The chopper circuit operates at a frequency of 200 Hz with on time of 4 msec, the load current is ripple free and constant. The average value of source current for the circuit shown below will be _______ A (up to 2 decimal place).

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Solution

$T = \frac{1}{200} = 5 m s e c$

Duty cycle,
$D = \frac{T_{o n}}{T} = \frac{4 m s e c}{5 m s e c} = 0.8$

For step down chopper,

$V_{0} = α V_{s}$

= 0.8 $\times$ 100 = 80 V

$V_{0} = E + I_{0} R$

$80 = 20 + I_{0} (5)$

$5 I_{0} = 60 \Rightarrow I_{0} = 12 A$

$V_{s} I_{s} = V_{0} I_{0}$

$I_{s} = \frac{80 \times 12}{100} = 9.6 A$

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The chopper circuit operates at a frequency of 200 Hz with on time of 4 msec, the load current is ripple free and constant. The average value of source current for the circuit shown below will be _______ A (up to 2 decimal place).

T=1200=5msec Duty cycle, D=TonT=4msec5msec=0.8 For step down chopper, V0=αVs = 0.8 × 100 = 80 V V0=E+I0R 80=20+I0(5) 5I0=60⇒I0=12A VsIs=V0I0 Is=80×12100=9.6A