Difference Amplifier

The difference amplifier circuit provides an output voltage that is proportional to the difference of its two inputs. This circuit configuration is shown on Picture 1. If we apply the KCL at the inverting terminal of the OA, we have:

i1 = (V1 - V-)/R1 = (V- - Vo)/R2 = i2

So, for the output voltage we can write:

Vo = (R1 + R2)/R1 * V- - R2/R1 * V1

Also, the input terminals voltages are equal to each other:

V- = V+ = R4/(R3 + R4) * V2

Picture 1: Difference Amplifier


Finally, for the output voltage we can write:

Vo = ((R1 + R2)*R4)/(R1*(R3 + R4)) * V2 - R2/R1 * V1

This relation is for general purpose. In case when we choose R1 = R3 and R2 = R4 (as it is in our circuit configuration example shown on Picture 1), for the output voltage we have now:

Vo = R2/R1 * (V2 - V1)

Picture 2: Output voltage of the Difference Amplifier (inputs at same frequency)

For the input signals at equal frequency, the output voltage waveform is shown on Picture 2. Here, the input signals have frequency of f1 = f2 = 500 Hz, and their amplitudes are V1 = +/- 100 mV and V2 = +/- 300 mV. As we can see from the plot the maximum of the output voltage is 1 V. This value can be also confirmed and with the last relation for the Vo:

Vo = R2/R1 * (V2 - V1) = 5K/1K * (V2 - V1) = 5 * (0.3 - 0.1) = 1 [V]

Picture 3: Output voltage of the Difference Amplifier (inputs at different frequency)

The output voltage waveform shown on Picture 3, is result of input signals at different frequencies. In this case, the amplitudes of the input signals remains the same as before V1 = +/- 100 mV and V2 = +/- 300 mV, but the frequencies now are f1 = 500 Hz and f2 = 1000 Hz. As we can see from the output voltage plot, the maximum values achieved are about 1.8 [V].