Help understanding a semi-controlled full rectifier

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SUMMARY

The discussion focuses on understanding the output waveform (Vout) of a semi-controlled full rectifier circuit involving controlled rectifiers T1 and T2, and diodes D1 and D2. The Vout waveform drops to zero when both T1 and T2 are reverse biased, as the inductor opposes changes in current by utilizing its stored energy. The participants clarify that during specific intervals, the diodes act as flyback diodes, allowing current to flow through the resistor and inductor, which affects the voltage readings. Additionally, discrepancies in waveform measurements from lab analysis are addressed, particularly regarding the voltage across T1 and D2.

PREREQUISITES
  • Understanding of semi-controlled full rectifier circuits
  • Knowledge of thyristors and their operation
  • Familiarity with inductor behavior in electrical circuits
  • Experience with waveform analysis and oscilloscopes
NEXT STEPS
  • Study the operation of phase-controlled rectifiers in detail
  • Learn about the characteristics and applications of flyback diodes
  • Research the impact of firing angles on thyristor performance
  • Examine waveform discrepancies in laboratory settings and their causes
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Electrical engineers, students studying power electronics, and anyone involved in the analysis and design of rectifier circuits will benefit from this discussion.

jendrix
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Hi, I'm analysing the the circuit here

http://imgur.com/a/23GUZ

And I'm having some trouble understand the Vout waveform and why it goes to 0 when when pi<wt<pi+a

If there is always an output current wouldn't there always be a voltage out? The only thing I could think would be the voltage across the inductor cancelling out the voltage across the resistor?
 
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I swear this looks exactly like an old professor of mine hand writing (uddin?)

To answer your question - the Vout waveform goes to zero because both controlled rectifiers (T1 and T2) are not conducting (reverse biased). Are you aware that inductors oppose changes in current by creating their own current flow by using their own stored energy? that is what is happening here.

The path that io is flowing is through the resistor and inductor and through diodes D1 and D2. And you are correct, the inductor is creating the voltage that is dropped across the resistor. think of the path through D1 and D2 as a wire in a conventional circuit and Vo is basically putting your meter across it.

FYI - diodes D1 and D2 are acting like "flyback" or "freewheeling" diodes during this part of the cycle.
 

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It could be :)I'm working through a similar problem and found the notes here

http://www.slideshare.net/maneesh001/phase-controlled-rectifiers

Thanks for clearing up the question, it's clearer now I'm a bit rusty with inductors.

I'm still a bit confused with the voltage across the thyristors, we analysed a similar circuit in the lab and I got a different waveform. I added it to the link, it is the pink line on the oscilliscope? Any idea why there is a discrepancy?
 
Not sure where the waveforms are
 
Hi, I added them here

http://imgur.com/a/i67f8

They bottom (pink) signal in each one is supposed supposed to be Voltage of T1, voltage of D2 and voltage of the inductor with a firing angle of 90 degrees - I think they may have been mislabeled

Yellow -Vin
Green -Vout
Blue -Iout

But I'm having trouble what was being measure on the pink signal in each
 
It would make sense to me if the pink was the voltage across T1 and the voltage across D2, since they conduct during the positive half cycle of Vs, and do not conduct during the negative half cycle of Vs.
 
FOIWATER said:
It would make sense to me if the pink was the voltage across T1 and the voltage across D2, since they conduct during the positive half cycle of Vs, and do not conduct during the negative half cycle of Vs.

Thanks, that sounds right to me

Do you know what the waveform would look like for the ac source current? I thought it would be the same as Iout but 0 from

0-pi

pi+a - 2pi

Or would it be negative during the period pi+a to 2pi?

Thanks
 
Only time the source current isn't flowing (is 0) is from $$0 - \alpha$$ $$\pi - (\pi+\alpha)$$ $$2\pi - (2\pi + \alpha)$$ $$\vdots$$ Unlike the output current, the source current flows in two directions depending on which rectifiers are on
 
Thanks for the help, all much clearer now
 
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Not a problem
 

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