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.

  • $\begingroup$ Something to do with the internal resistance of the power supply? $\endgroup$ – leongz Apr 12 '12 at 17:20

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:

$$V_R = \frac {VR}{R + R_{LED}}$$


$$R = \frac{V_R}{V - V_R} R_{LED}$$

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.


LED's are not like your standard circuit elements. Voltage current relation is highly nonlinear, so you don't generally speak the LED having a unique resistance.

The voltage vs current relation is almost vertical at the operating voltage, i.e. you will have a huge range of currents with voltage very close to say 3V. The resistance of the LED can be almost anything, depending on current. Choosing the resistor to make the voltage across the LED is getting it backwards. The LED will adjust its own internal resistance to keep its voltage drop near 3V (within some range), the purpose of the external resistor is to limit the current through the LED so that it does not damage itself. You start with the current Imax you want to limit to, assume 3V accross the LED, solve 5V = 3V + R*Imax.

  • $\begingroup$ The current for the LED is 0.2A@3v. With both of your methods, I get ~10. That sounds a bit low. Is it right though? $\endgroup$ – Prehistoricman Apr 12 '12 at 22:18
  • $\begingroup$ 10 ohms sounds good to me. Why do you think it's low? $\endgroup$ – user1631 Apr 12 '12 at 22:54
  • $\begingroup$ Not sure. I've done a few more calculations and they all sound right. I just hope they work in practice. I am trying to replace some green/red LEDs with some blue/red LEDs but when I got the new ones, I found out their required voltage was higher than the power supplied so I am using the 5v of the system to get a bit more power to the lights. I need a 68 ohm resistor for the blue light and 165 for the red. Thanks guys, this was really useful. $\endgroup$ – Prehistoricman Apr 13 '12 at 10:42

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