 # A noob resistance question

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

EDIT:

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?

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!

//Andrew

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.

Lefty

The moment I hit post I knew I screwed up, but the dog asked to go out and you caught it before I could fix it :blush:.

so it should be a 50? resistor... Do you know HOW I know I want to learn this? Because I keep on trying even though I have been wrong 100% of the time!

macweenie: The moment I hit post I knew I screwed up, but the dog asked to go out and you caught it before I could fix it :blush:.

Oh, the old "the dog eat my correct posting" excuse, go to the corner. ;)

so it should be a 50? resistor... Do you know HOW I know I want to learn this? Because I keep on trying even though I have been wrong 100% of the time!

Grumpy_Mike:
No with a forward voltage drop on the LED of 2.2V the drop across the resistor is 5 - 2.2 = 2.7 V

Are you sure?

macweenie:
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.

Now you’re getting good a reading datasheets, the second thing you need to look at is the graph of voltage vs. current and the min/max range of voltages that the LED needs. As you’ll see the graph shows that ass you approach maximum current, even minor variations in voltage (eg. 0.1V) can produce huge difference differences in current. The LEDs voltage tolerances are often more than 0.1V either way, so what voltage do you aim at…?

This is why controlling LEDs with resistors is a bad idea, and why they make special chips for driving LEDs - you need to control the current, not the voltage.

This is why controlling LEDs with resistors is a bad idea,

It is not a bad idea it is done all the time in a lot of professional applications and is quite satisfactory technique. It is not the technique to get the absolute exact control over the current but often this is not necessary.

The LEDs voltage tolerances are often more than 0.1V either way, so what voltage do you aim at...?

Well fungus, I would say that I would add the 0.1V to the 4V maximum forward voltage so 5V - 4.1V = 0.9V drop voltage, then to calculate the resistor needed (if I use a resistor vs a chip) it would be 0.9V / .02mA = a 45? resistor (or the next larger value), HOWEVER since I am always wrong about these things my gut is telling me to SUBTRACT 0.1V from the 4.0V BECAUSE it is the MAXIMUM forward voltage and if it is maximum then I can't go beyond it right? So 5V - 3.9V = 1.1V and 1.1V / .02mA = a 195? resistor (or the next larger value). I am aware that you may have been asking a rhetorical question, but I wasn't sure.

macweenie: I am aware that you may have been asking a rhetorical question, but I wasn't sure.

It was rehetorical. The correct answer is not to aim for a voltage.

Grumpy_Mike:

This is why controlling LEDs with resistors is a bad idea,

It is not a bad idea it is done all the time in a lot of professional applications and is quite satisfactory technique. It is not the technique to get the absolute exact control over the current but often this is not necessary.

If you're aiming at 10-12mA (or less) on a 20mA LED, sure. The power curve is usually quite flat there.

If you're aiming at 20mA on a 20mA LED then resistors simply don't work.

Since you brought it up, I was wondering, How does using an LED Driver IC differ from powering the LEDs through a PWM enabled pin? Couldn’t I use a shift register IC on a PWM pin to get the same result? Not that I have any immediate plans.

How does using an LED Driver IC differ from powering the LEDs through a PWM enabled pin

Rather the same way as driving a car differs from making a cake. No relationship whatsoever.

macweenie: How does using an LED Driver IC differ from powering the LEDs through a PWM enabled pin?

It's completely different.

Not all ICs do PWM.

The IC will clamp the LED current to a fixed value (eg. 20mA), independent of voltage.