# Calculating DC Output Current From Unregulated AC Transformer

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#### Phil Allison

Jan 1, 1970
0
"The Phantom"
I have a transformer with a nominal 24 volt secondary, rated at 8 amps.
It
has a measured series resistance (secondary plus reflected primary) of

** This is a fabricated falsehood - the numbers simply do not add up.

A 192VA rated transformer does not have 4% regulation - correctly rated
it has 8%.

Look up makers data if you doubt this.

The oft quoted ratio of 1.6 for AC amps to DC amps applies ONLY at full
load and for a correctly rated transformer.

BTW:

Percent voltage regulation and percent power loss with resistive load are
virtually the same numbers (ignoring I mag loss).

...... Phil

P

#### Phil Allison

Jan 1, 1970
0
"The Phantom"
"Phil Allison"
This one does.

** Cos it is not correctly VA rated.

This transformer has no maker's identification, so I can't look it up.

** So you also have no idea what its VA rating is.

However, the primary (it's a 60 Hz transformer) says 120 VAC and it
measures .961 ohms, cold.

** So it is a circa 360 VA transformer.

The *correctly* rated secondary load is not 8 amps - but more like 15.

The oft quoted ratio of 1.6 for AC amps to DC amps applies ONLY at full
load and for a correctly rated transformer.

...... Phil

P

#### Paul E. Schoen

Jan 1, 1970
0
The Phantom said:
I have a transformer with a nominal 24 volt secondary, rated at 8 amps.
It
has a measured series resistance (secondary plus reflected primary) of
about .125 ohms. I connected a bridge rectifier consisting of 4 80 amp
Schottky diodes, and a real 100,000 uF capacitor.

If you simulate this, a DC load which gives 8 amps RMS in the secondary
may
give a DC current of less than 4 amps. The ratio of secondary RMS
current
to DC load current will exceed 2 to 1 if the transformer is much larger
than this, with a series resistance less than this transformer has.

I posted a partial analysis over on ABSE in which I indicate that the
grid
waveform has a large effect on the RMS to DC current ratio in these
rectifier circuits.

I have not looked at the analysis, but I did find an error in my analysis
as stated above, although it does not change the essential fact that the
transformer will not be overloaded if you keep the DC current out to about
50% of the AC current rating.

My error was that I used the voltage and current out of the transformer as
a measure of the power it was delivering, and that is correct in a sense,
but the internal resistance sees an RMS current of about 1.8 amps, for a
power dissipation of 3.24 watts, and not 0.8. I found it easier to use an
external resistance for the simulation. This model would be for a 12 VAC
transformer rated at 2 amps (24 VA) with 2/12 = 16.7% regulation. Larger
transformers will generally have better regulation, partly because they do
not have as much surface area to volume, and cannot as easily get rid of
internal heat by convection.

Simulating your circuit with a 3.3 ohm load, I get Pin = 142W, Pout = 127W,
Iin = 8.14A, Iout=4.39A. The internal resistance of the tranny dissipates
8.7 watts, and the diodes 1.7 watts each. The peak current is 19.8 amps.
The Iin/Iout is 1.85. Using a transformer with less internal resistance, or
better regulation, will give a ratio over 2:1, but it will then be a
transformer with a much higher rating, or rated much more conservatively
than normal (as even this one seems to be). New ASCII file follows:

Paul

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P

#### Phil Allison

Jan 1, 1970
0
"The Phantom"
"Phil Allison"
The manufacturer rated it,

** Nevertheless, it is not correctly rated.

I am taking the manufacturer's word for it. Printed on the transformer is
the
designation 24 volts, 8 amps.

** Nevertheless, it is not correctly rated.

That would depend on the insulation system the manufacturer used, and the
resultant allowable temperature rise, wouldn't it?

** The 8% figure is for the lowest temp grade insulation in common use.

Using higher temp grade will only increase the figure.

....... Phil

P

#### Phil Allison

Jan 1, 1970
0
"John Popelish"
If all transformers were manufactured to a single regulation and
temperature rise standard,

** The vast majority on offer do.

The oft quoted ratio of circa 1.6 applies to stock lines transformers.

...... Phil

P

#### Phil Allison

Jan 1, 1970
0
"The Phantom"

For example, for the 24 volt, 8 amp transformer I've mentioned in this

** An incorrectly rated example.

Plucked out of his arse.

The
recommendation that John Fields made, to assume Irms/Idc = 2, which he
says will
always be safe, may not be safe if you are using a transformer with good
regulation and if your grid waveform is a good sinusoid.

** A transformer with unusually good regulation ALSO has unusually LOW temp
rise.

Which wipes you asinine case out.

Piss off.

....... Phil

P

#### Phil Allison

Jan 1, 1970
0
"The Phantom"
"Phil Allison"
A transformer with unusually good regulation was presumably rated that way
by
the manufacturer for some reason.

( snip even worse DRIVEL )

** How completely ASININE !!

A transformer with unusually good regulation = a transformer that is
being under utilised !!!

No need for exists for transformer makers to make special high regulation
models - the customer simply picks a model with more VA capacity than
strictly needed !!!

Piss off - you PITA fool.

....... Phil

P

#### Phil Allison

Jan 1, 1970
0
"The Phantom Congenital LIAR "

But there are still lower temperature grades available.

** No.

** No *we* are not - asshole.

Using an isolated example to prove a case is false logic.

Using an isolated example that only you know about is a classic debating
cheat.

Combining the two, as you have done, is the act of a desperate LIAR.

You have no case to put - **** off.

....... Phil

P

#### Phil Allison

Jan 1, 1970
0
"John Fields"
Explain that a transformer that might be seen as being under-utilized
while feeding a resistive load might have to be as big as it is in
order to drive reactive loads which pump current back and forth

So what is the context ?

Generally true, but there _are_ those exquisite ferroresonant
transformers

So what is the context ?

Is being a context shifter related to being a shirt lifter ?

....... Phil

P

#### Phil Allison

Jan 1, 1970
0
"The Phantom"

Piss off - you PITA BLOODY TROLL.

...... Phil

M

#### [email protected]

Jan 1, 1970
0
Sounds like you have a quibble with the manufacturer, not with me. I only
report the transformer's measured parameters and labelling.

Could well be that the manufacturer de-rated the transformer slightly
to reduce their liability in case of transformer failure. For
example, if the transformer could really take up to 100 VA, the
manufacturer said the transformer is good for up to 80 VA

Michael

P

#### Paul E. Schoen

Jan 1, 1970
0
The Phantom said:
If we assume Irms/Idc <= 2, and if it isn't, because the transformer has
low
internal resistance, then the secondary current will be higher than we
expected.
This may or may not cause a problem, but there is probably some reason
the
secondary current had a rated value. Perhaps the transformer was
originally
used in an enclosure without adequate ventilation, and cannot be allowed
to get
as hot as it might otherwise. Whatever the reason for the given
secondary
current rating, exceeding it is at the user's risk, and the user ought to
know
about it. If the transformer/rectifier combination will be used in a
better
environment than the transformer without rectifier was originally
designed for,
then there may be no problem. But if Irms/Idc could be > 2 in some
circumstances, the user should know what those circumstances might be.

Interestingly, with respect to your reply to whit3rd and the behavior he
described, the analysis I came up with seems to have a limit to the ratio
Irms/Idc. I allowed the transformer internal resistance to become orders
of
magnitude lower than would ever be possible in the real world, without
superconducting wire, and the calculated Irms/Idc never got over 4, no
matter
how peaky the waveform. I wonder what simulation would show. It might
be
numerically difficult for a simulator to correctly calculate the RMS
value of a
spike of current whose duration is one billionth of the period!
What will limit the narrow high current spikes you are talking about will
ultimately be the inductance of the other circuit components and their
internal resistance, and the diode characteristics. Also, the current
spikes are a function of the rate of rise of the waveform at the time of
onset of conduction (which also is not immediate), and the size of the
capacitor. For low values of capacitance, the conduction occurs on a faster
rising portion of the waveform, but the peak current is limited because the
capacitance is smaller. As you increase the capacitance, the spikes are
greater, and RMS current can be high enough while charging the capacitor to
exceed the transformer's specifications and overheat or even
instantaneously blow the windings, but that would be an extreme case.

For real life, reasonable circuits, after a large capacitor has charged to
its final value, the diodes will conduct only when the waveform is greater
than the diode forward drop and the minimum output voltage excursion (Max
DC - P-P ripple), and this will occur very close to the nearly flat top of
the sine wave, so the slew rate is very slow. Eventually the lower slew
rate is balanced by the high capacitance so the peak current will be
limited, and the Irms/Idc reaches a maximum value, such as the figure of 4
you mentioned.

Ordinary waveform distortion (and normal transformer saturation) will
usually cause a flattening of the peaks and an even slower slew rate. If
the waveform somehow has sharp peaks, like a triangle wave, this indicates
high frequency harmonics, and peak currents can become extreme. But that is
a very rare situation, and it may cause transformer overheating even with
an AC load, due to excessive saturation. We must assume reasonable power
quality.

Paul

P

#### Phil Allison

Jan 1, 1970
0
"Paul E. Schoen"
Ordinary waveform distortion (and normal transformer saturation) will
usually cause a flattening of the peaks

** Where in time are the current peaks from Imag ( or Isat) ???

Engage brain, before putting mouth in gear.

..... Phil

P

#### Paul E. Schoen

Jan 1, 1970
0
Phil Allison said:
"Paul E. Schoen"

** Where in time are the current peaks from Imag ( or Isat) ???

Engage brain, before putting mouth in gear.

The transformer will have current peaks that coincide with the peaks of
applied sinusoidal voltage. The output of the transformer will have the
voltage peaks flattened where the current peaks occur.

Paul

P

#### Phil Allison

Jan 1, 1970
0
"Paul E. Schoen"
"Phil Allison"
The transformer will have current peaks that coincide with the peaks of
applied sinusoidal voltage.

** No way - there is a 90 degree phase lag ( ie inductive load ..... )
that means currents peaks are at or near the supply voltage zero crossings.

The output of the transformer will have the voltage peaks flattened where
the current peaks occur.

** See above.

Hard to believe that someone who spends so much time * wrangling * with
some very serious transformers was unaware.

..... Phil

E

#### ehsjr

Jan 1, 1970
0
I'm sorry if I did not clarify this, but I do not need an introductory
course in sizing transformers.

Yes, you do. Your question reveals that you do not
understand the introductory material.

The whole point was moving beyond the
generic overestimates towards the real science in what is actually
necessary. I lecture towards the end of saying, ok, but this is
common knowlege, what else do you know? Maybe it takes a village,
maybe no one person has the entire answer but it seems to get there we
need to have some lecture, some dispelling of comfort zones and get
right down to the actual criteria necessary and have that proven
through real worl examples of success or failure, not just saying "use
a bigger hammer", until it is proven to be needed.

Yes, I will argue with an answer when I ask for an equation and
someone tells me otherwise. I have built plenty of PSU over the
years, if I needed to know what I have already done successfully then
I would have asked a different question? No offense intended, but you
need to focus on what I asked, as do others. If they are ignorant of

Sorry, but you seem unable/unwilling to understand the answers
you've been given.

In your case, you can try a different path. Rather than posting
and fighting, build the supply with your transformer. Construct
a variable load. Run the thing, set the current to wherever
you want, monitor the voltage to see what you can get. Watch the
temperature rise and keep it within safe limits - or test to
destruction if you have deep pockets and time to keep burning
out transformers. Arguing with the replies has gotten you nowhere,
so far, so the testing approach might be more productive for you.

Ed

P

#### Paul E. Schoen

Jan 1, 1970
0
Phil Allison said:
"Paul E. Schoen"
"Phil Allison"

** No way - there is a 90 degree phase lag ( ie inductive load
..... ) that means currents peaks are at or near the supply voltage zero
crossings.

** See above.

Hard to believe that someone who spends so much time * wrangling * with
some very serious transformers was unaware.

OK, you are right on this point, and the flattened voltage waveforms and
peaked current waveforms I have seen may be due to the fact that the
circuit breaker loads may be highly inductive, or contain current
transformers with non-linear diode and capacitor loads that draw more
current at the peaks.

For breaker testing, I'm usually looking at the current, and I often see a
peaked waveform, while the voltage appears to be flattened. But it could be
a reflection of distortion on the input waveform. When drawing several
hundred amps from a 480 VAC line, to pump 30,000 amps or more into a
breaker, strange things can happen.

The following waveforms show flattening as I have seen, but maybe it was a
clipped waveform to begin with:
http://www.bookcracker.com/transformer/

This paper shows saturation occurring near the zero crossing, when the
volt-seconds of the primary is reached:
http://www.rane.com/pdf/ranenotes/Unwinding Distribution Transformers.pdf

and this also shows current peaks just before each zero crossing, and no
apparent flattening.
http://www.ibiblio.org/kuphaldt/electricCircuits/AC/AC_9.html

Here is some good information (but mostly a copy of the above):

I seem to recall ferroresonant regulating transformers having flattened
waveforms.
http://www.alpha.com/files/documents/016538B0002_revB.pdf

I found this interesting:
http://www.ece.mtu.edu/faculty/bamork/FR_WG/Panel/WallingFerroPanel.pdf

Paul

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