Calculating DC Output Current From Unregulated AC Transformer

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.

Thanks, JC

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emailaddress
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** This is your error.

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.

** 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

Reply to
Phil Allison

This what? I was looking for details.

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.

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

e

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.

VA

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.

Reply to
emailaddress

"Phil Allison" >

This what? I was looking for details.

** Read the whole post before making silly replies.

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.

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

** Don't complain when you get free, good advice - f*****ad.

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.

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

Reply to
Phil Allison

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

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?

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

See above.

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.

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.

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.

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.

.
(

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 not answered the question, because I am not asking about 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.

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.

ng

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.

=A0for

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 that this was not what I asked about, not what the question was about, not what I care about. I care only about the equation. If you have the answer because of your computer application that was relevant.

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 asked that question instead of some other question.

Reply to
emailaddress

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

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(watch the line wrap)

Next, find out how much headroom your regulator needs, usually 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

Reply to
ehsjr

wrote in message news: snipped-for-privacy@e39g2000hsf.googlegroups.com...

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

Reply to
Paul E. Schoen

"Paul E. Schoen"

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

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

..... Phil

Reply to
Phil Allison

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

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.

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) when they are overloaded.

Paul

Reply to
Paul E. Schoen

ut -

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).

Reply to
whit3rd

snipped-for-privacy@insightbb.com 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 !

--
Best Reagrds:
                        Baron.
Reply to
Baron

"Paul E. Schoen"

** 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

Reply to
Phil Allison

"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.

** 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.

** 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 rectifier / capacitor loads.

..... Phil

Reply to
Phil Allison

Go to

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and download their Power Transformer Selection Guide - it is a one page .pdf that shows various circuit configurations and resulting voltage/current relations.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
 Click to see the full signature
Reply to
Peter Bennett

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 just about your maximum.

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

Reply to
Paul E. Schoen

"Peter Bennett"

** 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

Reply to
Phil Allison

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.

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.

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.

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.

Reply to
emailaddress

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!

Reply to
emailaddress

Thank you. This is a lot closer to what I was looking for. While it may not resolve all variables it begins to explain the discrepancies between what a certain someone insisted and what has been observed about existing designs field proven to work acceptibly.

Reply to
emailaddress
[snip]

The info you need can be found in any electronic text, or in numerous sources on the web. If I understand what you want transformer sizing and bulk filter cap "unregulated supply".

Here is just a couple sources.

Onsemis "Linear & Switching Voltage Regulator Handbook" see SECTION 8 " DESIGNING THE INPUT SUPPLY".

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Other free HB from onsemi

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Midcom also has an extensive collection of tech notes.

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More transformer info

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Similar to onsemis

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Reply to
Hammy

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