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Topic: A noob resistance question (Read 2 times) previous topic - next topic


Jan 03, 2013, 06:35 am Last Edit: Jan 03, 2013, 07:19 am by JimboZA Reason: 1
My tuppence worth....

I think what might not be becoming clear to messrs anorton and macweenie yet, is the notion that all of the voltage across the LED and its resistor has to be accounted for: the 5v in the 5V---LED---1K(ohm)---Gnd scenario has to disappear by the time we get to the right hand end.

So here's how it all hangs together....

1) What's the input voltage? We're saying 5v, so we have that.
2) What's the voltage we must account for over the LED? Manufacturer will tell us, let's go with the 2.2V mentioned earlier
3) So what voltage must we disappear over the resistor? We started with 5, accounted for 2.2, so we have 2.8 left to drop over the resistor
4) How are we doing with Ohm's Law for the resistor so far? Well we don't know R, that's what we need to calculate; we do know the voltage (2.8v); we don't know the current....
5) But we do know that in a series circuit like this, the current through the LED has to go through the resistor: there's no other path. And we do know (or manufacturer will tell us) that it's 20mA, or 0.020A... this is the value to which we need to limit the current, the object of the exercise.
6) So now we can apply Herr Ohm to the resistor with R=V/I giving R=2.8/0.02 = 140 Ohm


This is all a consequence of Kirchoff's Voltage Law which Wikipedia describes thusly:

The algebraic sum of the products of the resistances of the conductors and the currents in them in a closed loop is equal to the total emf available in that loop.

The products of the resistances of the conductors and the currents is simply the voltages (Ohm's Law).

Do we have a loop though? The 5V---LED---1K(ohm)---Gnd doesn't look like a loop... but it is if you think of the 5v and Gnd being eg the two terminals on a battery... then it would look like a loop: 5V---LED---1K(ohm)---Gnd---Battery---5v

So now we can see Kirchoff's Voltage Law at work in our simple circuit...

1) Total EMF available in loop: 5V
2) Sum of the voltages in the loop: Vled + Vresistor = 2.2 + Vresistor
3) So 5 = 2.2 + Vresistor or Vresistor = 5 - 2.2

Does that help?
Roy from ITCrowd: Have you tried turning it off an on again?
I'm on LinkedIn: http://www.linkedin.com/in/jimbrownza


No, V is actually the voltage "drop" across the resistor. Eg, if 1K is tied between 5V and
3.3V, the current I = Vdrop / R = (5V - 3.3V) / 1K = 1.7 mA, and not 5 mA or 3.3 mA.
In your example, you mean a 1K(ohm) resistor, right?  (just making sure--I'm really new, if you haven't figured it out yet.  )

So, in this LED (for example), it says "forward voltage drop" is 1.8V-2.2V (DC).  Let's say I have a circuit that has 5V---LED---1K(ohm)---Gnd.  I want to find the current.  I = Vdrop / R = 2.2V / 1K = 2.2mA.

Right?  Or am I still confused... :/

Yeah, the other guys are correct, and as I said you use "the voltage "drop" across the resistor"
to get the current through it. Ohm's Law is for resistors.


JimboZA's post really cleared it up for me--I needed that background information that is probably really basic to a lot of people here... :)

Thanks, everyone, for all your help!



Thanks JimboZA,

Although I got a great deal of information before your post, you gave me the piece of the puzzle I was lacking. I hadn't understood just WHERE the current I needed to power, in this case, an LED was coming from. Having looked briefly at the datasheets for different chips and sensors I wasn't making the now obvious connection that the items current requirements was in there. I was under the assumption that 15mA (right?) was the rule of thumb for ALL LEDs, but now it's so obvious I could face-palm.

So using the following datasheet:

I would subtract 4V (maximum forward voltage) from the 5V the Arduino is putting out to get a drop voltage of 1V which using V = r * I would be 1V = r * 20mA or 1 / .2 = a 5 ? resistor? I combed the datasheet for any other values that looked right but theses were the only ones that seemed to fit the bill.


or 1 / .2 = a 5 ? resistor?

No .2 is 200 ma, you need to use .02 in your calculation.


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