Choppers Questions and Answers

Answer: C

With \(R = 0\) the load is a pure inductance, and when the switch S1 closes at \(t = 0\) the constant source voltage \(V_s\) is applied across \(L\) (the diode \(D_1\) conducts, and \(D_m\) is reverse biased). The current rises linearly.

\[ V_s = L\frac{di}{dt} \Rightarrow i(t) = \frac{V_s}{L}t \]

\[ \begin{aligned} i(100\ \mu\text{s}) \\ &= \frac{300}{300 \times 10^{-6}} \times 100 \times 10^{-6} \\ &= 100\ \text{A} \end{aligned} \]

The answer is 100 A, option C. The value of 50 A would be at \(t = 50\ \mu\)s and 200 A at \(t = 200\ \mu\)s.

Answer: D

The chopping frequency is 100 Hz.

The chopping period is the on time plus the off time, and the frequency is its reciprocal:

\[ \begin{aligned} T &= T_{ON} + T_{OFF} = 9 + 1 = 10\ \text{ms} \\ f &= \frac{1}{T} = \frac{1}{10 \times 10^{-3}} = 100\ \text{Hz} \end{aligned} \]

The duty cycle is 9/10 = 0.9, so a step down chopper gives an average output of 0.9 times the input. The option 1000 Hz is \(1/T_{OFF}\), and 200 Hz and 500 Hz do not follow from the given times.

Answer: B

The correct code is (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii).

A phase-controlled converter (controlled rectifier) turns AC into variable DC (iv). A chopper switches a DC supply on and off rapidly to give a different average DC voltage (i). An inverter produces AC from DC (ii). A dimmerstat, a variable autotransformer, changes the magnitude of an AC voltage, AC to AC (iii).

The other codes give the chopper an AC role or treat the dimmerstat as a DC device, which does not fit how these circuits work.

Answer: C

The correct matching is a-iii, b-iv, c-ii, d-i.

• A chopper converts fixed DC to variable DC, used for PMDC motor speed control (iii).
• A dimmerstat (variable autotransformer) gives a variable AC voltage (iv).
• An inverter produces variable-frequency AC for variable speed AC drives (ii).
• A phase-controlled rectifier converts AC to variable DC for variable speed DC drives (i).

The other codes give the dimmerstat or the rectifier the wrong application, for example treating a phase-controlled rectifier as an AC drive.

Answer: B

The control action is to increase the chopper frequency keeping the duty cycle constant.

The average load voltage of a chopper is \(D \times V_{s}\), so the average load current is fixed by the duty cycle D. The peak-to-peak ripple in an RL load current is inversely related to the switching frequency relative to the load time constant: at a higher frequency the current has less time to rise and fall in each period, so the ripple shrinks.

Changing the duty cycle, alone or together with frequency, changes the average voltage and hence the average current, which the question forbids. Decreasing the frequency would increase the ripple, the opposite of what is wanted.