GATE Electrical Engineering (EE) Test: Thyristor Commutation Techniques

To turn-off a thyristor, anode current must be reduced below holding current and a reverse bias must be applied across thyristor for a finite period of time. If all these two conditions are not met simultaneously then forced commutation can be used to turn-off the thyristor.

The maximum value of capacitor voltage is V_c = 120 V after long time when T_A, T_A1 is ON

Then the maximum value of current I_{c max}

The circuit turn off time for thyristors are
t _c1 = R₁ C ln 2
t c₂ = R₂ C ln 2
t _c1 = t c₂
R₁ = R₂
safety margin = 1.5
for safe turn off
R₁ c ln 2 = 1.5 t _c1
t c₁ = 10 m sec

The turn-on time of an SCR depends on time constant (L/R) of the load.

It is necessary to keep the thyristor reverse biased for a finite period before a forward anode voltage can be reapplied because if a forward voltage is applied immediately after reducing the anode current to zero, it will not block the forward voltage and will start conducting again, although it is not triggered by a gate pulse. Reason is also a true statement but not the correct explanation of assertion.

In class D commutation turn off time for auxillary thyristor is 

tc₁ = 2.221 μs

• In forced commutation, the commutating components L and C do not carry load current continuously.
• Forced commutation is usually employed in dc choppers because natural current zero is not available in the circuits containing dc choppers.
• In load commutation. L and Care connected in series with the load or C in parallel with the load such that overall load circuit is underdamped.
• Load commutation is employed in series inverters.
•  In class-C commutation, the main thyristor that is to be commutated is connected in series with the load while the complementary thyristor is connected in parallel with the main thyristor as shown below.
  

Thus, statements 2 and 4 are only correct.

Given, L = 5μH, C = 20μF
Resonant frequency,

∴ Conduction time for auxiliary thyristor

Given, V_s = 200 V, R₁ = 10 Ω, and R₂ = 100 Ω
Given circuit represents a class-C commutation circuit.
The peak value of current through thyristor T₁ is


The peak value of current through thyristor T₂ is

Under steady-state condition, voltage across capacitor is V_s = 200 V and SCR conducts having a current through it equal to

When the SCR is forced commutated, the capacitor voltage at time t is


 ...(i)
Let C_j be the junction capacitance of SCR. Now, when SCR get turned-on, the

or, 
= 0.02 μF
Thus, the minimum value of C so that SCR does not get turn-on is 0.02 μF.
If C > 0.02 μF, SCR will be turn-on.