For ease of explanation, in the circuit shown in figure 3-25 all the resistors have a value of 1 kilohm, but any value could be used as long as the above ratio is true. For a subtractor circuit, the values of R1 and R3 must also be equal, and therefore, the values of R2 and R4 must be equal. It is NOT necessary that the value of R1 equal the value of R2.

Using figure 3-25, assume that the input signals are:

The output signal should be:

To check this output, first compute the value of R2 plus R4:

With this value, compute the current through R2 (I_R2):

(indicating current flow from left to right)

Next, compute the voltage drop across R2 (E_R2):

Then compute the voltage at point B:

Since point B and point A will be at the same potential in an operational amplifier:

Now compute the voltage developed by R1 (E_R1):

Compute the current through R1 (I_R1):

Compute the voltage developed by R3 (E_R3):

Add this to the voltage at point A to compute the output voltage (E _out):

As you can see, the circuit shown in figure 3-25 functions as a subtractor. But just as an adder is only one kind of summing amplifier, a subtractor is only one kind of difference amplifier. A difference amplifier can amplify the difference between two signals. For example, with two inputs (E1 and E2) and a gain of five, a difference amplifier will produce an output signal which is:

The difference amplifier that will produce that output is shown in figure 3-26. Notice that this circuit is the same as the subtractor shown in figure 3-25 except for the values of R3 and R4. The gain of this difference amplifier is:

Figure 3-26. - Difference amplifier.

Then, for a difference amplifier:

With the same inputs that were used for the subtractor, (E1 = + 3 V; E2 = + 12 V) the output of the difference amplifier should be five times the output of the subtractor (E_out = + 45 V).

Following the same steps used for the subtractor:

First compute the value of R2 plus R4:

With this value, compute the current through R2 (I_R2):

Next, compute the voltage drop across R2 (E_R2):

Then, compute the voltage at point B:

Since point A and point B will be at the same potential in an operational amplifier:

Now compute the voltage developed by R1 (E_R1):

Compute the current through R1 (I_R1):

Compute the voltage developed by R3 (E_R3):

Add this voltage to the voltage at point A to compute the output voltage (E_out):

This was the output desired, so the circuit works as a difference amplifier.

Q.28 What is the difference between a summing amplifier and an adder circuit?
Q.29 Can a summing amplifier have more than two inputs?
Q.30 What is a scaling amplifier?