I want to use 3-6 3mm led lights with resistors making them 12 v.
If I want to use batteries to power these rather than a transformer, what will I use?
Also how can I figure this out myself for the future please?
Thanks for answering such a basic question. Obviously this is a new area for me.
Brenda
LED's have a fixed voltage threshold that varies with color:
Color ---------- Voltage Drop
Red ------------ 1.8 V to 2.1 V
Orange --------- 2.2 V
Yellow ----------2.4 V
Green ---------- 2.6 V
Blue ----------- 3.0 V to 3.5 V
White ---------- 3.0 V to 3.5 V
You can't just hook up a power supply or battery and try to drive them at the above voltages directly though, because the conduction curve above the threashold is quite steep, so a resistor is normally required to limit and stabilize the current.
For 3mm LED's I would stay on the conservative side and use a resistor that limits the current to about 20ma to 30ma.
If your 12 volt battery voltage is fairly stable, you can string LED's in series until their voltage thresholds add up to a voltage about 2 or 3 volts below your battery voltage, then drop the rest of the voltage across a single series dropping resistor to control the current for the whole string.
White or blue LED's will drop just a little over 3 volts each, but other colors have lower voltage drops as shown above and you have to allow for this.
Let's go with a simple example of three white LED's in series with a single dropping resistor. The three LED's in series will drop a total of about 9.6 volts, so we then figure the drop needed in the dropping resistor based on the remaining voltage after the drop in all three LED's are subtracted from the 12 volt battery voltage.
For 3 white LED's on 12 volts:
3.2 volts per LED x 3 LEDs = 9.6 volts total drop in the LED series chain
12 volts - 9.6 volts = 2.4 volts remaining drop needed across the dropping resistor
Now to calculate the resistor value needed to limit the current to a safe 25mA with the 2.4 volts drop.
2.4volts / 0.025amps = 96 ohms
The resistor will dissipate the 2.4volts it drops times the 0.025amps passing through it = 0.06 Watts, so in this case a 1/4 watt 96 ohm to 100 ohm resistor would be a good choice.
You have to connect up the series LED's with the correct polarity, so the final hookup will look like this:
BATTERY_MINUS(-)-----(-)LED(+)-----(-)LED(+)-----(-)LED(+)-----RESISTOR-----(+)BATTERY_PLUS
The plus(+) side of the LED is called the ANODE in the data sheet and the minus(-) is called the CATHODE.
Usually the anode lead is longer, but sometimes it's the exact opposite, so you will have to get this info for your LED's.
The above three LED chain is pretty efficent for light output vs. power consumption, but will also be very sensitive to battery voltage fluctuations. For example if the voltage increases by only 2.4 volts the current through the LED's will double, and if the battery voltage drops by only 2.4 volts, the current through the LED's will drop to almost zero.
If your application is with a car electrical system, be aware that the voltage is anything but stable and can swing from less than 10 volts when the car is cranking to 15 volts when charging.
To make the circuit less sensitive to voltage fluctions we give up a little efficiency, by using fewer LED's in series and dropping more of the voltage in the dropping resistor.
For car applications with the unstable voltage available, and given that we are not usually concerned about getting really high efficiency anyway if we are driving small 3mm LED's, it's better to just limit the series chains to 1 or 2 white or blue LED's, 3 Green or Amber LED's, or 4 red LED's in series.
So, for a car applications (or any application with a 12 volt lead acid gel cell), I would calculate the max current at 30mA (0.030 Amps) at 15 volts to be on the safe side.
Here are a few examples for 0.030 amps @ 15 volts Max battery.
1 x White LED @ 3.2 volts in series with a dropping resistor:
15 volts - 3.2 volts = 11.8 volts drop required in the dropping resistor.
11.8 volts / 0.030 amps = 390 ohms dropping resistor (0.5 watt rating)
2 x White LED's @ 3.2 volts each in series with a dropping resistor:
3.2 volts x 2 = 6.4 volts total in the LED series chain
15 volts - 6.4 volts = 8.6 volts needed in the dropping resistor
8.6 volts / 0.030 amps = 286 ohms (270 ohm or 330 ohm 0.5 watt rating will be fine)
3 x Green LED's @ 2.6 volts each in series with a dropping resistor:
2.6 volts x 3 = 7.8 volts total in the LED series chain
15 volts - 7.8 volts = 7.2 volts drop needed in the dropping resistor
7.2 volts / 0.030amps = 240 ohms (240 or 270 ohm 0.5 watt rating)
4 x Red LED's @ 1.8 volts each in series with a dropping resistor:
1.8 volts x 4 = 7.2 volts total in the LED series chain
15 volts - 7.2 volts = 7.8 volts drop needed in the dropping resistor
7.8 volts / 0.030 = 260 (270 ohm 0.5 watt rating will be fine)
So the bottom line here is that for a series chain of 2 x White, 3 x Green (or Amber), or 4 x Red LED's in series we can use a simple 270 ohm 1/2 watt resistor and get a safe 25 to 30mA of current.
For more brightness, it's better to just get higher quality LED's with higher MCD ratings, but if that's not enough and you do want to drive the LED's harder (and are not so concerned that your LED's last forever), you can drop the series resistor down to half the above value by using two of the 270 ohm 1/2 watt resistors in parallel and then putting these two paralleled resistors in series with your 2 x White, 3 x Green (or Amber), or 4 x Red LED's in series.
This will double the worst case peak current up to twice the value calculated above (60mA at 15 volts).
This is much more than the data sheets typically allow for a 3mm LED but less than they are hit with in some short term applications like coin cell flashlights.
