Test: Thyristors - 2 - Electrical Engineering (EE) MCQ

During the turn on time, the voltage across the SCR is going down and the current through it is slowly rising as it is going into the conduction mode.

Initially for a fraction of a microsecond (delay time) after the gate signal is applied the anode current only flows near the gate terminal where the gate current density is maximum, as the gate current takes some time to spread all over the cross section of the device.

 The device will remain in the conducting state unless the anode current falls below the holding current value.

The losses are maximum during the rise time because both Ia & Va are high.

 The latching current is the value of current on which the device will remain in the on state even after removal of the gate signal. Whereas, the holding current is the threshold above which the device will work.

During the spread time the conduction starts spreading all over the SCR cathode cross-section structure, which depends upon the structure of the gate & cathode. Higher the cathode area more is the time required for the charges to spread all over.

 To enable the device to regain its reverse blocking capabilities, the stored charges in the junctions of the SCR must be removed.

If the thyristor turn off time is more than the circuit turn off time, the circuit will be turned off and the thyristor will keep conducting, which is not at all desirable.

It is the gate current versus the gate voltage plot and gives the minimum and maximum values of gate parameters.

As the gate characteristics is a plot of Ig vs Vg consisting of two curves one for the maximum values & other for the minimum the area between them gives the total average gate power dissipation. (A very important parameter in designing of the triggering circuits).

It gives the min gate to source resistance.

 Lesser time is required to inject the charges & turn on the device with higher gate pulse magnitude.

Vg.Ig = 0.5 W, the power dissipation mustn’t exceed the average power dissipation.

Vg/Ig = 130 .. (given)
Vg.Ig = 0.5 watts .. (given)
use both the given data & find the gate current.

The two transistor model consists of p-n-p and n-p-n transistors, of which the middle n-p layer is common in both the transistors.

 For L load, E = L di/dt
I = E/L t
Therefore, 0.100 = 200t/0.2
T = 100 μsec.

Load line is nothing but Rs
Es = 16V
Vg.Ig = 0.5
Rs = 128
We have Es = Ig x Rs + Vg.

The sum of both the collector currents of T1 and T2 forms the total anode current of SCR. Refer the model.

To turn on the device sum of both the current gains should approach unity value.

E = Ri + L di/dt
Solve the above D.E for I & substitute the above values.
t = 100.503 μsec.