# Re: Power Transformer Current

Just measure the voltage not the amperage. you are only supposed to measure amperage (current) when you have a load attached. Just attaching the
multimeter across the transformer to measure current is the wrong way and you will get useless results.You put the ammeter in series between the transformer and the load. The load you attach will draw only the current required for it to run. it will not draw the 4 or the 3.2 amps you measured. The current ratings on a transformer are only there to say "be sure the load you attach don't use more current than this".
Ideally the center tap should be exactly half the voltage as the two ends. But, in the real world the voltage between one side and the centertap will be slightly different than the other side to the centertap. But it is not much to make any real difference.
If the transformer has any markings or numbers on it you can do a search on yahoo or google with those to get its specifications on a data sheet. If it doesn't have these, see if it has any power ratings written on it . It might be something like 300W for 300 watts or maybe 100VA for 100 VoltAmps (different than watts). These specifications are saying that the transformer can only supply up to this much to the load. From this specification and a little math you can get the maximum current you can run through there. Which you might not even have to do if the load you attach uses only a little current.
If this is all new to you I recommend studying a little more before you proceed with your transformer project.
Josh

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remove the urine to answer

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| Again, the whole point of my exercise is to find a method of | determining the approximate current rating of different power | transformers (generally small transformers, so approximating by weight | is not too practical).
Sean. Further to the comments already posted by Josh and Don.
The power rating of most electrical devices is the maximum power you can get from it on a continuous basis, without the thing failing. There are margins on the figures quoted so that variations in ambient temperature and perhaps altitude, can be accomodated. Usually the Safety codes of individual countries also demand some margin as well.
Generally speaking, exactly what the limiting factor is, of any particular device, depends on the device and how it is constructed.
It could be the quality of the insulating material, It could be the size of the cable supplying it, or perhaps the rating of the input fuse ( if it has one) or the size of the actual conductors inside it.
The point is that no simple test will give you the answer because the limiting factor could be anything.
You might try looking for a change in voltage with an increasing load and find out how much current draw will drop the output voltage below your tolerence band, however you do not know if this amount of power could produce a hot spot somewhere, possibly leading to insulation catching fire.
Your best bet is to follow the advice given in the other posts such as looking for printed watt or VA or VoltAmps ratings on the casing.
Perhaps there is a fuse on the input side, it will have a curent rating printed on it. This may give you a pointer. From What I have seen, These fuses must be included, and are genreally non replaceable. You would probably have to pull apart the casing to find this so it might be self defeating proceedure.
Put simply, You need to find data on the device. Look it up on the web, Look in the radio shack catalog, or Look for printing on the outside.
Most sorts of simple tests will not give you the answer.
Tom Grayson
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On Sun, 10 Aug 2003 04:31:11 +0100, Sean Mathias wrote:

You're not just connecting an ammeter across each secondary and applying power, are you?
If you are, don't, either you'll blow a fuse or the transformer will saturate and get very hot.
If you want to determine a transformer's current rating, the only way to do it safely is apply a series of decreasing loads (resistors) to all the secondaries simultaneously until the voltage falls to the nominal (in this case 12.6V. V/R will then give you the current.
Open circuit, any transformer will show more voltage than loaded, due to iron and copper losses. The trick is to design so that the voltage is correct *on load*.
--
Then there's duct tape ...
(Garrison Keillor)
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"Fred Abse" wrote in message

Unlike inductors, transfomer core saturation is unrelated to current. Transformer cores saturate from applying too much voltage per turn for too long a period for the particular core material.
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While what you say is quite true for pulse transformers, where Volt-Second capacity and core saturation are directly related, it's for me a bit of a stretch to apply the same concept to an a.c. power transformer where entirely different issues are at play.
In designing a power transformer, core characteristic are more often determined by an analysis of the combined magneto-motive force of all its windings vis-a-vis the BH curves of the core material.
Harry C.
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snipped-for-privacy@yahoo.com (Harry Conover) wrote in message

So, why does a microwave oven power transformer go into saturation with no load? I tried one awhile back and it drew over 3 amps from the line with no load. I assume it was running into saturation since it was getting hot.
-Bill
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I don't know what's going on, it seems a little lightly built for the job. The core cross section area is 1.25 by 1.75 inch or .00141 sq meter. The laminations are all welded together on a base plate so the eddy currents are quite high and probably account for most of the heat. But I don't think it should draw 3 amps without only eddy currents as a load. The primary only has 160 turns of looks like 12 gauge wire which is a little over one turn per volt. Most power transformers are at least 2 turns per volt.
There is a transformer formula relating voltage, frequency, turns, area, and flux density: Erms =4.44*FNAB
and using B=1.1 for transformer steel, I get:
Erms = 4.44 * 60 * 160 * .00141 * 1.1 = 66 volts rms.
Looks like a 66 volt transformer running at 120. -Bill
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snipped-for-privacy@prosolve.com mentioned...

Just exactly how are you measuring the current? Short circuit? Have you considered that your methods are incorrect or your meter is misleading you?

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If you actually had about 4 amps flowing through out of a secondary that is only rated for 0.45 amps it's a wonder that you didn't overheat the transformer and burn it, or something else, out? Maybe you were just lucky! Overheating a transformer can lead to burnt up windings, burn insulation which in turn can cause contact between different windings! See Note 1.
To understand what you are doing maybe you should do some ohms law calculations. For example; 14.0 volts across a 140 ohm lamp is; 14/140 = 0.1 amps. 14.0 volts across a 70 ohm lamp or resistor is; 14/70 = 0.2 amps 14.0 volts across a 10 ohm resistor is; 14/10 = 1.4 amps 14.0 volts across an almost dead short circuit (or maybe the internal resistance of an ammeter if that didn't burn up the internal shunt resistor inside the meter!) could be 14.0 volts across 3 ohms = 4.7 amps. Don't do that; transformers (or ammeters either) don't like short circuits that's why good circuits should have small fuses in the transformer primary.
Note 1: I had a transformer that worked OK and delivered the rated voltage from its secondary; unfortunately it had been overheated at some time and the heat had caused the secondary to be in electrical contact with the primary. So the radio of which the transformer was a part was electrically 'hot' and when one touched the radio you got a shock from the AC mains input!
Note 2. I respectfully suggest, but this is only a supposition, that the total ohmic resistance in the secondary winding of such a transformer, the meter leads and the shunt resistor inside the meter may have been 'just sufficient' to limit the current to 4.5 amps and if not connected for too long may have prevented severe damage? 4.5 amps at 14 volts is approximately 63 watts! Imagine the heat of a 60 watt bulb that turns most of it's electrical input into heat going into the insides of a transformer; it is not mentioned if anything got warm/hot?
I don't understand what the original poster means by 28.15 at 3.2 amps 'between' secondaries? If it was voltage due to capacitance there wouldn't be any current flowing. But there is possibility of seeing voltage between any two parts of the transformer due to capacitance effect and depending on the AC impedance of his meter.
Unless the poster does not know how to connect and read the ammeter setting of the meter? It's all very confusing about what he is doing and trying do? From what I've understood it's very lucky that secondary windings didn't burn out with ten times their design current and/or the transformer/s didn't catch fire! Keep a fire extinguisher handy.
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Also don't forget the voltage and current your measuring with your multimeter is RMS not peak.
Josh

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On Sat, 09 Aug 2003 20:31:11 -0700, Sean Mathias

If, as others have surmised, you short circuited your transformer with your multimeter set to a high current setting - like 10 amps - then you have taken an awful risk of burning out your meter. On the other hand, if you read the current scale on an analog meter when set to volts, then the reading is meaningless.
The rating of the Radio Shack transformer is the maximum current it should be allowed to deliver for an extended period of time without overheating and risking burning out. All ratings are ultimately decided in this way. If you connect a load across the transformer that just draws the rated current, the voltage should drop from the open circuit voltage to 12.6 volts, due to internal resistance of the windings.
There are many simple books available that describe exactly how a transformer works. It might be worthwhile borrowing one from your local library to understand the principles. Also read up on the use of your multimeter. Depending on the setting of the range selector, you can read either volts or amps or possibly resistance in ohms, but only one at a time.
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Thanks to all for the replies. To clarify some points as requested:
I am familiar with Ohm's law. I did, in fact, short the secondaries through my DMM with current scale set to 10A, apparently a no-no, got it.
It is not possible to do a google search as there are no markings whatsoever on most of these transformers. Most of them were picked up as surplus or parts from various places.
What I am trying to figure out is how to even approximate what the 'safe' or rated current is. Finding the secondary no-load voltages is pretty straightforward, as is resistance, and determining the different windings.
As for gradually adding load, I have done this, but what still eludes me is that as resistance (load in this case) increases, current increases and voltage drops. At what point do I know that the load is at 'maximum' or design as I can continue to load it with a corresponding decrease in voltage and increase in current.
Thanks for any help.
Sean
On Sun, 10 Aug 2003 16:41:55 GMT, snipped-for-privacy@oanet.com (Rusty) wrote:

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030810 1609 - John Jardine wrote:

Or, you could give it the smoke test. Just keep adding load until it starts to smoke, then reduce the load a little.
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Yeah but once you've turned it into an SED how to you get the smoke back in? ;-)
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Chris wrote:

BMFH????
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BINGO! Finally someone has the easiest way. This is how we did it in the surplus business years ago.

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Well, your current measurement may not have been wasted after all. Take a look at the procedure at http://home.earthlink.net/~jimlux/hv/xfmrmeas.htm and give it a try. See what you come up with by using it on a transformer for which you know all the specs. Ferinstance, a RadioShack unit with ratings printed on the case. Then try it with some of your unknowns.
Before I found that site, I was thinking in terms of getting a physical measurement of the wire in each of the windings. Convert the diameter of the wires into circular mils, and use the typical design standard of 700 circular mils/amp to get a pretty close current rating for the secondaries. How to do that.... you'll have to get to a small piece of the wire from each secondary; a piece that's free of any coatings such as enamel, insulating varnish, extraneous debris, etc. You may have to tear a small patch of the outer layer of insulation away to get to the wire, but generally, a small piece of wire is visible where it attaches to the terminals. Of course, you'll have to go digging if the transformer doesn't have solder terminals. Using a caliper or micrometer, measure the diameter of the wire. Search the web for an AWG table (very easy to find) that lists the diameter and cross-sectional area (in circular mils), and find the circular mils for the wire diameter just measured. Divide that number by 700 to get the approximate current, in amps, that the secondary was designed for.
Double check it by the procedure laid out by the web site above. See how close they come to each other.
I would be interested in hearing back from you as to which method comes closest to the ratings of your known transformer.
Cheers!!!
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Tweetldee
Tweetldee at att dot net (Just subsitute the appropriate characters in the
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secondaries.
Depends also heavily, on the iron in the core. How much, what type, how thick the lams are.
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Transformer design is a tradeoff between many things. For example, the secondary current rating determines the wire size, sort of. Larger wire means less voltage loss (I*R) in the winding and less heating (I*I*R). Larger wire is more expensive and requires a larger transformer for the same ratings but has less loss and less heating. More current requires a larger core size in order to not saturate. Saturation is when the steel core can't be magnetized any stronger therefore no more energy can be "transformed" to the secondary. When this happens, the primary current jumps to large values and can overheat the transformer or worse!
A good rule of thumb and often a design spec for transformers, is the rule already stated by John: "The rated current causes a 10% voltage drop from open circuit voltage." So Johns trick is a good one to use for unknown transformers. You'll always be close. Zoramy
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