Question

Figure shows a chopper operating from a 100 V dc input. The duty ratio of the main switch S is 0.8. The load is sufficiently inductive so that the load current is ripple free. The average current through the diode D under steady state is

• A. 10.0 A
• B. 6.4A
• C. 8.0 A
• D. 1.6A

Answer 1

The correct option is D 1.6A

(a)



During the period $T_{ON}$ chopper is on and load voltage is equal to source voltage $(V_s=100V)$. During the interval $T_{OFF}$. chopper is off load current$I_0$ flows through the diode as result load voltage is zero during $T_{OFF}$
Average load voltage,
$V_0=\frac{T_{ON}}{T_{ON}+T_{OFF}}{V}_s=\alpha V_s$
$=\frac{0.8}{100}=80V$
Average load current,
$V_0=\frac{V_0}{R}=\frac{80}{10}=8A$
As load current is ripple free, so average diode current,
$I_D=\frac{1}{T}{\int }_{T_{ON}}^T{I}_{0}dt=\frac{(T-T_{ON})}{T}{I}_0$
$I_D=(1-\alpha )I_0$
$=(1-0.8)\times 8=1.6A$