What is needed to reduce a 9v battery to output 2mA?

Question

I am re-creating my own tDCS machine which requires the device to use a 9V battery reduced to output 2mA which is then hooked up to electrodes on your head. I supposedly increases focus and learning rate.

I'm having issues on what type of resistor is needed to reduce the 9V down to 2mA charge. If the line is hooked up directly with a simple resistor is a regulator needed for the charge to pull or is the regulator more for just safety concerns?(clearly anyone wanting to hook a battery up to their head doesn't need to worry with safety concerns.ha).

Thanks

biire2u · Accepted Answer

Amps is different than volts.

You have a 9V battery which has so many amps of electrical capacity. Just like if you have a 12V car battery that has 550 amps of current available is much different than a 12V charger for a computer printer that has .05 amps of current.

2mA is .002 amps. Using the equation V= IR, you have 9V = .002A * R
R = 9V / .002A = 4,500 ohms or 4.5 K resistor

The resistor is the regulator. It holds the voltage back so that only 2mA runs thru the circuit

? · Answer

Your question is unusual and not a typical question.  However, if you want to limit the current from a 9 volt battery to 2 mA put a 4.5 k ohm resistor in series with it.

Anonymous · Answer

current is dependent on load you will need a proper current limiter the 9v has sweet fa to do with anything, but the cos its a battery a limiter needs adding

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What is needed to reduce a 9v battery to output 2mA?

3 Answers