# ECE 220 Network Analysis I

## Lesson 13. Operational Amplifiers and Three-Phase Power

### Operational Amplifiers

Operational amplifiers are used frequently in many analog circuits, such as signal amplifiers, active filters, signal conditioners, and impedance matchers.  The circuit symbol is shown in Figure 1. Figure 1.  Operational amplifier circuit symbol

Although the inputs may be shown in either orientation, it is usually more convenient to show the inverting input on the top.

The operational amplifier (op amp) is usually supplied as a small integrated circuit with one or two op amps on the chip.  In addition to the signals shown in Figure 1, a practical op amp usually also has "offset null" connections for fine tuning.

The V+ and V- supply connections are required.  They provide power for the op amp and determine the maximum and minimum excursions of the output voltage.  As long as these excursions are not exceeded, V+ and V- need not be considered when making op amp calculations.  Based on this concept, a simplified op amp circuit can be used, as shown in Figure 2. Figure 2.  Simplified operational amplifier circuit symbol

Figure 3 shows a circuit model of the op amp.  vn is the voltage between the inverting input and ground.  vp is the voltage between the noninverting input and ground. Figure 3.  Operational amplifier circuit model

Ri is the input resistance (more generally, the input impedance would be used) of the op amp.  Typically, this resistance is very large - usually in the millions of ohms.  Therefore, the current flowing through it will be very small - usually a few microamps or even nanoamps.

The voltage-controlled voltage source in the circuit amplifies the difference voltage vd, which is  vp -  vn, by a factor A.  This A is called the gain of the op amp, and it is usually very large; a typical value for A is one million.  Later on, we will discuss the gain of the circuit.  Do not confuse the gain of the op amp with the gain of the circuit.

Ro is the output resistance (more generally, the output impedance would be used) of the op amp.  This will be relatively small, perhaps 100 Ω.  The output voltage vo is taken with respect to ground.

Let us use this model of the op amp to calculate the output voltage for the circuit in Figure 4. Figure 4.  Sample op amp circuit.

Let us suppose that this is a very poor op amp.  Its input impedance is rather low at 18 kΩ, its output impedance is rather high at 1 kΩ, and its gain is only 2000.  The result, using our circuit model, is shown in Figure 5. Figure 5.  Sample op amp circuit using op amp model.

Let's solve for vn using the node voltage method:

(vn - 1)/900 + (vn + 2000vn)/(9000 + 1000) + vn/18000 = 0

The first term represents the current through the 900 Ω resistor.  The second term represents the current from vn to the dependent source by way of the 9 kΩ resistor and the 1 kΩ resistor; note that the 9 kΩ resistor and the 1 kΩ resistor can be considered in series if no current is drawn from vo; also note that vd = vp - vd = -vn.  The third term represents the current through the 18 kΩ resistor.

To solve this equation, first multiply by 18000:

20vn - 20 + 1.8vn + 3600vn + vn = 0

vn = 0.00552 V

This is approximately 6 mV, which is very nearly zero.  It is negligibly small.

This should lead to a very small current in Ri.  It does:

IRi = .00552/18000 = 0.307 μA

This current is certainly negligibly small.

Now let's calculate the current in the 900 Ω resistor:

I900 = (1 - .00552)/900 = 1.1050 mA

The current in the 9 kΩ resistor is

I9k = (.00552 + 2000 × .00552)/(9000 + 1000) = 1.1046 mA

Notice that these two currents are very nearly the same.  For practical purposes, they can be assumed to be equal.

Now let's calculate vo:

vo = vn - v9k = .00552 - 9000I9k = .00552 - 9000 × .0011046 = -9.94 V

This voltage is very nearly -VsRf/Rs, which calculates to be -10 V.

Let us summarize:  Even for this very poor op amp,

vn is approximately 0.
IRi is approximately 0.
IRs is approximately IRf.
vo is approximately -VsRf/Rs

Since these relationships are nearly true, even for this very poor op amp, we can guess what the relationships will be for an ideal op amp circuit:

vn = 0.
IRi = 0.
IRs = IRf.
vo = -VsRf/Rs

#### Summary of Ideal Op Amp Properties

Ri = infinity.  Therefore Ii = 0 at either the n or p input.
Ro = 0.
A = infinity.  Therefore, vd = 0, (vp - vn) = 0, and vp = vn.

Using this ideal model for the op amp will greatly simplify our calculations, and will result in only very small errors in resulting voltages and currents.

The op amp is very rarely used "open loop" to amplify a tiny signal into a big one.  Instead, feedback is used.  The effect is to make the circuit insensitive to variations in A, Ri, and Ro, even rather large variations.  This makes it possible for the engineer to design op amp circuits (almost) without regard to which brand of op amp is selected.

#### Ideal Op-Amp Analysis Method

The following analysis will work for almost all op-amp circuits.  Even if the op-amp cannot be considered ideal, this works as a good approximation.

1)  Calculate vp.  This is the voltage at the positive input.  To make this calculation, assume that no current flows into the positive input.
2) vn = vp.  This is the voltage at the negative (inverting) input.  This is because the difference voltage vp - vn is infinitesimally small.
3) Solve Kirchhoff's Current Law at the vn node.  In the KCL equation, assume the current into the negative input is zero.

#### Ideal Op-Amp Example

We will now solve an op-amp circuit using the ideal op-amp analysis method.  Consider Figure 6 below. Figure 6.  Example circuit using ideal op-amp model.

1) Since no current flows into the positive input, no current flows in the 8.9 kΩ resistor.  Therefore, the voltage on both sides of the 8.9 kΩ resistor is the same (1 V).  So vp = 1 V.
2) vn = vp = 1 V.
3) KCL at vn is:
-IS - IF = 0
To find the I values, use this form of Ohm's Law:
IXY = (VX - VY)/R
This important equation says that the current between points X and Y is the voltage on the X side minus the voltage on the Y side divided by the resistance. Applying this to the two currents in the circuit, we get:
IS = (3 - 1)/10 k = 2/10 k
IF = (vo - 1)/80 k
Plugging these into KCL gives:
-2/10 k - (vo - 1)/80 k = 0
Multiplying by 80 k:
-16 - vo + 1 = 0
vo = -15 V

You can use this method to analyze almost any op-amp circuit.

There is a reason for the presence of the 8.9 kΩ resistor.  It is the parallel combination of the resistors connected to the negative input.  Real op-amp circuits have tiny "bias currents" that flow into the positive and negative inputs.  By balancing the resistances seen by these two terminals, the effects of the bias currents tend to cancel out.

Before going on, you should complete Tutorial 13A on operational amplifiers.

### Three Phase

Most electric power is transmitted over high voltage three-phase lines.  This method of power transmission is more efficient than single phase power, as you might use in your home.  As you might expect, three-phase power transmission uses three wires. The voltage between any wire and ground has the same magnitude as the voltage between any other wire and ground, but the phases are different.  Each voltage will be 120° out of phase with the other voltages, as shown in the figure below. Each color represents a different phase.  At any point in time, the sum of the three voltages is exactly zero.

#### Three-Phase Power Sources

Three-phase power is produced by specially designed three-phase electric generators.  When a wire moves through a magnetic field, a voltage is generated in the wire (Faraday's Law).  For a three-phase generator, three coils of wire are placed at angles of 120° to each other, and are rotated in a magnetic field 60 times per second.  This produces three voltasges which are 60 Hz sinusoids, each out of phase with the others by 120°.

There are two ways these three coils of wire can be connected together.  The two connections are called delta and Y.  For a delta-connected generator the coils are connected together as shown in the figure below. The coils of wire shown do not represent inductors; rather they represent windings of the generator.  Each such winding generates what is called a phase voltage, meaning the voltage of one phase of the source, shown in the figure as Vφ. Each winding also has a phase current, Iφ.  The phase voltage and phase current will produce a line voltage, VL, and a line current, IL.

To back up a bit, this discussion covers only balanced three-phase.  This means that all loads for all phases are exactly the same.  In practice, this is (usually) very nearly true. The analysis of unbalanced three-phase is much more difficult and is beyond the scope of this course. Given that this is balanced three-phase, each Vφ is exactly the same as every other, keeping in mind that these variables represent rms magnitudes.  The same is true for Iφ, VL, and IL.

To return to the delta-connected generator, you can easily see that any pair of lines is directly connected to one phase of the generator.  It should therefore be obvious that:
(1)  VL = Vφ

The relationship between IL and Iφ is not nearly so obvious.  One must sum the phasor currents in two phases of the source to find the line current. The result is shown below.
(2)  IL = √3Iφ

This peculiar result occurs because the currents are 120° out of phase with each other.

A Y connected generator has coils connected as shown in the figure below. It should be clear from this figure that the current in any phase must be the same as the current in its connected line.
IL = Iφ

The line voltage can be found by taking the phasor sum of the voltage in two phases of the source.  The result is shown below.
(4)  VL = √3Vφ

The analysis of three-phase circuits reduces to little more than deciding where to put the square root of three.

The loads for three-phase power can also be either delta- or Y-connected, as shown in the figure below. The voltage across each Z is a phase voltage, Vφ, for the load. We can reuse Equations (1) and (2) above for the delta-connected load.  We can reuse Equations (3) and (4) for the Y-connected load.  The same equations apply for sources and loads.  For example, the current in a Y-connected load is the same as the line current.

The analysis of a three-phase circuit is virtually the same as the analysis for a single phase AC circuit.  We can still use phasor analysis, power triangles, and all the rest.  There are a few specialized equations for total and phase power that may be helpful.  These equations work for both delta and Y loads.

The power consumed in one phase of the load can be found if the current resistance of the phase are known:
(5)  Pφ = Iφ2Rφ

By conservation of energy, the total power consumed by the load must be 3 times the power consumed in any phase:
(6)  PT = 3Pφ

To get the total power consumed in the load, the following equation is useful:
(7)  PT = √3VLILcosθ where cosθ is the power factor.

The voltage and currents in the above equations are rms values, not peak values.

#### Three-Phase Example

In the circuit below, a three-phase Y-connected generator is connected through power lines to a delta-connected load.  Each phase of the generator produces 277 V.  The load consumes 180 kW of power with a power factor of 66% lagging.  The task is to find the following:  (a) the line voltage, (b) the phase voltage of the load, (c) the line current, (d) the phase current of the generator, (e) the phase current of the load, and (f) the power consumed by one phase of the load. (a)  From Equation (4) for Y-connected sources or loads:
VL = √3Vφ = 480 V

(b)  From Equation (1) for delta-connected sources or loads:
VφLoad = VL = 480 V

(c)  From Equation (7):
PT = √3VLILcosθ
IL = PT/(√3VLcosθ)
IL = 180,000/(√3 × 480 × 0.66) = 328 A

(d)  From Equation (3) for Y-connected sources or loads:
IφGen = IL = 328 A

(e)  From Equation (2) for delta-connected sources or loads:
IφLoad = IL/√3 = 328/√3 = 189 A

(f)  From Equation (6):
Pφ = PT/3 = 180 kW/3 = 60 kW

#### Practical Considerations

Most commercial power plants use Y-connected generators with the center tap grounded.  If the phases are balanced (all phases have exactly the same current), no current will flow through the grounded center tap.  When single-phase power is needed from a three-phase system, it can be extracted from any one of the three phases.

Most large motors are three-phase delta-connected motors.  Three-phase motors have a distinct advantage over single-phase motors:  the power is constant.  In a single-phase motor, p(t) is a sinusoid, but in a three-phase motor, p(t) is a constant.

The line voltage and line current, VL and IL, are more useful than the phase voltage and phase current, Vφ and Iφ.  This is because VL and IL are much easier to measure.  Vφ and Iφ can only be measured by opening up the machine.

Most utilities will provide a customer with only one type of electrical service, either single-phase or three-phase.  (See http://www.federalpacific.com/university/transbasics/chapter6.html and http://en.wikipedia.org/wiki/Three-phase_electric_power for further explanation).  If the system is single-phase, it will probably be three-wire 120 V and 240 V.  For three-phase systems, more options may be available:  240 V three-wire, 480 V three-wire, 600 V three-wire, 208Y/120 V four-wire, 480Y/277 V four-wire.

#### Three-Phase Power Measurement

Power is measured in three-phase systems using the "two-wattmeter method," as shown below. The two wattmeters are connected to any two phases of the input, and then to the load as shown.  Each wattmeter measures the average power, the average of v(t) × i(t), not Vrms × Irms.  Wattmeter 1 senses the current in the top wire and the voltage between the top wire and the center wire; it calculates the average of the product of the voltage and the current.  If Wattmeter 1 measures P1 and Wattmeter 2 measures P2, the total power and total reactive power can be calculated:
PT = P1 + P2
QT = √3(P1 - P2)

Before going on to the homework, you should complete Tutorial 13 on three phase.

### Homework Problems

1.  Use a circuit similar to Figure 4 above to design an op amp circuit that will have an output of -20 V.  Available devices:  an ideal op amp, a 1 V battery, a 2 V battery, and the following resistors:  1 kΩ, 2 kΩ, 4 kΩ, 10 kΩ, 20 kΩ, 40 kΩ.  Show the calculations for your design and sketch your design.

2. Carefully show your analysis of the circuit, using the assumptions for an ideal op amp.   Find vo.

3. Carefully show your analysis of the circuit, using the assumptions for an ideal op amp.   Find vo.

4. Each Vs in the above figure has an rms value 220 V.  Each one is 120° out of phase with the other two.  R = 90 Ω, ZL = j110 Ω.  Find the magnitude (rms value) of the current in any line and the total power consumed by the load.

5. A Y-connected three-phase load has a phase voltage of 440 V and a phase current of 15 A. Each phase consumes 5.2 kW of power. What is the line voltage, the line current, and the total power consumed by the load?  What is the power factor of the load?

6. A delta-connected three-phase source has a phase voltage of 220 V. It feeds a delta-connected load consisting of three identical impedances of 10/42° Ω each. Find the magnitude of the phase current in the load, the line current, and the total power consumed by the load.

7. A balanced Y-connected three-phase motor uses 20 kW of power at 0.81 pf lagging when connected to 220 V lines. What is the line current? What is the current in one phase of a delta-connected generator?

Bonus (no partial credit). In Problem 6, solve if each wire in the transmission line has a resistance of 1 Ω. (Hint: Use delta-Y transformation.)