I have searched everywhere but I can't seem to find out how I can calculate the output voltage (or voltage over a component) from the resistance of a resistor (or other component) and the supply voltage. Everything seemed to be about Ohm's Law, which looks like it could help but I don't know how.
In the case I need to solve, there is a supply of 5v and I need to get the resistance needed to lower the voltage for an LED (the current and resistance at optimum voltage I can get if needed) to 3v. I will definitely need to do this operation again so if you could give something like an equation, rather than the necessary resistance.
Best Answer
You need to know the resistance of the LED: call this $R_{LED}$ and call the resistor $R$. The current flowing out of the power supply is just the voltage (5 volts) divided by total resistance, $I = V/R$, so:
$$I = \frac {5}{R + R_{LED}}$$
The voltage drop across your resistor, $R$, is just $V = IR$ and you want this to be two volts to leave a three volt drop across the LED, so:
$$2 = IR = \frac {5R}{R + R_{LED}}$$
and a quick rearrangement gives:
$$R = \frac{2}{3}R_{LED}$$
Reply to comment:
More generally, suppose the power supply voltage is $V$ and the voltage drop across the resistor is $V_R$, then you get:
$$V_R = \frac {VR}{R + R_{LED}}$$
so:
$$R = \frac{V_R}{V - V_R} R_{LED}$$
But as user1631 says, this assumes the LED can be treated as a simple resistor, and in practice this isn't true. A quick Google found http://www.oksolar.com/led/led_color_chart.htm, which gives some graphs of current against voltage for LEDs. In fact the same Google found http://en.wikipedia.org/wiki/LED_circuit, which describes exactly the problem you're asking about.