# ECE 252 Introduction to Electrical Engineering

## Lesson 13A. Operational Amplifiers

#### Operational Amplifier Basics

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 Kirchhoff's Current Law at the vn node:

The current out of vn through the 900 Ω resistor is (vn - 1)/900.
The current out of vn through the 9000 Ω resistor is (vn - 2000vd)/(9000 + 1000).
The current out of vn through the 18000 Ω resistor is vn/18000.

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 - vn; since the positive input is grounded, vp = 0 and vd = -vn.  The third term represents the current through the 18 kΩ resistor.

KCL at vn is therefore:

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

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.

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