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ehsjr...
Posted: Sun Jul 13, 2008 10:49 pm
Guest
John Fields wrote:
Quote:
On Sat, 12 Jul 2008 22:00:45 GMT, ehsjr
e.h.s.j.r.removethespampunctuation at (no spam) bellatlantic.net> wrote:


John Fields wrote:

On Sat, 12 Jul 2008 07:56:53 -0700 (PDT), emailaddress at (no spam) insightbb.com
wrote:



On Jul 11, 2:34 pm, ehsjr
e.h.s.j.r.removethespampunctuat... at (no spam) bellatlantic.net> wrote:



I don't want to argue - that seems your intent.
I know you want a magic formula - you have made that
clear. There is no single formula for you. Those who
have replied have made that clear.

We've tried to show you some of the factors that are
involved. You insist on ignoring them. So, we can't
help you, no matter how hard we try. Sorry.

A formula yes, but no it's not magic. It simply requires considering
all the significant variables which I'd hoped others would assist
with, but obviously nobody else wants to do more than argue instead of
putting thought into what such an equation would look like. If you
say "it depends", then that should be in an equation. If you say some
other thing depends too, then that too can be put into the equation.

There is an equation that could be made. How accurate the answer from
it would depend on how complete it was.


---
Funny, meteorologists feel the same way, and yet it seems that
detecting that last elusive flap of butterfly wings has been
perpetually beyond their grasp.

JF

:-) Nice analogy, wonder if he'll get it.

What is interesting to me is the apparent contradiction in
the op's thinking. He chooses to lecture respondents with
his thinking, rather than attempt to understand what they
are saying. Or, if not lecture, argue with points that are
made. I've seen this a few times here - someone comes here,
states that they don't know about X, and asks a question.
But then they argue with the answer! What kind of thinking
is that? Is it the "new way of thinking" or ???


---
I've seen it a few times also, and I think it's a way of trying to
save face by not admitting that the gift that was given incurred
emotional indebtedness.

That is, if you give me a gift of knowledge and I dis it, I'll still
have the gift but in my mind I won't owe you anything for it.

Make sense?

JF

Ah. Thanks.

Ed
ehsjr...
Posted: Sun Jul 13, 2008 10:51 pm
Guest
Phil Allison wrote:
Quote:
"John Fields"


I've seen it a few times also, and I think it's a way of trying to
save face by not admitting that the gift that was given incurred
emotional indebtedness.

That is, if you give me a gift of knowledge and I dis it, I'll still
have the gift but in my mind I won't owe you anything for it.

Make sense?



** If you imagine some little a brat throwing a toy that is not the one he
wanted back at Santa.



..... Phil



Thanks! That example makes the concept clear.

Ed
...
Posted: Wed Jul 16, 2008 8:13 am
Guest
On Jul 12, 2:29 pm, "Paul E. Schoen" <pst... at (no spam) smart.net> wrote:
Quote:
mrdarr... at (no spam) gmail.com> wrote in message

news:0631a23f-0a93-4a34-a703-b3bd4e22ba06 at (no spam) j22g2000hsf.googlegroups.com...



On Jul 12, 7:40 am, emailaddr... at (no spam) insightbb.com wrote:
..snip..

However, it is fairly beside the point that I am looking for
a universal equation that ignores all of this. Many people seem to be
saying the rest matters and yes of course it does - but it is still
theoretically expressable in an equation with the factors that change
as variables.

Universal equation. Here you go.

http://en.wikipedia.org/wiki/Heat_equation
http://en.wikipedia.org/wiki/Newton%27s_law_of_cooling#Newton.27s_law...

Also check out "Introduction to Heat Transfer" by Incropera and
DeWitt.

The game is to keep the transformer wires at only a moderate
temperature higher than ambient, or you will melt the wiring
insulation.

Knock yourself out.

Michael

These equations are only part of the overall solution. Transformers are
complex physical entities with many materials having different thermal
characteristics, and the temperature of any spot in the transformer depends
on the amount of power being applied, time, and the way in which heat is
radiated, conducted, or convected away from the source of the heat. The
winding insulation is rated at a certain temperature according to the
insulation class, and may vary from about 105 C (Class A) to 180 C (Class
H).

http://www.engineeringtoolbox.com/nema-insulation-classes-d_734.html

That is for motors (and I think it applies also to transformers), but here
is information more specific to trannies, and allows for hot spots:http://www.jeffersonelectric.com/cgi-bin/site.pl?3208&dwContent_conte...

This shows a wider range of classes up to "S", which is 250 C. Now, that's
hot!http://www.pleo.com/ulsystem/eis_apply.htm

This is a detailed document with more formulas for the OP to use when he
designs his transformer:http://www.superioressex.com/uploadedFiles/News/White_Papers/emcwa-ne...

Paul



Yep, my links were meant as a starting point. Boy is he going to have
fun if he takes heat transfer coefficients from natural convection
into account.

Did you know, Excel 2003 can solve Bessel functions? (useful for
solving partial differential equations in cylindrical coordinate
systems) Type BesselJ in the HELP toolbar, then follow the "How?"
link to install the Analysis ToolPak

Michael
...
Posted: Wed Jul 16, 2008 8:17 am
Guest
On Jul 16, 2:55 am, emailaddr... at (no spam) insightbb.com wrote:
Quote:
On Jul 12, 6:00 pm, ehsjr



e.h.s.j.r.removethespampunctuat... at (no spam) bellatlantic.net> wrote:
John Fields wrote:
On Sat, 12 Jul 2008 07:56:53 -0700 (PDT), emailaddr... at (no spam) insightbb.com
wrote:

On Jul 11, 2:34 pm, ehsjr
e.h.s.j.r.removethespampunctuat... at (no spam) bellatlantic.net> wrote:

I don't want to argue - that seems your intent.
I know you want a magic formula - you have made that
clear. There is no single formula for you. Those who
have replied have made that clear.

We've tried to show you some of the factors that are
involved. You insist on ignoring them. So, we can't
help you, no matter how hard we try. Sorry.

A formula yes, but no it's not magic. It simply requires considering
all the significant variables which I'd hoped others would assist
with, but obviously nobody else wants to do more than argue instead of
putting thought into what such an equation would look like. If you
say "it depends", then that should be in an equation. If you say some
other thing depends too, then that too can be put into the equation.

There is an equation that could be made. How accurate the answer from
it would depend on how complete it was.

---
Funny, meteorologists feel the same way, and yet it seems that
detecting that last elusive flap of butterfly wings has been
perpetually beyond their grasp.

JF

:-) Nice analogy, wonder if he'll get it.

What is interesting to me is the apparent contradiction in
the op's thinking. He chooses to lecture respondents with
his thinking, rather than attempt to understand what they
are saying. Or, if not lecture, argue with points that are
made. I've seen this a few times here - someone comes here,
states that they don't know about X, and asks a question.
But then they argue with the answer! What kind of thinking
is that? Is it the "new way of thinking" or ???

Ed- Hide quoted text -

- Show quoted text -

I'm sorry if I did not clarify this, but I do not need an introductory
course in sizing transformers. 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
the answer, there is no need to reply.


http://www.despair.com/incompetence.html
whit3rd...
Posted: Wed Jul 16, 2008 11:08 am
Guest
On Jul 13, 4:35 pm, John Fields <jfie... at (no spam) austininstruments.com> wrote:

Quote:
Use Irms = 2.0 Idc and you'll always be safe .

Well, not completely, no. Consider a 1A (RMS)
rated transformer, with a diode/capacitor output.
When attached to a 1A (DC) load, the transformer
actually forward-biases the diode for a brief
time at the peak voltage, and otherwise current
from the transformer is nil. If the conduction
period is one-tenth of the full cycle period, that
means 10A current from the transformer during
the active time.

Here's where it gets mathematical: the power dissipated
in a 1A transformer with (for instance) 1 ohm winding
resistance is 1 watt, when the load is taking a simple
1A (AC) current.

Heat = 1A **2 * 1 ohm = 1 watt

When the same transformer feeds the DC load as described above, the
power is

Heat = (0.9 * 0) + (0.1 * 10A**2 * 1 ohm) = 10 watt

So a perfectly good transformer can burn up feeding a
DC rectifier and load, when a similar AC load wouldn't
bother it. That isn't always covered by a 'factor of two'
or any other rule of thumb. I've seen manufacturers offer
tables of the permissible DC output ratings for their
transformers, but you can't count on that. The AC rating
looks better, so salesmen will feed you that info first.
Paul E. Schoen...
Posted: Wed Jul 16, 2008 12:28 pm
Guest
"John Fields" <jfields at (no spam) austininstruments.com> wrote in message
news:ninr74hjcudl6v8eaf69i3mpplk45rest9 at (no spam) 4ax.com...
Quote:
On Wed, 16 Jul 2008 02:55:03 -0700 (PDT), emailaddress at (no spam) insightbb.com
wrote:


I'm sorry if I did not clarify this, but I do not need an introductory
course in sizing transformers. 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
the answer, there is no need to reply.

---
You may have _built_ plenty of PSU over the years, but from your
question re. transformer sizing it appears that all you were doing was
grunt work, putting together someone else's design, with no real
insight as to the "why" of the transformer's capacity needing to be
larger than what was required to drive the load.

Instead of being an annoying little ass you might try thanking those
of us who have given you good advice instead of pretending that,
somehow, we did you a disservice.

Oh, and by the way, Fuck you.

I usually give someone the benefit of doubt and try to come up with various
ways of looking at a problem and proposing a reasonable solution, or
pointing out the need for more information. The OP has chosen not to
comment on those posts, but only those where he feels compelled to argue
and assume a self-righteous attitude. So I second your motion. All in
favor?

Paul
Dan Coby...
Posted: Wed Jul 16, 2008 1:05 pm
Guest
<emailaddress at (no spam) insightbb.com> wrote in message news:daad0917-cdf2-4ff5-8a75-
Quote:
I'm sorry if I did not clarify this, but I do not need an introductory
course in sizing transformers. The whole point was moving beyond the
generic overestimates towards the real science in what is actually
necessary.

The usual limiting factor for the current output of a transformer is heat dissipation.

The factors that need to be considered are:
1) The amount of heat being produced.
2) How the heat is removed from where it is produced.
3) The maximum desired operating temperature.

Heat is produced in a transformer in several different ways. There are electrical
losses in the windings due to the currents in the windings. There are electrical losses
in the core materials dues to induced currents in the core There are probably other
factors related to frequency and materials.

The easiest one of these to handle quantitatively is the power (heat) lost in the
windings' resistance. As the current increases, so does the power that is lost in the
resistance of the windings. For more information (and equations) about the power loss in
a resistance see: http://en.wikipedia.org/wiki/Root_mean_square#Average_electrical_power

Short of giving you the equations for RMS current we cannot answer your question
since we do not know that the winding currents are for your application. (You mentioned
DC currents but that is pretty much meaningless for a transformer.) Please note that the
RMS current is not the same as the average current. There can be a big difference in
RMS current (and thus power lost) between a current waveform that consists of pulses
with a low duty cycle and high peak current versus a constant DC with the same
average current. If you are building power supplies, the RMS current will depend upon
(among other things) the output current, the filter capacitance, and the internal resistance
of the transformer. You have given us no information about any of the details of what
you are trying to accomplish and we really do not want to repeat the contents of
textbooks on basic electrical engineering.


How the heat is removed from where it is produced depends upon the specific
details of the design of the transformer and the environment in which is being used.
You have given us no details on your application and we really do not want to repeat
the contents of the textbooks on heat flow.


The maximum operating temperature for a transformer depends upon the materials
and upon operating considerations. As the operating temperature increases the
rates of break down of the insulation increase. In general, the higher the operating
temperature of a transformer, the shorter its life will be. There can also be other
operating considerations. For instance, the materials that are used to build a
transformer, may allow for a long life while operating at 100 C but I would not like
to have that transformer in close proximity to my skin at that temperature.
Paul E. Schoen...
Posted: Wed Jul 16, 2008 5:27 pm
Guest
"whit3rd" <whit3rd at (no spam) gmail.com> wrote in message
news:6a58988a-5bd9-4c16-af89-69652a6d8f6c at (no spam) k30g2000hse.googlegroups.com...
Quote:
On Jul 13, 4:35 pm, John Fields <jfie... at (no spam) austininstruments.com> wrote:

Use Irms = 2.0 Idc and you'll always be safe .

Well, not completely, no. Consider a 1A (RMS)
rated transformer, with a diode/capacitor output.

When attached to a 1A (DC) load, the transformer
actually forward-biases the diode for a brief
time at the peak voltage, and otherwise current
from the transformer is nil. If the conduction
period is one-tenth of the full cycle period, that
means 10A current from the transformer during
the active time.

Quote:
Here's where it gets mathematical: the power dissipated
in a 1A transformer with (for instance) 1 ohm winding

resistance is 1 watt, when the load is taking a simple
1A (AC) current.

Quote:
Heat = 1A **2 * 1 ohm = 1 watt

When the same transformer feeds the DC load as described above, the
power is


Quote:
Heat = (0.9 * 0) + (0.1 * 10A**2 * 1 ohm) = 10 watt

So a perfectly good transformer can burn up feeding a
DC rectifier and load, when a similar AC load wouldn't

bother it. That isn't always covered by a 'factor of two'
or any other rule of thumb. I've seen manufacturers offer
tables of the permissible DC output ratings for their
transformers, but you can't count on that. The AC rating
looks better, so salesmen will feed you that info first.
-----------------------------------------------------------

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).

If you can show me a circuit where you will get these 10 ampere peaks at
10% duty cycle, then I will agree that the tranny will be overloaded. But
you will probably need to use some sort of PWM, and there will also be a
lot more power being dumped into the load. If you are talking about AC to
DC rectifier circuits, it's a safe bet to design the circuit so that the DC
output voltage under load is about the same as the nominal RMS AC voltage
of the transformer, and in this case the RMS input current is 1.81/1.08 =
less than twice the output current. So the 2:1 ratio that John proposed is
very reasonable.

Paul

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Paul E. Schoen...
Posted: Thu Jul 17, 2008 2:27 pm
Guest
"The Phantom" <phantom at (no spam) aol.com> wrote in message
news:8t1u74d5s811flfu5bj3s4q36lnhp86d2l at (no spam) 4ax.com...
Quote:
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

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The Phantom...
Posted: Thu Jul 17, 2008 3:11 pm
Guest
On Thu, 17 Jul 2008 12:36:21 -0400, John Popelish <jpopelish at (no spam) rica.net> wrote:

Quote:
The Phantom wrote:

I don't think it's that difficult for the reasons you cite.

Knowing the rated secondary current means we don't have to know the allowable
temperature rise.

Only if you assume the design environment (sealed container,
forced air velocity, ambient temperature range, etc. ).
Many transformers are designed for very specific (and
varied) environments.

I don't see what this has to do with it. Of course I am assuming that the
manufacturer's rating was determined (by the manufacturer) knowing the design
environment, and we don't need to do that ourselves, and I further assume that
we know what the manufacturer's rating is, and that the transformer will be used
in its design environment with the rectifier load.

I explain in the next paragraph that if the rectifier circuit causes the same
secondary RMS current as a pure AC load, then the heating will be the same.
Therefore, we don't need to know what the heating is; we already know it's safe.
We only need to know that the transformer will be safe operating at its rated
secondary current, whether that current is a result of a pure resistive AC load,
or a rectifier load. If we don't know that (the rated secondary current in its
intended environment), then we don't know enough to safely use the transformer
even with a resistive load.

There's no point in trying to calculate the safe DC current available from a
rectifier if we don't even know the safe AC secondary current; in that case we
must resort to characterizing the transformer thermal behavior ourselves.

Quote:

We need only see to it that the RMS secondary current when
used in a bridge rectifier circuit is the same as the rated secondary current.
Then we will get the maximum possible DC current with the nearly the same
temperature rise as when a pure AC load drawing rated current is applied. The
actual temperature rise may not be exactly the same because the higher crest
factor pulses of current drawn by the bridge cause a little extra IR drop in the
primary wire, and change the core loss slightly. I tried to measure this
effect, but it is so small as to be completely negligible.

Yes, that effect is more pronounced for cheap, low
efficiency transformers than it is for high efficiency units
that have plenty of iron and are rated to perform well with
+10% or more line voltage and possibly at 50Hz, ans well as
60 Hz.

So is a negligible effect which is, in some cases, more pronounced, still
negligible? :-)

I was unable to measure the effect using a cheap Radio Shack transformer; I
think it's negligible in all cases.

As a side note, I measured the core loss of the cheap Radio Shack transformer
with a wattmeter designed for accurate measurement of low power factor loads
(unloaded transformers, in other words). The loss was 2.65 watts cold and 2.42
watts hot. After make a series of measurements with the transformer hot, I
connected it, unloaded, to the wattmeter, and over the course of a couple of
hours watched the core loss drift back up from 2.42 watts to 2.65 watts.

Quote:

The regulation is easily measured and provides what would seem to be the
remaining needed information, but I found that the grid voltage waveshape has a
large effect on the result, and knowing and specifying that may be the most
uncertain part of the process.

And changes day to day and even time of day. These wave
shape effects are almost unnoticeable with a resistive
load., but can be major players for rectifiers with
capacitor input filters.

The saving grace is that the clipping of the grid waveform reduces the heating
of the transformer rather than increasing it. Therefore, if a formula is used
which assumes that the grid waveform is a good sinusoid, not flat-topped, the DC
current calculated to give the rated secondary current will be lower than the
true (measured) value, more so for transformers with good regulation.

For example, for the 24 volt, 8 amp transformer I've mentioned in this thread, a
calculation assuming an undistorted grid waveform gives a value of 3.6 amps for
the DC current to give an 8 amp RMS secondary current. But, the measured value
is 4.6 amps; the calculation is very conservative. The calculated Irms/Idc is
2.22, but the measured Irms/Idc is only 1.74.

If I change the calculation to use a grid waveform with a 2.1 millisecond flat
spot at the top of the waveform, then I get a result of about 4.6 amps. 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.

Quote:

Did you look at my post on alt.binaries.schematics.electronic? I show a method
of deriving a "formula", and discuss the results I got compared to actual
measurements.

I did a fly over and saved it, but do not wade through your
math.
Phil Allison...
Posted: Thu Jul 17, 2008 5:28 pm
Guest
"The Phantom"
"Phil Allison"
Quote:


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.



Quote:
** 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.


Quote:
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
Phil Allison...
Posted: Thu Jul 17, 2008 5:31 pm
Guest
"John Popelish"

Quote:
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
Phil Allison...
Posted: Thu Jul 17, 2008 5:39 pm
Guest
"The Phantom"


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

** An incorrectly rated example.

Plucked out of his arse.


Quote:
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
...
Posted: Thu Jul 17, 2008 6:14 pm
Guest
On Jul 17, 7:05 pm, The Phantom <phan... at (no spam) aol.com> wrote:
Quote:
On Fri, 18 Jul 2008 09:56:01 +1000, "Phil Allison" <philalli... at (no spam) tpg.com.au
wrote:





"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 !!!

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
The Phantom...
Posted: Thu Jul 17, 2008 6:40 pm
Guest
On Fri, 18 Jul 2008 08:39:18 +1000, "Phil Allison" <philallison at (no spam) tpg.com.au>
wrote:

Quote:

"The Phantom"


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. Whatever the temperature rise with rated
secondary current, it will be exceeded if the secondary current is greater than
its rating. If the user wants to do that, it's his choice. He should be aware
that under certain conditions, some transformers may give Irms/Idc greater than
2 in rectifier service if used at the stated rating, and this may or may not
cause a problem, depending on the enviroment and other factors.

Quote:
Which wipes you asinine case out.

Piss off.


...... Phil
 
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