Question

A type-A chopper is fed from a 200 V dc source. The chopper is connected to load RLE with R = 0, L = 2.5 mH and constant E, for a duty cycle of 0.25. The chopping frequency to limit the amplitude of load current to 8A is ______ kHz.

Answer 1

Average output voltage is given by,

$V_0=\alpha V_s$

$\alpha =dutycycle=0.25$

$V_0=0.25\times 200=50V$

As the average value of voltage drop across L is zero

$V_0=E=\alpha V_s=50V$

During $T_{ON}$, the difference in source voltage $V_s$ and load emf E appears across inductance L. Also during $T_{ON}$, the current through L rises from $I_{min}to{I}_{max}$.

$V_L=L\frac{di}{dt}$

$V_s-E=L\frac{\Delta I}{T_{ON}}$

$T_{ON}=\frac{L\Delta I}{V_s-E}=\frac{2.5\times 8\times {10}^{-3}}{200-50}$

$=1.33\times {10}^{-4}sec=133\mu sec$

and, duty cycle, $\alpha =\frac{T_{ON}}{T}$

$\frac{\alpha }{T_{ON}}=f$

Chopping frequency, $f=\frac{\alpha }{T_{ON}}=\frac{0.25}{133\times {10}^{-6}}=1.879kHz$