Infrared spotlight project.

I have a PCB BOARD set up for 150x 5mm IR LED 1.5-1.6 forward voltage. Forward current is 60 mA

150 LED’s are divided into 50 groups (3 LEDs each + 1 resistor)

What ohm resistors are needed for the 50 groups of three LEDs each? I was thinking of using x2 18350 Ultrafire 3.7v x2= 7.4 volts.

Suggestions on more or less voltage using increments of 3.7.

Using the online resistor LED calculator: for 150 LEDs grouped 3 LEDs + 1 resistor, the answer given was x50 1/2 W 56 Ohm resistor needed. Being a broke novice I don't want to order and wait for fifty wrong resistors. Do you think this is a safe bet? To better understand my confusion please refer to item #251216249922 on eBay or look up PCB board for 150 x LED 3 mm or 5 mm. The third photo completed tells a lot.

  • \$\begingroup\$ See my answer I already edit it with a suggestion about that using the same batteries. \$\endgroup\$
    – zzz
    Jan 22, 2013 at 9:11

5 Answers 5


This is an application of Ohm's law.

Assuming each set of three LED's are connected in series with a current limiting resistor:

Using a \$V_f\$ of 1.5 means the total \$V_f\$ for each diode group is 4.5 V.

The resistor value is \$R = \frac{E}{I}\$, or \$R = \frac{E}{0.060}\$

  • at 6V: \$R = \frac{(6 - 4.5)}{ 0.060} = \$ 25 ohm (90 mW, use 1/4 W resistor)
  • at 12V: \$R = \frac{(12 - 4.5)}{0.060} = \$ 125 ohm (450 mW, use 1/2 W resistor)

Be sure to use a resistor rated for the power dissipation (\$P = IE\$). Assuming these 50 groups are in parallel, that works out to be a total of 3 amperes. You'll lose less power by having smaller resistor from a lower supply voltage, so of these examples, 6 volts is preferable to 12 volts. To be more efficient you may want to research LED driver circuits.

Make sure your power supply/batteries can supply at least 3A!


Per your edit, using 14.8 V would require larger current limiting resistors and waste more power:

14.8V: \$R = \frac{(14.8 - 4.5)}{0.060} = \$ 172 ohm (618mW, use 1W resistor)

At this voltage, the power dissipation of the resistor is now 618 mW and you would therefore have to use 1W power resistors.

Edit 2:

Some of the math and resistor values were off. It was late at night apparently, and I was trying to get the TeX right. I've adjusted the math and values to be accurate.

  • \$\begingroup\$ Would 1/2w 56 ohms work for 7.4volts? I just don't want to end up with 50 wrong resisters! \$\endgroup\$
    – Tim
    Jan 22, 2013 at 15:38
  • \$\begingroup\$ 7.4 volts, minus the 4.5 forward voltage of the diodes leaves 2.9 V across the resistor. The resistor value should therefore be 48 ohms for 60mA. 56 ohms certainly is acceptable. It would dissipate 174mW so you could even use a 1/4 watt resistor. \$\endgroup\$
    – JYelton
    Jan 22, 2013 at 19:04

12 volt can be used to light up 3 of this LED but the practical problem is that this battery wont last long. When you connect LED in series, they share same amount of current, but for any additional string you need 60mA more. 50 x 0,060A = 3A which is quite high for battery. Resistor value can be found easily with Ohms law. Simply subtract the string voltage drop from power source and divide it with desired current - example: 12V - (2v + 2v +2v) = 6v / 0,06a = 100ohm resistor for each string. I can see demonstration of this but with only 10 LED strings here.

I've see you edit your post so I want to suggest something about your batteries. I will just assume they have 3000mA capacitance for the example.

When batteries are connected in series they make equivalent of battery that have more voltage but same capacity. If you connect all 4 batteries you will have 3000mA and 14,8V but you will use only 6 volts and other will be dissipated as heat in resistors. Which is simply loss for you. When batteries are connected in parallel they make equivalent of same voltage but with more capacity battery, which will fuel you device longer (6000mA, 3,7V). Since 2 of this can provide enough potential for 3 LED (3,7V x 2 = 7,4V). You can make 2 pairs in series and connect them in parallel.

Thus you will end with battery equivalent of 7,4V with 6000mA. Added value is that you don't need big resistors, since they will dissipate about 3 volt x 60 mA and this can be handled with easy with 1/4 watt resistors. They are cheap and prolific. Example of parallel and series battery and better example. Just one of the firsts google results.

  • 1
    \$\begingroup\$ Capacitance is the wrong word to use here. Capacitance is measured in farads, not mA. Batteries do have a limit to the current they can supply, usually modeled as a series resistance that increases as the battery is drained, and decreases with temperature. Putting batteries in series also puts these resistances in series, increasing the voltage and power lost at a given current. \$\endgroup\$
    – Phil Frost
    Jan 22, 2013 at 12:01
  • \$\begingroup\$ Very true, I just use a auto-correction as non-native English speaker and this is result. Feel free to edit it as appropriate. About ESR it is definitely worth mentioning, but I highly doubt that batteries can cope with load as high as 3A. Maybe a wet car battery. \$\endgroup\$
    – zzz
    Jan 22, 2013 at 13:06
  • \$\begingroup\$ some tests show the internal resistance for the batteries in question around \$187m\Omega\$. With two of those in series, and 3A, the expected voltage drop would be \$ 187m\Omega \cdot 2 \cdot 3A \approx 1.12V \$. Not nothing, but not enough to not work. \$\endgroup\$
    – Phil Frost
    Jan 22, 2013 at 13:26
  • 1
    \$\begingroup\$ 1,122 volts x 3A = 3,366W dissipated as heat in the battery. I believe that this is simply way too much and battery will fail due to excessive heating alone. In practice when something is pushed above it's limits diminishing results occur. 3A is way above intended use and tests will show even worse values if they was made with higher current. \$\endgroup\$
    – zzz
    Jan 22, 2013 at 13:59
  • \$\begingroup\$ Just noticed, there was a down-vote. I would like to get a reason for it, to find whatever mistake I made there. \$\endgroup\$
    – zzz
    Jan 22, 2013 at 14:07

The total voltage drop of 3 LEDs in series should be 3 x 1.5 V = 4.5 V. The difference between the DC power supply and the LEDs' voltage drop would be the voltage across the resistor.

Let's say you use a 14.8 V power supply. You'll then have 14.8 V - 4.5 V = 10.3 V voltage across the resistor, so your resistor will have to dissipate a lot of power as heat: 10.3 V * 0.060 A = 0.618 W. So you would need a resistor rated for at least 1 W. The value of the resistor can be calculated using Ohm's Law(R = V / I), so you'll need a resistor of 10.3 V / 0.060 A = 171.6 Ohm Also, make sure your DC power can supply at least 50 x 0.060 A = 3 A.

To minimize the wasted power in the resistor, you could do one of two things or both.

  • Put more LEDs in series
  • Lower your DC power supply

That way more of the power will go into the LEDs and less will be lost as heat in the resistor.

Here's a useful tool for calculating the LED's resistor.

EDIT: I see that you've made some changes to your original question and you decided to use a lower DC power supply, with only two 3.7 V batteries in series instead of three. Let's calculate the resistor value again: 7.4 V - 4.5 V = 2.9 V across the resistor, so R = 2.9 V / 0.060 A = 48.3 Ohm, which is pretty close to the 56 Ohm. You can go ahead and buy those resistors.

  • \$\begingroup\$ Would anyone agree that the answer is x 50 (1/2w 56 Ohms)? I don't want to order 50 resisters I can't use! \$\endgroup\$
    – Tim
    Jan 22, 2013 at 15:25
  • \$\begingroup\$ @Tim You should upvote the answers that you consider helpful. \$\endgroup\$
    – m.Alin
    Jan 22, 2013 at 18:49

What you really need to consider here is the batteries. The resistors are a secondary concern if you can't power the circuit. The internal resistance of your batteries is about \$187 m \Omega\$. Whatever current is going through the LEDs must also go through the batteries. The internal resistance of the battery will reduce the apparent voltage of the battery, and the power dissipated in this resistance heats the battery.

In your proposed configuration, you will have 50 parallel strings of LEDs, for a total current of \$60mA \cdot 50 = 3A\$. The internal resistance of each battery will experience a voltage drop of \$ 3A \cdot 187mA \approx 0.56V \$ and dissipate \$ 0.56V * 3A \approx 1.68W\$ of power. That power manifests as heat in the battery, and while it may be fine for short periods of time (seconds), it may overheat the battery if operated constantly. Also, you will want to minimize power in anything besides your LEDs to maximize your runtime. Every second you run this system represents 1.68 joules of energy you used to make the batteries warm instead of powering your LEDs.

It can be shown from Ohm's law that the power used in a resistor is given by \$ P = I^2R \$. We can't change \$R\$ without changing the battery, but we can minimize \$ I \$. Assuming we want the total power of the circuit to remain constant, we must raise the voltage to reduce the current, since power is the product of current and voltage, \$ P = IE \$.

If you use four batteries in series, each series LED chain can be about twice as long, since we have more voltage available. There will still be 60mA in each one, but there are half as many. So the total current required from the battery is now \$ 60mA \cdot 25 = 1.5A \$. Power losses in each battery will now be \$ (1.5A)^2 187m\Omega \approx 0.42W \$ per battery. This is still a bit much, so let's design for five batteries in series.

Five batteries will give you a voltage of \$ 5 \cdot 3.7V = 18.5V \$. We can put nine LEDs in series for a drop of \$ 9\cdot 1.55V = 13.95V \$. We want to leave more than 25% of the voltage to be dropped over the LED series resistor to get adequate current regulation. With 9 LEDs in each string, you will need 7 strings of them to get 153 LEDs. Total battery current:

\$ 60mA \cdot 7 = 420mA \$

Power and voltage lost in each battery:

\$ 420mA \cdot 187 m\Omega \approx 78mV \$
\$ (420mA)^2 \cdot 187 m\Omega \approx 33mW \$

I'll leave the calculation of the necessary resistor as an exercise. The other answers have covered that pretty well. Another interesting calculation is to calculate the total power of the circuit, the power in the LEDs, and the power in the current limiting resistors. This will give you some insight into the overall efficiency of your circuit.

  • 3 led per group = 3x1.5V = 4.5V voltage drop on your leds.
  • battery is 7.4V, you have to vaste 7.4V-4.5V = 2.9V on the resistor
  • you need 60mA, R=U/I = 48 ohm (you can get 47 ohm resistors).

But, as my colleagues highlighted above: * 60 mA x 50 group = 3000mA - you can never load your battery with that, unless you have a small car battery, or a SLA battery at your disposal * as you can see on the calculation, you use 4.5V x 60mA power for making light, and 2.9V x 60mA for making heat. You lose like 40% of power on the resistors.

You may consider:

  • making not 50 groups, but much less, say, 6 or 7
  • use a step up led driver to drive these led strings and convert your 7.2V battery to a higher voltage with only 10% of power loss.


The search at http://www.linear.com/products/step-up_(boost)_led_drivers tells that LT3754 is a good choice.

I agree that stepup converters are not a beginners topic - the other way of making a good IR flashlight is to use a high performance power IR led, instead of messing 150 standard leds.

Stackexchange does not allow us to answer anything else than your literal question, but I dare to suggest these:


The power leds are a way more effective than standard leds, i.e. from the same volt and amper (power) they produce more light.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.