Hello,

I have a transformer from a Commodore 1540/1541 drive. Attached is a picture of the transformer schematic.

I would like to know if this transformer is adequate for use in a +/-12V and +5V linear power supply with a common ground.

When connecting the white and black wires on the transformer to mains, the voltage on the orange to orange is 16.85V and blue to blue is 10.58V. I don't believe this configuration would work as the AC output of the bottom secondary seems too low.

When connecting the white and red wires on the transformer to the mains, the voltage on the orange to orange is 19.59V and blue to blue is 12.29V. I am thinking that if I use this configuration, the bottom secondary AC output of 12.29V is still too low. My concern is that the voltage dropped by diodes and then by the voltage regulator (7812 or 7912) would be too much. Is this correct?

Also according to the documentation I could find on line, it mentions that a 1A 250V slow blo fuse is used with this transformer. From this is it safe to conclude that this transformer is rated at 1A at the very least?

 

When connecting the white and black wires on the transformer to mains, the voltage on the orange to orange is 16.85V and blue to blue is 10.58V. I don't believe this configuration would work as the AC output of the bottom secondary seems too low.

Also according to the documentation I could find on line, it mentions that a 1A 250V slow blo fuse is used with this transformer. From this is it safe to conclude that this transformer is rated at 1A at the very least?

Let's see.
16.85V * 1.1414 - 0.7 * 2 = 17.83V
10.58V * 1.1414 - 0.7 * 2 = 10.68V
The dropout voltage for the Fairchild chips is 2V, so there is plenty of voltage for regulating the output.

 

Also according to the documentation I could find on line, it mentions that a 1A 250V slow blo fuse is used with this transformer. From this is it safe to conclude that this transformer is rated at 1A at the very least?

No. 1 amp is when it has failed. However, that is 1 amp on the primary side. The secondary side will have a lot more amps than the input side because the voltage output is a lot lower. You might be able to get 4 or 5 amps out of this one.

 

Let's see.
16.85V * 1.1414 - 0.7 * 2 = 17.83V
10.58V * 1.1414 - 0.7 * 2 = 10.68V
The dropout voltage for the Fairchild chips is 2V, so there is plenty of voltage for regulating the output.

So based on this I can connect the mains to the primary black and white wires, and then I would have the following voltages after rectification and after the voltage regulator drop:

16.85V * 1.414 - (0.7[diode drop] * 2) - 2[regulator drop]= 20.43V
10.58V * 1.414 - (0.7[diodes drop] * 2) - 2[regulator drop]= 11.56V

Since I want +/- 12V using the 7812 and 7912 I am guessing in the case of the 10.58V output from one of the secondary coils, the result will be cutting it too close, is this correct? Would it be recommended to use schottky diodes instead for the lower forward voltage drop?

Would this transformer be classified as having a certain tap? I am unsure as it has four wires instead of three. Could I connect the two secondary coils in series and would the following configuration then be possible (using 12volt regulators) as shown in this example:

http://dragonsoul.darkbb.com/t3-schematic-for-dual-rail-power-supply-15volt

 

16.85V * 1.414 - (0.7[diode drop] * 2) - 2[regulator drop]= 20.43V
10.58V * 1.414 - (0.7[diodes drop] * 2) - 2[regulator drop]= 11.56V

Since I want +/- 12V using the 7812 and 7912 I am guessing in the case of the 10.58V output from one of the secondary coils, the result will be cutting it too close, is this correct? Would it be recommended to use schottky diodes instead for the lower forward voltage drop?

Would this transformer be classified as having a certain tap? I am unsure as it has four wires instead of three. Could I connect the two secondary coils in series and would the following configuration then be possible (using 12volt regulators) as shown in this example:

There is a problem with your math; you should have 15.83V and 8.68V.

The circuit in the link has a higher voltage output then 30V. Using the higher voltage, I get 18.96V (still not enough).

I have never seen it done, and it would reduce the available current by half. A possibility is to have 2 bridges, use one for the +12V supply and use the other for a -12V supply. Would this work?

 

I get the same numbers as Van53.
However, this is the unloaded voltage of the transformer. It will be less under load, possibly by as much as 5% to10%.

 

I get the same numbers as Van53.
However, this is the unloaded voltage of the transformer. It will be less under load, possibly by as much as 5% to10%.

Silly me. I'm using 1.1414.

 

There is a problem with your math; you should have 15.83V and 8.68V.

The circuit in the link has a higher voltage output then 30V. Using the higher voltage, I get 18.96V (still not enough).

I have never seen it done, and it would reduce the available current by half. A possibility is to have 2 bridges, use one for the +12V supply and use the other for a -12V supply. Would this work?

Using two bridge rectifiers should work as you suggested. I noticed that in the attached schematic of the power supply from which I had extracted this transformer, this is how it was designed. It looks like it is just a matter of swapping the LM340-12 and LM340-5 for a 7812 and a 7912.

The design of the circuit I linked in my previous reply looked simpler. So I was curious if such a circuit would work using 12 volt regulators (instead of the 15 volt regulators shown in the circuit) by connecting the secondary's in series... I am wondering that since the secondary voltages are different in my transformer, one is high than the other, if I connect them in series, would the result be equal voltages relative to the center? The idea was that if this is the case, then an equal load on the +12 and -12V lines would result in each regulator dissipating the same amount of heat.

 

The diagram you attached produces +12V and +5V. Did you mean to attach a different diagram?

Edit: since the grounds are connected together, the 5V is between the ground and 12V.

 

The diagram you attached produces +12V and +5V. Did you mean to attach a different diagram?

Edit: since the grounds are connected together, the 5V is between the ground and 12V.

I meant to attach this diagram. This is the schematic for the unit from which the transformer was extracted. It shows two bridge rectifiers as I believe you were suggesting to connect it this way. It also shows a 12V and 5V output with a common ground (to my understanding). So I would conclude that for a +/-12V supply,all that would need to be modified in this diagram is to swap the LM340-12 and LM340-5 for a 7812 and a 7912. (Assuming the voltage is enough as poster #12 indicated the voltage will be 5-10% less under load).

The other diagram I was referring to was this :
http://dragonsoul.darkbb.com/t3-sche...-supply-15volt

Please ignore the 15V regulators (I intend to use 7812/7912). So in the above link they have a center tap transformer.

I just read at this site:

http://www.diyaudio.com/forums/parts/162991-center-tapped-transformer.html

Where it seems it is possible to create a center tap from two secondary's. I think I may try this as IF the voltage is equal relative to the center, and if the current drawn on the positive and negative 12 volt rails are the same during load conditions, then the heat dissipated from the regulators would be equal to my understanding.

 

Looking at my opening post, would anyone know why is it that when the red and white wires connected to the mains results in a higher voltage on the secondary as compared to when the white and black wires are connected to the mains?

edit: Is this because when the mains is connected to the red and white, the number of turns on the primary is reduced, and since the number of turns on the secondary remains the same, the voltage on the secondary is increased?

 

For a transformer (assuming no loss)
V1 / N1 = V2 / N2
The voltage divided by the number of turns on the primary side equals the voltage divided by the number of turns on the secondary side.

 

For a transformer (assuming no loss)
V1 / N1 = V2 / N2
The voltage divided by the number of turns on the primary side equals the voltage divided by the number of turns on the secondary side.

Excellent. Makes sense. Thanks!

 

I connected the two secondary's in series so they were out of phase. The voltages remained the same as when measured before they were connected in series (i.e. 16.8V RMS and 10.4VRMS).

I think I'll leave them separate and use two bridge rectifiers are previously suggested.

 

No. 1 amp is when it has failed. However, that is 1 amp on the primary side. The secondary side will have a lot more amps than the input side because the voltage output is a lot lower. You might be able to get 4 or 5 amps out of this one.

That's great, 4 to 5 amps would be more than enough as I plan to design for a maximum of 1 amp per rail (+12V@1A,-12V@1A,5V@1A)

I've attached an image of the transformer.

 

I connected the two secondary's in series so they were out of phase. The voltages remained the same as when measured before they were connected in series (i.e. 16.8V RMS and 10.4VRMS).

.

They will be identical across the individual winding ends, but when out of phase connected, there would be a cancellation effect across the outer ends.
When connecting identical windings in parallel, care has to be done to ensure in-phase connected.
Max.

 

They will be identical across the individual winding ends, but when out of phase connected, there would be a cancellation effect across the outer ends.
When connecting identical windings in parallel, care has to be done to ensure in-phase connected.
Max.

When I connected the secondary's in series, out of phase, measuring the outer ends showed 27.3V RMS.

When I connected the secondary's in series, in phase, measuring the outer ends showed 6.2V RMS.

From this it appears there is a cancellation effect across the outer ends when connected in phase. Above you mentioned this cancellation effect should occur when connected out of phase. Am I missing something here?

Thanks

 

In phase, out of phase...just connect them for the larger voltage. That way they won't be fighting each other.

 

When I connected the secondary's in series, in phase, measuring the outer ends showed 6.2V RMS.

From this it appears there is a cancellation effect across the outer ends when connected in phase. Above you mentioned this cancellation effect should occur when connected out of phase. Am I missing something here?

Thanks

You obviously had the in/out of phase reversed?
A decent transformer diagram comes with a dot marking the start of any secondary windings, these show automatically how to connect in phase.
The end of any winding goes to the start of the next, IOW wound in the same direction, in phase.
Max.

 

You obviously had the in/out of phase reversed?
A decent transformer diagram comes with a dot marking the start of any secondary windings, these show automatically how to connect in phase.
The end of any winding goes to the start of the next, IOW wound in the same direction, in phase.
Max.

I think I am looking at this the wrong way. Referring to the transformer diagram in my opening post (it does not have the phase dots), I connected one orange to one blue. Then on my scope I connected both channel ground pins to this center tap point. Next I connected the other orange wire to one channel, and the remaining blue wire to the next channel. On the scope both sine waves were being displayed "in phase". In this configuration, when measuring between the orange and blue extremes, I get 6.2VRMS. Since each point, relative to the joined center tap is showing in phase on the scope, I referred to this configuration as "in phase".

Starting over. When connecting the same orange with the other blue wire to make a center tap point, and connecting and measuring as described above, the sine waves on each channel of the scope appear out of phase, and I get 27.3V RMS across the extremes. I referred to this configuration as "out of phase".

If you could please let me know where I've gone wrong it would be appreciated.