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 Kirchhoff's Current Law at the v_{n} node:

The current out of v_{n} through the 900 Ω resistor is (v_{n}
- 1)/900.

The current out of v_{n} through the 9000 Ω resistor is (v_{n} -
2000v_{d})/(9000 + 1000).

The current out of v_{n} through the 18000 Ω resistor is v_{n}/18000.

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_{n}; since the positive input is grounded, v_{p} =
0 and v_{d} = -v_{n}. The third term represents
the current through the 18 kΩ resistor.

KCL at v_{n} is therefore:

(v_{n} - 1)/900 + (v_{n} + 2000v_{n})/(9000 + 1000) +
v_{n}/18000 = 0

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:

I_{900} = (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_{s}R_{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_{s}R_{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_{s}R_{f}/R_{s}.

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.

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 v _{p}**. This is the voltage at the
positive input. To make this calculation, assume
that no current flows into the positive input.

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 v_{p} = 1 V.

2) v_{n} = v_{p} = 1 V.

3) KCL at v_{n} is:

-I_{S} - I_{F} = 0

To find the I values, use this form of Ohm's Law:

**I _{XY} = (V_{X} - V_{Y})/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:

I

I

Plugging these into KCL gives:

-2/10 k - (v

Multiplying by 80 k:

-16 - v

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.

At this time you should complete
Tutorial 13A
on **operational amplifiers**.