I have just started learning the basics of electronics. I found a circuit which uses NPN as signal Amplifier and powered by 12V source. The signal source is a audio from Mobile headphone jack. Circuit schematic

I tested the circuit, it works as expected. Now I need this circuit powered by a 3V Button cell. So I can pack the whole circuit in tiny enclosure and I guess it's completely okay to power a IR LED by 3v battery. I know I need to replace the resistor by lower value resistors and I don't know what resistors to use.

As I'm reducing power source to 1/4th of 12v. Do I need to use 1/4th of resistor used? Also explain how to find out and Math behind it.

  • \$\begingroup\$ Can you provide a link (data sheet) to the IR LED? \$\endgroup\$
    – Andy aka
    May 3, 2015 at 10:34
  • \$\begingroup\$ radioshack.com/high-output-infrared-led/… Forward voltage (V): 1.28V, rated at 100mA \$\endgroup\$
    – Arun kumar
    May 3, 2015 at 10:37
  • 2
    \$\begingroup\$ Hate to rain on your parade, but coin cell batteries like the 2032 are not designed to deliver more than a mA or two for an extended period (actually specs for a continuous load are more like 100 µA). Your LED is rated at 100 mA; I imagine it is being driven at a current much lower than that, say 20 to 30 mA. Unfortunately because of the high internal resistance of the battery, this will drop the voltage well below 3v. \$\endgroup\$
    – tcrosley
    May 3, 2015 at 10:48

2 Answers 2


The original 12 volt powered circuit uses a 15 ohm current limiting resistor in series with the IR LED and because the LED has a forward volt drop of 1.28 volts at 100 mA, I believe that the 15 ohm resistor could be too low for 12 volts.

Basically it works like this - When the transistor is fully turned on it might drop 0.2 volts across collector and emitter leaving 11.8 volts across IR LED and 15 ohm resistor. Because the IR LED can be assumed to drop 1.28 volts, this leavesabout 10.5 volts across a 15 ohm resistor i.e. a current is implied of 700mA.

This is far too much for the LED and it will burn so we then have to assume that the voltage applied to the base via the 2.2kohm resistor is maybe about 1 volt max. This means that maybe 0.3 volts is dropped across the 2k2 base resistor implying a current into the base of 0.136 mA. If the transistor has a gain of (say) 100, the current flow in the collector is going to be about 14mA.

But this creates an issue in trying to get the circuit to work from a 3V button cell. With maybe 14mA flowing through the transistor, 15 ohm resistor and IR LED, the transistor is dropping about 10 volts across its terminals and this won't be as clear-cut on a smaller supply voltage.

The upshot of all my ramblings is that without knowing the precise details of the input voltage feeding the 2k2 resistor it will be guesswork to predict if the circuit will work from a 3V supply.

EDIT - as T Crosley mentions, a button cell will only be capable of supplying a handful of milli-amps (even the bigger ones). See the data sheets on this page for proof.

  • \$\begingroup\$ Nice link. For the CR2032 MFR datasheet, the "Standard Discharge Current" is 400 µA and the Max. Cont. Discharge Current is just 3.0 mA. A little higher than the numbers I quoted in my previous comment, but not much. \$\endgroup\$
    – tcrosley
    May 3, 2015 at 19:39

at three volts form a CR2032 the cell's internal resistance will be more than 15 ohms. try it with no resistor


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