how to assemble my own power supply

Need someone to guide me to make my own power supply from 115vac to 6vdc.
would appreciate someone to tell me what all i would require for this project and what board i need for such.

thanks in advance,

luis

Here you will find detailed description on a homemade power supply based on the very populard voltage regulator - LM317 ..
**broken link removed**

Regards,
IanP

My first question is why?

The Chinese make one way cheaper than you could ever make. However, the Chinese fudge their numbers too, so watch out. A 5v/5a supply might just be 20v/much less current supply that loads to 5v @ 5a. ;-)

But, to answer your question:

1. AC cord
1a. fuse
1b. snubber network (if you care)
2. Transformer
3. Rectifier
4. Filter capacitor (size determines load current capacity)
5. Regulator (6v can be had via the LM317 adj regulator)
6. Post regulaton capacitor (which you may not need if your load is nearby)

For a 6v supply, I'd go to Radio Shack and pick up their 120/12ish transformer. For the regulator, I'd use the 317, as it's easily set up (vo=(1+r2/r1)*1.25 or some such) .

If, per chance, you want large current output, you'll need a series pass transistor, where the regulator's output sets the voltage Vo, but the series pass transistor feeds in current to the regulator output.

Also, if you don't use the series pass, you should put a diode (a 1n4001?) from output to input of the regulator, so that when you kill the supply power your output capacitor's discharge bypasses your regulator's output circuitry.

(Incidentally, if you blow yourself up, remember 1) I told you to just buy one, and 2) there is no warranty expressed, implied, or written that the above information isn't a crock of hooey. ;-) )

ian p,
thanks for ur prompt repl and assistance.
i enjoy the site.

cheers

luispapas

Added after 2 minutes:

EULER's,
Thanks for ur assistance, will try and purchse the parts needed.

Thanks a Mill.

luispapas

I failed to mention something...

That 120/12 is vin=120v(rms) and vo=12v(rms).

As I remember it,

v(rms)

=

v(peak)

1
---
2^0.5)

Hence, if the transformer puts out 12v(rms) with a 120v(rms) vin

12

=

v(peak)

0.707...

then v(peak) is

v(peak)

=

12
----
0.707

=

16.97v

If you used a bridge rectifier (4 diodes, two for each cycle), you'd drop (roughly) 2(0.7)=1.4v.

This gives you roughly 16.97-1.4 = 15.57v into your regulator. However, this creates a 15.57-6 = 9.57 difference between input and output. Don't try to pull too much current (LM317 can do appprox 1.5a after all) or you're going to need a heat sink (chunk of metal to carry heat away from the device).

If you don't know what size heat sink you need, it's ok, as the LM317 has thermal protection. You'll know thermal protection has kicked in if you get negative spikes with your load attached. (Thermal protection means the current (your output!) will be reduced to reduce heat.) Just increase your heat sink's size until the spikes go away.

Although another trick (if you haven't a scope) is to just monitor the heat with your finger, and use a meter, on dc for your output v and ac for spike sensing. If you feel the temperature increasing, increase your heat sink's size. Increase until you get a temperature that remains fairly (it is your finger thermometer after all) steady.

If you don't like the heat sink idea, you can always reduce the input v via a series resistor. However, at all times (all loading) you need a 2.5 to 3v difference between output and input for your regulator to work right. Otherwise, you'll lose regulation (your output v).

The above is courtesy of SOP Engineering (seat-o-pants) ;-)