A.C. Amplifier

Build an a.c. voltage amplifier with a voltage gain of 10. A gain of 10 means that the ratio R_C/R_E should be equal to 10, as it was in the previous experiment. When no a.c. input voltage is applied, the transistor voltages and currents should be near the center of the operating range that you have determined earlier. Reasonable values would be V_C ~ 5 volts, and I_C ~ 10 mA, allowing for both a positive and a negative swing of the time­varying output signal. Given a supply voltage of 15 V, a collector voltage of V_C = 7.5 V at a current of I_C = 10 mA implies a 5 V drop across R_C. R_C should, therefore, be about 500 W. From R_C = 500 W, follows R_E = 50 W. The resistors used in the previous experiment, 470 and 47 W, are close enough. Using these values build the circuit of Fig. 11. The 47 W resistor is yellow­purple­black and the 470 W resistor is yellow­purple­brown. Try to be neat to avoid wiring errors and accidental short circuits between adjacent wires. After you have turned down both power supplies connect them to your circuit. Check the circuit carefully - the layout of your circuit should look something like this - then turn on the power. Set the collector supply to 15 V, then raise the bias voltage until the bias current is ~ 50 mA. (With the 47 W and the 470 W resistor in series with the transistor it is safe to exceed 10 V on the collector supply.)

When the d.c. voltages and currents have been set you are ready to apply an a.c. signal: Connect one channel of the oscilloscope to the output of the oscillator and set the oscillator output to an amplitude of ~ 0.2 V. Be sure the dc offset control on the oscillator is set to zero. After verifying this value on the scope, connect the oscillator to the amplifier input. Connect the other channel of the oscilloscope to the Output terminal, i.e. to the collector, so that you can simultaneously view both the input and the output waveforms, using the ALT mode of the scope. As you probably know, a d.c. current cannot flow through a capacitor, but an a.c. current can. By using the 0.1 mF capacitor, as shown in Fig. 11, you can introduce an a.c. signal without altering any of the d.c. levels. The amplitude of the output voltage should be 10 times larger than that of the input voltage but cannot be exceed about 6 volts for any input amplitude. Why?

Once you have both the input and the output waveforms on the screen, compare their amplitudes using the calibrated sensitivity setting of the scope. (Be sure the variable sensitivity knobs on the scope inputs are in the fully clockwise CAL position.)

• How big is the output to input ratio that you actually measure?
• Compare the phase of the output waveform relative to the phase of the input waveform. Is it what you expected?
• If you do not know what to expect click here

Now take another look at Fig. 10. As I_B gets bigger, so does I_C. However, a larger value of I_C means a smaller value of V_CE due to voltage drop across the series resistors, (47 W and 470 W). Indeed, I_C cannot possibly get bigger than I_CMAX = 15 V/(R_C+R_E) = 29 ma. Mark this value on your graph. Conversely, a small value of I_B implies a small value of I_C. A smaller I_C in turn means a large value for V_CE. The obvious limit is (V_CE)_max = 15 V, the voltage of the collector power supply. The intermediate values of I_C and V_EC one finds to lie along a straight line between the extremes. This straight line is the load line, shown in Fig. 10, and you should draw it on your graph.

The supply voltage V must equal the sum of the voltage V_CE across the transistor and the voltage drop across the two resistors R_C and R_E.

V = V_CE(I_C) + I_C(R_C + R_B)

We cannot readily solve that equation because we have only the measured curves for the characteristic V_C(I_C), not a mathematical expression. If we write the same equation in the form

V - I_C(R_C + R_B) = V_CE(I_C)

we have on the left the mathematical expression for the load line as a function of the collector current, and on the right the expression for the measured collector voltage as a function of the collector current. The equal sign means that both expressions must be true simultaneously, even if the second one is not known to us in a mathematical form. We conclude that for any given base current the collector current must be given by the intersection of the characteristic for that base current and the load line. In other words the load line lets us solve graphically the equation that we could not solve mathematically.

Following are some suggestions for further exploration. Be systematic and keep careful notes of your observations and measurements. Take enough time to do a good job and don't feel that you have to do everything. Be sure to leave the right power supply set at 15 V.