# Choppers

 Question 1
The chopper circuit shown in figure (i) feeds power to a 5 A DC constant current source. The switching frequency of the chopper is $100 \mathrm{kHz}$. All the components can be assumed to be ideal. The gate signals of switches $S_{1}$ and $S_{2}$ are shown in figure (ii). Average voltage across the $5 \mathrm{~A}$ current source is A $10 \mathrm{~V}$ B $6 \mathrm{~V}$ C $12 \mathrm{~V}$ D $20 \mathrm{~V}$
GATE EE 2023   Power Electronics
Question 1 Explanation:
Output Voltage Waveform : $\therefore$ Average voltage, $\mathrm{V}_{0}=\frac{1}{10}[20 \times 3]=6 \mathrm{~V}$
 Question 2
The steady state current flowing through the inductor of a DC-DC buck boost converter is given in the figure below. If the peak-to-peak ripple in the output voltage of the converter is 1 V, then the value of the output capacitor, in $\mu F$, is _______________. (round off to nearest integer) A 124 B 148 C 165 D 182
GATE EE 2022   Power Electronics
Question 2 Explanation:
We have for buck boost converter,
\begin{aligned} \Delta V_c&= \frac{I_o}{fC}\\ where, \; \alpha &=\frac{T_{ON}}{T}\frac{20}{20+30}=0.4 \\ I_{L(Avg)}&=\frac{16+12}{2}=14A \end{aligned}
For buck-boost converter,
\begin{aligned} \frac{I_{o(Avg)}}{I_{L(Avg)}}&=\frac{1}{1-\alpha }\\ \Rightarrow I_L&=I_o(1-\alpha )\\ &=14(1-0.4)=8.4A\\ Given: \; \Delta V_c&=1V\\ Therefore, \; 1&=\frac{0.4 \times 8.4 \times 50 \times 1^{-6}}{C}\\ C&=168\mu F \end{aligned}

 Question 3
Consider the buck-boost converter shown. Switch Q is operating at $\text{25 kHz}$ and 0.75 duty-cycle. Assume diode and switch to be ideal. Under steady-state condition, the average current flowing through the indicator is ____________A. A 15 B 18 C 30 D 24
GATE EE 2021   Power Electronics
Question 3 Explanation: $\alpha=0.75, \quad f=25 \mathrm{kHz}$
Assume continous conduction:
\begin{aligned} V_{0}&=\frac{\alpha V_{s}}{1-\alpha}=\frac{0.75 \times 20}{1-0.75} \\ V_{0}&=60 \mathrm{~V}\\ I_{0} &=\frac{V_{0}}{R}=\frac{60}{10}=6 \mathrm{~A} \\ I_{L} &=\frac{I_{0}}{1-\alpha}=\frac{6}{1-0.75}=24 \mathrm{~A} \\ \Delta I_{L} &=\frac{\alpha V_{s}}{f_{L}}=\frac{0.75 \times 60}{25 \times 10^{3} \times\left(1 \times 10^{-3}\right)}=1.8 \mathrm{~A} \\ I_{L \min } &=I_{L}-\frac{\Delta I_{L}}{2}=24-\frac{1.8}{2}=24-0.9 \\ \left(I_{L \min }=23.1 \mathrm{~A}\right) &>0 \end{aligned}
$\therefore$ Continous conduction assumption is correct.
$I_{L}=24 \mathrm{~A}$
 Question 4
Consider the boost converter shown. Switch Q is operating at $\text{25 kHz}$ with a duty cycle of 0.6. Assume the diode and switch to be ideal. Under steady-state condition, the average resistance $R_{\text{in}}$ as seen by the source is __________ $\Omega$.
(Round off to 2 decimal places). A 1.25 B 1.6 C 2.2 D 3.45
GATE EE 2021   Power Electronics
Question 4 Explanation:

Checking for continuous conduction mode
\begin{aligned} \Delta I_{L} &=\frac{\alpha V_{S}}{f L}=\frac{0.6 \times 15}{25 \times 10^{3} \times 1 \times 10^{-3}}=0.36 \mathrm{~A} \\ \frac{\Delta I_{L}}{2} &=0.18 \mathrm{~A} \\ I_{L, \min } &=I_{L}-\frac{\Delta I_{L}}{2}=I_{S}-\frac{\Delta I_{L}}{2} \\ &=(9.375-0.18)=9.195>0 \end{aligned}
As it is continuous conduction
\begin{aligned} V_{0}&=\frac{V_{S}}{1-\alpha}=\frac{15}{1-0.6}=37.5 \mathrm{~V} \\ I_{0}&=\frac{V_{0}}{R}=\frac{37.5}{10}=3.75 \mathrm{~V} \\ \frac{V_{0}}{V_{S}}&=\frac{I_{S}}{I_{0}}=\frac{1}{1-\alpha} \\ I_{S}&=\frac{I_{0}}{1-\alpha}=\frac{3.75}{1-0.6}=9.375 \mathrm{~A} \\ R_{\text {in }}&=\frac{V_{S}}{I_{S}}=\frac{15}{9.375}=1.6 \Omega \end{aligned}
 Question 5
In the dc-dc converter circuit shown, switch Q is switched at a frequency of 10 kHz with a duty ratio of 0.6. All components of the circuit are ideal, and the initial current in the inductor is zero. Energy stored in the inductor in mJ (rounded off to 2 decimal places) at the end of 10 complete switching cycles is ________. A 10 B 5 C 15 D 20
GATE EE 2020   Power Electronics
Question 5 Explanation:
Buck boost converter,
$D=0.6\rightarrow \text{store energy}$
$D=\frac{T_{ON}}{T}=0.6$
$T_{ON}=0.6\: T\rightarrow \text{store energy}$
$T_{OFF}=0.4\: T\rightarrow \text{releasing energy}$ For one cycle: Rise in current for $0.2T$
For 10 cycles: Find rise in current $(0.2T) \times 10 = 2T$
$i=\frac{50}{L}t$
$i=\frac{50}{L}(2T)=\frac{50\times 2}{LP}=\frac{100}{10\cdot 10^{-3}\times 10\cdot 10^{3}}=1\, A$
$\therefore \, \text{Energy stored}=\frac{1}{2}Li^{2}=\frac{1}{2}\times (10\cdot 10^{-3})(1)^{2}=5\: mJ$

There are 5 questions to complete.