pos 12 volts to split 5 volts supply?

dacflyer · Post by **dacflyer** » Mon May 22, 2006 8:47 pm

as you all can see from all my post lately, i am a real buzzzy body..many projects in the works..and with everyones help, i am making good headway..just wanted to say thanks to all..

now on to one more question..
i am wanting to make me a split 5 volts supply.

-5 volts - GND - +5volts
from a 12 volt DC single supply
i tried a pos and neg regulators,,but it will not work
they clash... and trying to series 2 pos regulators do not work either..
i also have a circuit using a 4049 IC and apair of dioads, but i cannot get this to work either.. the 4049 needs a osc. input of which i have, but i cannot get the darnthing to work.. i need something that will work

its no problem if i use batterys, but i want to use a old wal-wart

ok, i am tired now, i am going to bed,, thanks

dyarker · Post by **dyarker** » Mon May 22, 2006 9:43 pm

Assuming negative common of 12V supply:
+5V can be linear regulated from 12V, wasteful but can be done.
-5V will have to be a DC-DC converter.

If neither pole of 12V is common (floating), put two resistors in series across the 12V, call the point where the resistors connect to each other common, use 7805 and 7905.

Externet · Post by **Externet** » Mon May 22, 2006 9:47 pm

Hi.
It can be done if the Â±5V currents needed are not that high, with a '7805' and a '7905' of the LOW dropout type, whatever are their correct part numbers; and the 12V supply should be somewhat current 'beefy' compared to the Â±5V circuit currents.
7805 and 7905 part numbers cannot be used here!

Connect two resistors of equal value in series across the 12V, use a LOW resistance but that will not load the 12V source to less than ~11.5V with the circuit operating.

[That will allow only 0.25V dropout headroom per regulator, choose them with such low dropout or less.]

The junction of the resistors will be your common for the regulators and for the circuit to be powered.
The positive of your 12V source end to feed your +5V positive regulator input; the negative of your 12V source to feed the -5V negative regulator input.

Second method:
Connect two resistors of equal value in series across the 12V, use a LOW resistance but that will not load the 12V source to less than ~11.4V with the circuit operating.
Connect plain silicon diodes to each 12V terminal and they will drop each 0.7V to yield Â±5V.

Miguel

MrAl · Post by **MrAl** » Mon May 22, 2006 11:58 pm

Hello there,

I didnt see any mention of output current requirements...
this needs to be stated in order to reply properly, but...

did you consider using two 7805 type regulators?

Lower regulator connects to input, ground, and output is +5v.
Upper regulator connects to input, instead of ground +5v (the lower reg)
and output is +10v.
Relatively speaking, the +5v becomes ground and ground becomes -5v,
and +10v becomes +5v.

Does this make sense?

You'll have to test this configuration because i dont remember if
i've ever tested this or not

Sorry.

State your output current requirements next time ok? Thanks.

fripster · Post by **fripster** » Tue May 23, 2006 12:01 pm

hi there,

What you need is not a set of regulators, but a power-opamp. The output of this opamp will be your '0V' point and the others will be the +6 and -6 Volts. If you need precisely +5/0/-5 then regulate to 10V (one side!) before the opamp. Any power opamp will do that is stable for unity gain..

Happy hacking!

Da fripster

Edit: Look at the datasheet of the ST Micro L165 power opamp. A sampl e schematic is on page 7, fig. 13.

L. Daniel Rosa · Post by **L. Daniel Rosa** » Tue May 23, 2006 1:01 pm

One more option is a charge pump voltage inverter. They are only good for low current (maybe 20mA) and the voltage droops when loaded, but will run from a single +5 regulator (or higher, then you could put a -5 reg on the output).

Robert Reed · Post by **Robert Reed** » Tue May 23, 2006 1:10 pm

As MrAl said current demands are an important part of this equation. If you require a common ground to all supplys (12V included) and if your current demands are small (especially -5 V), then constuct a simple relaxation ocillator, capacitively couple and extract negative voltage thru proper rectification. From there filter and drive your negative regulator. And if the low current demand is small, use a 79L05 (TO-92 case). A very simple circuit if it fits your requirements.

jimandy · Post by **jimandy** » Tue May 23, 2006 3:26 pm

I believe Maxim makes a chip that will develop neg. 5V from a pos. voltage source.

haklesup · Post by **haklesup** » Tue May 23, 2006 3:48 pm

Starting from the +12V supply is what makes this unnessarily complicated. Just start with a center tapped transformer then run the output through a full wave bridge rectifier and you end up with a + and - ungegulated DC voltage WRT the center tap. Now you can filter and regulate those two supplies using 3 terminal regulators or any other.

Check out the LM723. It is a quite a versitile part.[/img]

dacflyer · Post by **dacflyer** » Tue May 23, 2006 6:33 pm

MrAl >> the current -5 current is minute...maybe 5ma. i need it for developing -5 for the Lm7107

haklesup >> i have no CT wal-warts..most all that i have are just 2 wire outputs.

robert reed >> i tried what your describing, using a 4049 and other parts you describe..but it will not function for some reason.. perhaps i have a bad chip. i need it test it more..

fripster >> thanks, i will look into this..

Robert Reed · Post by **Robert Reed** » Tue May 23, 2006 8:27 pm

Dac
Use one or two sections of your 4049 to make a simple square wave generator. Run this at 1000 Hz or higher to reduce demands on coupling and filtering. I believe you can parrelel the remaining inverters up, thereby increasing current source/sink levels. Capacitively couple your out put to a rectifier/ filter and directly drive the 79L05 with it. Make sure your capacitors are sized for the task at hand. The 79L05 requires 5 ma of operating current plus Your 5 ma load current current makes 10 ma total which this circuit should easily handle. Make sure you power the 4049 from 12 vdc, it can handle 18vdc.

ecerfoglio · Post by **ecerfoglio** » Wed May 24, 2006 6:12 am

Dacflyer's original post:

from a 12 volt DC single supply .....
its no problem if i use batterys, but i want to use a old wal-wart ...

If you donÂ´t need too much current and you have an AC wall wart you may try something like this:

One wire from the wall wart goes to ground.

The other wire goes to a half wave rectifier (anode |>| cathode), big filter cap and regulator (positive supply). It also goes to a second half wave rectifier (but reversed, cathode |<| anode), big filter cap and regulator (negative supply).

Being half wave it will need bigger filter caps, but if you start with something like 12 V AC and regulate to 5V you have plenty of headroom for the regilators to reject ripple. Just be sure that the filterÂ´s minimum voltage is higher than 7 or 8 volts (Vout + 2)

Newz2000 · Post by **Newz2000** » Wed May 24, 2006 6:42 am

I saw these in EPE - you can use an LM386 chip to create a +/- power source:

sorry for the eye strain with those images. I haven't tried it yet.

MrAl · Post by **MrAl** » Wed May 24, 2006 7:30 am

dacflyer wrote:MrAl >> the -5v current is minute...maybe 5ma. i need it for developing -5v for the Lm7107

Im quite sure they make chips just for this purpose... for a light duty -5v
regulator having only a +V supply available...

Want me to see what i can find?

dacflyer · Post by **dacflyer** » Wed May 24, 2006 1:45 pm

thanks , i think i can deal with it now...

T&L Publications

pos 12 volts to split 5 volts supply?

pos 12 volts to split 5 volts supply?

Who is online