-The transistor amplifier in Figure has the following values: hie = 2 kΩ, hfe = 120, Cbc = 5 pF, Cbe = 10 pF, RS = 300 Ω, Rth = 1.2 kΩ, rc = 4 kΩ, and Cin = 1200 pF. What is the voltage gain of the circuit?
A) 0.24
B) 2
C) 240
D) 480,000

Question

-The transistor amplifier in Figure  has the following values: hie = 2 kΩ, hfe = 120, Cbc = 5 pF, Cbe = 10 pF, RS = 300 Ω, Rth = 1.2 kΩ, rc = 4 kΩ, and Cin = 1200 pF. What is the voltage gain of the circuit?
A) 0.24 
B) 2 
C) 240 
D) 480,000

Quizplus · Accepted Answer

The voltage gain (Av) of a common-emitter transistor amplifier can be approximated by the formula Av = -hfe * (Rc || RL) / (hie + (1 + hfe) * Re), where Rc is the collector resistance, RL is the load resistance, hfe is the current gain, hie is the input impedance, and Re is the emitter resistance. In this scenario, since specific values for RL and Re are not provided, and assuming the load resistance is much larger than the collector resistance (a common assumption for simplification), the formula simplifies to Av ≈ -hfe * Rc / hie. Plugging in the given values: Av ≈ -120 * 4000 / 2000 = -240. The negative sign indicates a phase inversion which is typical for common-emitter amplifiers, but the magnitude of the gain is what's usually of interest, hence 240.

-The Transistor Amplifier in Figure Has the Following Values