I assembled a bridge rectifier circuit, and got an output voltage of 4V peak-to-

Question

I assembled a bridge rectifier circuit, and got an output voltage of 4V peak-to-peak. I am required to calculate the ripple voltage using the formula Vr = Vs - 2VD, where VD is the diode drop voltage. Vs is the amplitude of the ac signal.

My question is: my output voltage is peak-to-peak, so should I use 2V instead of 4V for my Vs, since Vs is th amplitude? or I should use the whole 4V for my Vs, and can you explain why and what the difference is plz , just a brief explanation.

Another question: I used three different capacitors (10-22-and 47 micro-farad) in the circuit and measured ripple voltage at each one, my question is why does the 47 micro-farad capacitor give smallest amount of ripple, AND largest mean value ?

thank you

Explanation / Answer

Use 4 volt not 2 volt

out put from rectifier is dc. But practically it is not pure pure DC like battery. It contains small AC components. Ripple is measure of ac component above dc value. Out put from bridge is 4 v peak to peak.In bridge rectifier only two diodes are conducting at a time, hence deduct 2Vd.

Capacitors used to reduce the ripple. More is value less will be ripple ( Variation in out put ac components above dc ). More is value of capacitor less will be variation. If variation is zero mean value is maximum. Capacitor stores energy at peak voltage and supplies energy when voltage reduces. If You will use a very very high capacitor ripple will redude to zero and out put voltage variation will be zero and output will be equal to peak voltage.

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