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Testing Amps on rewound microwave transformer

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RAV1965

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I need a max of 40v and 8a to power my drop down variable power supply. I have old microwave transformers so I removed the secondary and rewound it with 14 gauge wire. I now get 48vac and when I use a bridge rectifier it brings it down to 40vdc. I tried to test amps by hooking up my tester (rated at 20amps) and only succeeded in burning up some jumper wires luckily no other damage.

I don't have a clamp on tester and can't afford to buy one.

How can I determine the amperage, and if too high bring it down to my target?

Thanks
 

One way is to attach various resistors and measure AC volts. That gives you an idea when the secondary output voltage starts to drop, thus letting you know how much power you can get from your rewired transformer.

It's too risky to directly attach an ammeter. It has very low ohms. (My DMM's have a 10A range, made possible by a thick wire shunt about an inch long inside the case.) Hence it pulls output voltage very low. Even if you get a high Ampere reading, it's not a normal, valid use of the transformer.
 

Magnetron transformer is special design (with artificially increased leakage inductance) to achieve constant current output. It's no well suited for general transformer applications.
 

also the pri of a uWave Tx has fewer turns than a conventional Tx - it runs very high peak flux and high peak magnetising current - as a result they hum loudly

you cannot connect an ammeter to a rectified sec without a series load as the ammeter is essentially a short - so the current is limited by the Tx design and winding resistances

not surprising you melted the wires on the ammeter ( and probably buggered the ammeter internals for 20A - or blew the fuse - and the bridge tect ) if you didn't know this about basic measurements - it leaves us all wondering when you will kill your self with mains volts ...

some times a little knowledge is a dangerous thing -

you need to load up the DC o/p gradually and note the heat / temp rise in the Tx ...
 

One way is to attach various resistors and measure AC volts. That gives you an idea when the secondary output voltage starts to drop, thus letting you know how much power you can get from your rewired transformer.

I don't completely understand this. Please explain with more detail.

What about automotive fuses? What if I momentarily touch one to both wires? I would guess that if I leave it on too long then the current would burn it up just like it did my jumpers wires, but if only for a moment then not burn it up if the amp rating was high enough?

Thanks
 

The current needs to be limited by a suitable load on the output of the Tx - if you apply a too low ohm load you will overload the Tx, this is what fuses and polyfuses & circuit breakers are for - protection if you short the output or apply too low ohms ...
 

Simulation of transformer with various loads. Switches were closed in sequence. Scope trace travels to the left. Notice volt reading drops as current draw increases.

A resistive type load is recommended because it behaves in a predictable manner. You may use heating elements, or plain incandescent bulbs.

transformer step-down various loads switchable.png

Most likely your transformer will not perform the same as the simplified transformer model in my screenshot. It's hard to predict its current draw with different loads, or no load.
 

A resistive type load is recommended because it behaves in a predictable manner. You may use heating elements, or plain incandescent bulbs..

I didn't understand the graphic at all, and understood the rest a little bit. But I have pieced together things you said with other things I do understand and hopefully have a working plan.

So I take a resistive load like a heating element or light bulb and measure it's resistance using my multimeter. I hook that load up to my transformer and measure the voltage drop on both sides of the load. Then I use the formula v/r to find I. Will that work?

Thanks
 

You are not understanding how current ratings work.
Firstly, take note of what FvM wrote in post #3, it will work but that kind of transformer is not optimal.

A transformer produces voltage, not current. Current is what you draw from it.
The voltage will drop and the temperature will rise as more current is drawn. The current rating is decided by how much voltage drop is acceptable and how hot it can be allowed to run. These are things you have to work out for yourself based on the new windings you added.

The classic way to test a transformer rating is to see what it does under different loads. By connecting different resistors across the secondary, different load currents will flow into them and the voltage will drop accordingly. You have to verify the voltage does not drop below your requirement at full load current and at the same time the transformer doesn't overheat.

Referring to your first post, something is wrong with your measurements. If you have 48V AC from the transformer you should only get about 1.5V drop in the bridge rectifier and if filtered it should rise to almost 70V. I suspect your DVM is not being used correctly.

Brian.
 

You are not understanding how current ratings work.

A transformer produces voltage, not current. Current is what you draw from it.
The voltage will drop and the temperature will rise as more current is drawn. The current rating is decided by how much voltage drop is acceptable and how hot it can be allowed to run.

Brian.

Brian, Thank you very much for a very clear and understandable response.

I've been basing my questions on the specs of the drop down power supply which states the input max is 40vdc and 8 amps. Sounds like you are saying that only the volts matter, and the amps will take care of themselves? I do know that is true with other loads that only pull the current that they need.

Firstly, take note of what FvM wrote in post #3, it will work but that kind of transformer is not optimal.

I'm still pretty fuzzy on this part. If this is not a good way to do this, then what IS the best way?

I used a bridge rectifier that I scavenged from an old TV board. Maybe that would help to explain my readings. Otherwise I'm pretty sure I am using the dmm correctly. That rectifier is burned up so next time I will be using individual diodes to make my own.

Thanks
 

if you apply a too low ohm load you will overload the Tx, this is what fuses and polyfuses & circuit breakers are for - protection if you short the output or apply too low ohms ...
... ... ... .
 

amps will take care of themselves
Provided you have enough capability to provide the maximum current you need, that is true. Amps represent the load, all you have to do is make sure the source is capable of producing them safely.

If the maximum input to your power supply is 40V you have too much voltage from the transformer and you need to remove some turns. It works like this: The transformer produces AC, it is a waveform close to a sine wave that reverses polarity at your AC line rate, usually 50Hz or 60Hz depending on where you are. Because it is a constantly changing voltage we have a special way of measuring it to give a meaningful number, we call it RMS (Root Mean Squared) which is exactly what it is, the square root of the mean of the voltage squared. To measure it in real time would mean it was constantly changing and actually passing through zero twice as the polarity reversed back and forth again. Your meter measures in RMS so when you measured 48V it was the RMS value. The problem is that when you pass it through the bridge rectifier, it no longer reverses polarity, one or the other of the diode pairs conducts and 'steers' the voltage so it comes out all the same polarity. It isn't a sine wave any longer though so your meter gives a wrong reading.

To compound your problem, the RMS value isn't the same as the peak value, it is somewhat below it. That means when you measured 48V RMS, the peaks were quite a lot higher, in fact 1.414 times the RMS voltage. 48V RMS = (1.414 * 48) = 67.8V. In your power supply there are capacitors which charge up to that peak voltage so you have far more than is safe. You really want the power supply to see no more than 40V which means the RMS voltage must be (40/1.414)=28.3V or lower.

Don't worry too much about the comments on the type of transformer, it will work but not as well as one designed for use in a normal power supply. Microwave oven transformers are designed to run in a kind of constant overload condition, it has to do with the way Magnetrons work but that is a different topic altogether!

Brian.
 

To compound your problem, the RMS value isn't the same as the peak value, it is somewhat below it. That means when you measured 48V RMS, the peaks were quite a lot higher, in fact 1.414 times the RMS voltage. 48V RMS = (1.414 * 48) = 67.8V. In your power supply there are capacitors which charge up to that peak voltage so you have far more than is safe. You really want the power supply to see no more than 40V which means the RMS voltage must be (40/1.414)=28.3V or lower.

My Digital Multi Meter DMM (Extech 470 with True RMS) has settings for A/C power so when I measure A/C power it calculates the RMS and gives me the correct reading. When I measure D/C power I change the setting for that and so it gives the correct reading.

I live in the US and we have 120vac and when I check that known voltage it gives me the reading as expected.

I am very familiar with measuring volts. It's amps that I'm fuzzy about.

if filtered it should rise to almost 70V.

Can you tell me more about this comment? Are you talking about smoothing the D/C output with capacitors? Should I do that? I was thinking that I should. What kind of capacitors should I use?

Thanks
 

So I take a resistive load like a heating element or light bulb and measure it's resistance using my multimeter. I hook that load up to my transformer and measure the voltage drop on both sides of the load. Then I use the formula v/r to find I. Will that work?

Yes, if your load doesn't change resistance in use. Incandescent lamps increase in resistance when they glow. Normally this throws off your calculation. However if you apply only 48V instead of house voltage, it may not change filament resistance so much.

Try various loads. Lamps in parallel, lamps in series. A toaster or space heater can be suitable. (One rated 1000W with house voltage, draws a fraction of that amount at 48V).

At all times touch a finger to the transformer. Disconnect if it heats up alarmingly. It may get hot even with no load attached. A fuse offers predictable protection.

To limit current through the transformer, insert resistance inline with the primary. (Such as a space heater.)

It would be better if you had a variac so you could start applying low voltage, and increase it gradually.
 

What exactly are you trying to measure?
You cannot measure the 'Amps' because that is what you take FROM the transformer, not what it will just give to you. However, if you intend to draw 'X' Amps from the transformer it obviously must be rated to at least 'X' Amps at the voltage and temperature you have specified.

Thin in terms of the transformer being a gasoline engine. A 40cc grass cutter 'strimmer' runs at about 2,000 RPM, a 4,000cc car engine can also run at 2,000 RPM. But if you fitted the strimmer engine in a car it probably wouldn't get off your driveway. Substitute RPM for voltage and you see the problem, measuring voltage alone means nothing, it is the capability to produce power under load that matters. In a transformer where you have wound your own coil, the only way to discover it's performance is to try it under different load conditions. If you measure the voltage as you draw different currents and note how much it drops you can guage how well it performs. As it will get hotter under higher loads, you have to ensure it doesn't exceed safe levels too.

Regarding the higher voltage I mentioned, what is your transformer going to supply power to? You stated you were using a bridge rectifier so you seem to need a DC output. The voltage after the bridge is not steady DC though, each AC cycle going in will come out as two DC cycles. It will still go to 0V when the AC goes through zero crossing and it will still rise to the same peak value of the sine input. Remember what I said about RMS being used for AC measurment, it doesn't work for DC and the pulsating DC voltage will not measure correctly as DC on your testmeter range either. What you probably see is the meter trying to show the average DC voltage, what it shows will vary according to meter manufacturer, you need dedicated instruments to be able to measure the instantaneous voltage in real time as it rises and falls 120 times a second.

If you do add capacitors across the DC pulses, they will charge to the PEAK of the DC pulse waveform. This is 1.414 times the original RMS value (1.414 is an approximation of the square root of 2). If your transformer is feeding a power supply for example, it will almost certainly have input capacitors in it and they have to withstand that peak voltage. Your first post said the maximum input voltage is 40V so you have to drop the AC voltage to ensure the peaks don't go above that.

Brian.
 

Yes, you explained most of that earlier. Did you read my latest post?

Thanks
 

If you mean this:
My Digital Multi Meter DMM (Extech 470 with True RMS) has settings for A/C power so when I measure A/C power it calculates the RMS and gives me the correct reading. When I measure D/C power I change the setting for that and so it gives the correct reading.
It isn't correct. The RMS reading of the AC from the transformer will be fine but the DC range will not give an accurate result. It is accurate for steady DC voltages, such as from a battery or regulated PSU but after the bridge rectifier you don't have a steady voltage (unless you add capacitance), you have a voltage constantly changing between zero and peak and back again 120 times a second. The meter can't measure a moving target accurately and will show something approximating to the average of the waveform. To measure it accurately, add a capacitor across the output of the bridge rectifier and divide the DC reading by 1.414 then if you want absolute accuracy, add the voltage dropped across the two conducting diodes in the bridge (about 1.5V)

Brian.
 

If you mean this:
It isn't correct. The RMS reading of the AC from the transformer will be fine but the DC range will not give an accurate result. It is accurate for steady DC voltages, such as from a battery or regulated PSU but after the bridge rectifier you don't have a steady voltage (unless you add capacitance), you have a voltage constantly changing between zero and peak and back again 120 times a second. The meter can't measure a moving target accurately and will show something approximating to the average of the waveform. To measure it accurately, add a capacitor across the output of the bridge rectifier and divide the DC reading by 1.414 then if you want absolute accuracy, add the voltage dropped across the two conducting diodes in the bridge (about 1.5V)

Brian.

OK great. I didn't know that about the DC measurement with this transformer. To stay on the safe side I will just plan on the DC being 1.5v lower than the AC.

But I did plan on smoothing the output with capacitors. Any advice on which ones to use?

Thanks
 

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