromechanical starter is used, the motor coasts to a stop at a rate determined by the load connected to the motor.

Solid-state motor starters can be programmed for different stopping modes. This allows greater application flexibility and protection of the motor/load. Motor stopping modes include a soft stop, pump control, and brake stop. See **Figure 11.**

**Figure 11** Solid-state motor starters can be programmed for different stopping modes to allow greater application flexibility and protection of the motor/load.

The soft stop is the most common solid-state motor starter stopping method. Soft stops allow for an extended controlled stop. In a soft stop, the deceleration time is controlled by the starter, not the load. The soft stop mode is designed for friction loads that tend to stop suddenly when a voltage is removed from the motor.

The pump control mode is used to reduce surges that occur when centrifugal pumps are started and stopped. The pump control mode produces smooth acceleration and deceleration of motors and pumps.

Common motor and pump starting times range from a few seconds to 30 sec. Common motor and pump stopping times range from a few seconds to 120 sec, depending on the size of the motor and pump.

Some applications require a fast motor stop. The brake stop mode provides motor braking for a faster stop than a coast stops or soft stop. The amount of braking (and thus braking time) is programmed based on the application requirements. When using the brake stop mode, the longest time is set first and adjusted downward as needed.

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]]>The post Solid State Relay Vs Electromechanical Relay appeared first on Electrical A2Z.

]]>In addition, the physical features and operating characteristics of EMRs and SSRs are different. See **Figure 1.**

**Figure 1.** An Electromechanical relay provides switching using electromagnetic devices. A solid state relay depends on SCRs and triacs to switch without contacts.

An equivalent terminology chart is used as an aid in the comparison of EMRs and SSRs. Because the basic operating principles and physical structures of the devices are so different, it is difficult to find a direct comparison of the two.

Differences arise almost immediately both in the terminology used to describe the devices and in their overall ability to perform certain functions. See **Figure 2.**

**Advantages and Limitations**

Electromechanical relays and solid state relays are used in many applications. The relay used depends on the electrical requirements, cost requirements, and life expectancy of the application.

Although SSRs have replaced EMRs in many applications, EMRs are still very common. Electromechanical relays offer many advantages that make them cost-effective. However, they have limitations that restrict their use in some applications.

**Figure 2.** An equivalent terminology chart is used as an aid in the comparison of EMRs and SSRs.

Electromechanical relay advantages include the following:

- normally have multi-pole, multi-throw contact arrangements
- contacts can switch AC or DC
- low initial cost
- very low contact voltage drops, thus no heat sink is required
- very resistant to voltage transients
- no OFF-state leakage current through open contacts

Electromechanical relay limitations include the following:

- contacts wear, thus having a limited life
- short contact life when used for rapid switching applications or high-current loads
- generate electromagnetic noise and interference on the power lines
- poor performance when switching high inrush currents

SSRs provide many advantages such as small size, fast switching, long life, and the ability to handle complex switching requirements. SSRs have some limitations that restrict their use in some applications.

Solid state relay advantages include the following:

- very long life when properly applied
- no contact to wear
- no contact arcing to generate electromagnetic interference
- resistant to shock and vibration because they have no moving parts
- logic compatible with programmable controllers, digital circuits, and computers
- very fast switching capability
- different switching modes (zero switching, instant- on, etc.)

Solid state relay limitations include the following:

- normally only one contact available per relay
- heat sink required due to the voltage drop across switch
- can switch only AC or DC
- OFF-state leakage current when the switch is open
- normally limited to switching only a narrow frequency range such as 40 Hz to 70 Hz

**Input Signals**

The application of voltage to the input coil of an electromagnetic device creates an electromagnet that is capable of pulling in an armature with a set of contacts attached to control a load circuit. It takes more voltage and current to pull in the coil than to hold it in due to the initial air gap between the magnetic coil and the armature.

The specifications used to describe the energizing and de-energizing process of an electromagnetic device are coil voltage, coil current, holding a current, and drop-out voltage.

A solid state relay has no coil or contacts and requires only minimum values of voltage and current to turn it on and off. The two specifications needed to describe the input signal for an SSR are control voltage and control current.

The electronic nature of an SSR and its input circuit allows easy compatibility with digitally controlled logic circuits. Many SSRs are available with minimum control voltages of 3 V and control currents as low as 1 mA, which makes them ideal for a variety of current state-of-the-art logic circuits.

**Response time**

One of the significant advantages of a solid state relay over an electromechanical relay is its response time (ability to turn on and turn off). An EMR may be able to respond hundreds of times per minute, but an SSR is capable of switching thousands of times per minute with no chattering or bounce.

DC switching time for an SSR is in the microsecond range. AC switching time for an SSR, with the use of zero-voltage turn-on, is less than 9 ms. The reason for this advantage is that the SSR may be turned on and turned off electronically much more rapidly than a relay may be electromagnetically pulled in and dropped out.

The higher speeds of solid state relays have become increasingly more important as industry demands higher productivity from processing equipment. The more rapidly the equipment can process or cycle its output, the greater its productivity.

**Voltage and Current Ratings**

Electromechanical relays and solid state relays have certain limitations that determine how much voltage and current each device can safely handle. The values vary from device to device and from manufacturer to manufacturer. Datasheets are used to determine whether a given device can safely switch a given load.

The advantages of SSRs are that they have a capacity for arc-less switching, have no moving parts to wear out, and are totally enclosed, thus allowing them to be operated in potentially explosive environments without special enclosures.

The advantage of EMRs is that the contacts can be replaced if the device receives an excessive surge current. In an SSR, the complete device must be replaced if there is damage.

**Voltage Drop**

When a set of contacts on an electromechanical relay closes, the contact resistance is normally low unless the contacts are pitted or corroded. However, because an SSR is constructed of semiconductor materials, it opens and closes a circuit by increasing or decreasing its ability to conduct.

**Even at full conduction**, a solid state relay presents some residual resistance, which can create a voltage drop of up to approximately 1.5 V in the load circuit. This voltage drop is usually considered insignificant because it is small in relation to the load voltage and in most cases presents no problems. This unique feature may have to be taken into consideration when load voltages are small.

A method of removing the heat produced at the switching device must be used when load currents are high.

**Insulation and leakage**

The air gap between a set of open contacts provides an almost infinite resistance through which no current flows. Due to their unique construction, solid state relays provide a very high but measurable resistance when turned off. SSRs have a switched-off resistance not found on EMRs.

It is possible for small amounts of current (OFF-state leakage) to pass through an SSR because some conductance is still possible even though the SSR is turned off. OFF-state leakage current is not found on EMRs.

**OFF-state leakage current** is the amount of current that leaks through an SSR when the switch is turned off, normally about 2 mA to 10 mA. The rating of OFF-state leakage current in an SSR is usually determined at 200 VDC across the output and should not usually exceed more than 200 mA at this voltage.

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]]>The post Solid State Relay Switching Methods appeared first on Electrical A2Z.

]]>**Zero Switching**

A zero-switching relay is a solid state relay that turns on the load when the control voltage is applied and the voltage at the load crosses zero (or within a few volts of zero).

The relay turns off the load when the control voltage is removed and the current in the load crosses zero. **See Figure 1.**

The zero-switching relay is the most widely used relay. Zero switching relays are designed to control resistive loads. Zero switching relays control the temperature of heating elements, soldering irons, extruders for forming plastic, incubators, and ovens. Zero switching relays control the switching of incandescent lamps, tungsten lamps, flashing lamps, and programmable controller interfacing.

**Figure 1.** A zero-switching relay turns on the load when the control voltage is applied and the voltage at the load crosses zero.

**Instant-On Switching**

An instant-on switching relay is a solid state relay that turns on the load immediately when the control voltage is present. This allows the load to be turned on at any point on the AC sine wave.

The relay turns off when the control voltage is removed and the current in the load crosses zero. Instant-on switching is exactly like electromechanical switching because both switching methods turn on the load at any point on the AC sine wave. **See Figure 2.**

Instant-on relays are designed to control inductive loads. In inductive loads, voltage and current are not in phase and the loads turn on at a point other than the zero voltage point that is preferred.

Instant-on relays control the switching of **contactors**, magnetic valves and **starters**, valve position controls, magnetic brakes, small motors (used for position control), **1φ motors**, small **3φ motors**, lighting systems, **programmable controller interfaces**, and phase controls.

**Peak Switching**

A peak switching relay is a solid state relay that turns on the load when the control voltage is present and the voltage at the load is at its peak.

The relay turns off when the control voltage is removed and the current in the load crosses zero. Peak switching is preferred when the voltage and current are about 2° out of phase because switching at peak voltage is switching at close to zero current. **See Figure 3.**

Peak switching relays control **transformers** and other heavy inductive loads and limit the current in the first half period of the AC sine wave.

Peak switching relays control the switching of transformers, large motors, DC loads, high inductive lamps, magnetic valves, and small **DC motors**.

**Figure 2.** An instant-on switching relay turns on the load immediately when the control voltage is present.

**Figure 3.** A peak switching relay turns on the load when the control voltage is present and the voltage at the load is at its peak.

**Analog Switching**

An analog switching relay is a solid state relay that has an infinite number of possible output voltages within the rated range of the relay.

An analog switching relay has a built-in synchronizing circuit that controls the amount of output voltage as a function of the input voltage. This allows for a ramp-up function of the load. In a **ramp-up function**, the voltage at the load starts at a low level and is increased over a period of time. The relay turns off when the control voltage is removed and the current in the load crosses zero. **See Figure 4.**

**Figure 4.** An analog switching relay has an infinite number of possible output voltages within the rated range of the relay.

A typical analog switching relay has an input control voltage of 0 VDC to 5 VDC. These low and high limits correspond respectively to no switching and full switching on the output load.

For any voltage between 0 VDC and 5 VDC, the output is a percentage of the available output voltage. However, the output is normally nonlinear when compared to the input. Also, the manufacturer’s data must be checked.

Analog switching relays are designed for **closed-loop** applications. One closed-loop application is a temperature control with feedback from a temperature sensor to the controller.

In a closed-loop system, the amount of output is directly proportional to the amount of the input signal. **For example**, if there is a small temperature difference between the actual temperature and the set temperature, the load (heating element) is given low power. **However**, if there is a large temperature difference between the actual temperature and the set temperature, the load (heating element) is given high power. This relay may also be used for starting high-power incandescent lamps to reduce the inrush current.

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]]>An SSR uses a silicon-controlled rectifier (SCR), triac, or transistor output instead of mechanical contacts to switch the controlled power. The output is optically coupled to a light-emitting diode (LED) light source inside the relay. The relay is turned on when the LED is energized, usually with low-voltage DC power. **See Figure 1.**

**Figure 1.** A solid-state relay (SSR) is an electronic switching device that has no moving parts.

The industrial control market has moved to solid-state electronics. Due to declining cost, high reliability, and immense capability, solid-state devices are replacing many devices that operate on mechanical and electrical principles.

The selection of a solid-state relay is based on the electrical, mechanical, and cost characteristics of each device and the required application.

A solid state relay circuit consists of an **input circuit, a control circuit, and an output (load-switching) circuit**. These circuits may be used in any combination to provide many different solid-state switching applications. **See Figure 2.**

**Figure 2.** A solid state relay circuit consists of an input circuit, a control circuit, and an output (load-switching) circuit.

**1. Input Circuit**

An input circuit is the part of an SSR to which the control component is connected. The input circuit performs the same function as the coil of an EMR.

The input circuit is activated by applying a voltage to the input of the relay that is higher than the specified pickup voltage of the relay. The input circuit is deactivated when a voltage less than the specified minimum dropout voltage of the relay is applied.

Some SSRs have a fixed input voltage rating, such as 12 VDC. Most SSRs have an input voltage range, such as 3 VDC to 32 VDC. The voltage range allows a single SSR to be used with most electronic circuits.

The input voltage of a solid state relay may be controlled (switched) through mechanical contacts, transistors, digital gates, etc. Most SSRs may be switched directly by low-power devices, which include integrated circuits, without adding external buffers or current-limiting devices. Variable-input devices, such as thermistors, may also be used to switch the input voltage of an SSR.

**2. Control Circuits**

A control circuit is the part of a solid state relay that determines when the output component is energized or de-energized.

The control circuit functions as the interface between the input and output circuits. In a solid state relay, the interface is accomplished by electronic circuitry inside the SSR. In an EMR, the interface is accomplished by a magnetic coil that closes a set of mechanical contacts.

When the control circuit receives the input voltage, the circuit is switched or not switched depending on whether the relay is a zero switching, instant-on, peak switching, or analog switching relay.

Each relay is designed to turn on the load-switching circuit at a predetermined voltage point. **For example**, a zero-switching relay allows the load to be turned on only after the voltage across the load is at or near zero. The zero-switching function provides a number of benefits, such as the elimination of high inrush currents on the load.

**3. Output (Load-Switching) Circuits**

The output (load- switching) circuit of a solid state relay is the load switched by the SSR. The output circuit performs the same function as the mechanical contacts of an electromechanical relay. **However,** unlike the multiple output contacts of EMRs, SSRs normally have only one output contact.

Most solid state relays use a thyristor as the output switching component. Thyristors change from the OFF state (contacts open) to the ON state (contacts closed) very quickly when their gate switches on. This fast switching action allows for high-speed switching of loads.

The output switching device used depends on the type of load to be controlled. Different outputs are required when switching DC circuits than are required when switching AC circuits.

Common output switching devices used in SSRs include the following:

- SCRs are used to switch high-current DC loads.
- Triacs are used to switch low-current AC loads.
- Transistors are used to switch low-current DC loads.
- Anti-parallel Thyristors are used to switch high- current AC loads. They are able to dissipate more heat than a triac.
- Thyristors in diode bridges are used to switch low- current AC loads.

A solid state relay can be used to control most of the same circuits that an EMR is used to control. Because an SSR differs from an EMR in function, the control circuit for an SSR differs from that of an EMR. **This difference** is how the relay is connected to the circuit. An SSR performs the same circuit functions as an EMR but with a slightly different control circuit.

**Two-Wire Control**

A solid state relay may be used to control a load using a momentary control such as a pushbutton. **See Figure 3.** In this circuit, a push button signals the SSR, which turns on the load.

The **pushbutton** must be held down to keep the load turned on. The load is turned off when the pushbutton is released. This circuit is identical in operation to the standard two-wire control circuit used with EMRs, magnetic motor starters, and contactors. **For this reason**, the pushbutton could be changed to any manual, mechanical, or automatic control device for simple ON/OFF operation.

The same circuit may be used for liquid level control if the pushbutton is replaced with a float switch.

**Three-Wire Memory Control**

A solid state relay may be used with an SCR to latch a load in the ON condition. **See Figure 4.** This circuit is identical in operation to the standard three-wire memory control circuit.

An SCR is used to add memory after the start pushbutton is pressed. An SCR acts as a current-operated OFF-to-ON switch.

The SCR does not allow the DC control current to pass through until a current is applied to its gate. There must be a flow of a definite minimum current to turn on the SCR. This is accomplished when the start pushbutton is pressed.

Once the gate of the SCR has voltage applied, the SCR is latched in the ON condition and allows the DC control voltage to pass through even after the start pushbutton is released.

Resistor R_{1} is used as a current-limiting resistor for the gate and is determined by gate current and supply voltage.

**Figure 3.** A solid state relay may be used to control a load using a momentary control such as a pushbutton.

**Figure 4**. A solid state relay may be used with an SCR to latch a load in the ON condition.

**Tech Fact**

*When a motor drive is programmed, the control circuit must be programmed for two-wire or three-wire operation. *

*The term “two- wire” means that a switch can perform two functions. The term “three-wire” means that a switch can perform only one function. All three-wire switches require a second switch to control the load. *

**Equivalent NC Contacts**

A solid state relay may be used to simulate an equivalent normally closed (NC) contact condition. An NC contact must be electrically made because most SSRs have the equivalent of a normally open (NO) contact. **This is accomplished** by allowing the DC control voltage to be connected to the SSR through a current-limiting resistor (R). **See Figure 5.**

The load is held in the ON condition because the control voltage is present on the SSR. The selector switch is moved to turn off the load. This allows the DC control voltage to take the path of least resistance and electrically remove the control voltage from the relay. This also turns off the load until the pushbutton is released.

**Figure 5**. A solid state relay with a current-limiting resistor may be used to simulate an equivalent normally closed (NC) contact condition.

Solid state relays are also capable of being controlled by electronic control signals from logic gates and transistors. **See Figure 6.**

In this circuit, the SSR is controlled through an NPN transistor that receives its signal from IC logic gates, etc. Two resistors (R_{1} and R_{2}) are used as current-limiting resistors.

**Figure 6.** SSRs may be controlled by electronic control signals from logic gates and transistors

Solid state relays can be connected in series or parallel to create multi-contacts that are controlled by one input device. Multi-contact SSRs may also be used.

Three SSR control inputs may be connected in parallel so that when the switch is closed, all three are actuated. **See Figure 7.** This controls the 3φ circuit.

In this application, the DC control voltage across each SSR is equal to the DC supply voltage because they are connected in parallel. When a multi-contact SSR is used, there is only one input that controls all output switches.

SSRs can be connected in series to control a 3φ circuit. **See Figure 8.** The DC supply voltage is divided across the three SSRs when the switch is closed. For this reason, the DC supply voltage must be at least three times the minimum operating voltage of each relay.

**Figure 7.** Three solid state relays may be connected in parallel to control a 3φ circuit or a multi-contact SSR may be used.

**Figure 8.** Three solid state relays may be connected in series to control a 3φ circuit. When SSRs are connected in series, the DC supply must be three times the minimum operating voltage of each relay.

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]]>The failure rate of most solid state relays doubles for every 10°C temperature rise above an ambient temperature of 40°C. An ambient temperature of 40°C is considered standard by most manufacturers.

**Figure 1.** As temperature increases, the number of operations of a solid-state relays decreases.

Solid-state relay manufacturers specify the maximum relay temperature permitted. The relay must be properly cooled to ensure that the temperature does not exceed the specified maximum safe value.

Proper cooling is accomplished by installing the solid state relay to the correct heat sink. A heat sink is chosen based on the maximum amount of load current controlled.

**Heat Sinks**

The performance of a solid state relay is affected by ambient temperature. The ambient temperature of a relay is a combination of the temperature of the relay location and the type of enclosure used.

The temperature inside an enclosure may be much higher than the ambient temperature of an enclosure that allows good air flow.

The temperature inside an enclosure increases if the enclosure is located next to a heat source or in the sun. The electronic circuit and solid state relay also produce heat. Forced-air cooling is required in some applications.

- You May Also Read: Solid State Timers

**Tech Fact**

*Higher-than-rated heat is a common cause of SSR failures. High heat is difficult to locate and measure. *

*To determine if the solid state relay is overheated, the relay is operated at full load for at least one hour. An IR (infrared) noncontact temperature test instrument or thermal imager is used to measure the heat on the relay. *

*The heat should be no more than 75% of the relay’s maximum heat rating. A thermal imager works best because it can visually indicate the exact location of heat on the relay, heat sink, and the surrounding area.*

**Selecting Heat Sinks**

A low resistance to heat flow is required to remove the heat produced by an SSR. The opposition to heat flow is thermal resistance.

Thermal resistance (R

_{TH}) is the ability of a device to impede the flow of heat. Thermal resistance is a function of the surface area of a heat sink and the conduction coefficient of the heat sink material. Thermal resistance (R_{TH}) is expressed in degrees Celsius per watt (°C/W).

Heat sink manufacturers list the thermal resistance of heat sinks. The lower the thermal resistance value, the more easily the heat sink dissipates heat. The larger the thermal resistance value, the less effectively the heat sink dissipates heat.

The thermal resistance value of a heat sink is used with a solid state relay load current/ambient temperature chart to determine the size of the heat sink required. **See Figure 2.**

**Figure 2.** The thermal resistance value of a heat sink is used with an SSR load current/ambient temperature chart to determine the size of the heat sink required.

A relay can control a large amount of current when a heat sink with a low thermal resistance number is used. A relay can control the least amount of current when no heat sink (free-air mounting) is used. Heat conduction through a relay and into a heat sink can be maximized with the following recommendations:

- Use heat sinks made of a material that has a high thermal conductivity. Silver has the highest thermal conductivity rating. Copper has the highest practical thermal conductivity rating. Aluminum has a good thermal conductivity rating and is the most cost-effective and widely used heat sink.
- Keep the thermal path as short as possible.
- Use the largest cross-sectional surface area in the smallest space.
- Always use thermal grease or pads between the relay housing and the heat sink to eliminate air gaps and aid in thermal conductivity.

**Mounting Heat Sinks**

A heat sink must be correctly mounted to ensure proper heat transfer. A heat sink can be properly mounted using the following recommendations:

- Choose a smooth mounting surface. The surfaces between a heat sink and a solid-state device should be as flat and smooth as possible. Ensure that the mounting bolts and screws are securely tightened.
- Locate heat-producing devices so that the temperature is spread over a large area. This helps prevent higher temperature areas.
- Use heat sinks with fins to achieve as large a surface area as possible.
- Ensure that the heat from one heat sink does not add to the heat from another heat sink.
- Always use thermal grease between the heat sink and the solid-state device to ensure maximum heat transfer.

**Relay Current Problems**

The overcurrent passing through a solid state relay must be kept below the maximum load current rating of the relay. An overload protection fuse is used to prevent overcurrents from damaging a solid state relay.

An overload protection fuse opens the circuit when the current is increased to a higher value than the nominal load current. The fuse should be an instantaneous fuse used for the protection of semiconductors. **See Figure 3.**

**Relay Voltage Problems**

Most AC power lines contain voltage spikes superimposed on the voltage sine wave. Voltage spikes are produced by the switching of motors, solenoids, transformers, motor starters, contactors, and other inductive loads. Large spikes are also produced when lightning strikes the power distribution system.

The output element of a relay can exceed its breakdown voltage and turn on for part of a half period if overvoltage protection is not provided. This short turn-on can cause problems in the circuit.

Varistors are added to the relay output terminals to prevent an overvoltage problem. A varistor should be rated 10% higher than the line voltage of the output circuit. The varistor bypasses the transient current**. See Figure 4. **

**Figure 3.** An overload protection fuse opens the circuit when the current is increased to a higher value than the nominal load current.

**Figure 4.** Varistors are added to relay output terminals to prevent an overvoltage problem.

**Voltage Drop**

In all **series circuits**, the total circuit voltage is dropped across the circuit components. The higher the resistance of any component, the higher the voltage drop. The lower the resistance of any component, the lower the voltage drop. **Thus**, an open switch that has a meter connected across it shows a very high voltage drop because the meter and open switch have very high resistance when compared to the load.

**Conversely**, a closed switch that has a meter connected across it shows a very low voltage drop because the meter is closed and closed switches have very low resistance when compared to the load.

A voltage drop in the switching component is unavoidable in a solid state relay. The voltage drop produces heat. The larger the current passing through the relay, the greater the amount of heat produced. The generated heat affects relay operation and can destroy the relay if it is not removed. **See Figure 5.**

**Figure 5.** The voltage drop in the switching component of an SSR produces heat and can destroy the relay if it is not removed.

The voltage drop in an SSR is usually 1 V to 1.6 V, depending on the load current. For small loads (less than 1 A), the heat produced is safely dissipated through the relay’s case. High-current loads require a heat sink to dissipate the extra heat. **See Figure 6**.

For example, if the load current in a circuit is 1 A and the SSR switching device has a 2 V drop, the power generated is 2 W. The 2 W of power generates heat that can be dissipated through the relay’s case.

However, if the load current in a circuit is 20 A and the SSR switching device has a 2 V drop, the power generated in the device is 40 W. The 40 W of power generates heat that requires a heat sink to safely dissipate the heat.

**Figure 6.** For small loads (less than 1 A), the heat produced in a solid state relay is safely dissipated through the relay’s case.

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]]>The post Electric Power Distribution System Basics appeared first on Electrical A2Z.

]]>One-line diagrams use the most basic symbols because the intent of the drawing is to illustrate as clearly as possible the flow of current throughout the building distribution system and where each component or device connects into the system.

One-line diagrams are also used when designing large commercial and industrial installations to show the path of electric power throughout a building.

One-line diagrams are also used when troubleshooting distribution system problems such as loss of electric power, low voltage, blown fuses, tripped circuit breakers, and poor power quality.

- You May Also Read: Electrical Power: Transmission & Distribution | Distribution Substation Components

They are also used to determine electric power shut-off points, future expansion capacity, and where emergency backup generators or secondary power systems are connected into the system.

A one-line diagram is helpful when troubleshooting an electric power system and can show the entire distribution system or specific parts of a system. See **Figure 1.**

**Figure 1.** One-line diagrams use single lines and symbols to show system components and operation.

For example, a one-line diagram may show a 13.8 kV feed into a building and the **transformers** used for the distribution of specific voltages.

High voltages are used for the distribution of large amounts of electric power using small conductor sizes. The high voltage is then stepped down to low voltage levels and delivered to distribution panels. The distribution panels route power to individual loads such as industrial equipment, motors, lamps, and computers.

The electrical distribution system of a building must transport electric power from the source of power to the loads. In large buildings, the distribution may be over large areas with many different electrical requirements throughout the building. See **Figure 2**.

In many cases where it is common to shift production machinery, the distribution system must be changed from time to time.

A busway is a metal-enclosed distribution system of busbars available in prefabricated sections. Prefabricated fittings, tees, elbows, and crosses simplify the connection and reconnection of the distribution system.

When sections are bolted together, electric power is available at many locations and throughout the system.

A busway does not have exposed conductors. This is because the power in a plant distribution system is at a high level. To offer protection from the high voltage, the conductors of a busway are supported with insulating blocks and covered with an enclosure to prevent accidental contact.

A typical busway distribution system provides for fast connection and disconnection of machinery. Busways enable manufacturing plants to be retooled or re-engineered without major changes in the distribution system.

**Figure 2.** The electrical distribution system in a plant must transport the electric power from the source of supply to the loads.

The most common length of busways is 10′. Shorter lengths are used as needed. Prefabricated elbows, tees, and crosses make it possible for the electric power to run up, down, and around corners and to be tapped off from the distribution system. **This allows the distribution system** to have maximum flexibility with simple and easy connections as work is performed on installations.

The **two basic types** of busways are feeder and plug-in busways. See **Figure 3.**

**Feeder busways** deliver the power from the source to a load-consuming device. Plug-in busways serve the same function as feeder busways, but they also allow load-consuming devices to be conveniently added along the bus structure.

A **plug-in power panel** is used on a plug-in busway system. The three general types of plug-in power panels used with busways are fusible switches, circuit breakers, and specialty plugs such as duplex receptacles with circuit breakers and twist-lock receptacles.

The conduit and wire are run to a machine or load from the fusible switches and circuit breaker plug-in panels. Generally, power cords may be used only for portable equipment.

**Figure 3.** The two basic types of busways are feeder and plug-in busways.

The loads connected to the electric power distribution system are often portable or unknown at the time of installation. For this reason, the power distribution system must often terminate in such a manner as to provide for a quick connection of a load in the future. **To accomplish this**, an electrician installs receptacles throughout the building or plant to serve the loads as required. With these receptacles, different loads can be connected easily.

Because the distribution system wiring and protective devices determine the size of the load that can be connected to it, a method is required for distinguishing the rating in voltage and current of each termination. This is especially true in industrial applications that require a variety of different currents, voltages, and phases.

Equipment grounding is required throughout an entire electric power distribution system. All non-current-carrying metal parts including conduit, raceways, transformer cases, and switchgear enclosures must be connected to ground.

The objective of grounding is to limit the voltage of all metal parts to the ground and establish an effective ground-fault current path. See **Figure 4.**

**Figure 4.** Electronic grounding, equipment grounding, and building grounding are the three types of grounding needed to create a safe work environment for individuals.

Grounding is accomplished by connecting the non-current-carrying metal to a ground with an approved grounding conductor and fitting.

A ground bus is a network that ties solidly to grounding electrodes. A grounding electrode is a conducting object through which a direct connection to earth is established.

The ground bus must be connected to the grounding electrodes in several spots. The size of the ground bus is determined by the amount of current that flows through the grounding system and the length of time the current flows.

In addition to grounding all non-current-carrying metal, lightning arresters may be needed. A lightning arrester is a device that protects transformers and other electrical equipment from voltage surges caused by lightning.

**A lightning arrester provides a path over which the surge can pass to the ground before it has a chance to damage electrical equipment.**

**Tech Tip**

Never assume that a metal enclosure is properly grounded. An electrical shock can be caused by ungrounded enclosures when the metal handle is touched to turn off a circuit at a disconnect switch.

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