2. Connect the LED to the meter leads, verifying correct polarity
3. Meter will indicate forward drop (usually 1V-3V for most LEDs.) Note that the LED may glow.
Now that you have the LED's forward voltage drop you can figure out how much voltage everything else
in the "chain" will need to drop. For very simple circuits it may just be a limiting resistor. For more
complex circuits it may be a bipolar or field-effect transistor, or maybe even something more esoteric.
Either way: The voltage through a series circuit will be distributed through all the elements in the circuit.
Let's assume a very simple circuit with a red LED, a resistor and the supply.
If the meter indicated 1.2V Vf for the LED, you know your resistor will have to drop 5V - 1.2V or 3.8V.
Assuming you want about 10mA through the LED it's now a simple matter of applying Ohm's law. We
know that in a series circuit the current through all elements must be identical, so 10mA through the
resistor means 10mA through the LED. So:
R=V/I
R = 3.8V / 10mA
R = 380 ohms
If you connect your LED to your 5V supply with a 380 ohm resistor in series, you will find the LED glowing
brightly as you intended. Now can your resistor handle the power dissipation? Let's see:
P=V*I
P = 3.8V * 10mA
P = 38mW
38mW is well within the dissipation spec for any 1/4 or 1/8W resistor. Generally speaking, you want to
stay well under the power rating for a device unless you know what you're doing. It's important to
realize that a resistor that is rated for 1/4W will not necessarily be cool to the touch when dissipating
1/4W!
What if you wanted to drive that same LED with a 24V supply? Ohm's law to the rescue again:
R=V/I
R = (24V - 1.2V) / 10mA
R = 22.8V / 10mA
R = 2280 ohms (let's use 2.4k since it's a standard E24 stock value):
And a power check (using an alternate power equation just to change things up):
P = V^2 / R
P = 22.8V * 22.8V / 2400 ohms
P = 217mW
Now you'll notice that by driving the applied voltage up we have driven the voltage across the resistor
up, and that in turn causes the total power dissipated by the resistor to go up considerably. While
217mW is technically under the 250mW a quarter-Watt resistor can handle, it will get HOT. I'd suggest
moving to a 1/2W resistor. (My rule of thumb for resistors is to keep their dissipation to under half their
rating unless you're actively cooling them or have specific needs laid out in the specification).