Friday 22 November 2013

Don't Waste Power With L.E.D.s

I've been playing with a mixed bag of white LEDs to work out how many I need to illuminate BirdBox 2014.

As this project is battery powered, I want to minimise power consumption where possible to extend battery life.

Like all 5mm LEDs, there is no obvious way to determine what exactly they are (i.e. there are no part numbers). And as this is a mixed bag, I need to sort them by type before including any of them in my circuit.

So armed with a 9Volt battery, a multimeter, breadboard and a selection of resistors, I set about conducting a few simple tests.

Now call me old fashioned, but I like to run LEDs with a current of 20mA, because in the old days, this was normally considered optimum. Some devices can be driven much harder, but I've never had an LED torch for more than 12months before at least one of the LEDs fails.

At 20mA, any self-respecting LED should last for years, maybe decades.

LED Forward Voltage

When correctly polarised (i.e. the LED anode is more positive than the cathode) the LED voltage (known as the forward voltage, Vf) will depend upon the device type and the current flowing through it. So red and infrared LEDs may have a Vf of less than 2V, while other colours are higher, with white LEDs having a Vf typically over 3V.

If you are lucky enough to have the specification for your LED, you may find it quotes the forward voltage for 20mA drive. But with my little test rig, I set about finding the forward voltage by selecting series resistors until the current was approximately 20mA.

For 4 of my LEDs, the forward voltage varies with current like this:-
 18mA:    3.1V
 20mA:    3.15mA
 25mA:    3.2V

Checking My LEDs

So with my simple test rig I can also compare my mixed bag of LEDs, and it turns out I have 4 of one type (Vf=3.15V), 3 of another (Vf=3.6V) and one other (Vf=3.3V). The 3.15V devices appear to be super-bright, while the others are less bright.

Consider Circuit Values

If I want to use the super-bright LEDs with a 12Volt supply, I can calculate the required resistor as follows:-
 Volt drop across the resistor = battery - LED Vf = 12V - 3.15V = 8.85V approx.
 Resistor value(k Ohms) = 8.85/20 = 0.443k or 443 ohms

The nearest standard value is 470 ohms, so let's use this. The actual current will be:-
 8.85V/0.47 = 18.8mA (approx)

Now, if I want to use 6 LEDs to illuminate my bird box using this simple circuit, the current drain on the battery will be:-
 6 x 18.8mA = 113mA

Just these 6 LEDs running on their own would fully discharge my 4Ahr battery in:-
 4/0.113 = 35hours

Hmmm, looking again at this simple circuit, it is clear we are wasting power in the series resistor. But we could put several LEDs in series, and still only draw the same current as the single LED in our simple circuit, like this:-

The series resistor is now calculated as follows:-
 Volt drop across resistor = 12V - (3 x 3.15V) = 2.55Volts
 Resistor value (k ohms) = 2.55/20 = 0.128k or 128 ohms
 Nearest standard value is 120 ohms

Re-calculate the current:-
 2.55V/0.12k = 21.15mA

Now we can power 6 LEDs with a similar current as previously used for just 2:-
 2 x 21mA = 42mA

Other Possibilities

I suggest you also research LED driver/controllers as there may be suitable devices for your particular application which may reduce battery drain, albeit at the cost of component count.

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