0

I have a very simple circuit: 5v -> LED (5mm, forward voltage of 2 V and max. current of 20mA) -> 1k ohm resistor -> GND.

I want to know how much current is flowing in my circuit. I am using this formula:

U = R x I

Since U = 5, and R = 1000 the current is (according to the formula): 5/1000 = 0.005A.

So far so good. The problem is when I try to check the voltage drop over the resistor and the LED. For calculating the voltage drop in the resistor I would use the same formula where I is my current (0.005A) and R is the resistor 1k ohm. The result is: 5 V.

Something very weird is happening here cause the voltage drop around my resistor should not be 5 V cause I know there must be another voltage drop around the LED. If 5V is in the resistor it would mean that 0v is around the LED (and it should not light up but it does).

Thank you helping me with this basic calculus.

Improve this question

1 Answer 1

Reset to default
0

The problem is that if you're trying to measure the current directly, you're going to get the wrong answer. Whether it's due to burden voltage or the (in)accuracy of the meter, it's a hopeless situation. The 0.005A your meter is showing is actually closer to 3mA in reality, but your meter can't distinguish between the two well enough.

If you need to measure current this small then the only way to do so accurately is to use a very low-value shunt resistor and an amplifier. This is the method the µCurrent uses, and it's essentially the defacto gold standard for measuring current in small circuits.

Improve this answer
6
  • thank you ignacio. But my question is moer about this: if I use a higher or lower resistor, will always my LED drag 2 V? I thoght that if I used a different value of resistor I would be changing the current and voltage drop across the LED. Isnt that true?
    – Samul
    Commented Nov 22, 2014 at 16:44
  • It is. But unless you change the current to either almost nothing or to too high for the LED to handle, you may not be able to measure the change; a diode's ohmic region is very steep.
    – Ignacio Vazquez-Abrams
    Commented Nov 22, 2014 at 16:56
  • Leds have a voltage-current curve. If less current is going through the led, the voltage drop will decrease somewhat. Lower voltage drop means higher voltage across the resistor, leading to slightly higher current. So the resistor and led are (kind of) working together to limit the current. So e.g. doubling the voltage doesn't mean twice the current, but less than twice.
    – Gerben
    Commented Nov 22, 2014 at 17:02
  • Thank you :) So just to be clear: if I use any resistor value my LED will always drag 2V, no more no less than that? I am asking this to you cause I measured the voltage across my resistor and it was supposed to be 3 V (5 V is the source power and 2 V should be used in the LED) but it's 1 V. What is going on?
    – Samul
    Commented Nov 22, 2014 at 20:23
  • @Samul: Approximately 2V, until the current is low enough that the LED enters its flat region.
    – Ignacio Vazquez-Abrams
    Commented Nov 22, 2014 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.