# Choppers

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 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 $(0.2T) \times 10 = 2T$
$i=\frac{50}{L}t$
$i=\frac{50}{L}(2T)=\frac{50\times 2}{LP}=\frac{100}{10\cdot 10^{-3}\times 10\cdot 10^{3}}=1\, A$
$\therefore \, \text{Energy stored}=\frac{1}{2}Li^{2}=\frac{1}{2}\times (10\cdot 10^{-3})(1)^{2}=5\: mJ$
 Question 5
In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48 V. The input DC voltage is 24 V. The output power is 120 W. The switching frequency is 50 kHz. Assume ideal components and a very large output filter capacitor. The converter operates at the boundary between continuous and discontinuous conduction modes. The value of the boost inductor (in $\mu H$) is _______.
 A 12 B 24 C 28 D 14
GATE EE 2019   Power Electronics
Question 5 Explanation:
\begin{aligned} P_0&=120W,\\ V_s&=24V\\ V_0&=48V\\ V_0&=\frac{V_s}{(1-D)}\\ 1-D&=\frac{24}{48}\\ D&=0.5\\ P_0&=V_0I_0=120\\ I_0&=\frac{120}{48}=2.54A\\ V_sI_s&=V_0I_0\\ I_s&=\frac{120}{24}=5A \end{aligned}
At boundary of continuous and discontinuous,
\begin{aligned} I_L&=I_s=\frac{\Delta I_{L}}{2}\\ \Delta I_L&=\frac{DV_s}{fL_c}=2 \times 5\\ L_c&=\frac{0.5 \times 24}{50 \times 10^3 \times 10}=24\mu H \end{aligned}
 Question 6
A DC-DC buck converter operates in continuous conduction mode. It has 48 V input voltage, and it feeds a resistive load of 24 $\Omega$. The switching frequency of the converter is 250 Hz. If switch-on duration is 1 ms, the load power is
 A 6 W B 12 W C 24 W D 48 W
GATE EE 2019   Power Electronics
Question 6 Explanation:
Geven that,
\begin{aligned} & \text{Supply voltage, } V_s=48V\\ & \text{Load resistance, }R_L=24\Omega \\ & \text{Switch frequency, }f_s=250Hz\\ &\text{on time of switch } (T_{ON})=1ms\\ & \text{Time period, } T=\frac{1}{f_s}=\frac{1}{250}=4ms\\ & \text{Duty cycle, } \alpha =\frac{T_{ON}}{T}=\frac{1ms}{4ms}=0.25\\ & \text{Load power }=\frac{(V_0)^2}{R}=\frac{(\alpha V_s)^2}{24}\\ &=\frac{(0.25 \times 48)^2}{24}=6 \text{ Watt} \end{aligned}
 Question 7
A dc to dc converter shown in the figure is charging a battery bank, B2 whose voltage is constant at 150 V. B1 is another battery bank whose voltage is constant at 50 V. The value of the inductor, L is 5 mH and the ideal switch, S is operated with a switching frequency of 5 kHz with a duty ratio of 0.4. Once the circuit has attained steady state and assuming the diode D to be ideal, the power transferred from B1 to B2 (in Watt) is ___________ (up to 2 decimal places). . A 8 B 10 C 12 D 16
GATE EE 2018   Power Electronics
Question 7 Explanation: During $T_{on}$ the circuit behaves as,
\begin{aligned} V_s&=L\frac{di}{dt} \\ di&=\frac{V_s}{L}dt \end{aligned}
Integrating on both sides,
\begin{aligned} I_P&=\frac{V_s}{L}T_{on}\\ T_{on}&=\alpha T=\frac{\alpha }{f}=80 \times 10^{-6}s\\ I_P&=\frac{50}{5 \times 10^{-3}}\times (80 \times 10^{-6})\\ &=0.8A \end{aligned} During $T_{off}$, it is $T_{on}\leq t\leq \beta T$
\begin{aligned} \text{KVL, } V_L&=V_s-V_0 \\ (V_L)_{avg}&=0 \\ V_sT_{on}+&(V_s-V_0)(\beta T-T_{on}) =0 \\ V_s\beta T &=V_0(\beta T-T_{on}) \\ \frac{V_0}{V_s}&=\frac{\beta }{\beta -\alpha } \\ \beta &=0.6\\ &\text{Fromt the graph of } i_L, \\ I_{L(avg)}&=\frac{\frac{1}{2} \times b \times h}{T}\\ &=\frac{\frac{1}{2} \times \beta T \times I_P}{T}\\ &=\frac{1}{2} \times 0.6 \times 0.8=0.24A\\ &\text{Power transferred to } B_2 \text{ is,}\\ P&=V_s \times (I_L)_{avg}\\ \text{ where } I_s&=I_L\\ P&=50 \times 0.24\\ &=12W \end{aligned}
 Question 8
The figure shows two buck converters connected in parallel. The common input dc voltage for the converters has a value of 100 V. The converters have inductors of identical value. The load resistance is 1 $\Omega$. The capacitor voltage has negligible ripple. Both converters operate in the continuous conduction mode. The switching frequency is 1 kHz, and the switch control signals are as shown. The circuit operates in the steady state. Assuming that the converters share the load equally, the average value of $i_{S1}$, the current of switch S1 (in Ampere), is _____ (up to 2 decimal places). A 5.55 B 6.85 C 7.25 D 12.5
GATE EE 2018   Power Electronics
Question 8 Explanation:
Hence it is buck converter,
\begin{aligned} V_0&=\alpha V_s\\ V_0&=0.5 \times 100=50V\\ I_0&=\frac{V_0}{R}=\frac{50}{1}=50A\\ V_sI_s&=V_0I_0\\ I_s&=\frac{V_0I_0}{V_s}=\frac{50\times 50}{100}=25A\\ I_{s1}&=I_{s2}=\frac{I_s}{2}=12.5A \end{aligned}
 Question 9
In the circuit shown all elements are ideal and the switch S is operated at 10 kHz and 60% duty ratio. The capacitor is large enough so that the ripple across it is negligible and at steady state acquires a voltage as shown. The peak current in amperes drawn from the 50 V DC source is ________. (Give the answer up to one decimal place.) A 37 B 40 C 22 D 56
GATE EE 2017-SET-2   Power Electronics
Question 9 Explanation:
Buckboost converter,
\begin{aligned} V_0&=\frac{\alpha V_s}{1-\alpha }\\ V_s&=50V\\ \alpha &=0.6\\ V_0&=75V\\ \frac{V_0}{V_s}&=\frac{I_s}{I_0}=\frac{\alpha }{1-\alpha }\\ &=\frac{0.6}{1-0.6}=\frac{3}{2}\\ I_0&=\frac{V_0}{R}=\frac{75}{5}=15A\\ I_s&=\frac{\alpha }{1-\alpha }I_0=\frac{3}{2} \times 15=22.5A \end{aligned} Since capacitor is very large $i_c=0$ at steady state
\begin{aligned} (i_L)_{avg}&=(I_s)_{avg}+(i_0)_{avg}\\ I_L&=I_s+I_0\\ I_L&=22.5+15=37.5A\\ \Delta I_L&=\frac{\alpha V_s}{fL}\\ &=\frac{0.6\times 50}{10 \times 10^3 \times (0.6 \times 10^{-3})}\\ &=5A\\ (i_L)_{peak}&=I_L+\frac{\Delta I_L}{2}\\ &=37.5+\frac{5}{2}=40A \end{aligned}
Peak value of current drawn from source $=(i_L)_{peak}=40A$
 Question 10
The input voltage $V_{DC}$ of the buck-boost converter shown below varies from 32 V to 72 V. Assume that all components are ideal, inductor current is continuous, and output voltage is ripple free. The range of duty ratio D of the converter for which the magnitude of the steady state output voltage remains constant at 48 V is A $\frac{2}{5}\leq D \leq \frac{3}{5}$ B $\frac{2}{3}\leq D \leq \frac{3}{4}$ C $0 \leq D \leq 1$ D $\frac{1}{3}\leq D \leq \frac{2}{3}$
GATE EE 2017-SET-1   Power Electronics
Question 10 Explanation:
\begin{aligned} V_0&=\frac{\alpha V_s}{1-\alpha }\\ \text{when, } V_s&=32V \text{ and }V_0=48V\\ \alpha &=\frac{3}{5}\\ \text{when, }V_s&=72V \text{ and }V_0=48V\\ \alpha &=\frac{2}{5}\\ \frac{2}{5}\leq &\alpha \leq \frac{3}{5}\\ \therefore \; \text{This answer is }& \frac{2}{5}\leq D \leq \frac{3}{5} \end{aligned}
There are 10 questions to complete.