Re: How To: DC +/- Power Supply?

From: CFoley1064 (cfoley1064_at_aol.com)
Date: 11/08/04 

Date: 08 Nov 2004 21:00:44 GMT

>Subject: How To: DC +/- Power Supply?
>From: jeremycyr@gmail.com (Jeremy)
>Date: 11/8/2004 12:34 PM Central Standard Time
>Message-id: <8b3d9513.0411081034.4eb2eb6c@posting.google.com>
>
>Hello,
>
>Please excuse my stupidity here; I've checked the web, but can't seem
>to grasp the concepts.
>
>I'm working on a project that says it requires +/- 15volts DC to
>operate. It has a (common), (+), and (-) in.
>
>The 15 volt power supply that I have is the standard wall-wart style,
>and only shows a (+) and (-) out.
>
>How can I create the necessary supply for this device? I read
>something about needing to use a pair of voltage regulators, and other
>misc. parts to build a "rail splitter." Is this applicable in this
>situation of am I overcomplicating things?
>
>Any assistance in helping me understand what this means would be
>greatly appreciated! Thanks in advance,
>
>Jeremy

Hi, Jeremy. Good newbie question.

You're stuck with 15 volts potential between the two wires coming out of your
wall wart, and a +/-15V supply has 30V potential between the + and -.

One possible solution is a "rail splitter", which sets an artificial common
halfway between the two. As long as you didn't require very much current, you
might try this with two low ohm resistors splitting the difference like this
(view in fixed font or M$ Notepad):

 +o---o-------o +7.5V
      |
     .-.
     | |22 ohms 3 Watt
     | |
     '-'
 15V |
      o-------o"COM"
      |
     .-.
     | |22 ohms 3 Watt
     | |
     '-'
      |
 -o---o-------o -7.5V
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Of course this wastes a lot of power just to get a low output impedance for the
artificial GND. A better solution would be something like this:

 +o---o--------------o----------o+7.5V
      | |
     .-. .----|----.
     | |10K | | |
     | | | |\| |
     '-' '--|-\ |
 15V | | >---o-----oCOM
      o------------|+/
      | |/|
     .-. |
     | |10K |
     | | |
     '-' |
      | |
 -o---o--------------o----------o-7.5V
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

The op amp creates a low impedance for the virtual GND at the expense of 1 IC
and a couple of resistors.

These are the two basic ways to do a "rail splitter". But neither will get you
where you want to go, which is a +/- 15V supply. In order to do that, you can
do a bit of a trick with an oscillator and a few caps and diodes like this:

 +15V +15V +15V
  | + +
 .-. | |
 | |24K .---o----o---. C = 100uF 25V
 | | | 8 4 | D = 1N4002
 '-' | | C
  | | |
  o-----o7 | +|| D -14V
  | | 3o---||--o----|<--o----o
 .-. | | || | |
 | |56K | 555 | V C ---
 | | .--o6 | D - ---
 '-' | | | | +|
  | | | | | |
  o--o--o2 | === ===
  | | | GND GND
 --- | 1 5 |
 --- '---o----o---'
  | 3.3nF |
 === ===
 GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

This is called a voltage inverter, and isn't as tricky as it looks. Note that
a cap blocks DC, and lets AC through. The output of the 555 is AC coupled by
the first 100uF cap. The AC is clamped by the first diode so it doesn't exceed
0.6V or so. This 15V peak-to-peak signal, with to bound at +0.6V, is then
rectified by the second diode, and filtered by the second cap. It will provide
a -15V, less the voltage drop across the two diodes. This will give you a
-14VDC, which should be good for your op amps. Actually, this supply is
unregulated, and will have greater ripple as you increase the current draw. I
would limit the current from the negative supply to less than 20 mA with this
setup.

If you need more current, you'll have to go with a more complicated or
expensive setup. One thing you can do is buy a DC to DC converter which will
efficiently provide you more current. Or, you can just chalk it up to
experience and get the right power supply from Jameco, as another post has
suggested.

If you need more help, feel free to post again.

Good luck
Chris

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