How to know/measure Amperage of unknown transformers

[Whoops]

Hello,
I have 2 power transformers that I dont have the specs for.

Both are 220V volts input
and have 10V on the secondary output which is perfect for my next project

Although I need to know whats the power rating for the transformers,
is there anyway that I could measure the VA's of these transformers?
How to know that in an unknown specs power transformer?

Thank you

 

[emrr]

http://groupdiy.com/index.php?topic=28600.msg346444#msg346444

 

[Whoops]

http://groupdiy.com/index.php?topic=28600.msg346444#msg346444

Thank you so much Emrr, I did different searches here in the forum to try to find this answer but couldnt find any.

Thanks for that post.

So :

Here's another rule of thumb: Many power transformers are rated at the current which makes the voltage fall by 10% from unloaded. (And you shouldn't run them at higher current, because the voltage drop dissipates as heat, and could fry the tranny.)

So let's say a transformer puts out 200V when not loaded. If you load it down such that it's now putting out 180V, and discover that it takes 100mA of curent to do that, you have a transformer officially rated at 180V/100mA.

By the way, that lets you figure out the transformer's equivalent series resistance. When you pull 100mA out of it, you're dropping 20V, so the equivalent series resistance is 20V / 0.1A = 200 ohms. Which suggests that if you pull 65mA out of that transformer, the ESR will drop 0.065 * 200 = 13V from the unloaded figure, so your loaded voltage will be 200 - 13 = 187V.

YMMV if the manufacturer computes ratings differently, but that's how it works in classic parts like Triad or Thordarson.

Peace,
Paul

and

> Here's another rule of thumb: Many power transformers are rated at the current which makes the voltage fall by 10% from unloaded.

Depends on size. Under 25VA it is not uncommon to rate for 20% sag. Over 1,000VA it is unlikely to find 5% sag. Very small iron won't overheat, big iron will: Surface/Volume ratio. (The electrical rating runs nearly as Weight which is same-as Volume; the thermal rating runs as Surface Area.)

 

[Whoops]

So know I have to make a circuit that can have a variable load on the transformers down the transformers,
and I have to use 2 DMM , one to measure the the load (amps) and the other to measure the voltage drop on the transformer caused by that load.

Is that correct?

Thank you so much

 

[PRR]

> I have to use 2 DMM

Why? Use resistors. They are usually marked for Ohms. If not, measure them first. Watch the voltage no-load and some large arbitrary load. Compute the sag. One meter.

 

[Whoops]

> I have to use 2 DMM

Why? Use resistors. They are usually marked for Ohms. If not, measure them first. Watch the voltage no-load and some large arbitrary load. Compute the sag. One meter.

Thanks PRR.

I'm not understanding how do I do it.
So
I have the secondary, should I put a resistor between the secondary pins while measuring the voltage?
So how do I calculate the resistor value for 1 amp , and for 1.5 and 2 amps  for example?

thank you and sorry for my lack of knowledge

 

[analogguru]

First step:
with a ruler measure the length of the iron core.  (The result will be for example 54mm).
Then measure the thickness of the iron core. (for example 18,8mm).

Second step:
Connect to google - google is your friend (at least in this case).
Type as search item: EI[length value]/[thickness value]
In the case of the example above this would be: EI54/18.8
You will get a lot of results, intersting for you is a number followed by "VA".  (VA = Volt-Ampere)

In the example above this would be: 16 VA.

Now measure the voltage at the secondary without any load.  In your case this would be 10V.

Now divide the e.g. 16 VA with the 10V and you will obtain 1,6A.  That is the minimum "Amperege" your transformer should be able to deliver.  If the voltage breaks down from 10V to e.g. 8V under load your transformer should be able to handle 2A.  You could verify this by measuring the diameter of the copper wire of the secondary winding, but this would lead too far.

 

[Whoops]

Thanks PRR and thanks Analoguru

I will do a you advised.
As for the diameter of the copper wire in the secondary, I can do that in some transformers, but other are quite sealed and from the outside you can only see the Pins where the secondary copper wires attach to.



some simple questions:

1) How do I Load the transformer with resistor so I can measure the voltage drop? What do I need to do praticaly?
whats the resistor value equivalence to VA or amps?

2)what is the  the iron core and  thickness of the iron core?
can you please point that visually in the picture attached to this message?

 

[Whoops]

I was thinking and realized my question 1 was probably not clear enough.

I I need to measure the transformer secondary voltage, and present it a load. You should increase the load to the point where the secondary output voltage drops around 10%, then you convert that load to Amps, and you have an aproximate estimate of the output capabilities of the transformer.

So my questions are:
How do I setup a load?
How do I make it variable?
How do I convert that load (resistance value) to Amps?

Thanks

 

[PRR]

Start with a modification of analogguru's plan.

Weigh the lump.

Look at Hammond Mfg, Edcor USA, other power transformer sites for a lump of similar weight.

The VA will be similar.

Say your lump is 1.5 pounds.

Hammond's site is often easier to search, but they seem to be down tonight.

Edcor LVP2.5-3-120 is 1.4 pounds, very close. It is rated 2.5V at 3A which is 7.5VA. So your part is a bit over 7.5VA.

> whats the resistor value equivalence to VA or amps?

I = V/R

VA = V*A

(Somebody should print this on toilet-paper for extended study.)

If you have a transformer showing 7 or 8 Volts UN-loaded, it *may* be 6.3V nominal under load.

Say you have to feed six 12AX7. 6*0.3A is 1.8A. And 6.3V*1.8A is 11.34VA.

In simple heater work (tube heaters; or resistors), VA is Watts.

So the resistor must stand 11.34 Watts for long enough to read a meter. A 10W part may do. Bigger would be better. But we are not done.

The Ohms must be similar to 6.3V at 1.8A or 6.3/1.8= 3.5 Ohms.

In this range, I would get a baggie of 10 Ohms 10 Watt resistors. Three in parallel is 3.333 Ohms. IF we really sag to 6.3V, this will really be 6.3/3.333= 1.89 Amps. Close enough. Heat will be 11.9 Watts. Three 10W resistors sharing 11.9W is 4W each, will live for decades.

Clip your voltmeter (don't trust fingers when making close measurements, especially on higher voltage.)

Re-verify the NO-load voltage. Say 7.7V.

Figure a likely sag. If estimated under 20VA, use 20%. 20VA-50VA, 10%. Over 50VA, 5%.

Figure a likely loaded voltage. As 11.34VA is under 20VA, take off 20%. 0.80*7.7V= 6.16V.

Power-down, connect three 10r resistors. Power-up, read. If it reads 6.16V or more, the transformer is probably not suffering at 1.8A.

Since your resistors are ample, you can set it on a flame-proof board and touch the transformer at 2 minutes, 20 minutes, two hours, tomorrow, and see if it runs hot. Very-warm may be very-normal. Too hot to touch "can" be fine, but we generally do not like that.

There is one modern-age risk here. Many not-tiny transformers on the European market now have one-time Thermal Cutouts. If it is about to start a fire, the internal gizmo blows open first. In mild abuse, you will feel the outside get hot first. In gross abuse the inside may overheat before the outside gets warm. These fuses are often non-replaceable (without major and maybe illegal surgery).