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
10 \mathrm{~V} | |
6 \mathrm{~V} | |
12 \mathrm{~V} | |
20 \mathrm{~V} |
Question 1 Explanation:
Output Voltage Waveform :
\therefore Average voltage, \mathrm{V}_{0}=\frac{1}{10}[20 \times 3]=6 \mathrm{~V}
\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)
124 | |
148 | |
165 | |
182 |
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}
\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.
15 | |
18 | |
30 | |
24 |
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).
(Round off to 2 decimal places).
1.25 | |
1.6 | |
2.2 | |
3.45 |
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 ________.
10 | |
5 | |
15 | |
20 |
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
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.