Test: Power Electronics - 2 - Electrical Engineering (EE) MCQ

Latching current is the minimum value of anode current which it must attain during turn – ON process to maintain conduction when gate signal is removed.

Holding current is the minimum value of anode current below which it must fall for turning – OFF the thyristor.

1. Anode must be positive with respect to cathode.

2.  Gate pulse must be more than the turn – on time of an SCR. This will ensure that anode current exceeds the latching current before gate signal is removed.

3.  A finger voltage is that voltage below which an SCR cannot be turned on with a gate signal.

4.  Magnitude of gate current must be more than the minimum gate current required to turn – on a thyristor.

P.I.V = V_m × Voltage safety factor

700 = V_m × 2

V_m = 350 V

To turn off the SCR anode current should go less than holding current

⇒ V = 4 × 10^-3 × 4 × 10³

⇒ V = 16 V

Form the options, V = 15 V is correct. for the remaining options, holding current is less than anode current.

Derating factor = 1 – string efficiency

Let I and I_sb be the one – cycle and sub – cycle surge current ratings of the SCR respectively

t = 10 ms, T = 20 ms

⇒ I = 2121.32 A

I²t rating = I² × 1/2f

= 45,000 Amp².sec

V_gI_g = 1.2 W

Maximum fault current occurs when source voltage is at its peak = 240 √2 V

When terminal ‘A’ gets short – circuited to ground the resistance offered to the source

= 10+(20||20) = 20 Ω

⇒ t_c = 208.33 ms

Dynamic resistance of SCR with 200 A

= 2/200 = 0.01Ω = 10 mΩ
Dynamic resistance of SCR with 300 A

= 1.2/300 = 0.004Ω = 4 mΩ
Let Rs be the resistance inserted in series with each SCR.

Current shared by SCR₁ = 500 
Current shared by SCR₂ = 500  

⇒ 12 + 3 R_s = 20 + 2Rs

⇒ R_s = 8 m Ω

Here, n = 4