Transistor Amplifier

The amplifier is an analog circuit (Figure 2–15), and the calculations, plus the points that must be considered during the design, are more complicated than for a saturated circuit. This extra complication leads people to say that analog design is harder than digital design (the saturated transistor is digital i.e.; on or off). Analog design is harder than digital design because the designer must account for all states in analog, whereas in digital only two states must be accounted for. The specifications for the amplifier are an ac voltage gain of four and a peak-to-peak signal swing of 4 volts.

Figure 2–15. Transistor Amplifier

IC is selected as 10 mA because the transistor has a current gain (β) of 100 at that point. The collector voltage is arbitrarily set at 8 V; when the collector voltage swings positive 2 V (from 8 V to 10 V) there is still enough voltage dropped across RC to keep the transistor on. Set the collector-emitter voltage at 4 V; when the collector voltage swings negative 2 V (from 8 V to 6 V) the transistor still has 2 V across it, so it stays linear. This sets the emitter voltage (VE) at 4 V.

Use Thevenin’s equivalent circuit to calculate R1 and R2 as shown in Figure 2–16.

Figure 2–16. Thevenin Equivalent of the Base Circuit

We want the base voltage to be 4.6 V because the emitter voltage is then 4 V. Assume a voltage drop of 0.4 V across RTH, so Equation 2–35 can be written. The drop across RTH may not be exactly 0.4 V because of beta variations, but a few hundred mV does not matter is this design. Now, calculate the ratio of R1 and R2 using the voltage divider rule (the load current has been accounted for).

R1 is almost equal to R2, thus selecting R1 as twice the Thevenin resistance yields approximately 4 K as shown in Equation 2–35. Hence, R1 = 11.2 k and R2 = 8 k. The ac gain is approximately RC/RE1 because CE shorts out RE2 at high frequencies, so we can write Equation 2–38.

The capacitor selection depends on the frequency response required for the amplifier, but 10 µF for CIN and 1000 µF for CE suffice for a starting point.