Topic: Analog Circuits
Subsystem A requires two voltages $V_{1}$ and $V_{2}$ with equal power ratings. Power conversion scheme adapted for the power supply design is shown in the following figure. What is the overall power conversion efficiency?
(The primary converter with efficiency $\eta_1$ supplies $V_1$ directly to Subsystem A and also feeds a secondary converter with efficiency $\eta_2$ which supplies $V_2$ to Subsystem A).
- A. $\frac{\eta_1\cdot\eta_2}{\eta_1+\eta_2}$
- B. $\eta_1*\eta_2$
- C. $\frac{2\eta_1\cdot\eta_2}{1+\eta_2}$
- D. $\frac{2\eta_1\cdot\eta_2}{1+\eta_1}$
Answer: C. $\frac{2\eta_1\cdot\eta_2}{1+\eta_2}$
Explanation: Let Subsystem A require power $P$ for $V_1$ and power $P$ for $V_2$. Total output power = $2P$.
The secondary converter delivers $P$ with efficiency $\eta_2$, so it requires an input power of $P/\eta_2$.
The primary converter delivers $P$ directly and $P/\eta_2$ to the secondary converter, so its total output power is $P + P/\eta_2$.
The primary converter has efficiency $\eta_1$, so its input power is $(P + P/\eta_2) / \eta_1 = \frac{P(1+1/\eta_2)}{\eta_1}$.
Overall efficiency = $\frac{\text{Total Output Power}}{\text{Total Input Power}} = \frac{2P}{\frac{P(1+\eta_2)}{\eta_1\eta_2}} = \frac{2\eta_1\eta_2}{1+\eta_2}$.