I'm so sorry if this is a stupid question. I have read some of the questions previously asked but in afraid the maths and physics are way over my head.

I have made a wooden template and would like it to have light up eyes. I am hoping to use 1.5 V AA battery to power them. The eyes will be initially yellow LEDs. I don't want to fry them so if I wanted to power two separate 'eyes' which I believe are 2.1 V each, what resistor would I need to add to my circuit please? In the future, I may want to use other colours which are different voltages so if anyone can explain how to work this out in dummy terminology I would be incredibly grateful. Also, does it make any difference to the equation if you change the amount of LEDs being powered?

  • \$\begingroup\$ I don't understand the "1 x 5V AA Battery"; can you be more specific please? \$\endgroup\$ – BobU Sep 8 '16 at 19:19
  • \$\begingroup\$ The AA "batteries" I'm familiar with are 1.5V \$\endgroup\$ – brhans Sep 8 '16 at 19:21
  • \$\begingroup\$ I'm sorry. I meant 1.5v I was having a moment after reading all the other similar questions. My brain hurts!! \$\endgroup\$ – Jennymc101 Sep 8 '16 at 19:24

First of all, AAs are 1.5V, not 5V.

The equation is simple: V = I * R, where V is voltage, I is current, and R is resistance.

Start with your battery voltage and subtract the voltage drop of the LED itself. This is the amount of voltage that your resistor must use up. Then determine how much current you want to flow. Probably between 0.001 and 0.020 amps (i.e., 1-20 milliamps, but the equations are for amps).

Now, just plug the numbers into Ohm's Law and go!

So, if you have a 9V battery, an LED that eats 2.1V, and you wanted 1 milliamp (0.001 amp), you would do: V = 9 - 2.1 = 6.9. Ohm's Law:

V = I * R

6.9 = 0.001 * R

R = 6.9 / 0.001

R = 6900.

If you have two LEDs (let's say they are both 2.1, and are in series with each other), then you have to subtract them both. So V = 9 - 2.1 - 2.1 = 4.8. Now do Ohm's Law again:

V = I * R

4.8 = 0.001 * R

R = 4.8 / 0.001

R = 4800

If you need more current, just change out the current in the equations.

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  • \$\begingroup\$ So how do I decide how much current I need to flow through the circuit? \$\endgroup\$ – Jennymc101 Sep 8 '16 at 19:38
  • \$\begingroup\$ So if I have 1.5v battery that won't be string with to power 2 less? I did my wiring regs years ago and I have no idea how I passed!! I don't remember it being this complicated. I realise it is simple for you. I just can't get my head around it. If I have a 1.5v battery I'll be working in minus figures if I want to draw 4.2v from it? Am I getting this all wrong? \$\endgroup\$ – Jennymc101 Sep 8 '16 at 19:44
  • \$\begingroup\$ The current used by an LED is not critical, as long as you stay under its Absolute Maximum rating, which you can get from its datasheet, or just assume the maximum is 30 mA (OK for most common LEDs). Lower currents produce less light - I find 10 mA is bright enough for most uses (but I had to go below 1 mA for one particularly efficient green LED to get it as dim as I wanted...) \$\endgroup\$ – Peter Bennett Sep 8 '16 at 19:49
  • \$\begingroup\$ I'm getting 1.5 -2.1 -2.1 = -2.7. 2.7/0.001 = -2,700 \$\endgroup\$ – Jennymc101 Sep 8 '16 at 19:50
  • \$\begingroup\$ You will need to use at least two AA cells in series (= 3 volts) to light your 2.1 volt LED. (negative value resistors are rather rare :-) ) \$\endgroup\$ – Peter Bennett Sep 8 '16 at 19:52

For sure, you need more than 1.5V to do the job; easiest is 3 series AA for 4.5V. As Johnnyb said, then you have to lose 2.4V in the resistor; so that's (4.5V-2.1V)/10mA = 240 Ohm. My suggestion would be to use separate parallel branches for the two LEDs; battery life will be much longer with AA batteries than with 9V. The value of the resistor is really not that critical; anything in the several hundred ohm range should be good; could be an interesting experiment for you.

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  • \$\begingroup\$ So if I have 2 x 2.1v leds, and 4.5v worth of battery output and I wire them in parallel I will go. 4.5v -2.1-2.1=0.3 \$\endgroup\$ – Jennymc101 Sep 8 '16 at 20:20
  • \$\begingroup\$ If I want 10ma current. Is it 0.3 divded by 0.01 = 0.0003? I feel like this number should be in the hundreds at least? \$\endgroup\$ – Jennymc101 Sep 8 '16 at 20:23
  • \$\begingroup\$ No, hook up the first "eye" first with your 3 AA batteries and a 240 Ohm resistor in series with the LED. That should light up nicely. THEN, connect the second "eye", e.g. LED in series with ANOTHER 240 Ohm resistor to the 4.5V battery. Prepare to experiment a little; lower resistor values will make the lights brighter, but drain the battery faster. \$\endgroup\$ – BobU Sep 8 '16 at 20:24
  • \$\begingroup\$ Is 0.3v enough to power two leds in parallel circuit? \$\endgroup\$ – Jennymc101 Sep 8 '16 at 20:24
  • \$\begingroup\$ So do I need a 300 ohm resistor? \$\endgroup\$ – Jennymc101 Sep 8 '16 at 20:25

First off, you'll have trouble finding a 5V AA battery, but you could put four AA's in series to get 6V.

When a diode (including an LED) turns on, there is a voltage drop across it.

Let's say that you start with your 5V supply, a resistor, and your LED. When the LED is on, its voltage drop - AKA forward voltage (from the datasheet) could be something like 2.6V, with the positive side at the anode. The voltage drop across your resistor, added with your forward voltage is equal to the supply:

V_supply = V_r + V_LED


V_r = 5V - 2.6V = 2.4V

Using Ohm's Law: V=IR, you can find out what current you want to pump through that LED. That will allow you to select your resistance.

Say you want to pump no more than 12mA through that LED, because the datasheet says that over 20mA, and you'll fry it. R = V/I = 2.4V/(0.012A) = 200 Ohms.


simulate this circuit – Schematic created using CircuitLab

I hope that helps!

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  • 2
    \$\begingroup\$ When I put two AA cells in series, I usually get 3 volts, not 6. \$\endgroup\$ – Peter Bennett Sep 8 '16 at 19:44
  • \$\begingroup\$ Noted! Thanks for pointing out my mistake. I've changed that to 4 batteries. \$\endgroup\$ – Daniel Sep 8 '16 at 19:52

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