how change 5v 20Amper to 3v 20amper (help)

[alirezalahij]

hi

how i can give from 5volt 20Amper
3volt with 20 amper


 7803regulator  max outpot current 1 amper

but i need 20A

thank

[petarp]

Hi. You can use switching power supply from National Semiconductor like LM2737 (LM2727, LM2743, etc.): http://www.national.com/pf/LM/LM2737.html , http://www.national.com/cat/index.cgi?i=i//200 or you can use 7803 regulator with curren boost transistors(http://www.national.com/ds/LM/LM340.pdf page 13,14).
If you use linear regulator like 7803 power losses will be about 40W and need huge heat sink.

Best Regards.

[lillbear]

Try google with "current mirror"

yours

[Parmin]

mc34063 + external transistor banks can do it comfortably.

google for mc34063

[alirezalahij]

thanks

this cicurate working?

[insane4evr]

1N4007 is rated only at 1 ampere average when rectifying sine wave. The voltage drop at 1 ampere is typically 1 volt, not exact. Assuming you want each one share 1 ampere, you will need 20 in parallel for an average drop of 1 volt. You will need 2 of these 20 parallel diodes in series for a total average drop of 2 volts. Make heavy copper connections to pass 20 amperes without overheating. You will need forced air cooling (fan) to dissipate the 2x20=40 watts heat.

BUT don't do it. Read below.

Unfortunately, when you hook them up, with a 20 ampere load, they could self destruct since in reality each diode has a different voltage drop. If they don't, they will not last. The diode with the lowest voltage drop will be carrying the largest current among them and will be the first one to go. If it goes open, the rest of the diodes that are in parallel to it will each, in theory, carry 20/19 amperes which is higher than 1. The one diode among these 19 that has the lowest drop will again be carrying the highest current and the failure repeats. If the diode goes shorted then you no longer have 3 volts output but higher than 3, causing the load to draw more than 20 amperes. You destroy these diode array and you probably will detroy your load and if your power supply has no current limiter, destroy your power supply also.

Maybe your 5 volts 20 amps power supply has an internal trimmer to adjust its output down.

[alirezalahij]

Maybe your 5 volts 20 amps power supply has an internal trimmer to adjust its output down.

i use pc power

[mhemara]

i use pc power

New ATX PC power supply has a 3.3v output @ 30A
Have fun  

[Jehan]

You can use MIC5157 from micrel and use a mosfet mounted on a large heatsink (regulator mosfet will dissipate 30 Amp X 1.7 volt, that is 51 watt) this will be lowest cost option,

alternativly use a standerd stepdown POL converter from Power-One or any other manufacturer which will convert 5 to 3.3 at more than 90% efficiency ( PD will be less than 10 watt in this case)

if you want to build your own converter it will be difficult as this kind of designs will require a multilayer board with lot of thermal management and and use of SMD fets like Direct Fet from IR which will require PICK place and reflow oven to do it.

[fpgaguy]

Diode approach will be a problem since you can't match the diodes -

If you don't want to build and learn all about switching regulators I suggest a prebuilt module such as
http://focus.ti.com/docs/prod/folders/print/ptd08a020w.html

For what they do; they are quite inexpensive and compact.
I've used very similar TI parts and they are pretty reliable; but it's not your only option by far

Typically you need a few external caps and a programming resistor - and there is usually a way to turn on/off or sequence the module via a pin on the module.

[LabVIEWguru]

If you want to use a linear regulator, use old school with a regulator and 3 big pass transistors like the 2N3055 and a big heat sink. You could use one of the LM317 adjustable regulators, 2 resistors and 3 (8 amp each I think) transistors and a pot to adjust the LM317 for the voltage you want. Personally, I would just use a switching power supply out of a  computer (and probably some filter caps - they are really noisy) to give you your voltage.