Estimating transfomer current rating?

[Terry Pinnell]

John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that Terry Pinnell <terrypinDELETE@THES
Edial.pipex.com> wrote (in <7eujh012mquj4qarfp4pk7tm3fr1rj20m7@4ax.com&gt
about 'Estimating transfomer current rating?', on Wed, 11 Aug 2004:

But I still don't see how even a protracted series of such tests is
going to tell me with any accuracy what I can expect using DC loads at
various voltages in the range of my supply.

If you are using a series regulator following a bridge rectifier with a
large filter/reservoir capacitor, the a.c. secondary current will be
between 1.6 and 1.8 times the d.c. load current. It has a peaky
waveform, so you need a true r.m.s. meter to measure it, or use your
scope. The waveform is close to repeated half-cycles of a higher
frequency than 50 Hz, maybe 150 Hz, interleaved by zero-current periods.
So you can take the r.m.s. value as the peak value divided by sqrt(2)
and then divided by the ratio of the duration of the pulse to the whole
half-period, i.e. 3, if the pulse looks like 150 Hz.

The load voltage is irrelevant, because the rectifier always produces
the full voltage across the capacitor.

All you need to check is that the transformer doesn't get too hot with
the maximum current you want to draw from it.

Thanks. Must that be done with the transformer removed from the large
metal case of the power supply? That can get too hot to touch and
obviously heats up the transformer too. But if the 'hand test' is done
with the transformer removed, and a maximum current thus empirically
determined, then how reliable is it for regular use, when the
transformer will be heated by the hotter case? That's the 'snag' I
referred to.


--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Terry Pinnell]

legg <legg@nospam.magma.ca> wrote:

On Wed, 11 Aug 2004 11:56:43 +0100, Terry Pinnell
terrypinDELETE@THESEdial.pipex.com> wrote:

I've since implemented the alternative I suggested. Currently (sorry
g&gt I have the secondary delivering 2.2A (an arbitrary first choice),
while I monitor transformer case temperature with my DVM.

But I still don't see how even a protracted series of such tests is
going to tell me with any accuracy what I can expect using DC loads at
various voltages in the range of my supply.

The rms current values in the secondary winding are directly
substitutable. The 1.8 relationship of RMSin/DCout is typical of
capacitive rectifier filters with 15% ripple.

Theres no reason why you couldn't bodge a capacitive rectifier-filter
onto your winding as a test load.

When shopping for suitable junk transformers, it's often wise to take
a meter or two and scrap paper with you.

L values give turns ratios.
R values give first-order current-generated rises.

Otherwise, why not think of other uses for those extra windings? While
neccessity may be the mother of invention, opportunity or incident is
it's father.

RL

Thanks. Re the extra two windings, as mentioned in my opening post I
do have them accessible from the front panel. Three terminals, giving
me three AC voltages.

With the other couple of power supplies I've made over the years, I've
typically used extra secondaries for fixed DC voltages, usually 12V
and 5V (sometimes just smoothed, normally fully regulated with 78xx).

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[John Woodgate]

I read in sci.electronics.design that Terry Pinnell <terrypinDELETE@THES
Edial.pipex.com> wrote (in <3uokh0hr1tpjjmaatq378trl8mkcpcg467@4ax.com&gt
about 'Estimating transfomer current rating?', on Wed, 11 Aug 2004:

Terry Pinnell <terrypinDELETE@THESEdial.pipex.com> wrote:


BTW, I was thinking of sticking my temperature sensor flush against
the transformer and measuring temp with my DVM while using heavyish
currents. But the heat dissipated from the 2N3716 mounted onto the
rear of the case would grossly distort the result. To use that
approach I'd have to disconnect the secondary, which of course means I
can't use the unit for normal bench work.

I've had time for testing the isolated secondary at only two loads so
far, but here are the results. The temperature measurement was with
the miniature sensor that comes with my DVM, taped to top of
transformer. Case was open though.

Time Load (A) Temp (C) V (AC RMS)
---- -------- -------- ----------
11:03 0 25 25.4

11:03 2.20 25 22.0
12:00 2.18 42 21.9 OK to hold.

Allowing for it to get a bit hotter inside the (ventilated?) case,

that's about top whack then: 2 A, and about 22 V on-load.

12:03 2.92 42 20.6
12:50 2.87 53 20.4 Too hot to hold

So your d.c output is limited to 1.1 to 1.2 A.

--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[John Woodgate]

I read in sci.electronics.design that Terry Pinnell <terrypinDELETE@THES
Edial.pipex.com> wrote (in <gkpkh09q16hr9u8rv5jlj328iv9jl8aana@4ax.com&gt
about 'Estimating transfomer current rating?', on Wed, 11 Aug 2004:

Thanks. Must that be done with the transformer removed from the large
metal case of the power supply? That can get too hot to touch and
obviously heats up the transformer too.

What is heating it up, then? It certainly should NOT get 'too hot to
touch.'
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Rich Grise]

Terry Pinnell wrote:

"Henry Kolesnik" <kolesnik@sbcglobal.net> wrote:

Get a variable load like headlight bulbs and keep loading it up till it
gets
hot after several hours so the temp will stabilize.. If you can
comfortably hold your hand on it you've found a load that it'll be happy
with and should
work for years. Remember that half wave vs full wave recitifiers will
have double utilization factor.

Thanks, but I mentioned what IMO is the snag with that up-thread in
Message-ID: <an7hh0lsmino144ner517lu3uufc0ebjd5@4ax.com

I've since implemented the alternative I suggested. Currently (sorry
g&gt I have the secondary delivering 2.2A (an arbitrary first choice),
while I monitor transformer case temperature with my DVM.

But I still don't see how even a protracted series of such tests is
going to tell me with any accuracy what I can expect using DC loads at
various voltages in the range of my supply.

What this will tell you is the total power dissipation at the point
where you decided the temp. rise is just right. Whatever the output
VA is, regardless what secondaries you use in any combination, that's
the output VA. i.e., just see to it that the total doesn't go over
whatever the load is now.

That's actually pretty much all there is to it - it's about
transformer losses anyway, right?

Cheers!
Rich

 

[Terry Pinnell]

John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

Allowing for it to get a bit hotter inside the (ventilated?) case,
that's about top whack then: 2 A, and about 22 V on-load.

12:03 2.92 42 20.6
12:50 2.87 53 20.4 Too hot to hold

So your d.c output is limited to 1.1 to 1.2 A.

Thanks, John. What seemed a relatively small reduction in bulk was
plainly rather more significant than I first thought. I'd guessed at
least 2A, but will have to be content with much less. Not that I often
need such heavy currents (and I have other supplies I could use in
most cases). But it was comfortable to be able to largely ignore the
setting.

Anyway, I needed to recalibrate the current limiting and current meter
sections, so this removes any excuse for postponing it.

Thanks to all for the advice on this.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Terry Pinnell]

John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that Terry Pinnell <terrypinDELETE@THES
Edial.pipex.com> wrote (in <gkpkh09q16hr9u8rv5jlj328iv9jl8aana@4ax.com&gt
about 'Estimating transfomer current rating?', on Wed, 11 Aug 2004:

Thanks. Must that be done with the transformer removed from the large
metal case of the power supply? That can get too hot to touch and
obviously heats up the transformer too.

What is heating it up, then? It certainly should NOT get 'too hot to
touch.'

As mentioned in
Message-ID: <an7hh0lsmino144ner517lu3uufc0ebjd5@4ax.com>
it's the series power transistor, a 2N3716 mounted on the rear of the
case.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[John Woodgate]

I read in sci.electronics.design that Terry Pinnell <terrypinDELETE@THES
Edial.pipex.com> wrote (in <jnemh012sgk4eeb8m58ejqdt2tekct7b7j@4ax.com&gt
about 'Estimating transfomer current rating?', on Thu, 12 Aug 2004:

As mentioned in
Message-ID: <an7hh0lsmino144ner517lu3uufc0ebjd5@4ax.com
it's the series power transistor, a 2N3716 mounted on the rear of the
case.

I'd fit a bigger heat sink to it, then.
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Robert Baer]

John Woodgate wrote:

I read in sci.electronics.design that Robert Baer
robertbaer@earthlink.net> wrote (in <4119F1D4.1E0462C0@earthlink.net&gt
about 'Estimating transfomer current rating?', on Wed, 11 Aug 2004:
I have hand-wound hundreds of power transformers for personal use, and
*NEVER* had to worry or compensate for "copper loss". Never ran out of
window space, either.

Your windings have zero resistance? Maybe you are using a different
definition of copper loss than I^2R?

Now core loss is someting that cannot be avoided, but can be safely
ignored for 50W and higher power transformers (small percentage of core
rating).

This depends on the ratio of window height to limb width. For scrapless
and even semi-scrapless laminations, the copper loss considerably
exceeds the iron loss if proper allowance on maximum induction is made
fro high mains voltage. With modern silicon iron, an iron (hysteresis)
loss of 5 W/kg at 1.5 T is typical.
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

Not zero resistance; impossible. The total I^2R losses were small
compared to core magnetization losses.

 

[John Woodgate]

I read in sci.electronics.design that Robert Baer
<robertbaer@earthlink.net> wrote (in <411A6FE3.875150E2@earthlink.net&gt
about 'Estimating transfomer current rating?', on Thu, 12 Aug 2004:

John Woodgate wrote:

I read in sci.electronics.design that Robert Baer
robertbaer@earthlink.net> wrote (in <4119F1D4.1E0462C0@earthlink.net&gt
about 'Estimating transfomer current rating?', on Wed, 11 Aug 2004:
I have hand-wound hundreds of power transformers for personal use, and
*NEVER* had to worry or compensate for "copper loss". Never ran out of
window space, either.

Your windings have zero resistance? Maybe you are using a different
definition of copper loss than I^2R?

Now core loss is someting that cannot be avoided, but can be safely
ignored for 50W and higher power transformers (small percentage of core
rating).

**********************************************
This depends on the ratio of window height to limb width. For scrapless
and even semi-scrapless laminations, the copper loss considerably
exceeds the iron loss if proper allowance on maximum induction is made
fro high mains voltage.
**********************************************
With modern silicon iron, an iron (hysteresis)
loss of 5 W/kg at 1.5 T is typical.

Not zero resistance; impossible. The total I^2R losses were small
compared to core magnetization losses.

I'd like to know how you achieved that, in view of my comments
highlighted above. Ye canna change the laws o'physics, Cap'n.
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk