power supply

[abdullahseba]

dose any one know how to make a high voltage regulator?
i have a 250v : 50v 20A transformer that i need to produce a stable 50 - 52v 20A DC.

 

[KrisBlueNZ]

Hi there and welcome to Electronics Point

50~52V at 20A is 1 kW. That's a lot of power! I think your only option would be a large switching power supply. Your cheapest option might be four 12V, 250W switching supplies with their output voltages cranked up a bit, and their outputs connected in series. I don't think your 250V/50V transformer will be of much use; you would waste a huge amount of money and space in smoothing capacitors, for a start!

 

[abdullahseba]

Hi there and welcome to Electronics Point

50~52V at 20A is 1 kW. That's a lot of power! I think your only option would be a large switching power supply. Your cheapest option might be four 12V, 250W switching supplies with their output voltages cranked up a bit, and their outputs connected in series. I don't think your 250V/50V transformer will be of much use; you would waste a huge amount of money and space in smoothing capacitors, for a start!

i have already got the transformer so do you have a regulation circuit?
though i might do as you say.

 

[KrisBlueNZ]

Converting AC to DC at mains frequency, and regulating with a linear regulator, at 20A output current, are both very inefficient and require some very large and heavy duty components.

Your bridge rectifier needs to be rated for an average current of at least 25A. I would use a 35A one such as http://www.digikey.com/product-detail/en/GBJ3504-BP/GBJ3504-BPMS-ND/3191596 which is cheap but will need a heatsink.

Your mains transformer will have a peak output voltage of around 70V. Subtract 4V to allow a 2V drop across each diode in the bridge and your peak voltage into the smoothing capacitors would be about 66V. Allow 4V dropout for the regulator, from a 52V output voltage, and the trough voltage at the smoothing capacitors must be at least 56V. That leaves (66 - 56) = 10V maximum ripple depth.

At 50 Hz mains frequency the ripple drop period will be about 9 ms. So the smoothing capacitance has to drop 10V maximum in a 9 ms period at 20A load current. You can calculate the smoothing capacitance:
dV / dT = I / C

C = I dT / dV
= 20 × 0.009 / 10
= 18,000 µF

So your smoothing capacitor would need to be at least 18,000 µF (I would use 27,000 µF or 33,000 µF) and rated for 100V. Something like http://www.digikey.com/product-detail/en/LNR2A333MSE/493-7699-ND/3930079 which will set you back USD 57. But that capacitor is rated for an RMS ripple current of 10A and the actual ripple current will be about 53A!

Now the mean voltage across a series regulator will be about (61 - 50) = 11V. At 20A that's a power dissipation of 220W. It would be best to split this across several transistors or MOSFETs; I would use a bank of about four 2N3055s since they're cheap. But with 220W dissipation you'll need something pretty big, and/or forced air cooled (i.e. with a fan). Something at least as big as http://uk.mouser.com/ProductDetail/Wakefield/395-2AB/?qs=sGAEpiMZZMttgyDkZ5WiulP/BNP%2bTZRYdZPVZTJEIlM= which would set you back about USD 43 plus the cost of the fan(s).

I hope you see that this is a lot more impractical than using switching power supplies!

 

[abdullahseba]

Converting AC to DC at mains frequency, and regulating with a linear regulator, at 20A output current, are both very inefficient and require some very large and heavy duty components.

Your bridge rectifier needs to be rated for an average current of at least 25A. I would use a 35A one such as http://www.digikey.com/product-detail/en/GBJ3504-BP/GBJ3504-BPMS-ND/3191596 which is cheap but will need a heatsink.

Your mains transformer will have a peak output voltage of around 70V. Subtract 4V to allow a 2V drop across each diode in the bridge and your peak voltage into the smoothing capacitors would be about 66V. Allow 4V dropout for the regulator, from a 52V output voltage, and the trough voltage at the smoothing capacitors must be at least 56V. That leaves (66 - 56) = 10V maximum ripple depth.

At 50 Hz mains frequency the ripple drop period will be about 9 ms. So the smoothing capacitance has to drop 10V maximum in a 9 ms period at 20A load current. You can calculate the smoothing capacitance:
dV / dT = I / C

C = I dT / dV
= 20 × 0.009 / 10
= 18,000 µF

So your smoothing capacitor would need to be at least 18,000 µF (I would use 27,000 µF or 33,000 µF) and rated for 100V. Something like http://www.digikey.com/product-detail/en/LNR2A333MSE/493-7699-ND/3930079 which will set you back USD 57. But that capacitor is rated for an RMS ripple current of 10A and the actual ripple current will be about 53A!

Now the mean voltage across a series regulator will be about (61 - 50) = 11V. At 20A that's a power dissipation of 220W. It would be best to split this across several transistors or MOSFETs; I would use a bank of about four 2N3055s since they're cheap. But with 220W dissipation you'll need something pretty big, and/or forced air cooled (i.e. with a fan). Something at least as big as http://uk.mouser.com/ProductDetail/Wakefield/395-2AB/?qs=sGAEpiMZZMttgyDkZ5WiulP/BNP%2bTZRYdZPVZTJEIlM= which would set you back about USD 43 plus the cost of the fan(s).

I hope you see that this is a lot more impractical than using switching power supplies!

thank you very much.
the power supply is for an underground train at a museum in London
http://www.walthamstowpumphousemuseum.org.uk/filming.html

 

[12vdc]

Hi
. I would offer the suggestion using the rectification and regulation circuits from an automotive alternator. Some units are factory rated from 100 amp to 200 amps.(GM uses a sense wire feedback which could be fooled into increasing output voltage) From Canada

 

[abdullahseba]

Hi
. I would offer the suggestion using the rectification and regulation circuits from an automotive alternator. Some units are factory rated from 100 amp to 200 amps.(GM uses a sense wire feedback which could be fooled into increasing output voltage) From Canada

I decided to use 2x 24v 8A power supply's in parallel

 

[KrisBlueNZ]

You'd better make sure they're designed to be paralled. Standard power supplies aren't. That's why I suggested connecting some lower-voltage ones in series.

 

[abdullahseba]

You'd better make sure they're designed to be paralled. Standard power supplies aren't. That's why I suggested connecting some lower-voltage ones in series.

I asked the seller he said you could

 

[(*steve*)]

I asked the seller he said you could

Just as long as he's not an eBay seller...

 

[KrisBlueNZ]

I asked the seller he said you could

He may have misinterpreted your question as being "I have a technical question that you don't understand and can't answer truthfully, but if you answer yes, you will make a sale".