diode bridge rectifier

Come on. Draw a diagram, and get two different colored pencils, and trace the current path for each half-cycle, and apply Ohm's and Kirkhoff's laws.

Sheesh! We're happy to help if you get stuck, but you do have to do your own homework.

Good Luck! Rich

(BTW, are you any relation to Harry? ;-P )

The bridge could be configured for pos. or neg. output with respect to common. You can't get help with homework until you understand the problem and then you won't need any help.

SMOKE !

Hi,

Actually I got stuck at a point in drawing the output voltage waveform of diode bridge rectifier with load consisting of capacitor in parallel with diode. Let me first explain that.

The diode(connected in parallel with the capacitor in load) becomes forward biased whenever the voltage across the load diode(nothing but Vo) is greater than 0.7V. Here comes the appearance of violation of KVL. Assume the input voltage source to be ideal. When the diode is on, voltage across diode(Vo) is nearly 0.7V. But the output voltage(Vo) of a full bridge rectifier should be positive sine wave with a peak of Vm. Please tell me where I went wrong in this analysis?

Thanks, Krishna

Hi,

Actually I got stuck at a point in drawing the output voltage waveform of 1-phase diode bridge rectifier with load consisting of capacitor in parallel with diode(cathode down). Let me first explain that.

The diode(connected in parallel with the capacitor in load) becomes forward biased whenever the voltage across the load diode(nothing but Vo) is greater than 0.7V. Here comes the appearance of violation of KVL. Assume the input voltage source to be ideal. When the diode is on, voltage across diode(Vo) is nearly 0.7V. But the output voltage(Vo) of a full bridge rectifier should be positive sine wave with a peak of Vm. Please tell me where I went wrong in this analysis?

Thanks, Krishna

I guess this is homework. Kinda odd tho.

During one half cycle the cap will charge as usual, but bearing in mind its charging frmo zero, not just on peaks. so what does i into the cap depend on?

On the other half cycle youve got a near short. Now what does i depend on?

Not doing it all for you

NT

Most diodes drop 1v - 1.5v when running power. 0.6v is the knee voltage, not the drop at largeish currents. This is something remarkably many confuse.

If you were using a 35A diode to conduct 1A, you may be right with

NT

KVL is very difficult to violate ;>) The diode load approaches a short circuit for much of each alternation. Assuming an ideal source, something will open (burn out) in the real world as the current will be extremely high. With a non-ideal source of moderate to high impedance, the voltage across the diode-capacitor will reach a steady-state value of 0.7 volts or so.

By neglecting the internal resistance of the transformer, and by failing to look up "Kirkhoff's law".

Good Luck! Rich

It will be the same on every half cycle. Bridge rectifier!