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.