Power Electronics Test 3

Result

Power Electronics Test 3

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Weekly Quiz Competition

Question 1
1 / -0

For a SCR the maximum rms on-state current is ‘I’. If the SCR is used in a resistive circuit for a rectangular wave with conduction angle of 90°. Calculate the average and rms currents, respectively.

A
I/4, I/2

B
I/2, I/√2

C
I/4, I

D
I/4, I/√2

Solution

Conduction angle (α) = 90°
For rectangular wave,
The average current, \({I_{avg}} = \frac{\alpha }{{360}} \times I = \frac{{90}}{{360}}I = \frac{I}{4}\)
The rms value of current, \({I_{rms}} = \sqrt {\frac{\alpha }{{360}}} \times I = \sqrt {\frac{{90}}{{360}}} \times I = \frac{I}{2}\)
Question 2
1 / -0

An SCR is connected in series with a 50 Ω resistor and a 0.5 H inductor, across a 250 V, DC supply. If the width of the firing pulse is 50 μs, at what value of the latching current will the SCR be triggered? (Assume e^x ≈ 1 + x)

A
10 mA

B
15 mA

C
20 mA

D
25 mA

Solution

The current in the circuit should reach to the latching current during the firing pulse for the thyristor to switch ON.
The equation for current is,
\(i\left( t \right) = \frac{V}{R}\left( {1 - {e^{ - \frac{R}{L}t}}} \right)\)
Given that R = 50 Ω, L = 0.5 H
At t = 50 μs,
\(i\left( t \right) = \frac{{250}}{{50}}\left( {1 - {e^{ - 100 \times 50 \times {{10}^{ - 6}}}}} \right)\)
\( = 5\left( {1 - {e^{ - 5 \times {{10}^{ - 3}}}}} \right)\)
= 5 (1 – (1 – 5 × 10^-3)) = 25 mA
Therefore, the latching current = 25 mA
Question 3
1 / -0

An SCR has half cycle surge current rating of 2400 A for 50 Hz supply. It’s one cycle surge current and I²t rating respectively are

A
3394 A, 115200 Amp².sec

B
3394 A, 230 Amp².sec

C
1697 A, 28,800 Amp².sec

D
1697 A, 57600 Amp².sec

Solution

Let I and I_sb be the one – cycle and sub – cycle surge current ratings of the SCR respectively.
\({I^2}T = I_{Sb}^2.t\)
t = 10 ms T = 20 ms
\(\begin{array}{l}{I^2}\left( {20} \right) = {\left( {2400} \right)^2}\left( {10} \right)\\ \Rightarrow {I^2} = \frac{{{{\left( {2400} \right)}^2}\left( {10} \right)}}{{20}}\\ \Rightarrow I = \frac{{2400}}{{\sqrt 2 }} = 1200\sqrt 2 = 1697A\end{array}\)
I²t rating
\(\begin{array}{l} = {I^2} \times \frac{1}{{2f}}\\ = {\left( {\frac{{2400}}{{\sqrt 2 }}} \right)^2} \times \frac{1}{{100}} = 28,800Am{p^2}.sec\end{array}\)