# Can't understand this circuit (diode/capacitor)

• Engineering

## Homework Statement

What is the name given to this type of circuit? Sketch output voltage waveform for a sine wave input across nodes A-B.

## Homework Equations

Please see attachment for circuit.

## The Attempt at a Solution

Well im inclined to say it is a full wave rectifier with a smoothing capacitor. I just cannot see how the current will move through the circuit.

As v_in goes positive, D1 and D2 conduct. D3 and D4 will not conduct. I assume the current will split and so we will have current entering the capacitor through both terminals??

It seems as if once the current has gone in there is no way for it to come out? I really dont understand this diagram! (unless I dont' understand AC properly..)

Thanks.

#### Attachments

• 23.2 KB Views: 448

## Answers and Replies

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berkeman
Mentor
You are correct, that is exactly what it is. If there were no load resistor, then the DC voltage across the capacitor would be the max peak-to-peak AC input voltage, minus how many diode drops?

And with the load resistor, what happens to the capacitor voltage between the AC input peaks?

ranger
Gold Member
Yup it is a rectifier :) But more like a bridge rectifier.

t seems as if once the current has gone in there is no way for it to come out? I really dont understand this diagram! (unless I dont' understand AC properly..)
Gone in? You mean from the capacitor's perspective? Well thats what the resistor is there for. Do you know how it would discharge through the resistor and what the wave form will look like?

EDIT: aww honey mustard berkeman. You're quick!

Last edited:
You are correct, that is exactly what it is. If there were no load resistor, then the DC voltage across the capacitor would be the max peak-to-peak AC input voltage, minus how many diode drops?

And with the load resistor, what happens to the capacitor voltage between the AC input peaks?
Gone in? You mean from the capacitor's perspective? Well thats what the resistor is there for. Do you know how it would discharge through the resistor and what the wave form will look like?
The capacitor will discharge and then recharge on the negative input wave, so on and so forth. Essentially the output as seen from the load will be a DC voltage (almost a Dc voltage - as far as i know this will depend on how well the output is smoothed) The output DC voltage will be the input volatge minus 2(?) diode drops.

I still don't fully understand how the circuit works. The questions tells us its a sine wave, so when A goes positive, what is B at? .

If we assume its 6V peak to peak, then when A goes positive, 3V will enter into the circuit, do these 3 volts split between the branch containing D2 and the branch containing D1? Surely not?

Thanks for your help so far.

AlephZero
Homework Helper
Compared with a bridge rectifier that actually works, this one has two of the diodes the wrong way round.

That might explain why you are having problems understanding it!

Flip D1 and D4 the other way round, then it should make more sense.

berkeman
Mentor
Compared with a bridge rectifier that actually works, this one has two of the diodes the wrong way round.

That might explain why you are having problems understanding it!

Flip D1 and D4 the other way round, then it should make more sense.
Ack! Good catch. I only looked at the other two diodes

No wonder the OP was confused!

Thanks for your help all.