Why output waveform look like this in bridge rectifier

  • Thread starter George317
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  • #1
George317
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in our laboratory class we made a simple bridge rectifier circuit like this
diode20.gif


then we replaced D1 and D2 with resistors, and when we connected the probes of the oscilliscope on the load resistor it showed a DC output something like this(sorry for the bad drawing)

output waveform when D1 and D2 are resistors.png


can you guys explain why it produces that output voltage? like what happens in the postive cycle to produce that waveform and what happens in the negative that would result with that wave.

may attempt in answering this:

in the positive cycle D4 and D3 are both reversed bias the current would have to go through the 2 resistor(previouslyD1 & D2) and RL hence the low value in the load.

while in the negative cycle it appears bigger because current does not want to go to the 2 resistors(?) because of "the current goes to the least resistance thing" hence its only goes to D2 -> RL -> D4 -> source.
 
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  • #2
sophiecentaur
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Best not to use the term "wants to" because circuits are dumb and have no thought processes. It is always possible to come up with a different form of words to explain mechanisms in a way that sounds you know what you're talking about. :smile:
The way your modified circuit works is similar to the way the original bridge circuit works. Current will only pass through forward biased diodes so the current through the load will always be the same way for both halves of the input AC cycle.
The waveform you obtained could have different peak heights for several reasons. Most likely is the possibility that the two resistors have different values so current would be less through the higher resistance, giving a lower voltage (IRL) across the load. Two different diode types could produce different peak voltages but they would have to be of very different types and be operating at either very high current or very low (around 1V) AC supply volts. IF the AC supply is not totally separated from the rest of the circuit ('a floating supply') then the current on each half cycle could be different.
Different Rs is my favourite solution to the problem, though.
 
  • #3
George317
Best not to use the term "wants to" because circuits are dumb and have no thought processes. It is always possible to come up with a different form of words to explain mechanisms in a way that sounds you know what you're talking about. :smile:
The way your modified circuit works is similar to the way the original bridge circuit works. Current will only pass through forward biased diodes so the current through the load will always be the same way for both halves of the input AC cycle.
The waveform you obtained could have different peak heights for several reasons. Most likely is the possibility that the two resistors have different values so current would be less through the higher resistance, giving a lower voltage (IRL) across the load. Two different diode types could produce different peak voltages but they would have to be of very different types and be operating at either very high current or very low (around 1V) AC supply volts. IF the AC supply is not totally separated from the rest of the circuit ('a floating supply') then the current on each half cycle could be different.
Different Rs is my favourite solution to the problem, though.
thanks for answering.

we used same resistors and the same diodes



this is what we made
https://imgur.com/TxIoYuq
output waveform when D1 and D2 are resistors.png
 
  • #4
sophiecentaur
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There’s clearly something in the schematic that is different from the practical realization. There is something not quite and we need to know some details about the equipment and the components.
One way to resolve this would be to swap the resistors then swap the diodes then change the connections to the transformer. (This latter action would need the scope to be triggered properly; you would need help from a teacher perhaps. Is the AC source a transformer or a signal generator with a ‘balanced ‘ output?
 
  • #5
Tom.G
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The schematic in post #1 does not match the schematic in post #3.

Different circuits, different explanations. Please clarify.

Oops, I misread the schematic. :sorry: Sorry.

Thanks to @willem2 for pointing this out.
 
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  • #6
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A possible cause for your problem is that you are putting the reference of the oscilloscope probe in the bottom part of the resistor (load). However, the osciloscope probe is grounded.
So, use a diferenttial probe or remove the osciloscope connection to the ground pin of your AC outlet (maybe using an adaptor).
Tell us the news.
 
  • #7
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I see no problem here. The output is what you would expect.
Note that the two circuits are exactly the same apart from the naming of the diodes. If you replace D1 and D2 by resistors in the first circuit, you'd expect that D3 and D4 remain.
If the upper side of the voltage source is positive, the diodes won't conduct and you get 2x2kΩ and 500Ω in series. The output voltage you'd see at the peak voltage across the 500Ω resistor is 12 * √2 * 500/4500 = 1.88 V
If the upper side of the voltage source is negative, the diodes will conduct, and you'll see 12*√2 - 2 diode drops, or about 15.6V. There's some current through the resistors as well, but this has no influence on the output voltage. (the input is an ideal voltage source)
 
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  • #8
sophiecentaur
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Yep. That's the answer. :smile:
 

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