Question

The gate to source voltage in an SCR is 16 V and load line has a slope of 128 V/A. The trigger voltage and current respectively for an average gate power dissipation of 0.5 W are

Answer 1

The circuit will be

By applying KVL in the loop, we get,
$-E_s+I_g{R}_s+V_g=0$

$-16+128I_g+\frac{0.5}{I_g}=0$

$-16I_g+128I_g^2+0.5=0$

$or,I_g=0.0625A=62.5mA$

$V_g=\frac{0.5}{I_g}=\frac{0.5}{62.5mA}=8V$