# Calculating DC Output Current From Unregulated AC Transformer

E

#### [email protected]

Jan 1, 1970
0
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?

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?

Thanks,
JC

P

#### Phil Allison

Jan 1, 1970
0
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?

The 1.67 amp ( ie 20VA) rating of the transformer is only for AC output -
not some derived DC voltage. It is also a " free air " rating - meaning
there must be plenty of ventilation and no heat sources next to the
transformer.

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?

** It would be overloaded delivering that amount of DC current and
overheat - unless perhaps you blew cold air over it with a fan.

Normally, one applies a de-rating figure to the AC current rating of a
transforemer when it is feeding a bridge rectifier and capacitor filter.
The figure depends somewhat on the design of the particular transformer (
toroidal, E-core ) but is in the range of 0.5 to 0.7.

So, if you really do need 1.67 amps DC at 16 volts - then pick a 40 VA
transformer.

....... Phil

E

Jan 1, 1970
0

This what? I was looking for details.

The 1.67 amp ( ie 20VA) rating of the transformer is only for AC output -
not some derived DC voltage.

Ok, but I'm not asking what it's not, I'm asking what it is. what I
am asking is what is the like-equivalent DC it should be capable of
based on the AC rating. Is 20VA really the answer, since we're
talking about winding ratios? It would seem not, since we do see
things like peak output around 16V instead of ( picking some number
out of air) 40V @ 0.5A for 20VA.

It is also a " free air " rating -  meaning
there must be plenty of ventilation and no heat sources next to the
transformer.

Yes, I am looking only for a free air rating. It all starts there.

**   It would be overloaded delivering that amount of DC current and
overheat -  unless perhaps you blew cold air over it with a fan.

I am not interested in what overloads it, I am interested in what
current it can supply at 12.3V or what voltage it would be at 1.67A
DC, what the math is to arrive at that free air rating for DC current
and DC volts give either or the other would be a variable.

Only after I know that *overloaded* maximum can I begin to derate to
account for heat.
Normally, one applies a de-rating figure to the AC current rating of a
transforemer when it is feeding a bridge rectifier and capacitor filter.
The figure depends somewhat on the design of the particular transformer (
toroidal, E-core )  but is in the range of  0.5 to 0.7.

So, if you really do need 1.67 amps DC at 16 volts  -  then pick a 40VA
transformer.

As mentioned I need to know the transformer's capability, not 1.67A at
16V. Those were example figures, I need the actual equations for any
transformer, even the 40VA one must have that.

Beyond that, I will use the actual equations to achieve at least
12.3VDC and a current depending on cost, transformer volume, etc.

I can see I should not have given any details about the project
because I only want to know how to convert a AC transformer's voltage
and current rating into a DC voltage at same current, or the resulting
current at a specific DC voltage.

P

#### Phil Allison

Jan 1, 1970
0

This what? I was looking for details.

** Read the whole post before making silly replies.

The 1.67 amp ( ie 20VA) rating of the transformer is only for AC output -
not some derived DC voltage.

Ok, but I'm not asking what it's not, I'm asking what it is.

** Read the whole post before making silly replies.

what I
am asking is what is the like-equivalent DC it should be capable of
based on the AC rating.

** Read the whole post before making silly replies.

Is 20VA really the answer, since we're
talking about winding ratios? It would seem not, since we do see
things like peak output around 16V instead of ( picking some number
out of air) 40V @ 0.5A for 20VA.

** Read the whole post before making silly replies.

It is also a " free air " rating - meaning
there must be plenty of ventilation and no heat sources next to the
transformer.

Yes, I am looking only for a free air rating. It all starts there.

** It would be overloaded delivering that amount of DC current and
overheat - unless perhaps you blew cold air over it with a fan.

I am not interested in what overloads it,

** You need to be - for safety and functional reasons.

Read the whole post before making silly replies.

I am interested in what
current it can supply at 12.3V or what voltage it would be at 1.67A
DC, what the math is to arrive at that free air rating for DC current
and DC volts give either or the other would be a variable.

** There is no possible math based only on the figures you gave. The whole
topic is about ACTUAL temp rise of the windings of the particular
transformer under particular conditions.

Best way to find THAT is to *measure* it.

Only after I know that *overloaded* maximum can I begin to derate to
account for heat.

** Read the whole post before making silly replies.

Normally, one applies a de-rating figure to the AC current rating of a
transforemer when it is feeding a bridge rectifier and capacitor filter.
The figure depends somewhat on the design of the particular transformer (
toroidal, E-core ) but is in the range of 0.5 to 0.7.

So, if you really do need 1.67 amps DC at 16 volts - then pick a 40 VA
transformer.

As mentioned I need to know the transformer's capability, not 1.67A at
16V. Those were example figures, I need the actual equations for any
transformer, even the 40VA one must have that.

** I just supplied the derating figures and the reason why it cannot be
precise.

Read the whole post before making silly replies.

I can see I should not have given any details about the project
because I only want to know how to convert a AC transformer's voltage
and current rating into a DC voltage at same current, or the resulting
current at a specific DC voltage.

** Neither is possible with simple math and only the VA rating to go on.

I have written a ( non simple) program that gets fairly close to predicting
the results ( ie the DC output voltage at any specified current ) -
providing you have much more data on the transformer like the primary and
secondary resistances, the size of the filter cap and a figure for
transformer leakage inductance.

You are WAY over-analysing the problem.

You only need to be sure the voltage is high enough, but not too high, for
you charger to work and the transformer is not overloaded under any
operating condition.

...... Phil

E

#### [email protected]

Jan 1, 1970
0
This what?  I was looking for details.

** Read the whole post before making silly replies.

** Phil, isn't this a bit of irony since I clearly asked for something
specific and you replied throughout your entire prior post with

Ok, but I'm not asking what it's not, I'm asking what it is.

**  Read the whole post before making silly replies.

See "** Phil" above.
I recall you're always a bit cranky but that's why I wrote what I did
two lines up, because I'm "asking what it is", for an equation, which
makes it seem as though you either choose not to provide one, or you
don't know. If you don't know it's not a problem, but isn't it
important to clarify what I am looking for?

what I
am asking is what is the like-equivalent DC it should be capable of
based on the AC rating.

**  Read the whole post before making silly replies.

Because if I do, I'll see the equation(s) I was asking about?

Is 20VA really the answer, since we're
talking about winding ratios?  It would seem not, since we do see
things like peak output around 16V instead of ( picking some number
out of  air) 40V @ 0.5A for 20VA.

**  Read the whole post before making silly replies.

See above.
Yes, I am looking only for a free air rating.  It all starts there.

Phil, how do I put it so you understand? If you really want to build
this entire project for me then go right ahead, I'll email you the
parameters. Until then, all I needed was what was asked, not for you
to second-guess something that was even a first-guess since you have
no idea what current I plan to run from any particular transformer,
you only know that I asked how to convert a spec mathematically and
used a few numbers as random examples of a starting point.
I am not interested in what overloads it,

** You need to be  -  for safety and functional reasons.

This is not a fixed plan for a one-off design Phil, this is a generic
question about transformers. The numbers were just random, forget the
numbers because they have nothing to do with the topic. I am not
interested because I have no trouble building within margins once I
know what those are - which requires having the equations.

Because I am not dead set on using any particular transformer for any
particular thing, it is valid and correct to say I am not interested
in what overloads any one particular transformer, only in how to
mathematically determine the free air DC ratings from the free air AC
ratings.

Read the whole post before making silly replies.

I am interested in what
current it can supply at 12.3V or what voltage it would be at 1.67A
DC, what the math is to arrive at that free air rating for DC current
and DC volts give either or the other would be a variable.

** There is no possible math based only on the figures you gave.

False. Given all the parameters it can be solved. Some of those
paramaters I do not have like transformer core saturation values,
others can be resolved experimentally but that can still be expressed
in an equation.

The whole
topic is about ACTUAL temp rise of the windings of the particular
transformer under particular conditions.

This is getting amusing. The whole topic is not about building one
thing with one transformer and what that one transformer's rating
should be. The whole topic which I started (least you forget?) was
asking for equations. If you don't know those, join the club- neither
do I. I could guess, have some theories but it would be better to
hear from others who can stay focused on the topic.
As mentioned I need to know the transformer's capability, not 1.67A at
16V.  Those were example figures, I need the actual equations for any
transformer, even the 40VA one must have that.

** I just supplied the derating figures and the reason why it cannot be
precise.

You used the term "derating", are you saying it is not a derating
figure but rather a conversion figure? I ask because you have gone on
and on about derating for temp, when I am very specifically looking
for only a mathematical equation for maximum transformer capability, a
theoretical one in a model that ignores temperature because the answer
is not going to be directly used to build something!

So are you claiming that if I short circuited the rectified output of
a transformer rated for 1.67A @ 12VAC, that the output would be 20VA *
0.5 to 0.7? I do not think that is the correct answer, I vaguely
recall it should be higher than that.

If the shorted output power is not 20VA * 0.5 to 0.7V, then you have
implementation details for any transformer, only how to convert
mathematically to actual peak, not prudent design values. Perhaps I
did not initially describe the question well enough. If so that is my
fault, but I hope by now it is clear I am not looking to directly
apply any info directly towards choosing the mentioned transformer for
anything in particular based only on shorted power.
I can see I should not have given any details about the project
because I only want to know how to convert a AC transformer's voltage
and current rating into a DC voltage at same current, or the resulting
current at a specific DC voltage.

** Neither is possible with simple math and only the VA rating to go on.

If it's complex math so be it. If it were simple I expect I'd have
known already, which was why it was asked regardless of how complex
the answer would be, but perhaps to the wrong group, perhaps it is not
a sci.electronics.basics level question. Ultimately if I find a spare
rectifier board lying around I'll hook that up and short it to get an
answer experimentally, at least for one transformer but it still
doesn't provide the sought after equation, I'd have to speculate about
whether similar E-Cores (which the example transformer was) behave the
same and collect more data about other E-cores in the same scenario,
hoping all the while that their manufacturers rated them equivalently,
accurately, then try to find some relationship.
I have written a ( non simple) program that gets fairly close to predicting
the results ( ie the DC output voltage at any specified current )  -
providing you have much more data on the transformer like the primary and
secondary resistances, the size of the filter cap and a figure for
transformer leakage inductance.

Then the equation(s) used in that would be much closer to what I was
after. Variables unresolvable could be assumed as constants then
experimentally compared to determine the level of deviation.

You are WAY over-analysing the problem.

You only need to be sure the voltage is high enough, but not too high,  for
you charger to work and the transformer is not overloaded under any
operating condition.

Ok, back to the charger. I have too many transformers to count that
could be used there. The transformer heat is not a problem. Using a
fan would be fine, having it shut down during regular use would even
be fine through thermal cutoff circuit(s). I would rather a warmish
running transformer with a fan cooling it than a burnt out regulator,
a bigger case for a giant heatsink plus added weight, or hot project
case.

With all due respect Phil, you have no idea of the project
requirements beyond what I had told you. Three manditory ones are low
temp density, small size, light weight. I'd MUCH rather put a thermal
cutoff or two in than dump a lot of heat on a regulator. A 40VA
transformer is not going to work, unless it's one rated slightly below
12V which is an odd figure. I guess I could strip some windings off
of one but that is such an ugly solution.

Ultimately for that particular project I will most likely use a SMPS
instead, but that project was only an example not the question asked
which is what the equation is to determine AC to DC spec conversion.
I do not need an answer based on the limited specs of the transformer
info I had given, I only need the equation as complete as it can be
for future uses.

Perhaps we got off on the wrong foot Phil, but please believe someone
when they state plainly what they don't care about and focus on what
they asked. This is NOT a topic about building one project, this is
about actual vis-a-vis rating conversion. I wouldn't have ran a 12VAC
1.67A transformer at 1.67A, AC output either, I know about derating
and temperature issues already which is why I stated several times
not what I care about. I care only about the equation. If you have

I hope I've made things clearer. I do not care about anything except
an equation to determine exactly what I asked about. It was why I

E

#### ehsjr

Jan 1, 1970
0
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?

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?

There is no "magic formula". You need to learn a whole
lot more about power supplies to calculate with precision
and to understand what needs to be considered and why.

I'll mention some general things.

First, with a bridge rectifier and capacitive filter,
figure the current you can supply to the load at
a little over 1/2 the transformer rating, in your
case, a bit over .9 amps.

Next, figure the ripple voltage, Vr. To figure Vr, see
http://ocw.mit.edu/NR/rdonlyres/Ele...EC80-48C8-A502-61CDC632D5B7/0/ripplevolts.pdf

(watch the line wrap)

available from the datasheet. Then, the maximum voltage at
the output of the regulator will be
(Vsec*1.41) - 2Vf - Vr - Vheadroom = Vmax where Vsec is the voltage
at the secondary. Vf is the diode drop, which you can find
on the curve of Vf vs I on the datasheet. You should figure Vsec
based on worst case line voltage and worst case Vsec sag under load.

By the way, the 12.3 volt figure is odd - how did you arrive
at it?

Ed

P

#### Paul E. Schoen

Jan 1, 1970
0
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?

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?

I find it easier to just simulate the circuit using LTSpice and adjust
values until they are close. I used a voltage source with 18 V peak at 60
Hz, and internal resistance of 0.8 ohms, which is probably about right for
a 20 VA transformer with about 10% regulation at full load of 1.67 A. I
added four rectifiers in FWB, a 2000 uF capacitor, and a 12 ohm load. I got
1.71 amps RMS from the voltage source, which dropped to 11.5 VRMS under
load. The output power is 13.2 watts, with 12.5 VDC as you require, and a
little over one amp. So in this case the 20 VA tranny is derated by
13.2/20, or about 65%. That's right in the range of 0.5 to 0.7 suggested by
Phil. Some of this is due to rectifier losses, which I estimate as about 1
or 2 watts. Schottkys will help a little.

The ASCII file follows. You can take it from there.

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 diode 288 160 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL polcap 400 192 R0
SYMATTR InstName C2
SYMATTR Value 2000µ
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=.8
SYMATTR InstName V1
SYMBOL res 512 144 R0
SYMATTR InstName R1
SYMATTR Value 12
SYMBOL diode 352 272 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D3
SYMATTR Value MUR460
SYMBOL diode 288 384 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL diode 352 496 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D4
SYMATTR Value MUR460
TEXT 184 528 Left 0 !.tran 1

P

#### Phil Allison

Jan 1, 1970
0
"Phil Allison" >

Phil, isn't this a bit of irony since I clearly asked for something
specific

** And impossible.

Phil, how do I put it so you understand?

** Do not ever try that shit on - fuckhead.

** You need to be - for safety and functional reasons.

This is not a fixed plan for a one-off design Phil, this is a generic

** You still have not read the whole post - have you ?

What an utter ass you are.

** There is no possible math based only on the figures you gave.

False.

** You have no clue whatever.

Given all the parameters it can be solved.

** So what did I write above ?

The whole
topic is about ACTUAL temp rise of the windings of the particular
transformer under particular conditions.

This is getting amusing.

** You are one, colossally stubborn fool.

** I just supplied the derating figures and the reason why it cannot be
precise.

You used the term "derating", are you saying it is not a derating
figure but rather a conversion figure?

** The figure was fully explained.

It derates the AC current figure to give the available DC current while not
exceeding the VA rating.

I ask because you have gone on
and on about derating for temp, when I am very specifically looking
for only a mathematical equation for maximum transformer capability, a
theoretical one in a model that ignores temperature because the answer
is not going to be directly used to build something!

** You Q is just plain absurd.

Transformer current and VA ratings are ALL about temp rise !!!!!!

The regulation curve ( whether AC or DC plus rectifier /filter ) is a
function of VA and a number of other parameters plus the type of
construction used.

So are you claiming that if I short circuited the rectified output of
a transformer

** Never do that and I have never mentioned any such idiocy.

You are hee hawing like some barnyard ass now - pal.

If the shorted output power is not 20VA

** The power delivered into a short is always zero - fuckhead.

But massive heat build up in the transformer will destroy it - pronto.

implementation details for any transformer, only how to convert
mathematically to actual peak, not prudent design values.

** Utterly stupid question, only asked by fools.

I can see I should not have given any details about the project
because I only want to know how to convert a AC transformer's voltage
and current rating into a DC voltage at same current, or the resulting
current at a specific DC voltage.

** Neither is possible with simple math and only the VA rating to go on.

If it's complex math so be it.

** You do not need any complex math to size a tranny to do a given job.

Problem here is, you cannot even specify what the job is.

I have written a ( non simple) program that gets fairly close to
predicting
the results ( ie the DC output voltage at any specified current ) -
providing you have much more data on the transformer like the primary and
secondary resistances, the size of the filter cap and a figure for
transformer leakage inductance.

Then the equation(s) used in that would be much closer to what I was
after.

** I don't give a shit what garbage you are after - fuckhead.

What you NEED is a damn good kick up the arse.

You are WAY over-analysing the problem.

You only need to be sure the voltage is high enough, but not too high, for
you charger to work and the transformer is not overloaded under any
operating condition.

With all due respect Phil, you have no idea of the project
requirements beyond what I had told you.

** So you are now making then up as you go along - right ?

A 40VA transformer is not going to work, unless it's one rated slightly
below 12V which is an odd figure.

** Totally insane crap.

Ultimately for that particular project I will most likely use a SMPS
instead, but that project was only an example not the question asked
which is what the equation is to determine AC to DC spec conversion.
I do not need an answer based on the limited specs of the transformer
info I had given, I only need the equation as complete as it can be
for future uses.

You HAVE to measure individually them after you BUY a transformer !!!

Just as quick to run a couple of bench tests if you have the damn tranny
right there in front of you.

Perhaps we got off on the wrong foot Phil, but please believe someone
when they state plainly what they don't care about and focus on what

** Sorry - pal.

But I do not answer ridiculous Qs as specified by arrogant, bullshitting
novices

No-one on usenet EVER has to do that.

I hope I've made things clearer.

** What you have made very clear is that you are an utterly autistic, rabid
nut case.

Who is just not able to accept new information.

...... Phil

P

#### Phil Allison

Jan 1, 1970
0
"Paul E. Schoen"
I find it easier to just simulate the circuit using LTSpice and adjust
values until they are close.

** Fine - but the OP was after a math formula and will never forgive you
for offering him a LTSpice solution.

I used a voltage source with 18 V peak at 60 Hz, and internal resistance
of 0.8 ohms, which is probably about right for a 20 VA transformer with
FWB, a 2000 uF capacitor, and a 12 ohm load. I got 1.71 amps RMS from the
voltage source, which dropped to 11.5 VRMS under load. The output power is
13.2 watts, with 12.5 VDC as you require, and a little over one amp. So in
this case the 20 VA tranny is derated by 13.2/20, or about 65%. That's
right in the range of 0.5 to 0.7 suggested by Phil.

** Amazing ......

But you will have 3 volts p-p ripple with that puny 2000uF cap.

Means the DC voltage falls well below 12, a lot.

Formula: I = C dv/dt ( where dt = 6mS )

Plus a 20 VA tranny has typically 15% regulation - not 10 %.

Worse if it heats a lot.

The ASCII file follows. You can take it from there.

** Never assume folk have LT Spice available or the ability to use it.

...... Phil

P

#### Paul E. Schoen

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

** Fine - but the OP was after a math formula and will never forgive
you for offering him a LTSpice solution.

Well, Spice is really a math formula, if you dig deep enough...
** Amazing ......

But you will have 3 volts p-p ripple with that puny 2000uF cap.

Means the DC voltage falls well below 12, a lot.

Formula: I = C dv/dt ( where dt = 6mS )

You're right. Schottky diodes and a 2200 uF capacitor are just barely
adequate. And then one must figure lowest line voltage as well. That's
where switchers are often the way to go.
Plus a 20 VA tranny has typically 15% regulation - not 10 %.

Worse if it heats a lot.

** Never assume folk have LT Spice available or the ability to use it.

Why not? It's free! And fairly easy. But I'd like a version that actually
shows components smoking or blowing up (with appropriate sound effects)

Paul

W

#### whit3rd

Jan 1, 1970
0
None of the information given is useful in making a determination on
this issue. Phil is telling you that the AC
rating is for 1.67A RMS, which amounts to 2.3A peak, but
that your rectifier will only conduct for a fraction of the duty
cycle, so
the 1.67A average output of the supply is all delivered in very
short time intervals (when the transformer output exceeds
the capacitor charge plus the diode forward drop). The
rectifier current could be 18A repeating peaks (even though
the average is lower), because the rectifier is almost always
turned off.

18A can melt the copper windings.

The ratings for rectifier transformers are different than those
for AC transformers. If you used a choke filter instead of a
capacitor, the rectifier would have a larger conduction angle (and you
can get closer to the AC current rating, safely).

B

#### Baron

Jan 1, 1970
0
[email protected] Inscribed thus:

Phil is giving you good advice !

Watts in equals watts out less losses.

Its not his fault you don't understand what Phil is telling you !

P

#### Phil Allison

Jan 1, 1970
0
"Paul E. Schoen"
Why not? It's free! And fairly easy. But I'd like a version that actually
shows components smoking or blowing up (with appropriate sound effects)

** ROTFL !

Have to agree with that one - as I has mused over the same idea myself.

Be far more entertaining and instructive than boring old numbers.

Maybe it could keep score of the \$ value of the damage too !!

..... Phil

P

#### Phil Allison

Jan 1, 1970
0
"whit3rd"
Phil Allison

None of the information given is useful in making a determination on
this issue. Phil is telling you that the AC rating is for 1.67A RMS,
which amounts to 2.3A peak, but that your rectifier will only conduct
for a fraction of the duty cycle, so the 1.67A average output of the
supply is all delivered in very short time intervals (when the transformer
output exceeds the capacitor charge plus the diode forward drop).

** True enough.
The rectifier current could be 18A repeating peaks

** No way.

The ratio of average DC output current to AC current cap charging peaks is
more like 2 or 3 to 1 in practice.

The ratings for rectifier transformers are different than those
for AC transformers.

** No true at all.

AC supply transformers are NOT made in two versions - one for AC and one
for rectifier input.

Though ones with low leakage inductance ( like toroidals) work the best with

...... Phil

P

#### Paul E. Schoen

Jan 1, 1970
0
None of the information given is useful in making a determination on
this issue. Phil is telling you that the AC
rating is for 1.67A RMS, which amounts to 2.3A peak, but
that your rectifier will only conduct for a fraction of the duty
cycle, so
the 1.67A average output of the supply is all delivered in very
short time intervals (when the transformer output exceeds
the capacitor charge plus the diode forward drop). The
rectifier current could be 18A repeating peaks (even though
the average is lower), because the rectifier is almost always
turned off.
18A can melt the copper windings.

The only way copper windings could melt is if those 18 amp peaks lasted a
lot longer than one cycle. It is all related to temperature, which is
related to power and energy, and generally fusing current is given by I^t.
It also depends on how quickly the heat can get out of the hottest spot in
the windings and be dissipated by conduction, convection, and radiation.
Best bet is to put a thermistor in the deepest part of the windings and
plot temp vs time at the worst case conditions, and see if the maximum
temperature gets close to the rating of the insulation (usually about 130
C), but 90 C is safer. Another way to measure the temperature is by
measuring the winding resistance change after it has stabilized. The tempco
of copper is about 0.4% / Deg C. So a 40% increase in winding resistance is

For transformers, it is all about RMS current, and duty cycle. A
transformer rated at 1.67 amps RMS will handle 2.3 amps RMS for
intermittent duty, with 50% duty cycle, with ON times no longer than 10-20
minutes or so. It will also handle overloads of 2x for 25%, and even 10x
for 1% duty cycle, with ON times of a few cycles. We "abuse" the trannies
in our circuit breaker test sets in this way all the time. The outputs are
essentially short-circuits (several hundred microhms of breaker and
connections), and we adjust the input to get the current needed to trip the
breaker. We sometimes need 40,000 amperes to trip a 4000 amp breaker (in
onee or two cycles), and the tranny is rated at 4000 amps continuous.

Solid state devices, like the diodes in this circuit (and the SCRs we use
to control our test sets) have a sharper derating curve, and are based more
on I^t, so a 500 amp SCR or diode will usually be limited to maybe 3x its
rating before it reaches the area where the ON time is very short, as it is
for capacitive charging peaks.

In the simulation I did, with Schottkys and 2200 uF, the peak rectifier
current is 3.8 amps. Even with 22,000 uF it is not quite 4 amps. Lowering
the internal resistance of the source to 0.4 ohms makes these peaks about 5
amps. Phil is correct on this as well.

Paul

#### neon

Oct 21, 2006
1,325
You Have The Info As Stated 12v @1.6a Forget You 1.41 Unless You Have No Load You Will Never See That Voltage. The Rms Is Value Is .639 Of Peak So Use That. The Output Voltage Depends On Load Caps Diodes Are .7v For Calculation Get Them Hot Or Cold That Will Change Too So Unless You Do Stress Analyisis It Means Nothing.

P

#### Phil Allison

Jan 1, 1970
0
"Peter Bennett"
Transformer Selection Guide - it is a one page .pdf that shows various
circuit configurations and resulting voltage/current relations.

** There is an *unfortunate error* in several of the circuits shown, all the
ones where a cap is the filter.

The figure for " V (Avg) D.C. " is given as 64 % of " V (Peak) D.C " -
which is complete bollocks !!!

That is only the case where there is NO filter cap used.

With suitable filter cap values, the average and peak DC voltages can be a
close as you like.

...... Phil

E

#### [email protected]

Jan 1, 1970
0
There is no "magic formula". You need to learn a whole
lot more about power supplies to calculate with precision
and to understand what needs to be considered and why.

Ok, but this should be resolvable to a reasonable level given the
example context of it being a typical E-core transformer and basic
silicon bridge rectified and capacitive filtered circuit. The output
at that point is the question. Even if some formula has an unknown
variable, or two or one hundred, the start would be to resolve those
which requires a working equation in which to place those variables.

I'll mention some general things.

First, with a bridge rectifier and capacitive filter,
figure the current you can supply to the load at
a little over 1/2 the transformer rating, in your
case, a bit over .9 amps.

This seems to counter the majority of small transformer examples out
there, does it not? Consider products powering just about anything
that uses a wall wart.
Consider a sub-20VA transformer rated in a wall wart for 12VDC, 1A
output. Granted, we could call that a little over 1/2 the rating but
just how much or little is the crucial issue, what variables determine
how much or little and how to express those mathematically.

Next, figure the ripple voltage, Vr. To figure Vr, seehttp://ocw.mit.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-S...

Thank you for the link, though since I am not mass producing equipment
and so not overly concerned about small differences in component cost
or size, I'm essentially going to consider ripple current a constant
resolved later as would apply to any one design, while I am asking
about a general formula for conversion without regard to variables
that would change in different designs beyond the basic assumption of
a capacitor large enough to achieve acceptibly low ripple for example.
available from the datasheet. Then, the maximum voltage at
the output of the regulator will be
(Vsec*1.41) - 2Vf - Vr - Vheadroom = Vmax where Vsec is the voltage
at the secondary. Vf is the diode drop, which you can find
on the curve of Vf vs I on the datasheet. You should figure Vsec
based on worst case line voltage and worst case Vsec sag under load.

By the way, the 12.3 volt figure is odd - how did you arrive
at it?

This is what the example circuit would need at a bare minimum by
calculating similarly to what you have above plus other drops in the
circuit @ expected current levels, to still retain the necessary
minimum input voltage, plus or minus a margin of error as median
datasheet values where used.

However, I am not concerned about the entire circuit, not about drop
over a regulator, I am only concerned about resolving how the AC
transformer rating relates to DC output before anything further in a
circuit beyond a typical bridge rectifier and capacitor sufficient to
smooth to a hypothetical 0V ripple, and I am also accounting for the
(Often negligable) different in forward voltage over the rectifier(s)
at different current levels, but this too can be expressed
mathematically.

Essentially, I don't want information more applicable to one project
than to another. Only what remains true mathetically for all projects
which employ AC spec'd transformers of typical design and through
bridge rectification by the most common silicon diodes. In other
words, as you'd see in the vast majority of electronics already if
they're not using switching PSU.

E

#### [email protected]

Jan 1, 1970
0
Phil, isn't this a bit of irony since I clearly asked for something
specific

** And impossible.

Apparently not, because plenty of products out there use wall warts,
are you suggesting they all have to build prototypes for even a simple
AC-DC wall wart? You gave a figure of 0.5 to 0.7 times the rating but
clearly darn near every wart here (quite a big box full) don't adhere
to this, they're all using transformers smaller than your suggestion
would imply and they aren't all bursting into flames or melting
things.

Phil, how do I put it so you understand?

** Do not ever try that shit on - fuckhead.

You need a pet Phil, or something. If we were in church then the
cursing might seem dramatic but on usenet does it really have any
useful purpose? I suppose you're trying to incite me to call you a
fuckhead too. Will it make you happy, can we move on then or will you
still be stewing for no good reason? I suspect the latter.

This is not a fixed plan for a one-off design Phil, this is a generic

** You still have not read the whole post - have you ?

Yes

Wasn't it obvious when I replied in a way related to what you had
written throughout the whole post?

What an utter ass you are.

Thank you Phil, but that's not quite enough. You hassle so many
people on usenet that calling me an ass just isn't special or rare
enough to be satisfying.

** You are one, colossally stubborn fool.

Yes, I generally keep at it until I achieve the desired result. I'll
have an equation even if I have to invent one of my own and
experimentally test it. I was just hoping someone out there had more
related info than I did besides generalizations of jumping up in
transformer size to create some margin, which is what I'd been doing
all along and is not so desirable as knowing exactly what that margin
is instead of just a ballpark figure, or at least to get much closer
in accuracy. Even if "use a bigger hammer" is a popular phrase there
has to be a method others are using when they start out with AC
transformer specs, as they manage to produce gear that works fine but
is not rated as high as your conservative estimate suggested would be
necessary. If real world examples and my wanting to find the
discrepancy between them and what you claim, is upsetting, all i can
say is whatever you want to insist won't change the box of
transformers I'm staring at. I suppose we could call the entire wall
wart industry a massive fraud for being overrated but I doubt it's
that simple.

You used the term "derating", are you saying it is not a derating
figure but rather a conversion figure?

** The figure was fully explained.

It derates the AC current figure to give the available DC current while not
exceeding the VA rating.

Thank you for clarifying, but I'm still in disbelief. Take this
example-

I have two typical looking E-core transformers rated for 32VA.
One of which was in a cordless drill quickcharger, it charged a pack
in an hour and also has an appropriate spec on it to accomplish that
per the battery pack cell capacity, 1.8A @ 12V Output (after going
through the control circuit).

This charger was used frequently for over 10 years. No failure on it,
no discolorations or smell like you'd have from excessive heat, looks
new inside except for a little bit of sawdust from the environment it
was occasionally used in. 'Twas only scrapped because it became less
viable to rebuild the battery packs another time on a drill after many
years of wear.

Based on your suggestion, even the upper end of getting 0.7 * AC
capacity, 0.7 * 32VA = 22.4VA. Output from the entire charger after
charge control board losses was 12V * 1.8A = 21.6. I think we both
know an old charger board is not 21.6/22.4 = 96.4% efficient!

I ask because you have gone on
and on about derating for temp, when I am very specifically looking
for only a mathematical equation for maximum transformer capability, a
theoretical one in a model that ignores temperature because the answer
is not going to be directly used to build something!

** You Q is just plain absurd.

Not in the context of using it as a starting point for an equation
upon which other variables are plugged in per the transformer and
application.

Transformer current and VA ratings are ALL about temp rise !!!!!!

ALL? Phil you seem to be stuck in infinite loop mode or something.
Regardless of whether you feel that way, I am talking about like
specs, there had to be a constant involved in the inital AC spec so
that remains the constant in the DC spec with regards to temperature.
What DC voltage and current is there, resolved mathematically, if we
allow the transformer to rise to the temp it would attain operating at
the AC voltage and current but not hotter than that? Wouldn't you
concede that in this case, a fair comparison would be if they are the
same temp so it isn't about temp rise at all? Some of the example
wall warts I have would tend to run quite a bit hotter than their free
air rating once placed in a sealed (or barely, passively ventilated)
plastic shell, and yet they don't have the thermal issues you imply
they should.

The regulation curve ( whether AC or DC plus rectifier /filter ) is a
function of VA and a number of other parameters plus the type of
construction used.

Agreed. Since the transformer itself is a constant, including it's
parameters and construction, I am looking for the remaining equation
of the AC voltage & current to DC voltage and current. I am looking
for the basic equation in which things that don't have to be
different, that can be held as constants like temperature and the
transformer being the same, are held as constants.

So are you claiming that if I short circuited the rectified output of
a transformer

** Never do that and I have never mentioned any such idiocy.

You are hee hawing like some barnyard ass now - pal.

Phil, I have a box of transformers and it's not a big deal to short
one of the smaller ones. Yes it'll smoke, maybe blow an internal fuse
or something but until it gets up to the high temps you fear it might
just last long enough to get a current reading. I was hoping to avoid
doing such things when I came here asking for an equation, but it
looks like testing is the only way to resolve this. Dead shorts
aren't necessary of course, but plotting data on several transformers
would seem useful if you were the only source of info on this topic.

If the shorted output power is not 20VA

** The power delivered into a short is always zero - fuckhead.

Are you saying it won't create heat, at least until something blows?
It sure seems like all that energy is going somewhere, and that it's
measurable.

But massive heat build up in the transformer will destroy it - pronto.

I'm sure it won't run long like that, but nevertheless you got the
jist of what I meant so who is being stubborn now?

implementation details for any transformer, only how to convert
mathematically to actual peak, not prudent design values.

** Utterly stupid question, only asked by fools.

then I would have known what you wanted me to ask.
Unfortunately you failed to do so.

If it's complex math so be it.

** You do not need any complex math to size a tranny to do a given job.

Ah, but I do if I don't want to be half ignorant about it, if I have a
need to use one only large enough for the job, or if I need to know
whether one I had available is suitable, at least long enough to get a
prototype running or whatever the purpose.

There are lots of devices out there in the world that don't appear to
just size a transformer by throwing in one that's definitely larger
than it needs to be to supply the current and (your biggest concern)
stay cool enough not to fail or cause thermal stress to anything in
the vicinity), to have an acceptibly long service life without any
issues, unless the load itself causes excessive current and it blows
the thermal fuse.

Problem here is, you cannot even specify what the job is.

I especially do not want an equation that is only applicable to one
job. That would defeat the whole purpose.

Then the equation(s) used in that would be much closer to what I was
after.

** I don't give a shit what garbage you are after - fuckhead.

What you NEED is a damn good kick up the arse.

Pot and Kettle, Phil?

With all due respect Phil, you have no idea of the project
requirements beyond what I had told you.

** So you are now making then up as you go along - right ?

I am declaring them irrelevant constants because I am looking for an
equation that is independent of any one project. That's why I have
mentioned VARIABLES quite a few times, that it is not a problem to
have them to plug in later per any specific project.

A 40VA transformer is not going to work, unless it's one rated slightly
below 12V which is an odd figure.

** Totally insane crap.

Not work as-in, not meet the other criteria, I'm sure it'll supply the
minimal current in the example and then some but you still aren't
seeing that I am not looking for a recommendation for one project, it
was only an example posed towards coming up with a generic equation.
Again I will state that I should not have mentioned any specifics
about any project because it has distracted you too much.

Ultimately for that particular project I will most likely use a SMPS
instead, but that project was only an example not the question asked
which is what the equation is to determine AC to DC spec conversion.
I do not need an answer based on the limited specs of the transformer
info I had given, I only need the equation as complete as it can be
for future uses.

You HAVE to measure individually them after you BUY a transformer !!!

I think you are right, to resolve some variables that will be
necessary. Even then, these seem to be less relevant variables as
there is a typical ratio of size to current rating that is above what
your 0.5 to 0.7 derating suggesting would dictate.

Just as quick to run a couple of bench tests if you have the damn tranny
right there in front of you.

Yes, I could run a test on every transformer but at some point one has
to pick the transformer to test, make a primary selection of
candidates and based on what I've seen there has to be a more accurate
way. Consider the example I posed above about a drill battery
recharger (which is not the same example as the other battery charger
project I initially posed), where the 32VA transformer is producing
enough power that if you were right, it should be too hot but it
isn't.

But I do not answer ridiculous Qs as specified by arrogant, bullshitting
novices

I'm starting to wonder if you've built nearly as many power supplies
as I have Phil. I have done similar to what you suggest in the past,
using vague ranges to derate as rules of thumb but I know there has to
be a better way and I observe far too many real world products that
contradict your claims, even if they are not so different than my
prior beliefs, and that they contradict my prior beliefs as well as
yours is part of what brought me here in the newsgroup to ask in the
first place.

No-one on usenet EVER has to do that.

I hope I've made things clearer.

** What you have made very clear is that you are an utterly autistic, rabid
nut case.

Are you sure it's transformers that are overheating?

Have a nice day Phil. It's not worth getting upset over.

E

#### [email protected]

Jan 1, 1970
0
[email protected] Inscribed thus:

Phil is giving you good advice !

Watts in equals watts out less losses.

Its not his fault you don't understand what Phil is telling you !

I agree, and understand what Phil was saying, but that's a generic
guesstimation that I'm looking to move beyond. Given enough time most
generic guesstimations can be resolved mathematically and there are
plenty of products out there that don't go to overkill Phil suggests
is necessary. Yes there are unresolved variables but this is a
science after all, and equations can have variables in them and be
valid expressions. Remember I asked for an equation, if Phil can't
provide one that's not something to be ashamed of, neither can I, so I
asked hoping for someone who knows rather than someone who thinks I
don't need to know.

If Phil insists he is right and the rest of the world is wrong -
having produced working electronics that seem an aweful lot like
evidence, it would be nice to have some of the pixie dust they must be
sprinking on these things to keep them working.

There HAS to be a better way!

Replies
4
Views
870
Replies
1
Views
444
Replies
7
Views
964
Replies
3
Views
3K
Replies
23
Views
2K