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jbasanth
modified 1 year ago

Bridge rectifier with Filter

3
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438
16:42:59
Given: Ac source→Vrms=120V Transformer ratio=5:1 Frequency=60Hz Calculation: Secondary Emf=Vrms*(turns ratio) ^-1=120/5 Secondary emf=V2=24V V2(peak) =24*√2=33.9V Let's take each diode's voltage drop=0.8v And remember only two diodes are active in each cycle, So, The output peak Voltage=33.9V-2(0.8) =32.3V The output frequency=2*F(in) =2*60Hz=120Hz Since, My Load=1kΩ , Load current=Vo/1kΩ= 32.3/1kΩ=32.3mA The average dc output=Vo/π=10.28V Now for the capacitor: τ=RC, Since R=V/I and τ=1/f, V(ripple) =I/fC; I want a ripple of 1ΔV around, this will just give approximate value. C=32.3mA÷(120*1V) C=270μF Notice that the ripple voltage is around 800mV to 1.5 V, this because of the exponential growth and we used an approximation
published 1 year ago

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