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Hello guys i am really confused about the whole resistor issue. I have visited so many sites that tells you what resistor you need but i don't know the volt drop that asks to complete or the current.

I want to power 3 RED LEDs with a 9 Volt battery

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  • \$\begingroup\$ To calculate the resistor you need to know the desired LED current, the source voltage (9V) and the forward voltage of the LEDs. If you don't know any of the above then you could approximate the values. If you provided more info on the LEDs we could tell you what approx values to use in your calculation. You can then adjust the resistance value to get the brightness you want. Also, depending on the voltage drop you will need to think about whether or not you can have 3 LEDs in series with 9V source. \$\endgroup\$ – I. Wolfe Feb 21 '15 at 18:58
  • \$\begingroup\$ Yes i know this but some guys here are pretty expirienced with these stuff so i assumed they maybe know \$\endgroup\$ – Dionysis Nt. Feb 21 '15 at 19:01
  • \$\begingroup\$ We do know, but "red LED" doesn't provide enough information. I've seen red LEDs with vastly different voltage drops. I've used LEDs with a voltage drop large enough that 9V will not be enough voltage to power 3 LEDs in series. We aren't psychic so we can't tell you what voltage to use in the calculations if you don't tell us what LED you're using. \$\endgroup\$ – I. Wolfe Feb 21 '15 at 19:03
  • \$\begingroup\$ Yes i see. So the only solution is to buy new LEDs and read the label that provides me with all these information. The formula is: R=(Power supply volt - led drop)/current? Am i right? \$\endgroup\$ – Dionysis Nt. Feb 21 '15 at 19:05
  • \$\begingroup\$ Why not measure the drop yourself using multimeter -> forum.arduino.cc/index.php?topic=106994.0 \$\endgroup\$ – Plutonium smuggler Feb 21 '15 at 19:10
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Red LEDs usually have a 1.7V drop. If the LEDs are wired in series, they will drop about 5.1V. 9V - 5.1V leaves about 3.9V across the current limit resistor.

I'm going to further assume that you want to have the LEDs run at 20 mA max. So: 3.9V / 0.02 Amps = 195 Ohms. The closest standard (E12) resistors are 180 or 220 Ohms. I'd choose 220 Ohms.

Now let's see what happens as the battery dies. A standard Alkaline battery is considered to be dead when its' terminal voltage drops to about 1V under load. A 9V battery contains 6 cells. 6 * 1V = 6V. (6V - 5.1V) / 220 Ohms ~= 4 mA. The LEDs will be lit but dim.

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  • \$\begingroup\$ will the Led be light enough with 20mA ? I have a 120 Ohms resistor. If i connect it to the circuit it will give me a current of 40mA (am i right?) . Will this damage the leds? \$\endgroup\$ – Dionysis Nt. Feb 21 '15 at 19:17
  • \$\begingroup\$ Probably. Here's something you need to know about ratings. When (for instance) an LED is rated for a maximum of 20 mA, that means that as long as you keep the current below 20 mA, the LED will work as advertised. If you run it at more than 20 mA, the manufacturer says you're on your own, and if the LED stops working it's your problem, not theirs. Running a 20 mA LED at 40 mA will probably not kill it immediately, but it won't double the brightness, either (efficiency decreases above the rated max current). Will it be bright enough? I haven't a clue - how could I possibly know what "enough" is? \$\endgroup\$ – WhatRoughBeast Feb 21 '15 at 20:30
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    \$\begingroup\$ I recently used some green LEDs that were excessivly bright at 10 mA - I had to reduce the current to less than 1 mA to get the brightness (dimness?) that I wanted. \$\endgroup\$ – Peter Bennett Feb 21 '15 at 20:33
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The red LEDs I've used have had a forward voltage of about 1.8 volts (but there may be some newer technologies with higher voltages). Typical 5 mm LEDs usually have a recommended maximum current of 20 - 30 mA, but do produce ample light at lower currents.

So, three LEDs will drop 3 x 1.8 volts = 5.4 volts, which leaves 3.6 volts across the series resistor. I usually aim for 10 mA current, so the resistor is R = E/I = 3.6/.010 = 360 ohms. The resistor value is not critical - a higher value will reduce the current, and make the LED dimmer.

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  • \$\begingroup\$ what if the current reaches 40 mA? \$\endgroup\$ – Dionysis Nt. Feb 21 '15 at 19:18
  • \$\begingroup\$ If you exceed the maximum rated current for the LED, the LED will die. \$\endgroup\$ – Peter Bennett Feb 21 '15 at 19:49
  • \$\begingroup\$ @PeterBennett - not necessarily true - see my comment on the other answer. \$\endgroup\$ – WhatRoughBeast Feb 21 '15 at 20:32
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Lots of "ballpark" figures and guessing here. Red LED's are typically ABOUT 1.8V [as others have mentioned] but the actual voltage varies for each device. The current you are pumping through it will also effect the voltage slightly. A quick Google search for "red LED datasheet" got me devices with voltages ranging from 1.65V to 2.25V so 1.8 is about average.

USUALLY, the limiting resistor will average out these variations. If you put them in series, however, you COULD get one LED running at 1.9V and another at 1.6V... which impacts the voltage across the resistor... and your overall brightness is lower. It might not be a big change, but it CAN happen. If this isn't an issue, put the LEDs in series: you use less current, and batteries last longer.

For greater control [at the expense of using more current] you can have the LEDs in parallel, with one limiting resistor on each LED. 9v - 1.8V = 7.2V, which at 10mA is 720 ohms [or 360 ohms for 20mA]. 390, 470, and 560 ohm resistors are common values which should work. If you find one LED is brighter/dimmer, you can change its resistor: especially useful if you ever decide to mix LEDs with different colors.

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