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8 volt from 12 volt battery

Hello, i have a wireless camera that wants 8 (or 9) volt input. That's a little weird i think.. Anyway, i have a 12 volt battery, is there a way i can get the 8 volt?

I have a spare 3V regulator and a 5V regulator (the classic L7805 something). Is there a way i can use them together to get the 8 volt?

Schematics appreciated :)


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all the explanations here


(from http://www.cpemma.co.uk/diodes.html )


7805 with potential divider
Output adjusted by resistors

If the voltage at the "common" pin is raised above ground, this is added to the 5V regulator output, up to a limit about 2V below the input voltage – so a 12V supply can give 5-10V output to fans.

Text-book method shown is with a two-resistor potential divider on the output. Voltage across R1 is 5V, which produces a current of 5/R1 amps. This, plus the regulator's current from the "common" terminal (quiescent current Iq), flows through R2 producing a voltage across R2 of R2(5/R1 + Iq) volts. This voltage across R2 is added to the 5V across R1 to give the raised output voltage.

Formula for the output voltage Vo is

Vo = 5 + R2(5/R1 + Iq)

which rearranges to

Vo = 5(1 + R2/R1) + (Iq * R2)

Iq is typically about 4mA (.004A), so if R1 is 240 ohms and R2 is 150 ohms,

Vo = 5(1 + 150/240) + (.004 * 150) = 8.7V

To minimise the effect of Iq variations between regulators, the divider current needs to be significantly higher than Iq, so R1 should preferably be under 330 ohms.

Check the output with a multimeter, as small changes in Iq from one regulator to another have a significant effect. Or use a 220-ohm preset potentiometer for R2 and adjust as required.






The regulator still puts out 5V between its output and it's ground but the voltage divider made up of the 120ohm and 75 ohm resistors cause another 3.125V to be dropped across the 75ohm resistor.

I'm okay if you multiply your resistors by 100...

because you are sinking 25mA in this voltage divider

Regulators need a certain amount of current to flow through the ground pin. The LM805 needs 5-8mA. Read the data sheet.


From what I remember,  You can control the output of a 7805 by potting its GND


I found this to confirm:




 You can see here that a simple voltage divider controls the 7805's outpout