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Chapter 8. Differential Amplifier

Typical example

In the picture above, you see a typical schematic of a differential amplifier.

The DC current source I1 provides a continuous 1mA current flow.

Transistors T1 and T2 have the same electrical characteristics, e.g. hFE1=hFE2=100. Therefore, the quiescent emitter currents of T1 and T2 are the same: IE1=IE2=0.5mA. The voltage across R1 (and R2) will be: VR1=10k∙0.5mA=5V. So the voltage at the OUT terminal equals V1-VR2=9V-5V=4V.

If we inject a 1uA current in IN1, IE1 will raise by 1uA∙100=0.1mA, so IE1=0.6mA and IE2 will be 0.4mA since the sum must be 1mA. VR2=0.4mA∙10k=4V and VOUT=9V-4V=5V. So we can write down the following (DC) formula for VOUT: VOUT=V1-VR2=9V-(0.5mA-IIN1∙hFE)∙R2=9V-5V+IIN1∙hFE∙R2=4V+IIN1∙hFE∙R2. When we omit the DC component we get this AC equation: vOUT=iIN1∙hFE∙R2.

Of course, we can also inject a current in IN2, resulting in: vOUT=-iIN2∙hFE∙R2 (Note the minus sign). Combine both equations and you get:


Hence the name differential amplifier.

All so called operational amplifiers are based on a differential amplifier. We'll take a closer look on operational amplifiers in the next chapter. First, we have to find out how to create a DC current source.