**Table of Contents**

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.
h_{FE1}=h_{FE2}=100. Therefore,
the quiescent emitter currents of T1 and T2 are the same:
I_{E1}=I_{E2}=0.5mA. The voltage
across R1 (and R2) will be: V_{R1}=10k∙0.5mA=5V. So
the voltage at the OUT terminal equals
V1-V_{R2}=9V-5V=4V.

If we inject a 1uA current in IN1, I_{E1} will
raise by 1uA∙100=0.1mA, so I_{E1}=0.6mA and
I_{E2} will be 0.4mA since the sum must be 1mA.
V_{R2}=0.4mA∙10k=4V and
V_{OUT}=9V-4V=5V. So we can write down the following
(DC) formula for V_{OUT}:
V_{OUT}=V1-V_{R2}=9V-(0.5mA-I_{IN1}∙h_{FE})∙R2=9V-5V+I_{IN1}∙h_{FE}∙R2=4V+I_{IN1}∙h_{FE}∙R2.
When we omit the DC component we get this AC equation:
v_{OUT}=i_{IN1}∙h_{FE}∙R2.

Of course, we can also inject a current in IN2, resulting in:
v_{OUT}=-i_{IN2}∙h_{FE}∙R2
(Note the minus sign). Combine both equations and you get:

**v _{OUT}=(i_{IN1}-i_{IN2})∙h_{FE}∙R2**

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.