An ideal chopper operating at a chopping period of 2 ms supplies a load of 4 $Ω$ having an inductance of 8 mH from 80 V battery. Assuming a load is shunted by a perfect commutating diode, and battery to be lossless. If duty ratio is 0.2, maximum and minimum value of load current (A) respectively are

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Solution

Given,
Duty ratio, $α = 0.2$

Time constant $τ = \frac{L}{R} = \frac{8}{4} = 2$ ms
$\frac{T_{o n}}{T_{o n} + T_{o f f}} = \frac{1}{5}$

$T_{o n} = \frac{2}{5} = 0.4$ ms

$\frac{T_{o n}}{T_{o f f}} = \frac{1}{4}$

$T_{o f f} = 1.0$ ms

$I_{0, m a x} = \frac{E_{d c}}{R} [\frac{1 - e^{- T o n / τ}}{1 - e^{- T / τ}}] = \frac{80}{4} [\frac{1 - e^{- 0.4 / 2}}{1 - e^{- 2 / 2}}] = 5.73 A$

$I_{0 m i n} = I_{0 m a x} e^{- T_{o f f} / τ}$
$= 5.73 e^{- 1.6 / 2}$ = 2.57 A

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An ideal chopper operating at a chopping period of 2 ms supplies a load of 4 Ω having an inductance of 8 mH from 80 V battery. Assuming a load is shunted by a perfect commutating diode, and battery to be lossless. If duty ratio is 0.2, maximum and minimum value of load current (A) respectively are

Given, Duty ratio, α=0.2 Time constant τ=LR=84=2 ms TonTon+Toff=15 Ton=25=0.4 ms TonToff=14 Toff=1.0 ms I0,max=EdcR[1−e−T on/τ1−e−T/τ]=804[1−e−0.4/21−e−2/2]=5.73 A I0 min=I0maxe−Toff/τ =5.73e−1.6/2 = 2.57 A