A step up chopper is used to deliver 3 A to the 10 - load- The input battery voltage is 12 V- L- 20 - H- C -100 - H and the chopper frequency is 50 kHz- The minimum battery current required is

V_0 =I_0R = 3 × 10 = 30V V_0 =V_s/(1 D) (1 D) =12/30 D =(1 0.4) = 0.6 Δ I =DV/fL =0.6 × 12/50 × 10^3 × 20 × 10^ 6 = 7.2 A Average battery current, I_s =I_o ...

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A step up chopper is used to deliver 3 A to the $10 Ω$ load. The input battery voltage is $12 V, L = 20 μ H, C = 100 μ H$ and the chopper frequency is 50 kHz. The minimum battery current required is

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Solution

$V_{0} = I_{0} R$
$= 3 \times 10 = 30 V$
$V_{0} = \frac{V_{s}}{(1 - D)}$

$(1 - D) = \frac{12}{30}$
$D = (1 - 0.4) = 0.6$

$Δ I = \frac{D V}{f L} = \frac{0.6 \times 12}{50 \times 10^{3} \times 20 \times 10^{- 6}} = 7.2 A$

Average battery current,

$I_{s} = \frac{I_{o}}{(1 - D)} = \frac{3}{(1 - 0.6)} = 7.5 A$
$I_{s m i n} = I_{s} - \frac{Δ I}{2} = 7.5 - \frac{7.2}{2} = 3.9 A$

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A step up chopper is used to deliver 3 A to the 10 Ω load. The input battery voltage is 12 V, L=20 μ H, C=100 μH and the chopper frequency is 50 kHz. The minimum battery current required is

V0=I0R =3×10=30V V0=Vs(1−D) (1−D)=1230 D=(1−0.4)=0.6 ΔI=DVfL=0.6×1250×103×20×10−6=7.2 A Average battery current, Is=Io(1−D)=3(1−0.6)=7.5A Is min=Is−ΔI2=7.5−7.22=3.9 A

A step up chopper is used to deliver 3 A to the $10 Ω$ load. The input battery voltage is $12 V, L = 20 μ H, C = 100 μ H$ and the chopper frequency is 50 kHz. The minimum battery current required is