# Real Vdrops

Discussion in 'General Electronics Chat' started by Rosbife, Jun 3, 2013.

1. ### Rosbife Thread Starter New Member

Jan 17, 2013
9
0
Hey guys, my doubt is pretty straightforward.

Basically, when you're building, say, a very simple source-resistor-LED circuit, and you use Ohm's Law to calculate resistor / current values depending on your source, how exactly do you go about the precise voltage drop?

I ask this because LED datasheets will usually give you the typical voltage drop for a forward current of 20mA, and, if you're lucky, throw in a graphic of the I(Vd). So to get the 20mA you can simply apply Ohm's.
But now imagine you want to get, say, ~60mA to flow through your LED. Even though a very slight increase in Vd translates into a very steep increase in the current, there is still some difference in the drop you will have to subtract to the source in order to calculate the resistor value. Which might end up in an appreciably lower current flow.

So my question is, how do you go about these calculations? Especially if you have no chart/graphic, and only the typical VD at If=20mA, and you want a current value significantly higher (50mA and upward). Is the difference in Vd negligible?

2. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Whether it is negligible depends on what constitutes negligible for your application.

If you are trying to get 50mA and you end up with 49mA instead, is that tolerable? If so, then is 45mA still acceptable? If not, then is 49.9mA still to far off?

The current in the LED is going to be

I = V/R

where V is the voltage across the current limiting resistance and R is its resistance.

What you need to consider is how much of an uncertainty you pick up if your V changes by some amount due to the change in voltage drop across the LED.

If

Inom = Vnom/R

and

V = Vnom + ΔV, then

I = (Vnom + ΔV)/R = Inom + (ΔV/R)

ΔI = ΔV/R

ΔI/Inom = (ΔV/R)/(Vnom/R)

ΔI/Inom = ΔV/Vnom

With a bit of practice, you will be able to spot that relationship by inspection.

What this tells you is that the greater the absolute voltage dropped across the resistor, the less effect small changes in that voltage will have. If you have a 12V supply and your LED nominally drops 2V, then if you use that in your calculation but it turns out that the LED actually drops 2.1V, then you will have a 0.1V/10V or 1% error. If you are using 5% resistors, then the uncertainty in the actual value of R outstrips the effect of the change in LED voltage. But if you are running from a 3.3V supply, then your uncertainty due to the increase in LED drop is 0.1V/1.3V which is 7.7%. For many applications, that would still be quite acceptable, but for others it wouldn't.

And don't forget that the uncertainty in the LED voltage drop is only one source of uncertainty. You also have the actual value of the resistor and the actual supply voltage, both of which have tolerances that likely overshadow the change in LED voltage drop.

3. ### Rosbife Thread Starter New Member

Jan 17, 2013
9
0
Thanks for the thorough explanation.
I also understand that variation in the resistor and/or source will likely overshadow the variation in the LED voltage drop. I suppose a circuit with, say:
-12V source
-100Ohm resistor
-3x Series Red LEDs (typical 2.1 Vd@20mA, max 2.6)

would produce no greater variance than 54-60 mA, perhaps even 55-59.

4. ### JMac3108 Active Member

Aug 16, 2010
349
66
Rosbife,

If you have an LED circuit where its important for the LED current to be more accurate, a current source is a good way to get there.

5. ### mik3 Senior Member

Feb 4, 2008
4,846
63
You can easily make a current source with the LM317 regulator.

6. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
So which is "better"? The current limiting resistor or the current source?

It depends.

The current source performs better but costs more. If the current limiting resistor performs "good enough", then it makes no sense to spend the extra time, effort and money on a current source. But if the current limiting resistor does not provide adequate performance, then the cost savings are immaterial because the design doesn't meet the performance requirements.

A huge part of engineering is determine what constitutes a solution that is "good enough" and then designing a circuit that meets that level of performance but does not needlessly exceed it by a significant amount.