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If I want to supply power to a device that is rated for about 2V using a DC source of 12V. How can I find the resistor o should use.

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closed as too broad by Finbarr, Steve G, Elliot Alderson, evildemonic, JYelton Oct 18 at 18:07

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    \$\begingroup\$ What is the device? \$\endgroup\$ – Bruce Abbott Sep 5 at 1:50
  • \$\begingroup\$ LW1ECP's answer is good BUT it does NOT answer your question with any certainty - not because there is anything vastly wrong with the answer but because you have not told us what you REALLY want and it is easy for you to form an incorrect opinion. IF you describe what you are trying to do (rather than how to do it) it is likely that we can give you a much more useful answer. \$\endgroup\$ – Russell McMahon Sep 5 at 12:07
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It depends on the current required for your device.

If the current required is next to nothing, then a simple voltage divider will do it.

If you're not too fussy about the voltage and the current requirement is modest (1-5 mA), a shunt regulator made of three 1N4148 type diodes in series with a 5-10K resistor to 12V can make approximately 2V (each diode has about 0.66V Vf drop at that current.)

If you need more current draw than that, you will need to consider using a voltage regulator to set the voltage output for your system. 2V isn't a common standard voltage, but you can use an adjustable type to make that output.

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There is something missing: how much current does the device drain. Once known, you can calculate R = (12V - 2V)/current

That's in theory. In practice, many devices don't have a constant current consumption. So, the voltage that will get to them won't be constant. Suppose it drains 1A maximum, so R = (12-2)/1= 10 ohm. But if sometime it drains only 0.1A, then the 10 ohm resistor will drop just 10 ohm * 0,1A = 1V, and your device will receive 12-1=11V!!! Chances are it will get fried. If this is the case you'd better use an electronic regulator. And even if consumption does not vary so much, some circuits can break into self oscillation if applied voltage changes with current.

What is your device?

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  • \$\begingroup\$ I think that answers .. thanks.. \$\endgroup\$ – user221698 Sep 5 at 2:06

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