Calculating DC Output Current From Unregulated AC Transforme

On Jul 12, 6:00 pm, ehsjr
<e.h.s.j.r.removethespampunctuat...@bellatlantic.net> wrote:

John Fields wrote:
On Sat, 12 Jul 2008 07:56:53 -0700 (PDT), emailaddr...@insightbb.com
wrote:

On Jul 11, 2:34 pm, ehsjr
e.h.s.j.r.removethespampunctuat...@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.

 

On Jul 12, 2:29 pm, "Paul E. Schoen" <pst...@smart.net> wrote:

mrdarr...@gmail.com> wrote in message

news:0631a23f-0a93-4a34-a703-b3bd4e22ba06@j22g2000hsf.googlegroups.com...



On Jul 12, 7:40 am, emailaddr...@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

 

On Jul 16, 2:55 am, emailaddr...@insightbb.com wrote:

On Jul 12, 6:00 pm, ehsjr



e.h.s.j.r.removethespampunctuat...@bellatlantic.net> wrote:
John Fields wrote:
On Sat, 12 Jul 2008 07:56:53 -0700 (PDT), emailaddr...@insightbb.com
wrote:

On Jul 11, 2:34 pm, ehsjr
e.h.s.j.r.removethespampunctuat...@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]

On Jul 13, 4:35 pm, John Fields <jfie...@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.

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.

 

[The Phantom]

On Wed, 16 Jul 2008 18:27:11 -0400, "Paul E. Schoen" <pstech@smart.net>
wrote:

"whit3rd" <whit3rd@gmail.com> wrote in message
news:6a58988a-5bd9-4c16-af89-69652a6d8f6c@k30g2000hse.googlegroups.com...
On Jul 13, 4:35 pm, John Fields <jfie...@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.

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.
-----------------------------------------------------------

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.

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

============================================================

Version 4
SHEET 1 880 680
WIRE 240 144 128 144
WIRE 288 144 240 144
WIRE 384 144 352 144
WIRE 416 144 384 144
WIRE 512 144 416 144
WIRE 528 144 512 144
WIRE 528 160 528 144
WIRE 128 192 128 144
WIRE 416 192 416 144
WIRE 240 256 240 144
WIRE 288 256 240 256
WIRE 416 256 352 256
WIRE 416 304 416 256
WIRE 528 304 528 240
WIRE 528 304 416 304
WIRE 640 304 528 304
WIRE 640 336 640 304
WIRE 128 368 128 272
WIRE 288 368 128 368
WIRE 384 368 384 144
WIRE 384 368 352 368
WIRE 128 480 128 368
WIRE 288 480 128 480
WIRE 416 480 416 304
WIRE 416 480 352 480
FLAG 640 336 0
FLAG 512 144 V+
SYMBOL schottky 288 160 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D2
SYMATTR Value 1N5818
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL polcap 400 192 R0
SYMATTR InstName C2
SYMATTR Value 100000ľ
SYMATTR Description Capacitor
SYMATTR Type cap
SYMATTR SpiceLine V=63 Irms=2.51 Rser=0.025 MTBF=5000 Lser=0 ppPkg=1
SYMBOL voltage 128 176 R0
WINDOW 3 -11 133 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 24 44 Left 0
SYMATTR Value SINE(0 18 60 0 0 0 100)
SYMATTR SpiceLine Rser=1
SYMATTR InstName V1
SYMBOL res 512 144 R0
SYMATTR InstName R1
SYMATTR Value 12
SYMBOL schottky 352 272 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D3
SYMATTR Value 1N5818
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky 288 384 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D1
SYMATTR Value 1N5818
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky 352 496 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D4
SYMATTR Value 1N5818
SYMATTR Description Diode
SYMATTR Type diode
TEXT 184 528 Left 0 !.tran 1

 

[The Phantom]

On Thu, 17 Jul 2008 19:56:15 +1000, "Phil Allison" <philallison@tpg.com.au>
wrote:

"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
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.

With 121.7 VAC applied to the primary, the secondary voltage measures 26.9
VAC, unloaded.

With 121.7 VAC applied to the primary, and rated load of 8 amps drawn from
the secondary with a pure resistive load, the secondary voltage measures
25.85 VAC.

This is a change of 1.05 volts for a load of 8 amps; 1.05/8 = .13125 ohms.

Look up makers data if you doubt this.

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

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

The secondary (24 volts, 8 amps nominal) measures .080 ohms.

With 121.7 VAC applied to the primary and no load, the secondary voltage
measures 26.9 VAC, for an approximate turns ratio of 4.524:1. The .961 ohm
primary, divided by the turns ratio squared, gives .04695 ohms reflected to
the secondary. Adding the measured .080 ohm secondary gives a total .12695
ohms, which is *about* .125 ohms.

This is what I would expect to measure at the secondary if the primary were
shorted. What I actually measure is .131 ohms; the transformer is still a
little warm from having been under load a few hours ago. See the picture
of the impedance meter's display over on ABSE.

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

 

[The Phantom]

On Thu, 17 Jul 2008 21:53:47 +1000, "Phil Allison" <philallison@tpg.com.au>
wrote:

"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, and printed the rating on the outer paper.

Look up makers data if you doubt this.

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.

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

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?

Maybe it's only class O. Maybe they wanted to be very conservative.

The *correct* rating depends on the allowable temperature rise, which,
presumably, the manufacturer knew.

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

Wouldn't this ratio depend somewhat on the regulation of the transformer?

A transformer with poorer regulation would have a wider conduction angle, and a
smaller ratio of AC amps to DC amps.

For example, I have a cheap Radio Shack transformer rated at about 15 VA, and
its measured regulation is about 15%. The AC amps to DC amps ratio at rated
secondary RMS current is about 1.53.

The 192 VA transformer discussed above, with its better regulation has a ratio
of AC amps to DC amps of about 1.7, at the manufacturer's rated secondary
current.

That's not much variation, and the 1.6 figure would be a good rule of thumb,
although the actual number varies with the regulation of the transformer.
Perhaps slightly different ratios could be specified for different values of
regulation.

I varied the load current with the 192 VA transformer discussed above, and
measured slightly varying ratios at the various load levels:

DC current ratio of AC amps to DC amps

4.03 1.799
4.49 1.781
5.95 1.744
7.17 1.719
12.33 1.633

..... Phil

 

[John Popelish]

The Phantom wrote:

On Thu, 17 Jul 2008 21:53:47 +1000, "Phil Allison" <philallison@tpg.com.au
wrote:
(snip)
** 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?

Maybe it's only class O. Maybe they wanted to be very conservative.

The *correct* rating depends on the allowable temperature rise, which,
presumably, the manufacturer knew.

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

Wouldn't this ratio depend somewhat on the regulation of the transformer?

A transformer with poorer regulation would have a wider conduction angle, and a
smaller ratio of AC amps to DC amps.

For example, I have a cheap Radio Shack transformer rated at about 15 VA, and
its measured regulation is about 15%. The AC amps to DC amps ratio at rated
secondary RMS current is about 1.53.

The 192 VA transformer discussed above, with its better regulation has a ratio
of AC amps to DC amps of about 1.7, at the manufacturer's rated secondary
current.

That's not much variation, and the 1.6 figure would be a good rule of thumb,
although the actual number varies with the regulation of the transformer.
Perhaps slightly different ratios could be specified for different values of
regulation.

I varied the load current with the 192 VA transformer discussed above, and
measured slightly varying ratios at the various load levels:

DC current ratio of AC amps to DC amps

4.03 1.799
4.49 1.781
5.95 1.744
7.17 1.719
12.33 1.633

If all transformers were manufactured to a single regulation
and temperature rise standard, the formula sought by the
O.P. would be a lot closer to practical, but your two
examples span only a part of the considerable range of such
specifications. This range is what makes a universal
formula either inaccurate if simple, or so detailed that it
is impractical to fill in all the variables if complete.

--
Regards,

John Popelish

 

[The Phantom]

On Thu, 17 Jul 2008 11:12:19 -0400, John Popelish <jpopelish@rica.net> wrote:

The Phantom wrote:
On Thu, 17 Jul 2008 21:53:47 +1000, "Phil Allison" <philallison@tpg.com.au
wrote:
(snip)
** 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?

Maybe it's only class O. Maybe they wanted to be very conservative.

The *correct* rating depends on the allowable temperature rise, which,
presumably, the manufacturer knew.

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

Wouldn't this ratio depend somewhat on the regulation of the transformer?

A transformer with poorer regulation would have a wider conduction angle, and a
smaller ratio of AC amps to DC amps.

For example, I have a cheap Radio Shack transformer rated at about 15 VA, and
its measured regulation is about 15%. The AC amps to DC amps ratio at rated
secondary RMS current is about 1.53.

The 192 VA transformer discussed above, with its better regulation has a ratio
of AC amps to DC amps of about 1.7, at the manufacturer's rated secondary
current.

That's not much variation, and the 1.6 figure would be a good rule of thumb,
although the actual number varies with the regulation of the transformer.
Perhaps slightly different ratios could be specified for different values of
regulation.

I varied the load current with the 192 VA transformer discussed above, and
measured slightly varying ratios at the various load levels:

DC current ratio of AC amps to DC amps

4.03 1.799
4.49 1.781
5.95 1.744
7.17 1.719
12.33 1.633

If all transformers were manufactured to a single regulation
and temperature rise standard, the formula sought by the
O.P. would be a lot closer to practical, but your two
examples span only a part of the considerable range of such
specifications. This range is what makes a universal
formula either inaccurate if simple, or so detailed that it
is impractical to fill in all the variables if complete.

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. 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.

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.

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.

 

[John Popelish]

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.

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.

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.

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.

--
Regards,

John Popelish

 

[The Phantom]

On Thu, 17 Jul 2008 12:36:21 -0400, John Popelish <jpopelish@rica.net> wrote:

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.

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.

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.

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.

 

[The Phantom]

On Fri, 18 Jul 2008 08:31:49 +1000, "Phil Allison" <philallison@tpg.com.au>
wrote:

"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.

So what ratio should be used for transformers that aren't stock lines?

..... Phil

 

[The Phantom]

On Fri, 18 Jul 2008 08:28:05 +1000, "Phil Allison" <philallison@tpg.com.au>
wrote:

"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.

It's not up to you to determine the rating. The manufacturer *correctly* rated
it for its intended purpose.

Your argument is entirely false

It's not my argument; it's the manufacturer's.

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.

But there are still lower temperature grades available. We're talking about
*this* transformer, which may not be using the most common insulation system.

Using higher temp grade will only increase the figure.

It would still depend of the insulation system, as I said, wouldn't it?

...... Phil

 

[The Phantom]

On Thu, 17 Jul 2008 15:27:01 -0400, "Paul E. Schoen" <pstech@smart.net> wrote:

"The Phantom" <phantom@aol.com> wrote in message
news:8t1u74d5s811flfu5bj3s4q36lnhp86d2l@4ax.com...
On Wed, 16 Jul 2008 18:27:11 -0400, "Paul E. Schoen" <pstech@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

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!

(as even this one seems to be). New ASCII file follows:

Paul

 

[ehsjr]

emailaddress@insightbb.com wrote:

On Jul 12, 6:00 pm, ehsjr
e.h.s.j.r.removethespampunctuat...@bellatlantic.net> wrote:

John Fields wrote:

On Sat, 12 Jul 2008 07:56:53 -0700 (PDT), emailaddr...@insightbb.com
wrote:

On Jul 11, 2:34 pm, ehsjr
e.h.s.j.r.removethespampunctuat...@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.

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
the answer, there is no need to reply.

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

 

[The Phantom]

On Fri, 18 Jul 2008 02:00:59 -0400, "Paul E. Schoen" <pstech@smart.net>
wrote:

"The Phantom" <phantom@aol.com> wrote in message
news:urmv74l94pn1n49mjsijnevvjdb5tlohtp@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.

I thought I ought to consider the effect of the transformer's leakage
inductance in the analysis. I measured the 192 VA transformer I've been
using and it has 80 uH leakage inductance as seen from the secondary with
primary shorted. So, in your simulation, you should make the transformer's
equivalent impedance .125 ohms in series with 80 uH. Also, I used 80 amp
Shottky diodes in the bridge, so the peak voltage drop across the diodes is
about .4 volts at the peak current of about 20 amps.

Going further to determine the effect of the flattening of the grid
waveform, I got a high power audio amplifier out of the basement and drove
it with a low distortion 60 Hz sine wave. I used the bridged power amp
output to drive the transformer. This way I could see the effect of
flattening of the grid waveform. I was able to get a clipped sine by
slightly overdriving the power amp. I took 3 sets of measurements with
varying degrees of clipping. I've posted scope photos of the current and
voltage waveforms over on ABSE.

You can see the effect of the leakage inductance in the fact that the peak
current is slightly delayed in time with respect to the peak of the applied
voltage.

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

 

[The Phantom]

On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddress@insightbb.com wrote:

I have an application in which I need to determine if a transformer is
suitable.

Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?

I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?

Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.

How much rectified DC voltage can that transformer maintain at 1.67A?
How much current can it maintain while staying at or above 12.3V, or
if there is another voltage to consider because we don't know the
other properties of this transformer, what voltage would that be and
at that voltage what is the DC current capability?

Please I'm asking to learn how to calculate this myself instead of
only the numerical answer.

I've posted a derivation of a formula over on alt.binaries.schematics
..electronic

Thanks,
JC