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Dominique 25th April 2005 12:05 AM

power supply diodes voltage rating?
 
hi,

I was just wondering about using schottky diodes in my next 12V powersupply, now I wonder about the security margin I should consider...

If I use 40V diodes (they got a lower dropout) the datasheet tells me that they are good for 28V RMS and 40V peak voltages.

Now, the 12V transformer should produce 16,92V peak, but at low load the voltage is always higher.

And well, sorry, I don't know coils well, they are always a bit misterious to me... how high could the transformer's peak voltages eventually be and should I rather choose higher rated diodes?
I often read about coils producing high voltages when a circuit is switched on, for example.

Thanks for listening!
Cheers,
Dominique

Sch3mat1c 25th April 2005 06:13 AM

I'd be more worried about switching off than on.

If you must, you could add a power zener or something across the winding, possibly with a crowbar to a latching relay or something.

Shottkies are within ratings for 12VAC.

Tim

AndrewT 25th April 2005 08:48 PM

Hi,
the diodes must survive peak inverse voltage (PIV) which is twice the open circuit peak voltage of the transformer.
The regulation of your transformer has to be added to the 12Vac & then multiplied up by 2 * sqrt2. Then add on 6% for power company over voltage.
Starting with 40V and working back gives a max of 13.3 Vac from the transformer open circuit when on 240Vac input.
However if you are using a full bridge then you have 2 diodes working for you, giving maximum of 26.6Vac from the transformer.

Dominique 25th April 2005 11:53 PM

Thank you both for helping!
Unfortunately I don't understand everyting you say - partly because I'm no geek yet, also because I'm not much used to that english electronics jargon!

Andrew,
"the diodes must survive peak inverse voltage (PIV) which is twice the open circuit peak voltage of the transformer." sounds easiest to understand for me of all the statements to me, so I'll stick with that.
If I add your other statement "However if you are using a full bridge then you have 2 diodes working for you, giving maximum of 26.6Vac from the transformer."

the result for a full bridge rectifier would be:
* The diodes rating must be at least the open circuit peak voltage of the transformer *

Is this correct?

I checked my needs and for the next power supply I'll need a 15V transformer.
The worst open circuit voltage of (rather cheap) 15V transformers I found was 25V!
So the open circuit peak should be: 1,41 x 25V = 35,25V!

So a 40V diode should be ok?
Or did I misinterpret the given info?

Cheers,
Dominique

Sch3mat1c 26th April 2005 05:53 AM

You should get out a pad of paper and pencil and draw the circuit and waveforms and see where the voltage comes from. More instructionally, you should draw a progression of half wave to FWCT to FWB bipolar to FWB, noting the electrical behavior of each. In fact I might just have to draw that anyway, and post it.

A FWCT circuit (using two diodes) has PIV = sqrt(8) * ACV (note sqrt(8) = 2 * sqrt(2)). It also requires a centertapped winding (as you may've guessed from the acronym) and as such needs a greater winding voltage for a given output. Not used much anymore.

FWB uses four diodes, the voltage drop is double (at low voltages and high currents, a good application for shottky diodes!) and PIV = 1.414 * ACV. This can be seen because, at no point during the cycle does the AC waveform drop below ground (ground being defined by the negative cycle rectifiers) nor does it rise above the positive output (for the inverse reason). It's kind of hard to see so you have to imagine the middle of the winding grounded, which then reverts it into a positive and negative FWCT type arrangement.

Tim

AndrewT 26th April 2005 03:29 PM

Hi,
Is this correct - YES.
Your calculation of 35.25V is correct for normal mains voltage. Now you must also add on the 6% overvoltage (in the UK) the power company is allowed to supply.

You are asking about discrete diodes and all the previous applies.

If you buy a 4 diode encapsulated rectifier the voltage rating on that has already allowed for the two diodes in series, you cannot double the rating on this one, just accept the marked voltage.

Dominique 27th April 2005 02:45 PM

2 Attachment(s)
thank you both again!
To get my head clear again, I made a sketch, looking at the moment of one peak of the sine wave.
cheap 15V Transformer:
* Open circuit voltage (rms) = 25V
* Open circuit peak voltage = 35,25V
* 50V Schottky diode with 0,4V forward voltage at 100mA (e.g. SB150)
(Ps will only be used up to this current)
* Voltages are referenced to ground.

Ok, now I can see that each diode will have to block the peak voltage minus the diode forward voltage!

I'll have to get info on the possible voltage variation in my country now!

Still I wonder if there is a way to estimate a voltage peak from the transformer when switching the circuit off.
Because the diode voltage rating is meant for the "repetitive peak voltage", but I couldn't find a maximum (non-repetitive) voltage in the datasheets. Maybe this value would be much bigger anyway, but who knows!

Cheers,
Dominique


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