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Full Wave Bridge Rectifier Circuit

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gurse

gurse

Electrical
Jun 27, 2019
17
I have a full wave rectifier thyristor control circuit on a 10 Ohms resistor coil which produced the following results with 240V AC rms.

DC Current through coil (A) DC Voltage across Coil (V) AC Voltage across coil (V rms)

1.15 10.8 92.0
2.43 22.5 153.9
3.44 31.9 189.7
4.63 43.4 211.4


If I reduce the mains input from 240V rms to 110V rms to thyristor full wave rectifier circuit, will I still be able to pull 4.63 Amp DC as the output of a full rectifier circuit is Vrms = Vp / √2, which works out as 110 * 1.414 / √2 = 110 V rms.
 
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Sketchy math... lol.

You are multiplying by a number, then dividing it by the same number expressed in a different way.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
 
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Hi itsmoked,
My sketchy maths check,
The Vrms of a full bridge rectifier output is Voltage peak / √2 and the peak voltage of an Vrms = Vrms * √2.

Therefore, 110V Ac rms input to a full bridge rectifier will yield 110V rms at the output, 110 * √2 / √2.

Hi IRstuff,
The reason why raised the question is that when measuring the voltages across the coil when pulling 4.63 A DC is 211.4 V AC and 43.4 V DC.

So reducing the input voltage from 240V rms to 110 V AC rms, the output will only yield 110 V rms so it should be enough for DC current but what about the Vrms AC.
 
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Are you trying to say that the RMS voltage of a rectified sinewave with no filtering is the same as the RMS of the AC input sinewave that is applied to the rectifier? If so, then yes. Just actually thinking what RMS voltage means would confirm that. If not, then your math still doesn't check. Either way, I'm not sure what use knowing that is.

The Dc output voltage of a rectifier supplied with 1-phase input and feeding and inductor (a coil) is approximately Vrms * 0.89 when the inductor has enough inductance to cause a constant DC current to flow. So, 110VAC input could get you approximately 98VDC maximum. However, I can't say you will be successful since I have no idea if your coil acts as an inductor or how you made those measurements. I also can't figure out what you did with your measurements to get the a DC voltage so low and a AC voltage so high.
 
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DC current was measured in series with the coil and the AC and DC voltages across the coil. In fact, Also, I have measured the AC current and it was 2.54 A. Both were measured with a multimeter changing modes between AC and DC.
I was only interested in DC current to the coil but also measured AC as a matter of interest and trying to figure out correlation between AC and DC values.
The measured resistance of the coil was 10 ohms so I assume DC voltage and DC current relates to resistance of the coil and AC voltage and current to AC impedance of the coil as the rectifier output has AC ripple.
 
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The measured resistance of the coil was 10 ohms so I assume DC voltage and DC current relates to resistance of the coil and AC voltage and current to AC impedance of the coil as the rectifier output has AC ripple.
Click to expand...

From EE101, the source of the DC is the rectified AC, so if you cut your AC by more than half, you'll cut your DC by more than half.


TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
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So what is the correlation between AC and DC voltages that I measured across the coil in your opinion.
 
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Metering errors.
The meter may be sampling 1/2 wave rectified DC.
This would be average rather than RMS.
The meter indicates a value of twice the average further multiplied by a form factor to convert from average to RMS.
Some old analogue multi-meters gave strange results when DC was applied to an AC input.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
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thyristor full wave rectifier circuit
Click to expand...

Thyristor strongly implies a phase controlled rectifier. So, since no-one seems to have any idea what phase angle this thing is running at, it's likely time to stop the stupid comments about not remembering EE101 and the effect halving the voltage had.

I already posted what I believe is a sufficient answer, but apparently I have to spell it out. If you want to put 4.63A through a 10 ohm coil then you need 46.3VDC. Assuming this coil has enough inductance, the 110VAC should be capable of producing about 98VDC. You do get that that the possible voltage you can get or 98VDC is > than the voltage you need or 46.3VDC?

Failing to provide enough information for anyone to make a more precise determination, you'll have to try it to see if it works.
 
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Why do you need the coil to have enough inductance please, is it because of its filtering effect.
 
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The measurements don't make any sense to me. Simple theory tells me I could easily cause 4.63A to flow through a 10 ohm coil powering it from a phase controlled rectifier with a 110VAC input.

But, you can continue to say it's not possible if you want. With the limited info provided, the only way to know for sure is to try it.

gurse - yes with enough inductance the filter effect causes a constant DC current to flow.
 
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The OP continues to not show what his circuit actually looks like. I have no doubt that a blank sheet design could easily meet that requirement, as you sa, but that wasn't the OP's question.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
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If the OP ALREADY has a 1-phase, thyristor or phase controlled rectifier AS HE HAS INDICATED HE HAS, then why would him using what he already has require him to do a blank sheet new design? Do you have some specific reason to doubt he has the system he posted? He's already indicated he could control the voltage and current to the coil with his rectifier, which can't be done using diodes.
 
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The reason why I calculated the Vrms (Vp / √2) of the rectifier output rather than the Vav (2Vm /π) is to give the DC equivalent for the same power as the rectifier output is a complex waveform.
 
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So, have you tried reducing the voltage yet? Seems easier than speculation.
 
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Not yet, it is not speculation it is based on well established formulas.
 
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