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Problem 7—In the cascade amplifier shown below, determine the values of R1, R2 and R such that the quiescent current through the transistors is 1mA and the collector voltages V_C1 = 3 V and V_C2 = 6 V. Take V_BE = 0.7 V and assume the transistor current gain factor h_FE to be high and base currents to be negligible.

The voltage across the 18 kohm resistor, V_B1 = V_BE + the drop across the 200 ohm emitter resistor, or V_B1 = 0.7 V + 200 ohms ý 1 mA = 0.9 V. Therefore, the current through the 18k resistor is V_B1 / 18k = 0.9 V / 18 kohms = 0.05 mA. This current flows through both R1 and R2 since the base currents are assumed to be negligible.

V_B2 is simply V_BE + V_C1 = 0.7 V + 3 V = 3.7 V.

The voltage across R2 is V_B2 ý V_B1 = 3.7 V ý 0.9 V = 2.8 V. Given that the current is 0.05 mA, then R2 must be 2.8 V / 0.05 mA = 56 kohm.

The voltage across R1 is 9 V ý V_B2 = 9 V ý 3.7 V = 5.3 V. Again, given that the current is 0.05 mA, then R1 must be 5.3 V / 0.05 mA = 106 kohm.

If the h_FE is 100, then each transistor requires a base current of about 1 mA / 100 = 10 ýA. The base voltages are the same as before, so it's simply a matter of recalculating R1 and R2 to account for the higher current through them.

The current through R2 is now 50 ýA + 10 ýA = 60 ýA, so now its value needs to be 2.8 V / 60 ýA = 47 kohm.

The current through R1 is now 60 ýA + 10 ýA = 70 ýA, so now its value needs to be 5.3 V / 70 ýA = 75 kohm.

Contributor: Naveen PN

03-02 — NEXT Q&A