Experiment: Active Loads

The circuit in Figure 6.2 is a differential amplifier. The only difference between this circuit and the one you designed is that the R_C's have been replace by an active load. The active load is actually a current source that provides a very large effective load effective resistance for R_C.

1.

Build the circuit. By adjusting the 1K potentiometer, set the DC level between Q2 and Q4 at approx. $\frac{V_{CC}}{2}$ volts. This eliminates offset voltages due to transistor mismatches.

2.

Using a very small input (approx. 1mV) determine the gain at 100Hz, 1KHz, 10KHz and 50kHz. (You may have to use an inverting op-amp circuit to get the required small signal.)

3.

What happens if the magnitude of the input is too big?

4.

Try to determine a method to measure the input resistance of the diff-amp.

5.

By AC coupling 50K, 100K and 200K loads to your circuit, try to measure its output resistance. (Use a $0.1\mu f$ cap. for AC coupling the load).

  

Figure 6.2: Differential Amplifier