A transformer-coupled class A amplifier has a transformer with a 3:1 turns ratio and values of RL = 100 Ω, VCEQ = 6 V, and ICQ = 8 mA. What is the value of vceoff) for the circuit?
A) 12 V
B) 6 V
C) 13.2 V
D) Cannot be determined from the information given.

Question

Quizplus · Accepted Answer

The value of $V_{CE(off)}$ can be calculated using the given parameters and understanding the operation of a transformer-coupled class A amplifier. The turns ratio of the transformer affects the load seen by the transistor. The effective load ($R_{L(eff)}$) seen by the transistor is the actual load ($R_L$) multiplied by the square of the turns ratio ($n^2$), where $n$ is the turns ratio. Here, $n = 3$, so $R_{L(eff)} = R_L \times n^2 = 100 \times 3^2 = 900 \Omega$.The maximum swing in collector current ($I_C$) occurs from $0$ to twice the quiescent current ($2I_{CQ}$), so the peak-to-peak change in $I_C$ is $2I_{CQ} = 2 \times 8mA = 16mA$.The peak-to-peak voltage swing across the effective load is $V_{pp} = I_{C(pp)} \times R_{L(eff)} = 16mA \times 900\Omega = 14.4V$.The $V_{CE(off)}$ is the voltage at which the transistor is cut off, which occurs at the peak of the positive half-cycle of the output. This is $V_{CEQ} + \frac{V_{pp}}{2}$, where $V_{CEQ} = 6V$ and $\frac{V_{pp}}{2} = \frac{14.4V}{2} = 7.2V$.Thus, $V_{CE(off)} = V_{CEQ} + \frac{V_{pp}}{2} = 6V + 7.2V = 13.2V$.

A Transformer-Coupled Class a Amplifier Has a Transformer with a 3:1