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Chopper numerical: 220 V, 10 Ω, 50% duty

Short answer

For Vs=220V, Vch=2V, duty cycle 0.5, ON voltage is 218V. Average output Va = 0.5×218 = 109V. RMS output = √0.5×218 = 154.15V. Efficiency = (218/220)×100% = 99.09%.

For the chopper shown in figure below with a resistive load R = 10Ω, the input voltage is Vs = 220V, when the chopper remains ON its voltage drop Vch = 2V and the chopping frequency is f = 1kHz with duty cycle of 50%. Determine: a) The average output voltage b) The RMS value of output voltage c) The chopper efficiency. [Figure: dc source Vs (+ at top) connected through the chopper switch SW to a resistive load R; the output voltage vo appears across R and the load current io flows through it.]

A chopper is a DC-to-DC converter that switches a fixed DC supply ON and OFF to give a variable average DC output.

Given: Vs = 220 V, R = 10 Ω, Vch = 2 V, f = 1 kHz, duty cycle k = 0.5.

During ON time the load voltage is Vs − Vch = 218 V; during OFF time it is 0.

vo
218 |----+    +----+    +---
    |    |    |    |    |
  0 +----+----+----+----+----> t
    |<kT>|
    |<--T=1ms->|

a) Average output voltage

Va = (1/T)∫₀^(kT) (Vs − Vch) dt = k(Vs − Vch)
   = 0.5 × (220 − 2) = 0.5 × 218
Va = 109 V

b) RMS output voltage

Vo,rms = [ (1/T)∫₀^(kT) (Vs − Vch)² dt ]^(1/2)
       = √k · (Vs − Vch)
       = √0.5 × 218
Vo,rms = 154.15 V

c) Chopper efficiency

Output power:

Po = (1/T)∫₀^(kT) vo²/R dt = k(Vs − Vch)²/R
   = 0.5 × 218² / 10 = 2376.2 W

Input power (source current = (Vs − Vch)/R during ON):

Pi = (1/T)∫₀^(kT) Vs·(Vs − Vch)/R dt
   = k·Vs(Vs − Vch)/R
   = 0.5 × 220 × 218 / 10 = 2398 W

Efficiency:

η = Po/Pi = (Vs − Vch)/Vs = 218/220 = 0.9909

Answer: Va = 109 V, Vo,rms = 154.15 V, η = 99.09 %

✓ Answer checked by Claude Opus 5.5 (high effort)

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