transformer

[christos]

i have a transformer that get output 52 volts
its 10x10cm  very heavy
how do i know how many amps it can get at the output?
its    26-0-26

if you need pic i can show you

[Rudane]

Do you have any identifying numbers from the device? I'm not sure you can determine the maximum current out without knowing more information about the transformer.

[christos]

Do you have any identifying numbers from the device? I'm not sure you can determine the maximum current out without knowing more information about the transformer.

nothing..it got 2 secondaries 26v each

[Simon123]

Do you know, what it was connected to before?

[IonizedGears]

Do you have any identifying numbers from the device? I'm not sure you can determine the maximum current out without knowing more information about the transformer.

nothing..it got 2 secondaries 26v each

Assuming that your input is 120v, the ratio is 52/120(.4333) for the two combined secondaries  for V, or 120/52(2.3) for I. We could calculate this if you had the impedance of the main coil or the current through the main coil.

[christos]

Do you know, what it was connected to before?

a 40watt mic amplifier ..it used one more transformer at the output for the speakers

[christos]

Do you have any identifying numbers from the device? I'm not sure you can determine the maximum current out without knowing more information about the transformer.

nothing..it got 2 secondaries 26v each

Assuming that your input is 120v, the ratio is 52/120(.4333) for the two combined secondaries  for V, or 120/52(2.3) for I. We could calculate this if you had the impedance of the main coil or the current through the main coil.

input is 230v

[IanB]

i have a transformer that get output 52 volts
its 10x10cm  very heavy
how do i know how many amps it can get at the output?
its    26-0-26

if you need pic i can show you

The rating of a transformer is measured in VA (more or less the power handling).

Two ways to estimate the VA of a mains transformer are from the overall weight, or the dimensions of the core (height x width x thickness).

Given either or both of those, there are rules of thumb that will give you a ball park estimate.

[christos]

if i conect a  10ohm 10w res and measure the current..at the same time measure the tempp from the xformer ,,should it work?

[IonizedGears]

Well then, 52/230(.23)  for V, 230/52(4.2) for I. If you have the input impedance, then we could give you an exact answer. One way to do it is keep loading the secondaries until the voltage starts to drop and you could use E/R to find the max amperage for the secondaries this way.

[IanB]

if i conect a  10ohm 10w res and measure the current..at the same time measure the tempp from the xformer ,,should it work?

Item: the transformer provides 50 V
Item: the current through a 10 ohm resistor will therefore be 5 A
Item: the power dissipated in the resistor will therefore be 250 W

Conclusion: smoke will occur, if not flames

[IanB]

if i conect a  10ohm 10w res and measure the current..at the same time measure the tempp from the xformer ,,should it work?

As noted above the 10 ohm resistor is not a suitable load. But in general, increasing the load on a transformer until it starts to get warm is a good way to find the upper limit on the power handling.

[christos]

i have a transformer that get output 52 volts
its 10x10cm  very heavy
how do i know how many amps it can get at the output?
its    26-0-26

if you need pic i can show you

The rating of a transformer is measured in VA (more or less the power handling).

Two ways to estimate the VA of a mains transformer are from the overall weight, or the dimensions of the core (height x width x thickness).

Given either or both of those, there are rules of thumb that will give you a ball park estimate.

the core is 4x4x4

[rdl]

Do you know, what it was connected to before?
a 40watt mic amplifier

This might be a clue... or am I wrong?

[christos]

if i conect a  10ohm 10w res and measure the current..at the same time measure the tempp from the xformer ,,should it work?

As noted above the 10 ohm resistor is not a suitable load. But in general, increasing the load on a transformer until it starts to get warm is a good way to find the upper limit on the power handling.

so it may work?

[kg4arn]

If this helps get you in the ball park
About 70VA/kg is what I have read.

I have 2 transformers here that I have weighed:
Hammond mfg rated at 30VA weight = 0.522kg --> 58VA/kg
Signal mfg rated at 100 VA weight = 1.22kg --> 82VA/kg

[christos]

ill post some pics

[christos]

well the core is 7 height 5 width 7 length
its 2.5 kg te metal is 9 height 11 width and 3 leght

[IanB]

the core is 4x4x4

well the core is 7 height 5 width 7 length
its 2.5 kg te metal is 9 height 11 width and 3 leght

In what units? mm? cm? inches?

When you give measurements you must always give the units of measure.

[christos]

the core is 4x4x4

well the core is 7 height 5 width 7 length
its 2.5 kg te metal is 9 height 11 width and 3 leght

In what units? mm? cm? inches?

When you give measurements you must always give the units of measure.

oups..read the new measurements if you didnt and its in cm

[IanB]

I cannot make sense of your measurements. The core is the metal part with the laminations that has the coils wound around the middle of it. The height and width will usually be as big as the transformer. The thickness will usually be less than the height and width.

[christos]

I cannot make sense of your measurements. The core is the metal part with the laminations that has the coils wound around the middle of it. The height and width will usually be as big as the transformer. The thickness will usually be less than the height and width.

oups at the core i mean the windings..ill post a pic

[tasos987]

IMHO reply #9 helps always to stay on the safe side....
Other bulk solutions i think include a minimal risk of smoke poduction

[christos]

I cannot make sense of your measurements. The core is the metal part with the laminations that has the coils wound around the middle of it. The height and width will usually be as big as the transformer. The thickness will usually be less than the height and width.

oups at the core i mean the windings..ill post a pic

http://postimg.org/image/cfu7djqjf/

[christos]

IMHO reply #9 helps always to stay on the safe side....
Other bulk solutions i think include a minimal risk of smoke poduction

taso ellada edw

[christos]

Well then, 52/230(.23)  for V, 230/52(4.2) for I. If you have the input impedance, then we could give you an exact answer. One way to do it is keep loading the secondaries until the voltage starts to drop and you could use E/R to find the max amperage for the secondaries this way.

its 16 ohm

[IvoS]

I know it's possible to calculate VA ratings from cross section of the core in cm. I used to do that when I was young, but forgot it  .....

[christos]

I know it's possible to calculate VA ratings from cross section of the core in cm. I used to do that when I was young, but forgot it  .....

so is it posible to measure VA from the dimensions? if yes pls remeber how you did

[IanB]

http://postimg.org/image/cfu7djqjf/

You have measured the windings. The core is the big square silvery lump of metal surrounding the windings that you didn't measure.

Measure the thickness of it, then post a side view of the transformer with height and width measurements.

The size of the core is what mainly determines the power rating of the transformer.

[mrkev]

I know it's possible to calculate VA ratings from cross section of the core in cm. I used to do that when I was young, but forgot it  .....

so is it posible to measure VA from the dimensions? if yes pls remeber how you did

It is, but you have to know what to measure... Dimensions of the core are selected in the way that they don't have too much magnetic flow (is flux the propper english word? not sure here).
When you are calculating the transformer, you calculate the input power first. Then select the magnetic flux density (B) you want to have (usually B=1T - tesla). The area of the cross section of core should be S=sqrt(P/B), with that 1T it's just S=sqrt(P).
I made a quick image so you'd know what I mean by area of the cross section of core. It's S = l*d, just the middle grey area (dimensions of whole transformer as you posted won't help that much).

So power that you want to get from the output should be about 85% of input, because you are just calculating it, i would go with 70%...
P_o=70%*S²=0,7*(l*d)²

[IonizedGears]

Well then, 52/230(.23)  for V, 230/52(4.2) for I. If you have the input impedance, then we could give you an exact answer. One way to do it is keep loading the secondaries until the voltage starts to drop and you could use E/R to find the max amperage for the secondaries this way.

its 16 ohm

Well,  4.2(Turns ratio) = ?(16?/x?) so 17.64x=16, X = .907?

Since the secondary has .907?,  52V/.907? = 57.33A .

This is going by if you shorted it directly so I doubt this the actual max the wires could handle but this gives you theoretical max current capable at the secondary(I'm sure it is lower taking physical limitations and the huge voltage drop from 52V into consideration)

Edit: The first "?" was a square root symbol and the rest were ohm symbols I found online, turns out they don't work here although they looked fine before I posted.

[christos]

http://postimg.org/image/cfu7djqjf/

You have measured the windings. The core is the big square silvery lump of metal surrounding the windings that you didn't measure.

Measure the thickness of it, then post a side view of the transformer with height and width measurements.

The size of the core is what mainly determines the power rating of the transformer.

http://postimg.org/image/n34bam54t/

[christos]

I know it's possible to calculate VA ratings from cross section of the core in cm. I used to do that when I was young, but forgot it  .....

so is it posible to measure VA from the dimensions? if yes pls remeber how you did

It is, but you have to know what to measure... Dimensions of the core are selected in the way that they don't have too much magnetic flow (is flux the propper english word? not sure here).
When you are calculating the transformer, you calculate the input power first. Then select the magnetic flux density (B) you want to have (usually B=1T - tesla). The area of the cross section of core should be S=sqrt(P/B), with that 1T it's just S=sqrt(P).
I made a quick image so you'd know what I mean by area of the cross section of core. It's S = l*d, just the middle grey area (dimensions of whole transformer as you posted won't help that much).

So power that you want to get from the output should be about 85% of input, because you are just calculating it, i would go with 70%...
P_o=70%*S²=0,7*(l*d)²

thats nice..can you post any site if you find that they explain how to measure transformers or make ?im very interested now on transformers

[IanB]

http://postimg.org/image/n34bam54t/

That's good. Now we just need the front/back view as shown by mrkev above in post #29. I think I said "side view" earlier when perhaps I meant front/back view. We need a picture where we can estimate the measurement "l" shown in mrkev's picture (or you can measure it directly).

[christos]

http://postimg.org/image/n34bam54t/

That's good. Now we just need the front/back view as shown by mrkev above in post #29. I think I said "side view" earlier when perhaps I meant front/back view. We need a picture where we can estimate the measurement "l" shown in mrkev's picture (or you can measure it directly).

you want the middle iron?
if you need that..i cant get that on pic..cant see it but i can measure it if you want

[IanB]

you want the middle iron?
if you need that..i cant get that on pic..cant see it but i can measure it if you want

Yes, we need that measurement. If you take a picture of the front of the transformer at a slight downward angle it will be easy to estimate.

[johansen]

thats nice..can you post any site if you find that they explain how to measure transformers or make ?im very interested now on transformers

use my spreadsheet:
http://johansense.com/bulk/spreadsheets/

[liquibyte]

http://ludens.cl/Electron/trafos/trafos.html

The rest of the site has good info on transformers as well.

[christos]

you want the middle iron?
if you need that..i cant get that on pic..cant see it but i can measure it if you want

Yes, we need that measurement. If you take a picture of the front of the transformer at a slight downward angle it will be easy to estimate.

http://postimg.org/image/8klkrq2k7/   here are the real measurements

[IvoS]

My guess is that your transformer can deliver around 170 VA.

[IonizedGears]

Well then, 52/230(.23)  for V, 230/52(4.2) for I. If you have the input impedance, then we could give you an exact answer. One way to do it is keep loading the secondaries until the voltage starts to drop and you could use E/R to find the max amperage for the secondaries this way.

its 16 ohm

Well,  4.2(Turns ratio) = ?(16?/x?) so 17.64x=16, X = .907?

Since the secondary has .907?,  52V/.907? = 57.33A .

This is going by if you shorted it directly so I doubt this the actual max the wires could handle but this gives you theoretical max current capable at the secondary(I'm sure it is lower taking physical limitations and the huge voltage drop from 52V into consideration)

Edit: The first "?" was a square root symbol and the rest were ohm symbols I found online, turns out they don't work here although they looked fine before I posted.

Wait a second, I think i may have forgotten that for max power transfer the load on the secondary should be equal to the impedance of the secondary. So 52V/1.814Ohm = 28.67 Amps max on the secondary. Correct me if I'm wrong, but if I am wrong I would like to see the actual max secondary amperage.


When calculated another way, 230V/16Ohm = 14.375A on the primary and then 4.2*14.375 = 60.375 possible amps on the secondary, so my original estimate with the 57.33 A is pretty spot on for max current possible.

[christos]

Well then, 52/230(.23)  for V, 230/52(4.2) for I. If you have the input impedance, then we could give you an exact answer. One way to do it is keep loading the secondaries until the voltage starts to drop and you could use E/R to find the max amperage for the secondaries this way.

its 16 ohm

Well,  4.2(Turns ratio) = ?(16?/x?) so 17.64x=16, X = .907?

Since the secondary has .907?,  52V/.907? = 57.33A .

This is going by if you shorted it directly so I doubt this the actual max the wires could handle but this gives you theoretical max current capable at the secondary(I'm sure it is lower taking physical limitations and the huge voltage drop from 52V into consideration)

Edit: The first "?" was a square root symbol and the rest were ohm symbols I found online, turns out they don't work here although they looked fine before I posted.

Wait a second, I think i may have forgotten that for max power transfer the load on the secondary should be equal to the impedance of the secondary. So 52V/1.814Ohm = 28.67 Amps max on the secondary. Correct me if I'm wrong, but if I am wrong I would like to see the actual max secondary amperage.


When calculated another way, 230V/16Ohm = 14.375A on the primary and then 4.2*14.375 = 60.375 possible amps on the secondary, so my original estimate with the 57.33 A is pretty spot on for max current possible.

i dont know how its done..no idea..but can it have output 4 amps??http://postimg.org/image/8klkrq2k7/

[IanB]

From the information we have I would estimate that transformer can probably handle a 100 W load without overheating. It may be able to handle a higher load towards 200 W, but you would have to test it to find out.

If you can find a way of creating test loads (light bulbs?), then load it up and watch the temperature. As long as it can run continuously without getting too hot to touch it should be within its capabilities.

[johansen]

too hot is relative,
 and heat follows electrical loading squared.

also, rectifier loads double or triple copper losses due to very poor power factor.

my spreadsheet suggests 125 watts under the following conditions:
copper losses heat the wire at 1 degree C per 30 seconds.
1.5T flux@ 90% iron fill factor, primary wound with 23 gauge wire(230 volts), secondary with 16 gauge.
copper fill factor is limited to 50%,  actual fill is 36% for the primary half and 41% for the secondary half, averaging about 38%, and the copper should weigh just over half a kilogram.

Increasing the load to 150 watts results in the copper wire heating up at the rate of 1 degree C per 21.6 seconds.
Neglecting leakage inductance which my spreadsheet doesn't make any attempt to calculate, voltage drop under load is below 5% for both examples and efficiency is about 88%

iron loss is of no concern, copper losses almost always dominate.

[christos]

From the information we have I would estimate that transformer can probably handle a 100 W load without overheating. It may be able to handle a higher load towards 200 W, but you would have to test it to find out.

If you can find a way of creating test loads (light bulbs?), then load it up and watch the temperature. As long as it can run continuously without getting too hot to touch it should be within its capabilities.

thx alot..ill test it and see how much it can handle..ill post the results

[christos]

too hot is relative,
 and heat follows electrical loading squared.

also, rectifier loads double or triple copper losses due to very poor power factor.

my spreadsheet suggests 125 watts under the following conditions:
copper losses heat the wire at 1 degree C per 30 seconds.
1.5T flux@ 90% iron fill factor, primary wound with 23 gauge wire(230 volts), secondary with 16 gauge.
copper fill factor is limited to 50%,  actual fill is 36% for the primary half and 41% for the secondary half, averaging about 38%, and the copper should weigh just over half a kilogram.

Increasing the load to 150 watts results in the copper wire heating up at the rate of 1 degree C per 21.6 seconds.
Neglecting leakage inductance which my spreadsheet doesn't make any attempt to calculate, voltage drop under load is below 5% for both examples and efficiency is about 88%

iron loss is of no concern, copper losses almost always dominate.

can you tell me how many degree shoud a transformer work on max output..ill make a test whille measureing the temp and the current..

[peter.mitchell]

put it in a cardboard box just a little bit bigger than the transformer and tape a thermocouple to  the transformer, seal the box, load the transformer gradually until it runs at 45oC - 50oC without variation for an extended period of time.