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2007-01-31 10:41:25 · 3 answers · asked by Fred A 1 in Science & Mathematics Engineering

3 answers

Best way is use a DC-DC converter (LM317T Positive *cross* NTE 956 *cross* Radio Shack 276-1778) It's a very simple circuit that requires 2 resisters (or one resister and a pot if you want it adjustable) and 2 capacitors.
It will convert the 12 VDC in to an adjustable output of 0.5 - 11 VDC.
I use these all the time to install new (and inexpensive) 1 1/2 VDC quartz clock mechanisms in older, mechanical 12 VDC clocks in vintage and classic cars.

2007-01-31 10:58:22 · answer #1 · answered by LeAnne 7 · 1 0

A three terminal linear regulator would work well for this if you need less than 1A of current. Have a look at the LM7812 data sheet for the application circuit. These are fairly inexpensive and easy to find.

http://www.fairchildsemi.com/ds/LM/LM7812.pdf

A more simple, but less precise way would be to add a zener diode, or three forward biased silicon power diodes (1N4001 or similar) in series. This would give you approximately 12V (+/-1.5V), depending on the load current, characteristics of the diodes, and the temperature.

2007-01-31 18:59:41 · answer #2 · answered by DrewD 3 · 0 0

the downright easiest and laziest way is:

put 3 1 volt diodes on the positive terminal.

2007-01-31 18:45:15 · answer #3 · answered by rickpeet 3 · 0 0

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