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Transformer secondary current rating

R!f@@

Thread Starter

R!f@@

Joined Apr 2, 2009
9,918
Hi Guys,
I was just digging through my junk yard, when I came up with this transformer.
Now it is Huuuuuuuuuge. :D
I like to know is there anyway we can guess the secondary current capacity by measuring the coil diameter.
This former has a secondary coil diameter of around 2.7mm. (the width of the coil used)
What do you think the continues current rating would be.
Oh the secondary voltage is 35VAC
 
R!f@@

Thread Starter

R!f@@

Joined Apr 2, 2009
9,918
C'mon guys, we all know now about showing links to info's.
I know rod, from
http://sound.westhost.com/xfmr.htm
I use to email him, it was because of him that I have such a wonderful sound system.
one of a kind, every body wants it :cool:. But I ain't gonna sell it. :D.

I like you members opinion, and based on your experience I like to know what would be the current capacity you think would be, Just by looking at it or measuring the coil gauge.
that's all! Just give me what you think the current would be.:rolleyes:
 
kkazem

kkazem

Joined Jul 23, 2009
160
Hi,
Yes, certainly you can estimate the current rating of a transformer winding if you measure the wire diameter, which you say is 2.7mm = 0.1063" which is AWG #10 Heavy Film magnet wire. The area is 10,400 circular mils, and for a large transformer (although you never said how large, large is), is likely designed at between 500 Circular Mils/Amp (C.M./A) to 600 C.M./A, making the current rating likely between roughly 17 Amps to 21 amps. Now, having said that, the actual current rating will depend heavily on the total DC resistance of that winding, which I do not know. If you can, please measure it, not with an ohmmeter, but with a good DVM or two. Procedure for measuring low-resistances with a DVM. Using a lab power supply, or a D-cell battery and a 1-Ohm, 5 watt resistor, if no lab power supply is available, connect the source positive (lab supply or D-cell) in series with the 1-ohm resistor, then connect the other side of the 1-Ohm resistor to one-side of the winding to be measured, and connect the other side of the winding to be measured back to the source (lab supply or battery) minus side. Connect DVM-1, set to the 2 VDC scale, across the 1-Ohm resistor and connect DVM-2, set to 200 mVDC Scale (or lower if it has a lower scale), across the terminals or wire ends of the winding to be measured. If you only have 1 DVM, that will do, but you'll need to work quickly by having it initially setup for DVM-1 above, take a reading as quickly as possible if using a battery (since it will almost immediately start to go down in voltage), then as quickly as possible, connect it as DVM-2 above and take that reading (don't forget to change the voltage scale unless you have an autoscale DVM). If you are using a lab supply, there is no need to hurry up. Compute the unknown winding resistance as follows: Take the DVM-2 voltage divided by the DVM-1 voltage and this equals the unknown winding resistance. Once you have the winding resistance, you can compute the power dissipated in that winding by Power=I^2*R, with I=Current in the winding, and R=the measured/computed resistance of the winding. From there, it really depends on how big the transformer is. In order to set a maximum output current spec for that winding, knowing the transformer size and the lamination (core) dimensions, I could make a calculated guess at how much power loss that winding could safely tolerate, then calculate the current (I) from the formula above: I = Sqrt(Power/Resistance). Since I don't know how large, large is but you do, I can't make an estimate that way. Even if I did know or had a picture, I would still need the winding resistance measurement. You may ask why you can't simply measure the winding resistance using your DVM. The answer is that it is an low-Ohm measurement and your DVM leads may very well have more resistance than the winding you are trying to measure. Therefore, you need a more accurate method. This method works well for most DVMs down to a few milliOhms with a 3.5 digit DVM and down to less than a milliOhm for a 4.5 digit or greater DVM. Good luck and if you get that data and would like me to tryh and estimate the winding current rating a bit more accurately than I did at first, please send me a private email so that I'll know you're ready. I don't always get to AAC very often as I'm usually very busy at work.
Regards,
Kamran Kazem
 
R!f@@

Thread Starter

R!f@@

Joined Apr 2, 2009
9,918
kkazem said:
Hi,
Yes, certainly you can estimate the current rating of a transformer winding if you measure the wire diameter, which you say is 2.7mm = 0.1063" which is AWG #10 Heavy Film magnet wire. The area is 10,400 circular mils, and for a large transformer (although you never said how large, large is), is likely designed at between 500 Circular Mils/Amp (C.M./A) to 600 C.M./A, making the current rating likely between roughly 17 Amps to 21 amps. Now, having said that, the actual current rating will depend heavily on the total DC resistance of that winding, which I do not know. If you can, please measure it, not with an ohmmeter, but with a good DVM or two. Procedure for measuring low-resistances with a DVM. Using a lab power supply, or a D-cell battery and a 1-Ohm, 5 watt resistor, if no lab power supply is available, connect the source positive (lab supply or D-cell) in series with the 1-ohm resistor, then connect the other side of the 1-Ohm resistor to one-side of the winding to be measured, and connect the other side of the winding to be measured back to the source (lab supply or battery) minus side. Connect DVM-1, set to the 2 VDC scale, across the 1-Ohm resistor and connect DVM-2, set to 200 mVDC Scale (or lower if it has a lower scale), across the terminals or wire ends of the winding to be measured. If you only have 1 DVM, that will do, but you'll need to work quickly by having it initially setup for DVM-1 above, take a reading as quickly as possible if using a battery (since it will almost immediately start to go down in voltage), then as quickly as possible, connect it as DVM-2 above and take that reading (don't forget to change the voltage scale unless you have an autoscale DVM). If you are using a lab supply, there is no need to hurry up. Compute the unknown winding resistance as follows: Take the DVM-2 voltage divided by the DVM-1 voltage and this equals the unknown winding resistance. Once you have the winding resistance, you can compute the power dissipated in that winding by Power=I^2*R, with I=Current in the winding, and R=the measured/computed resistance of the winding. From there, it really depends on how big the transformer is. In order to set a maximum output current spec for that winding, knowing the transformer size and the lamination (core) dimensions, I could make a calculated guess at how much power loss that winding could safely tolerate, then calculate the current (I) from the formula above: I = Sqrt(Power/Resistance). Since I don't know how large, large is but you do, I can't make an estimate that way. Even if I did know or had a picture, I would still need the winding resistance measurement. You may ask why you can't simply measure the winding resistance using your DVM. The answer is that it is an low-Ohm measurement and your DVM leads may very well have more resistance than the winding you are trying to measure. Therefore, you need a more accurate method. This method works well for most DVMs down to a few milliOhms with a 3.5 digit DVM and down to less than a milliOhm for a 4.5 digit or greater DVM. Good luck and if you get that data and would like me to tryh and estimate the winding current rating a bit more accurately than I did at first, please send me a private email so that I'll know you're ready. I don't always get to AAC very often as I'm usually very busy at work.
Regards,
Kamran Kazem
Click to expand...
I haven't read the whole post yet, but I like you. :D
You are my kind of a guy
 
R!f@@

Thread Starter

R!f@@

Joined Apr 2, 2009
9,918
I got a coil resistance of 0.0685Ω.
1.5V applied, winding drops 100mV. R drops 1.4V.
So what do you think is the VA rating.
220VAC in put with a 35VAC out put
 
P

power_speed

Joined Jan 20, 2009
3
Could you not also just measure the current on the 220 side. If I remember correctly most transformers are very close to 100% transfer rate. So If 220VAC at 10amps in should be about 2200 watts on the output. Divided by 35VAC, I get about 60 amps. Of course my transformer theory is a bit rusty, so I may be wrong.
 
Z

zhgart

Joined Dec 28, 2009
19
coil diameter of around 2.7mm will be about 4mm^2, this will endure 10 A to 12A.
and your transf power will be 350w to 420w.
very heavy.
 
R!f@@

Thread Starter

R!f@@

Joined Apr 2, 2009
9,918


This is a screen print of a schema of the PSU I was planning to build with this transformer.
Simulation is OK and line plus load regulation is quite good I must say. :cool:
Problem is I cannot seem to get the output above 30V. :confused:
I like to get at least up to 40VDC.
Can any one of you give me a hint on how to increase the output.
I tried the feed back and V adjust. Any change from this config makes it worse. :mad:
I was thinking of trying a MOSFET as a pass element.
Will this change increase the output.
 

Attachments

SgtWookie

SgtWookie

Joined Jul 17, 2007
22,230
Your schematic is too small, cluttered and fuzzy for me to read.

You must have exported the image as a .jpg. That is not good.
Use .png format. It will be much more clear. And post a larger image of the schematic, without all of the instrumentation windows. Crop the edge so all we see is the schematic itself.

Look in National Semiconductor's datasheet for the LM723. They have lots of application examples shown.
 
R!f@@

Thread Starter

R!f@@

Joined Apr 2, 2009
9,918
SgtWookie said:
Your schematic is too small, cluttered and fuzzy for me to read.

You must have exported the image as a .jpg. That is not good.
Use .png format. It will be much more clear. And post a larger image of the schematic, without all of the instrumentation windows. Crop the edge so all we see is the schematic itself.

Look in National Semiconductor's datasheet for the LM723. They have lots of application examples shown.
Click to expand...
Find the attached pdf.
I tried the data sheets, that info is not enough.
Only way is to float the regulator, but I think it will worsen the load regulation. At least I like to squeeze 35 out of it. the best I can do is 30VDC tops. The main pass tr is not the actual one since I will be using 2SC3998 which can handle 25A at 250W, the NI did not have the model, so to simulate the 2N3055 is used. Other components are the default ones that I intend to use.
As for current limit, I will be implementing it once I am satisfied with the voltage. I can use a PIC for CC adjust and also after my new PIC's arrive, I will upgrade it to programmable PSU both with CV and CC :D

Shema is attached on a later post as I have made changes to it
 
Last edited:
eblc1388

eblc1388

Joined Nov 28, 2008
1,542
R!f@@,

You have told us you want to have 0~40V output from the power supply. Here are a few points to keep your mind focused.

1. Have you decided on how much current you want the power supply to give out? 5A, 10A or higher?

2. Do you want to maintain the same current even at low output voltage of 1 or 2V?

3. Do you want the power supply to endure possible output short circuit? For how long?

4. How much you can allow the output voltage to rise/drop/change say when the power supply output is at say 20V between no load and full load current? 1mV, 10mV, 100mV or 1 volt?

The last point is of vital important. Many hobbyist have built high current(>5A) *regulated* power supplies but very few have actually measured how its output performs between no load and loaded.

I can say for sure that 90% of them would drop more than 0.5V if not more. The performance is unrelated to the design of the schematic but to how one physically constructs the power supply.
 
R!f@@

Thread Starter

R!f@@

Joined Apr 2, 2009
9,918
First and for most, thank you L Chung for your time, I really appreciate it that you checked my diagram :)

You have told us you want to have 0~40V output from the power supply. Here are a few points to keep your mind focused.
Click to expand...
That was the idea, theoretical value gave me 49V from that transformer, so after building the rectifier and filter it measured 42V :mad:
So now my aim is at least 35V.

1. Have you decided on how much current you want the power supply to give out? 5A, 10A or higher?
Click to expand...
Now I know the transformer can handle 12A, so I will go for 12A max.

2. Do you want to maintain the same current even at low output voltage of 1 or 2V?
Click to expand...
May be, but I know the amount of power the pass transistor will be dissipating, and if I want it to handle 8 to 10A at 3.3VDC, I will add more transistors to keep the them below SOA of the device

3. Do you want the power supply to endure possible output short circuit? For how long?
Click to expand...
Definitely. that is why CC will be implemented later, Short circuit state will be time monitored by a PIC to cuttoff the out put (short circuit current is 15Amps)

4. How much you can allow the output voltage to rise/drop/change say when the power supply output is at say 20V between no load and full load current? 1mV, 10mV, 100mV or 1 volt?
Click to expand...
Load regulation should as low as posiible

The last point is of vital important. Many hobbyist have built high current(>5A) *regulated* power supplies but very few have actually measured how its output performs between no load and loaded.
Click to expand...
Very true indeed

I can say for sure that 90% of them would drop more than 0.5V if not more. The performance is unrelated to the design of the schematic but to how one physically constructs the power supply.
Click to expand...
What can you recommend to minimize load regulation.
 
eblc1388

eblc1388

Joined Nov 28, 2008
1,542
R!f@@ said:
What can you recommend to minimize load regulation.
Click to expand...
Sufficient component rating margins, proper wiring size and length and attention to every detail.

Don't expect to get a 5A variable power supply with a single pass transistor. It seldom works. On a "proper" commercial 5A 0-30V supply, no less than 9 off 2N3055 are being used to get the performance required.
 
t06afre

t06afre

Joined May 11, 2009
5,934
If your transformer is rated for 12 amp. You can not load your power supply with 12 amp on the output. How much will depend on your rectifier configuration. This has been discussed many times in this forum.
 
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