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Old 20th Apr 2018, 11:18 am   #1
Diabolical Artificer
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Default Quick transformer1/2 wave rectification question.

If I have a tfmr secondary 10v 0v 10v with a current capacity at 50mA, with 0v to ground, two diodes, half wave rectified. Does this mean the current capacity doubles, EG I get 10v @ 100mA?

Andy.
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Old 20th Apr 2018, 12:12 pm   #2
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Default Re: Quick transformer1/2 wave rectification question.

Hey Andy ,

Sounds like you’re talking about the full wave configuration with two diodes doing the work instead of a bridge with four — where each pair conducts alternately .

But no, regardless of what goes after it , the max current that can flow without causing problems is *limited* in practice by the ampacity of your transformer (heating losses included )

Also, each diode has to take turns to conduct and so each diode sees the full current in its loop, by turns.

Nice if it could be true, though !
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Last edited by Al (astral highway); 20th Apr 2018 at 12:26 pm. Reason: Clarity
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Old 20th Apr 2018, 12:14 pm   #3
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Default Re: Quick transformer1/2 wave rectification question.

Don't understand half wave rectified. If you have a diode in each 10v lead, 0v to ground, with the other ends connected together, it full wave. but only at 50mA.

Last edited by Boater Sam; 20th Apr 2018 at 12:15 pm. Reason: Added
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Old 20th Apr 2018, 1:00 pm   #4
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Default Re: Quick transformer1/2 wave rectification question.

In a word, yes.

Assume the diodes are connected cathode-to-cathode, with each anode connected to one end of the centre-tapped secondary winding. On the "crest", current flows up the top half of the winding, through one of the diodes and the load. The diode connected to the bottom half is reverse-biased and so does not conduct. On the "trough", current flows down the bottom half of the winding, through the second diode and the load. The first diode is not conducting, so no current flows in the top half.

Since each half of the winding is passing current only half the time, it contributes half the energy (= voltage * current * time) going into the load. If the current in the load is 100mA, then the top half of the winding is supplying 50mA and the bottom half of the winding is supplying 50mA. So there is still only an average current of 50mA flowing in the windings!
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Old 20th Apr 2018, 1:30 pm   #5
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Default Re: Quick transformer1/2 wave rectification question.

Yes, it's full wave, but not bridge rectified. I was thinking of something else when I typed.

Thanks Julie and Al, that clarifies things. Been thinking myself into a corner.

A.
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Old 20th Apr 2018, 2:05 pm   #6
G8HQP Dave
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Default Re: Quick transformer1/2 wave rectification question.

Each secondary only conducts for half the time, so only generates half the heat. Taken on its own this means that you can go up to 70mA. However, the primary and the core also have losses. 70mA on half the secondary means less primary and core losses than 50mA on the whole secondary, so you can go a bit higher but not as high as 100mA.

To get 100mA you need to parallel the secondaries and use a bridge rectifier.

Quote:
Since each half of the winding is passing current only half the time, it contributes half the energy (= voltage * current * time) going into the load. If the current in the load is 100mA, then the top half of the winding is supplying 50mA and the bottom half of the winding is supplying 50mA. So there is still only an average current of 50mA flowing in the windings!
Heat in the windings (which is usually the limiting factor) goes like current^2. Twice the current for half the time therefore gives twice the heat.

Last edited by G8HQP Dave; 20th Apr 2018 at 2:07 pm. Reason: extend
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Old 20th Apr 2018, 2:21 pm   #7
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Default Re: Quick transformer1/2 wave rectification question.

General convention with a full wave (not bridge) rectifier is that the DC output current (with some smoothing) is the same as the RMS current rating of the whole winding, in this case 50mA.
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Old 21st Apr 2018, 6:51 am   #8
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Default Re: Quick transformer1/2 wave rectification question.

That clear's things up even more Dave, thanks.

Thanks Barry.

A.
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Old 21st Apr 2018, 8:12 am   #9
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Default Re: Quick transformer1/2 wave rectification question.

The output voltage and the maximum current depends on the rectifier and smoothing configuration. Attached is a datasheet detailing the rating of the various rectifier/smoothing configurations.

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File Type: pdf rectifier smoothing data.pdf (259.2 KB, 75 views)
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Old 22nd Apr 2018, 6:03 pm   #10
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Default Re: Quick transformer1/2 wave rectification question.

That Hammond sheet has been criticised on other websites for containing some errors and ambiguities. It has been reported that when asked about it, Hammond declined to comment apart from saying that they found it on the internet somewhere.
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Old 22nd Apr 2018, 7:12 pm   #11
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Default Re: Quick transformer1/2 wave rectification question.

The RS version: https://www.vintage-radio.net/forum/...4&postcount=28
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Old 22nd Apr 2018, 7:21 pm   #12
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Default Re: Quick transformer1/2 wave rectification question.

I'm sure I have a copy of that RS one...somewhere!

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Old 24th Apr 2018, 5:41 am   #13
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Default Re: Quick transformer1/2 wave rectification question.

"The output voltage and the maximum current depends on the rectifier and smoothing configuration" Yes, I was talking in rough terms.

Thanks both for those links.

Andy.
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