So Ohm's law says I=V/R pretty simple, say I had a 5 volt source and a load I wanted to limit to 20 mA, so I would use a 250 ohm resistor which would limit the current to 20 mA at 5 volts. (I'm correct so far right?) (And presumably if the device tried to draw more the voltage would drop right?)

Now the calculation for an LED is R = Vsource - Vled/I.

So the equation is, simple if I want to power an LED with a forward voltage of 3.4v(which is basically the voltage the LED requires right? Sort of like my phone requires 5v? Not sure if Im understanding this) And a maximum current of 20ma then I would need an 80 Ohm resistor.

But then what would that make in the voltage that you're dividing (the V or Vsource-Vled in in ohms law, or the top of the fraction), the voltage you want to drop across resistor?

No that couldn't be right. The voltage going through the resistor?

As you can tell I'm a little confused and I might just misunderstanding this completely.

Also say I want the power to of those LEDs in series with one 5 volt source, how would I calculate that?

  • 3
    \$\begingroup\$ Your question is not clear. This should be well explain in your books. Would this help \$\endgroup\$ – engineer Jan 16 '18 at 5:05
  • \$\begingroup\$ If the voltage difference is constant between Supply and LED, Vf then the Resistor limits the current. I=V/R \$\endgroup\$ – Sunnyskyguy EE75 Jan 16 '18 at 6:05
  • \$\begingroup\$ Edited to (hopefully) make it clearer what I'm asking \$\endgroup\$ – David Scheiber Jan 16 '18 at 6:17

I don't like the phrase 'current-limiting resistor', as if there is a single current at which the resistor 'limits'. It's only words, but, words shape how we think, which is important when we're just setting out on the journey.

Unfortunately, current-controlling is not much better, neither is current-defining.

Current-dominating is about the least worst I have come up with to date. It has the great advantage that it sounds like it doesn't control the current completely, but allows other components to have an influence, which is exactly true. As the LED forward voltage changes, the current changes slightly because the remaining voltage across the resistor has changed, even though the resistor dominates the current calculation.

Anyhow, the voltage across the resistor, divided by its resistance, sets the current flowing through the LED+resistor combination. Hey, how about current-setting resistor?

  • \$\begingroup\$ I think this is the most relevant answer, maybe because it doesn't get bogged down into one specific example. I also remember having problems understanding how a resistor limits the current - it doesn't. \$\endgroup\$ – pipe Jan 16 '18 at 7:55
  • \$\begingroup\$ I now see why a LED would require a lower resistance resistor because of the voltage drop of the LED thanks \$\endgroup\$ – David Scheiber Jan 16 '18 at 8:01
  • \$\begingroup\$ @pipe I was a bit nerdy as a kid, starting to try to understand resistors when I was 8 or 10 or something. My uncle was an electronics engineer, and I remember behaving very badly, getting really cross with him, when he wouldn't tell me what resistor would drop the 12v from our model car transformer down to 9v for a radio. \$\endgroup\$ – Neil_UK Jan 16 '18 at 8:04

You are pretty right in your logic. If you draw it up like this I think it becomes much more clear.

As you state you already have a 3.4V forward voltage drop over the LED, which will leave you with a voltage over the resistor down to ground of 1.6V.

And then it gets really straightforward to select a proper value of the resistor to limit current according to Ohm's law U=R*I.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Touché - I think this "leave you with a voltage" explanation is more appropriate than my KVL explanation. I hope the OP finds it useful. +1 \$\endgroup\$ – Heath Raftery Jan 16 '18 at 7:34
  • \$\begingroup\$ Oh I think I'm getting it a bit more looking at it with the resistor after the led. So basically every diode has a voltage drop voltage goes in slightly lower voltage comes out. That simply leaves the rest of the voltage to be taken care of by the resistor. \$\endgroup\$ – David Scheiber Jan 16 '18 at 7:43
  • \$\begingroup\$ But what if for example I wanted to limit the current of a load like my phone (just as example because phones usually check to see how much current they can draw before causing the voltage drop ) to 20ma while keeping 5v, according to Ohm's law I would need a 250ohm resistor, but isn't that a higher resistance? \$\endgroup\$ – David Scheiber Jan 16 '18 at 7:50
  • \$\begingroup\$ @DavidScheiber A resistor is not a current limiter. If you think it is, you will keep having these problems while analyzing circuits. \$\endgroup\$ – pipe Jan 16 '18 at 7:51
  • \$\begingroup\$ Oh wait just read the answer under this one and I'm kind of getting it? I'm not really sure TBH \$\endgroup\$ – David Scheiber Jan 16 '18 at 7:54

This is a common stumbling block for those new to electronics. In addition to I=U/R (which is only simple on the surface), you also need to know two more things:

  1. voltages are measured across components (as opposed to currents which are measured through components).
  2. voltages sum to zero around a circuit (called Kirchhoff's Voltage Law).

Consider the following circuit, as described in your example.


simulate this circuit – Schematic created using CircuitLab

Let's go around the circuit clockwise, starting with Vsource. Vsource is a source so we add its voltage: \$+5V\$. Next is R1, which is a sink so we subtract its voltage: \$-V_{R1}\$. Last there's the LED, which is a sink again so we subtract its voltage: \$-3.4V\$. We're back where we started, so it all has to sum to zero:

$$+5V -V_{R1} -3.4V = 0$$ $$\implies V_{R1} = 5-3.4 = 1.6V$$

Great, so now we have the voltage across the resistor we can finally use I=U/R:

$$20mA = 1.6V / R1$$ $$\implies R1 = 1.6/0.02 = 80\Omega$$

giving us the value we expect. Now put the two equations together and see what happens:

$$ R1 = 1.6/0.02 = (5-3.4)/0.02 = (V_{source} - V_{LED}) / I$$

and there you have your expression for the resistor in a LED circuit.

Work through that example a couple of times and the logic for where the voltages come from, where they apply and when to use I=U/R should start to become a bit clearer.


The calculation for the current-limiting resistor used with an LED is R = (Vsupply-Vled)/A.

The voltage of interest here is the voltage across the resistor, which you get by subtracting the LED forward voltage from the power supply voltage..


If the LED drops 3.4V and you have a 5V supply then the voltage across your resistor is simply 5-3.4 = 1.6V.

You have calculated your resistor values correct but seem to be struggling to grasp an understanding of what is going on. Try building this up on a breadboard and measuring the current flowing through the LED as you change the resistor value (make sure not to exceed the maximum LED current rating - or purposely exceed it for learning purposes).

Your last question is poorly worded so I will answer both possibilities... The power dissipated by your LED is its forward voltage (3.4V as stated by you) x LED current. P=VI.

If you were asking how to power 2 LED’s in series then you need to subtract the TWO forward voltage drops (2 x 3.4 = 6.8V) from your supply voltage. In this case clearly a 5V supply is not enough for a series connection and you will need to increase this.


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