# Bridge rectifier with capacitor

## Homework Statement

For the following bridge rectifier you are asked to find the dc output voltage for the following errors:
a)the capacitor is an open circuit
b)one of the diodes is open
c)the electrolytic capacitor is connected with reverse polarity

## The Attempt at a Solution

My attempt is in the pdf below.If i should have added images of my work as well with the pdf,tell me so that i will know in the future.Thank you in advance for your time.

#### Attachments

• New Doc 30.pdf
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gneill
Mentor
If the load resistor were not present the capacitor voltage would follow the rectified voltage waveform to a peak and then remain there (current can't flow back through the rectifiers). So the DC output would then be equal to the peak voltage of the rectified voltage.

During normal operation the capacitor will charge up to the peak voltage at the peaks of the rectified waveform. At off-peak times the capacitor alone supplies current to the load. But the capacitor's voltage is going to drop as it does so. What curve will it follow? What's the time constant? How low can the voltage "sag" before the next rise-to-peak of the rectified waveform intercepts it and starts recharging the capacitor? A sketch will help to make it clear.

You can ballpark the sag ("ripple") by using the time between rectified peaks as the decay time for the capacitor voltage and calculate the change. The DC output can be estimated to lie between the peak and trough of the ripple if the ripple is a small fraction of the output. Think of it as a DC value with a small AC component superimposed on it. If the ripple is large compared to the DC component then it becomes time to wheel out the larger math machinery :)

NascentOxygen
Staff Emeritus