Author Topic: transformer  (Read 11456 times)

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christos

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Re: transformer
« Reply #25 on: January 11, 2014, 08:04:56 pm »
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

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Re: transformer
« Reply #26 on: January 11, 2014, 10:21:46 pm »
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

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Re: transformer
« Reply #27 on: January 11, 2014, 10:35:45 pm »
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

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Re: transformer
« Reply #28 on: January 11, 2014, 11:05:51 pm »
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.
I'm not an EE--what am I doing here?

mrkev

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Re: transformer
« Reply #29 on: January 11, 2014, 11:27:36 pm »
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%...
Po=70%*S2=0,7*(l*d)2

IonizedGears

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Re: transformer
« Reply #30 on: January 11, 2014, 11:35:58 pm »
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.
« Last Edit: January 12, 2014, 12:35:02 am by IonizedGears »
I am an EE student with interests in Embedded, RF, Control Systems, and Nanotech.

christos

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Re: transformer
« Reply #31 on: January 12, 2014, 12:10:48 am »
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

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Re: transformer
« Reply #32 on: January 12, 2014, 12:12:40 am »
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%...
Po=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

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Re: transformer
« Reply #33 on: January 12, 2014, 12:17:43 am »
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).
I'm not an EE--what am I doing here?

christos

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Re: transformer
« Reply #34 on: January 12, 2014, 12:35:31 am »
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
« Last Edit: January 12, 2014, 12:38:41 am by christos »

IanB

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Re: transformer
« Reply #35 on: January 12, 2014, 12:55:20 am »
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.
I'm not an EE--what am I doing here?

johansen

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Re: transformer
« Reply #36 on: January 12, 2014, 01:00:47 am »
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

liquibyte

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Re: transformer
« Reply #37 on: January 12, 2014, 02:44:28 am »
http://ludens.cl/Electron/trafos/trafos.html

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

christos

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Re: transformer
« Reply #38 on: January 12, 2014, 02:21:42 pm »
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

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Re: transformer
« Reply #39 on: January 12, 2014, 02:33:08 pm »
My guess is that your transformer can deliver around 170 VA.

IonizedGears

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Re: transformer
« Reply #40 on: January 12, 2014, 05:40:35 pm »
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 am an EE student with interests in Embedded, RF, Control Systems, and Nanotech.

christos

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Re: transformer
« Reply #41 on: January 12, 2014, 08:49:32 pm »
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

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Re: transformer
« Reply #42 on: January 12, 2014, 10:44:50 pm »
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.
I'm not an EE--what am I doing here?

johansen

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Re: transformer
« Reply #43 on: January 12, 2014, 11:58:31 pm »
too hot is relative,

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.

« Last Edit: January 13, 2014, 12:00:44 am by johansen »

christos

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Re: transformer
« Reply #44 on: January 13, 2014, 09:50:48 am »
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

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Re: transformer
« Reply #45 on: January 13, 2014, 09:53:50 am »
too hot is relative,

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

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Re: transformer
« Reply #46 on: January 13, 2014, 12:48:45 pm »
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.

Smf