Practice Questions: Parallel Operation of Thyristors - Electrical Engineering (EE) MCQ

Derating factor = 1 – string efficiency

Concept

Matched-pair SCR’s are generally available for parallel connection, but they are very expensive. With unmatched SCR’s equal current sharing is enforced by adding a low-value resistor or inductor in series with each SCR. Forced current sharing using equal-value resistors is shown in Figure 2. The basic requirement is to make current I₁ close to l₂, a maximum difference of 20% is accept across SCR₂, the value of R can be obtained from.

I₁R + V₁ = I₂R + V₂

R = (V₁ – V₂)/(I₂ – I₁)

Calculation:

Voltage across 0.05 ohm resistor  = 120 × 0.05 = 6 V
⇒ I₂ = 6/0.06 = 100 A
Current through T1 will be nearly equal to 100  A

Concept

Matched-pair SCR’s are generally available for parallel connection, but they are very expensive. With unmatched SCR’s equal current sharing is enforced by adding a low-value resistor or inductor in series with each SCR. Forced current sharing using equal-value resistors is shown in Figure 2. The basic requirement is to make current I₁ close to l₂, a maximum difference of 20% is accept across SCR₂, the value of R can be obtained from.

I₁R + V₁ = I₂R + V₂

R = (V₁ – V₂)/(I₂ – I₁)

Calculation:

In parallel anode to cathode voltage drops are same

1.8 + 300 R = 1.0 + 400 R

⇒ 100 R = 0.8

⇒ R = 0.008 = 8 mΩ

Concept

Matched-pair SCR’s are generally available for parallel connection, but they are very expensive. With unmatched SCR’s equal current sharing is enforced by adding a low-value resistor or inductor in series with each SCR. Forced current sharing using equal-value resistors is shown in Figure 2. The basic requirement is to make current I₁ close to l₂, a maximum difference of 20% is accept across SCR₂, the value of R can be obtained from.

I₁R + V₁ = I₂R + V₂

R = (V₁ – V₂)/(I₂ – I₁)

In parallel anode to cathode voltage drops are same

1.5 + 200 R = 1.2 + 300 R

⇒ R = 0.003 Ω

Dynamic resistance of 150 A. SCR1 = 1.2/150 = 8 mΩ

Dynamic resistance of 250 A SCR 2 = 0.8/250
= 3.2 mΩ

Let R_S be the resistance inserted in series with each SCR with this 
Current shared by SCR1 =
Current shared by SCR2 = 

⇒ 16 + 5 Rs = 3 Rs + 24
⇒ Rs = 0.004