Step-down chopper with switch drop: efficiency

When the chopper switch has an ON-state drop $V_{ch}$, the load gets $(V_s - V_{ch})$ during $T_{on}$, and the switch dissipates power. Efficiency is the ratio of load power to input power.

Given: $R = 20\ \Omega$, $V_s = 220$ V, $V_{ch} = 1$ V, $f = 2$ kHz, $k = 0.8$.

(i) Average output voltage

$$\begin{aligned} V_a &= k\,(V_s - V_{ch}) = 0.8\times(220-1) = 0.8\times219 \\ &= 175.2\ \text{V} \end{aligned}$$

(The frequency 2 kHz only fixes $T = 0.5$ ms and $T_{on} = 0.4$ ms; it does not change average values for a resistive load.)

(ii) Efficiency

Output power (load voltage is $219$ V for a fraction $k$ of the time):

$$\begin{aligned} P_o &= \frac{1}{T}\int_0^{kT}\frac{(V_s - V_{ch})^2}{R}dt = k\,\frac{(V_s-V_{ch})^2}{R} \\ &= 0.8\times\frac{219^2}{20} = 1918.44\ \text{W} \end{aligned}$$

Input power (source supplies current $(V_s - V_{ch})/R$ during $kT$):

$$\begin{aligned} P_i &= \frac{1}{T}\int_0^{kT}V_s\,\frac{V_s - V_{ch}}{R}dt = k\,\frac{V_s(V_s-V_{ch})}{R} \\ &= 0.8\times\frac{220\times219}{20} = 1927.2\ \text{W} \end{aligned}$$ $$\begin{aligned} \eta &= \frac{P_o}{P_i} = \frac{V_s - V_{ch}}{V_s} = \frac{1918.44}{1927.2} = \frac{219}{220} \\ &= 0.99545 = 99.55\% \end{aligned}$$

Switch loss $= 1927.2 - 1918.44 = 8.76$ W.

Answer: $V_a = 175.2$ V; efficiency $\eta = 99.55\%$.