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I have a submersible water pump that requires 2.5-6V. I want to power it from a 2A, 12V supply. I assume that's powerful enough, but the pump doesn't specify its current. How can I determine the current required, in order to find the right resistor to drop 6V?

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  • \$\begingroup\$ A resistor is not the way to regulate voltage, the drop across it is proportional to the current through it. Look for a voltage regulator. \$\endgroup\$
    – Rev
    Jul 20 '14 at 11:03
  • \$\begingroup\$ Measure the current under normal operation, that will give you the requirement and will (indirectly) tell you if your supply is adequate and will help with the decision as to what kind of regulator to use. Then make or buy a suitable regulator. A 2A 12V supply could supply as much as 3-4A at 5V if you use a switching regulator as @ShannonStrutz suggests. OTOH, a 7805 is super-simple to use but would not be suitable for more than a fraction of an ampere and would need a big heatsink and/or a fan for higher currents. \$\endgroup\$ Jul 20 '14 at 13:41
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You don't.

Resistors cannot be used like that

The voltage drop across a resistor is proportional to the current through it. The current drawn by a pump depends on many factors and varies during its normal operation.

The device you are looking for is called a Voltage Regulator

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That would have to be like a weapons grade quality resistor that would cost almost as much as the weapon itself probably.

You need a voltage regulator. While you could use a simple buck low dropout regulator (LDO), for an application that pulls 2A from a 12V supply, I would suggest going with a switching mode power supply. They are much more efficient. There are many ready made regulators on eBay if you are unwilling to create it yourself, which is not to be ashamed of as switching mode power supplies are not for the faint of heart.

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You should really get a supply whose voltage matches the pump's requirements - trying to use a resistor to drop the voltage definitely won't work.

If you have a meter that can measure current, you should connect the meter in series with the pump when it is operating with a normal load, and on a suitable power supply.

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As others said, resistor is bad idea to do that.

You could use voltage regulator (like LM7805) but you will loose a lot of power on it.

Better way would be switching regulator, but it complicates everything.

I think best way to do this would be just changing your power supply to another that provices 6V.

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