Question

A 1 : 1 Pulse transformer (PT) is used to trigger the SCR in te figure. The SCR is rated at 1.5 kV, 250 A with $I_L=250mA,I_H=150mA,and{IG}_{max}=150mA$$with{I}_L=250mA,I_{Gmin}=100mA.$ The SCR is connected to an inductive load, where L = 150 mH is series with a small resistance and the supply voltage is 200 V dc. The forward drops of all transistors/diodes and gate-cathode junction ON state are 1.0 V

The resistance R should be

Answer 1

When the pulses are applied to the base of the transistor. Tranisistor operates in ON state. So, the forward voltage drop in transistor $V_{CE}$ = 1 V.

$V_1=10-V_{CE}=10-1=9V$

$V_2=V_1\left(\frac{1}{1}\right)=V_1=9V$ [turn ratio 1 : 1]

$D_1$ is forward biased and voltage drop in diode $V_{D1}$ = 1 V.

$D_2$ is reversed biased and acts as open circuit.
Capacitor behaves as open circuit for dc voltage.
Forward voltage drop of gae cathode junction
$V_{gk}$ = 1 V
Voltage drop across resistor R,
$V_R=V_2-V_{D1}-V_{gk}$
= 9 - 1 - 1 = 7 V
To ensure turn-ON of SCR,
$R=\frac{V_R}{I_{g(max)}}=\frac{7}{150mA}\approx 47\Omega$