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'[EE]:Power transformer question'
2001\09\09@161243 by David VanHorn

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Rusty neurons here:

How do you calculate peak diode current in a full wave or bridge power supply?

I know it's based on the cap size, which determines the ripple voltage, but
I don't remember the diode peak current part.
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2001\09\09@180742 by Olin Lathrop

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> How do you calculate peak diode current in a full wave or bridge power
supply?

Fastest dV/dt of the AC waveform times the  capacitance after the
rectifier, plus maximum cicuit current draw.


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2001\09\09@191011 by Spehro Pefhany

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At 03:12 PM 9/9/01 -0500, you wrote:
>Rusty neurons here:
>
>How do you calculate peak diode current in a full wave or bridge power
supply?
>
>I know it's based on the cap size, which determines the ripple voltage, but
>I don't remember the diode peak current part.

For the recurring peak current... you can calculate it from the known point in
the cycle where the diode starts to conduct (from the p-p ripple voltage), and
the source impedance of the transformer winding (which you can measure by
loading it and measuring the voltage change) and the diode resistance and
capacitor ESR. You know the dv/dt of a perfect sine wave at the point where
the diode starts to conduct (dv/dt of  Vo*sin(wt) = w*Vo*cos(wt)). Typical
AC waveforms are a bit flattened these days, though. This is typically not
limiting, however it's worth noting that the RMS diode current is much higher
than the output DC average current, so the power dissipation is higher than
you would have with DC. Dividing the average current by fraction of the cycle
that the diode is conducting will get you a ball-park figure (though it will
be optimistic).

The inrush current at startup is much worse- worst case there is not much
limiting the current to a reasonable value other than the source impedance-
if the voltage is switched on at the peak of the cycle.

In one device I designed recently I used a spare PIC output (yeah, I did have
one) to control a small relay which shorted out a series resistor. The PIC
was also controlling the load so I could charge the capacitor (about the size
of a can of Coke) at a reasonable 4A peak to 300V for a second or so, before
shorting out the series resistor (waiiiiit for the relay to pull in) and
then cranking the power devices up. Or you can just use an NTC, but there
is a conflicting requirement in that the NTC stays hot longer than it takes
for the voltage on the capacitor to discharge.

Best regards,

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2001\09\09@191141 by Bob

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I believe the surge current is calculated thusly

I = w*C*Vs

Vs = Voltage coming off the centertapped secondaries
of the transformer

Which in a supply I did recently, looked like this:

2*pi * 60 Hz * 104uF * 35V = 1.37A

I used 1n4004 which were rated for 30A (!)
for non-repetitive peak surge current, so this
is obviously well within safety limits

Note: this was a Full Wave not a Bridge Rectifier

Hope this helps.

{Original Message removed}

2001\09\09@201154 by Russell McMahon

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An add on to the answers so far.
If you have any headroom then a small series resistor in series with the
transformer secondary can, rather unintuitively,  improve some aspects of
performance and reduce required component specifications. This reduces peak
currents, and spreads diode conduction angle thereby reducing diode spec,
switching noise, capacitor heating and more.  Reducing capacitor peak
currents can improve lifetime in temperature limited designs.

This resistor arguably worsens load affected regulation but is generally
insignificant with an active regulator and a smidgin of free head room
available.


       Russell McMahon
       apptechspamKILLspamclear.net.nz


> Rusty neurons here:
>
> How do you calculate peak diode current in a full wave or bridge power
supply?
>
> I know it's based on the cap size, which determines the ripple voltage,
but
> I don't remember the diode peak current part.

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2001\09\09@201742 by David VanHorn

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At 12:12 PM 9/10/01 +1200, Russell McMahon wrote:
>An add on to the answers so far.
>If you have any headroom then a small series resistor in series with the
>transformer secondary can, rather unintuitively,  improve some aspects of
>performance and reduce required component specifications. This reduces peak
>currents, and spreads diode conduction angle thereby reducing diode spec,
>switching noise, capacitor heating and more.  Reducing capacitor peak
>currents can improve lifetime in temperature limited designs.
>
>This resistor arguably worsens load affected regulation but is generally
>insignificant with an active regulator and a smidgin of free head room
>available.

I plan to.
I have some cap lifetime equations that I've done before, and I'll put that
in as well.
I developed them for a thermal printer I did. Most illuminating.


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2001\09\10@130457 by Roman Black

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David VanHorn wrote:
>
> Rusty neurons here:
>
> How do you calculate peak diode current in a full wave or bridge power supply?
>
> I know it's based on the cap size, which determines the ripple voltage, but
> I don't remember the diode peak current part.


Put a 0.1 ohm resistor in series and look on the CRO.
With all my high current DC supplies I do this anyway,
using a good sized resistor in series between the
transformer sec and the bridge. This also tends to
improve long term reliability when you turn the thing
on (and the cap is basically a short circuit). Even
dissipating a couple of percent in heat the resistor
is always worth the cost. Always try and use a resistor
to take the punishment, keep your semiconductors for
more subtle roles... :o)
-Roman

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2001\09\10@160846 by David VanHorn

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At 03:03 AM 9/11/01 +1000, Roman Black wrote:
>David VanHorn wrote:
> >
> > Rusty neurons here:
> >
> > How do you calculate peak diode current in a full wave or bridge power
> supply?
> >
> > I know it's based on the cap size, which determines the ripple voltage, but
> > I don't remember the diode peak current part.
>
>
>Put a 0.1 ohm resistor in series and look on the CRO.

I'd snap a current probe on it if it existed yet.
Unfortunately, I'm in the position of having to develop a spec.
It's further complicated by some design variables that take the required
current output over a pretty wide range.

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I would have a link to http://www.findu.com/cgi-bin/find.cgi?KC6ETE-9 here
in my signature line, but due to the inability of sysadmins at TELOCITY to
differentiate a signature line from the text of an email, I am forbidden to
have it.

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2001\09\10@231703 by David Huisman

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I(peak) = T/T1 * Idc

Where T = 1/f (1/50 for 1/2 wave and 1/100 for full wave) - or 1/60 and 1/20
depending where you are.
T1 = conduction angle of diodes
Idc = average current.

Also, Xc = 1/(2*pi*f*C) and average current = Vrms/Xc

Regards

David Huisman

{Original Message removed}

2001\09\11@103939 by Roman Black

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David VanHorn wrote:
>
> At 03:03 AM 9/11/01 +1000, Roman Black wrote:

> >Put a 0.1 ohm resistor in series and look on the CRO.
>
> I'd snap a current probe on it if it existed yet.
> Unfortunately, I'm in the position of having to develop a spec.
> It's further complicated by some design variables that take the required
> current output over a pretty wide range.

It's gonna get messy. :o) Power diodes Vf forward
drop is not fixed, and unless you know the conduction
time and impedances of the xformer and cap and secondary
filter (if there is one) it's a nightmare. And the
cap impedances will vary from part to part and with
device heat.

For example a common 1N4007 1A type rectifier diode
might drop 0.6v for low If, but at 1A constant they
can be as bad as 1.4v Vf. Now in a supply that had
0.8A constant load with a large cap the diode on-period
was very short, peak current was a few amps and I
think from memory the Vf during the period of actual
conduction was a few volts! Adding a tuned series
resistor helped a lot, and a second series resistor
from C1 to C2 (pi filter) with the bulk of the
capacitance in C2 gave much less losses and heat on
the diodes, allowing me to use the cheap diode instead
of going to larger diodes, and of course improved
the reliability of the circuit a lot. Most
manufacturers will "cheat" and use a lossy cheap
trnasformer which has very low current overhead.
In my case that was not possible.
-Roman

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