"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

"The Phantom"
"Phil Allison"


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.


** 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%.

This one does.


** Cos it is not correctly VA rated.

The manufacturer rated it,


** Nevertheless, it is not correctly rated.



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

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.

Your argument is entirely false

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.

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

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

** 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.

A transformer with unusually good regulation was presumably rated that way
by
the manufacturer for some reason.

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

But there are still lower temperature grades available.

We're talking about *this* transformer,

"The Phantom" <phantom at (no spam) aol.com> wrote in message
news:8t1u74d5s811flfu5bj3s4q36lnhp86d2l at (no spam) 4ax.com...
On Wed, 16 Jul 2008 18:27:11 -0400, "Paul E. Schoen" <pstech at (no spam) smart.net
wrote:


I beg to differ with your analysis, if you are talking about an ordinary
rectifier and capacitor circuit. As an example, I simulated a FWB with a
12
VAC nominal output transformer with 1 ohm series resistance, and a load
of
12 ohms, and a capacitor of 100,000 uF, which should produce the highest
possible current peaks. The simulation shows peak currents of 3.7 amps.
With 1000 uF, the peaks are 3.2 amps. Now, during the charging period,
with
100,000 uF, the peaks start at 14.6 amps and then diminish to 4.3 amps at
0.5 seconds. In the first 200 mSec, the tranny is supplying 34 watts, but
then settles down to 15.8 watts when the capacitor is fully charged. At
that time, the load is essentially pure DC, and the resistor dissipates
13.9 watts. So only about 2 watts is left, and that is shared among the
rectifiers (305 mW each), and the tranny (about 0.8 watts).

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

"The Phantom"
"Phil Allison"

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

** 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.

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 !!!

** How completely ASININE !!

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


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
instead of just forth.

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 !!!


Generally true, but there _are_ those exquisite ferroresonant
transformers

"The Phantom"
"Phil Allison"

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

** 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.

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

"The Phantom" <phantom at (no spam) aol.com> wrote in message
news:urmv74l94pn1n49mjsijnevvjdb5tlohtp at (no spam) 4ax.com...

SNIP

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

"Paul E. Schoen"
"Phil Allison"

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.

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.