# Question regarding the output of a half wave rectifier

Discussion in 'General Electronics Discussion' started by saadishere, Aug 18, 2012.

6
0
Aug 18, 2012
hello ,

could you please first have a look at my screenshot.
I do not really understand how the output of the resistor (A in the oscilloscope) has been displayed, when the input voltage across the ac power is positive , D1 is forwared biased , but D2 is then reverse biased as the anode is 0 and the cathode is positive ? right ? so no current can flow accross D2 , D2 is an open circuit.
when the input voltage is negative , D2 is forwared biased ,and D1 is reverse biased ,D1 is an open circuit.
how can we have this display on the oscilloscope for A , many thanks.

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2. ### Electrobrains

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Jan 2, 2012
You have no (or almost no) current flowing through R1. The only current flowing, is through your measuring probe. It's very high impedance, usually 1M or 10M Ohm. Compared to that, your 1k Ohm will act as a short circuit, keeping the two cathodes on the same potential.

Thus the resulting waveform is positive when D1 is conducting and 0V when your supply changes polarity.

In fact, D2 is hardly active at all in the circuit. Once it is reversed polarized and will block the current and in the next moment it's just (through the "short circuit" R1) connecting to the same potential as the ground of your oscilloscope.

It looks as if you have a small negative signal in the latter case. That's probably because D1 has a (negative) leakage current, that will flow through your high impedance probe (and a little through D2).
Because D2 is not ideal, it will allow for that negative voltage (theoretically up to approx. 0.6V) to hang around on the probe side.

3. ### davennModerator

13,785
1,936
Sep 5, 2009
welcome to the forums

for a fullwave rectifier using 2 diodes that circuit is incorrect
the AC would/should be coming from a centretapped source --- normally a transformer. the centretap is the 0V rail and the other 2 sides have the diodes in them and they will rectify every half cycle.
There only 2 ways to reftify a single ended AC source as you have shown, 1 is to use a single diode ( say your top one) and have a half wave rectifier, or use a bridge rectifier ( 4 diodes) for a fullwave rectifier.

Then when used in either of those 2 ways the 1k resistor will be the load across the DC output of the diodes.
have a look at these 3 circuits.....

Fullwave bridge rectifier

Fullwave 2 diode rectifier

Halfwave rectifier

for the first 2 rectifiers the waveform seen on the oscilloscope will be....

for the halfwave rectifier it will be....

Now in each example, the oscilloscope should be across the resistor load so you can view the waveform correctly

cheers
Dave

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4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Actually there's a third way...

But it's generally a poor option.

You can use 2 capacitors to divide the AC signal and create a "centre tap" You then do what you would normally do to full wave rectify. Redrawing it will make it look like the bridge rectifier but with 2 diodes replaced by capacitors.

I would consider this of technical interest only.

5. ### davennModerator

13,785
1,936
Sep 5, 2009
hey steve

havent seen that way done .... I assume, maybe incorrectly, non polarised caps ?

D

6. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
The only time I've seen it was with valve rectifiers. It can obviously be a little tricky to have all your directly heated cathodes in parallel when some are at ground potential and others at V+.

This was a solution, and was made more viable by the fact that currents were generally small and the voltage drop across a diode (tube) rectifier was often rather huge.

I believe you can get away with polarised caps, but you're stretching my memory. It certainly seems at first sight that you could not.