It's one of those "stick anywhere" lights that look like a household switch.
It's powered by 4 AAA batteries and has two-2 inch LED strips connected in series with a 3 ohm resistor(looks like a 1 watt).
I want to cannibalize the parts and reassemble into a ceiling fixture I bought at the thrift store. That way I can use it in my 12 volt system I'm installing in my house.
I'm having a heck of a time figuring out what resistance to use so I can up the voltage to 12 volts. I'm probably just looking at it wrong, I normally have the specs on the LEDs.
Thanks in advance for any help with this.
Electronic – I have an LED with no specs, but it does have a resistor. I want to double the voltage
ledresistors
Related Solutions
The LED forward voltage drop will remain (roughly) the same, but the current can change, so the calculation becomes (same equation solving for I):
$$I_{LED} = {(V_s - V_f)\over{R}}$$
So for a 3V \${V_f}\$ and a 5V supply, the \$100\Omega\$ resistor would give \${(5V - 3V)\over{100 \Omega }} = 20 mA\$.
So if you know what current you want, just plug the values in, e.g. for 10mA:
$$R = {(5V - 3V)\over{0.01 A}} = {200 \Omega}$$
Basically, the fact that the supply and the LED forward voltage can be relied upon to be pretty static, means that whatever value resistor you put in will also have a static voltage across it (e.g ~2V in this case), so it just leaves you to find out that voltage and select a resistance value according to the current you want.
Below is the V-I curve of a diode (from the wiki LED page), notice the current sharply rises (exponentially) but voltage stays roughly the same when the "on" voltage is reached.
For more accurate control of the current you would use a constant current, which is what most LED driver ICs provide.
An extremely simple and low cost constantish current source is shown in the circuit below.
Q1 is initially turned on by R2.
I_LED flowing in R1 produces a voltage per V = IR
of V = I_LED x R1
When V_R1 reaches about 0.6V Q2 is turned on and shunts base drive to Q1, thereby limiting any current increase.
ILED max ~= I_R2 x Beta_Q1 where Beta = current gain.
For say 5V and R2=1k IR2~= (5-0.6)/1k = 4.4 mA.
For Q2 Beta = 100 Imax = 440 mA - ample for most purposes.
This circuit drops a minimum of ABOUT 0.6V
Cost, size and complexity are minimal.
Results are far better than using just a series resistor.
ie Current is not "constant" but is far closer than with a resistor alone.
simulate this circuit – Schematic created using CircuitLab
Aka:
E&OE.
Best Answer
$$ R = \frac {V}{I} = \frac {6}{i} $$.
Pick the nearest standard value.
You might want to check the power rating required.
$$ P = VI = 6i $$
Chose a power rating greater or equal to that value.
As an alternative, buy a second lamp and just wire the two 6 V units in series in your new fitting. You'll have twice as much light for the same current draw on your battery.