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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?

pipe
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Jennymc101
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3 Answers3

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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.

johnnyb
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  • So how do I decide how much current I need to flow through the circuit? – Jennymc101 Sep 08 '16 at 19:38
  • 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? – Jennymc101 Sep 08 '16 at 19:44
  • 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...) – Peter Bennett Sep 08 '16 at 19:49
  • I'm getting 1.5 -2.1 -2.1 = -2.7. 2.7/0.001 = -2,700 – Jennymc101 Sep 08 '16 at 19:50
  • 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 :-) ) – Peter Bennett Sep 08 '16 at 19:52
  • 1.5 -2.1 -2.1 = 2.7. 2.7/0.010 = 270 ohms? – Jennymc101 Sep 08 '16 at 19:52
  • How many volts are in a smoke alarm battery? Are they 9? Or will I need 3xAA – Jennymc101 Sep 08 '16 at 19:54
  • Have I written current of 10ma correctly in the equation? – Jennymc101 Sep 08 '16 at 19:55
  • Yes, 10 mA is 0.01 Amp. – Peter Bennett Sep 08 '16 at 19:57
  • Could I get away with less AA'S if I wired in parallel? – Jennymc101 Sep 08 '16 at 19:57
  • The total battery voltage must be greater than the total voltage drops of the LEDs, plus some room for the current-limiting resistor. For two 2.1 volt LEDs in series, I'd suggest four AA cells (6 volts). That leaves 1.8 volts for the resistor. For 10 mA, you would want a 180 Ohm resistor. – Peter Bennett Sep 08 '16 at 20:06
  • I think I have my head around this now. Thank you so much for your advice and patience Johnnyb xxx – Jennymc101 Sep 08 '16 at 20:07
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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

therefore,

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.

schematic

simulate this circuit – Schematic created using CircuitLab

I hope that helps!

Daniel
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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.

BobU
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  • 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 – Jennymc101 Sep 08 '16 at 20:20
  • 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? – Jennymc101 Sep 08 '16 at 20:23
  • 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. – BobU Sep 08 '16 at 20:24
  • Is 0.3v enough to power two leds in parallel circuit? – Jennymc101 Sep 08 '16 at 20:24
  • So do I need a 300 ohm resistor? – Jennymc101 Sep 08 '16 at 20:25
  • Think about it one "eye" at a time; the first resistor gets 4.5V minus the 2.1V LED voltage, so that's the 2.4V/10mA = 240Ohms. OK, that's done. NOW got to the second "eye" and do the same thing; there is is no 0.3V anywhere involved. – BobU Sep 08 '16 at 20:35
  • Two 2.1 LEDs in parallel need the same voltage as a single LED. Each parallel LED should have its own series resistor, as the advertised 2.1 volts is only a nominal value, and the actual LEDs may very a little from that (and the voltage will vary with current as well). With a single current-limiting resistor for parallel LEDs, one LED may have a slightly lower voltage than the other, so will draw more current, and get brighter and hotter, which will lower the voltage more, make more heat... – Peter Bennett Sep 08 '16 at 20:38