Topic: Power Electronics
A high speed digital Subsystem requires three voltages $V_1$, $V_2$ and $V_3$ with 1:2:1 power ratings respectively. The power supply is designed with a distributed power conversion scheme. The primary converter has efficiency $\eta_1$ and feeds $V_1$ directly, while also feeding two secondary converters (with efficiencies $\eta_2$ and $\eta_3$) that supply $V_2$ and $V_3$. What is the overall power conversion efficiency?
- A. $\frac{4 \times \eta_1 \times \eta_2 \times \eta_3}{\eta_2 \eta_3 + 2 \times \eta_3 + \eta_2}$
- B. $\frac{\eta_1 \times \eta_2 \times \eta_3}{\eta_2 \eta_3 + 2 \times \eta_3 + \eta_2}$
- C. $\frac{4 \times \eta_1 \times \eta_2 \times \eta_3}{\eta_2 \eta_3 + \eta_2 + \eta_3}$
- D. $\frac{2 \times \eta_1 \times \eta_2 \times \eta_3}{\eta_2 \eta_3 + \eta_2 + \eta_3}$
Answer: A. $\frac{4 \times \eta_1 \times \eta_2 \times \eta_3}{\eta_2 \eta_3 + 2 \times \eta_3 + \eta_2}$
Explanation: Let the output power values be $P_1 = P$, $P_2 = 2P$, and $P_3 = P$. The total useful output power is $4P$. The input power to the primary converter is $P_{in} = \frac{P_1 + P_2/\eta_2 + P_3/\eta_3}{\eta_1} = \frac{P(1 + 2/\eta_2 + 1/\eta_3)}{\eta_1}$. The overall efficiency is $\eta_{overall} = \frac{\text{Total Output Power}}{P_{in}} = \frac{4P}{\frac{P(1 + 2/\eta_2 + 1/\eta_3)}{\eta_1}}$. Simplifying this expression gives $\frac{4\eta_1\eta_2\eta_3}{\eta_2\eta_3 + 2\eta_3 + \eta_2}$.