ECE 220 Network Analysis I

Supplementary Notes

Lesson 13. Operational Amplifiers and Three-Phase Power

Last edited: 04/25/07

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. v_n is the voltage between the inverting input and ground. v_p is the voltage between the noninverting input and ground.

Figure 3. Operational amplifier circuit model

R_i 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 v_d, which is v_p - v_n, 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.

R_o 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 v_o 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 v_n using the node voltage method:

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

The first term represents the current through the 900 Ω resistor. The second term represents the current from v_n 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 v_o; also note that v_d = v_p - v_d = -v_n. The third term represents the current through the 18 kΩ resistor.

To solve this equation, first multiply by 18000:

20v_n - 20 + 1.8v_n + 3600v_n + v_n = 0

v_n = 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 R_i. It does:

I_Ri = .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

I_9k = (.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 v_o:

v_o = v_n - v_9k = .00552 - 9000I_9k = .00552 - 9000 × .0011046 = -9.94 V

This voltage is very nearly -V_sR_f/R_s, which calculates to be -10 V.

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

v_n is approximately 0.
I_Ri is approximately 0.
I_Rs is approximately I_Rf.
v_o is approximately -V_sR_f/R_s.

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:

v_n = 0.
I_Ri = 0.
I_Rs = I_Rf.
v_o = -V_sR_f/R_s.

Summary of Ideal Op Amp Properties

R_i = infinity. Therefore I_i = 0 at either the n or p input.
R_o = 0.
A = infinity. Therefore, v_d = 0, (v_p - v_n) = 0, and v_p = v_n.

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, R_i, and R_o, 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.

Before going on, click on this tutorial.

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.

Three-Phase Power Sources

Three-phase power is produced by specially designed three-phase electric generators. Basically, 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 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 V_φ. Each winding also has a phase current, I_φ. The phase voltage and phase current will produce a line voltage, V_L, and a line current, I_L.

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 V_φ is exactly the same as every other, keeping in mind that these variables represent rms magnitudes. The same is true for I_φ, V_L, and I_L.

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:
V_L = V_φ

The relationship between I_L and I_φ 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.

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.
I_L = I_φ

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.

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

Three-Phase Loads

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, V_φ, for the load. We can reuse the equations above. 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.
P_T = 3P_φ
P_T = √3V_LI_Lcosθ where cosθ is the power factor.
P_φ = I_φ²R_φ

Naturally, the voltages and currents in the above equations are rms values, not peak values.

Before trying the Homework Problems, click on this tutorial.

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.

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

Each V_s in the above figure has an rms value 220 V. Each one is 120° out of phase with the other two. R = 90 Ω, Z_L = 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.)