# Misunderstanding Ohm's law

#### joachym

Joined Feb 21, 2022
14
Hi all,

I have an USB connector powered by 5V source. To indicate readiness I added a LED (3.2V, 0.02A). Using Ohm's law I calculated a resistor of 90Ω. LED was too bright so I used 160Ω... too bright - 220... 1k... 33k... 100k - and hardly reached acceptably low brightness. In my understanding, there should be 0.000018 Amps so why does the LED so happily shine? Why would anyone calculate resistance when even 1000 times higher resistor makes no difference? What am I missing here?

j.

#### Ya’akov

Joined Jan 27, 2019
7,026
You may be confused about Ohm's Law but I can't tell because you didn't provide the value of the resistance you finally used.but I can tell that you are confused about LEDs.

First, the 20mA rating of an LED is its nominal continuous current, not its ideal current for your use. If the LED is designed to be a bright white LED then at 20mA, it will be bright. Since you can't use more current without ultimately damaging it, it must be as bright as it can get at 20mA so it can be used in things like flashlights and other illumination, not indication uses.

Second, LEDs are non-linear with respect to current. That is, as oyu apply voltage the current rises much faster than the voltage does, following a non-linear curve.

This figure is a typical curve for an LED with a nominal continuous current of 20mA. Note how from the time it turns on to the time it is at its maximum 20mA current, it dramatically rises, and of course beyond that its very clear. LEDs can be deliberately overdriven, given more current than ideal for maximum life in order to get more light. This is often done but it means the heat will kill the LED long before its rated lifetime. If you give it too much voltage, it will burn up immediately.

So, 20mA isn't the current that will make the LED right for your use. It's the current that will maximize output and lifetime together. In your application you may need to provide more or less depending on your goals. You took an LED (I am assuming based on the rated forward voltage, a white one) intended for illumination and put it to work as an indicator. Nothing wrong with that but if you give it 20mA, you should expect it to be blinding.

In tests here I find that a typical white LED will turn on visibly in a normally lighted room at only 70μA. And as expected according to the curve it rapidly becomes much brighter with even small adjustments.

So if you were using a 277.8K resistor with 5V (I don't know where you would get one) your Ohm's Law calculation is fine but really that's not the issue. You need to understand how LEDs are rated and why. If you are concerned with matching LEDs subjective brightness to a particular application you won't find that information on the datasheet or in that abbreviated rating, you have to experiment with values, like you did. So you did that right.

#### ScottWang

Joined Aug 23, 2012
7,326
If your purpose is for indicates and doesn't for lighting then you should buy the LED as 2V/20mA, and adjust the values of current limits resistor to suit what you need.

#### joachym

Joined Feb 21, 2022
14
Hi and thanks for explanations.

I actually used some blue led with "not exactly known" specifics. The point is that in every tutorial they PRECISELY calculate resistance - and then it actually make no difference - I calculated 90 and finally used 100k...

So to conclude, in this very simple circuit it does not matter while I do not exceed the nominal current and wattage (not sure about this word ) of the LED, right?

Can I somehow "spoil" the circuit with such high resistance (like "loosing" power or whatever).

Thanks again!
j.

#### Alec_t

Joined Sep 17, 2013
13,324
Can I somehow "spoil" the circuit with such high resistance
It depends what function the resistor has in any given circuit. It's too little series resistance that spoils LEDs.

#### Ya’akov

Joined Jan 27, 2019
7,026
Hi and thanks for explanations.

I actually used some blue led with "not exactly known" specifics. The point is that in every tutorial they PRECISELY calculate resistance - and then it actually make no difference - I calculated 90 and finally used 100k...

So to conclude, in this very simple circuit it does not matter while I do not exceed the nominal current and wattage (not sure about this word ) of the LED, right?
νμμ
Can I somehow "spoil" the circuit with such high resistance (like "loosing" power or whatever).

Thanks again!
j.
Exceeding the specified maximum continuous current will kill the LED sooner than otherwise ranging from a snap and a bad smell to some long period of operation but a fraction of the rated life.

Reductions the current below this maximum, as you have done, will not harm anything and you can expected a much longer life from the LED. Choosing the proper resistor so the LED brightness matches the application is exactly the right thing to do. The only concern is if you choose to drive the LED harder than the manufacturer’s specifications, then you are dividing to trade light for life to whatever extent that turns out to be.

As far as the calculations, they are designed not to do that latter—exceed specifications—but they can’t tell you which resistance you need because you can’t predict how much current will give you the desired luminosity. If you did know that, say, 100μA was what would give you the luminosity you need you could calculate based on that rather than the generic 20mA but you can’t know that without testing, at least initially.

Let’s say that 100μA turned out to be the right number after experimenting with 5V and several resistors. At that point it would be possible to change the supply voltage to, say, 12V and recalculate using that and the 100μA to get a new value for the resistor because you know that’s what you want, so Ohm’s Law would be useful there.

You might even have three lights on a panel indicating voltages like 5V, 12, and 24V. Once you know that 100μA is needed to make the LED the right brightness you can calculate the resistors needed to make all the lights right and the same using the varying voltages, each with the proper current.

Choosing some other resistance than the one you find works best for you, understanding the implications of overdriving won’t “spoil” the circuit, it is the right thing to do.

#### ScottWang

Joined Aug 23, 2012
7,326
If you don't know the spec of LED then you can in series with an 1K resister and power it with 5V, and then from the other side of LED to see the brightness, you can't see the LED directly, from the brighness then you will know what kind of LED and which one is for lighting or indicates.

And you can using a Voltage Meter to measure the LED, mostly 1.8V~2V are for indicates, 3V~3.2V are for lighting.

#### Tonyr1084

Joined Sep 24, 2015
7,247
Going off what I personally have in MY stock, my Blue Super Bright LED's have an average Vf (forward voltage) of 2.82Vf. To calculate the needed resistance to run MY LED's at 20mA (0.02A) I use this formula:

(V - Vf) ÷ I = Ω

Using 5 volts as the source and subtracting the Vf I get 2.18 volts. That 2.18 volts needs to be dropped by the resistor. The LED drops 2.82V, leaving that 2.18 volts. Here's my numbers:

(5V - 2.82Vf) ÷ 0.02A = 109Ω.

So I'd use a 100Ω resistor to achieve 20mA (0.02A). However, 20mA is quite bright for my SB-LED's. I'd opt to run them on no more than 5mA (0.005A)

(5V - 2.82Vf) ÷ 0.005A = 436Ω. I don't have a 436Ω resistor, so I'd use a 470Ω resistor resulting in:
(5V - 2.82Vf) ÷ 470Ω = 4.6mA (0.0046A). Even then that's going to be pretty bright.

If I wanted to run it at 2mA then (5V - 2.82Vf) ÷ 0.002A = 1090Ω. Not having a 1090Ω resistor I'd use a 1KΩ resistor resulting in (5V - 2.82Vf) ÷ 1000Ω = 2.18mA (0.00218A)

Super Bright LED's are that - "Super Bright". You could probably get away running it on just 1mA. That'd likely be around a 2KΩ resistor. But using the 100KΩ you mention, I don't think the LED would produce any visible light. So I have to challenge you to check the resistor you're using. If you are mis-reading the color code, and if you have an ohm meter, check it to be sure. I do that all the time, mis-read the colors on my resistors. That's why I keep mine segregated by value. Even then, when I pick one up I always verify its resistance.

#### Tonyr1084

Joined Sep 24, 2015
7,247
Here's a video I did two years ago on using parallel LED's with a single resistor. Definitely NOT the preferred method, but it demonstrates what can be done. The reason for showing you this video is that at 2 minutes 40 seconds I plug in my first SB blue LED. Notice at 100Ω it's very bright. That was running at about 20mA.

#### Ya’akov

Joined Jan 27, 2019
7,026
Going off what I personally have in MY stock, my Blue Super Bright LED's have an average Vf (forward voltage) of 2.82Vf. To calculate the needed resistance to run MY LED's at 20mA (0.02A) I use this formula:

(V - Vf) ÷ I = Ω

Using 5 volts as the source and subtracting the Vf I get 2.18 volts. That 2.18 volts needs to be dropped by the resistor. The LED drops 2.82V, leaving that 2.18 volts. Here's my numbers:

(5V - 2.82Vf) ÷ 0.02A = 109Ω.

So I'd use a 100Ω resistor to achieve 20mA (0.02A). However, 20mA is quite bright for my SB-LED's. I'd opt to run them on no more than 5mA (0.005A)

(5V - 2.82Vf) ÷ 0.005A = 436Ω. I don't have a 436Ω resistor, so I'd use a 470Ω resistor resulting in:
(5V - 2.82Vf) ÷ 470Ω = 4.6mA (0.0046A). Even then that's going to be pretty bright.

If I wanted to run it at 2mA then (5V - 2.82Vf) ÷ 0.002A = 1090Ω. Not having a 1090Ω resistor I'd use a 1KΩ resistor resulting in (5V - 2.82Vf) ÷ 1000Ω = 2.18mA (0.00218A)

Super Bright LED's are that - "Super Bright". You could probably get away running it on just 1mA. That'd likely be around a 2KΩ resistor. But using the 100KΩ you mention, I don't think the LED would produce any visible light. So I have to challenge you to check the resistor you're using. If you are mis-reading the color code, and if you have an ohm meter, check it to be sure. I do that all the time, mis-read the colors on my resistors. That's why I keep mine segregated by value. Even then, when I pick one up I always verify its resistance.
I tested using a white LED with a 2.6V rating and was able to see it clearly as on in a normally lighted room with 70μA. So at 5V about 71.4KΩ. So, it‘s close. I did get light earlier but it wasn’t enough to overcome the general lighting, i didn’t note when it turned on in the dark.

#### Tonyr1084

Joined Sep 24, 2015
7,247
I tested using a white LED with a 2.6V rating and was able to see it clearly as on in a normally lighted room with 70μA. So at 5V about 71.4KΩ. So, it‘s close.
Maybe one day I'll test mine at that high a resistance and see what I get.

#### Audioguru again

Joined Oct 21, 2019
5,560
In the video one red LED produced a certain brightness then a second red LED was connected causing the current in the first red LED to suddenly be half. Did you see it dim to half the brightness? NO, because like our hearing, the sensitivity of our eyesight is logarithmic then 1/10th the current is seen as half as bright.

Also, the iris in our eyes adjusts to changing brightness.

#### Ya’akov

Joined Jan 27, 2019
7,026
Just a data point:

I took the same white LED, which my Fluke 287 says has a 2.65344V forward bias and adjusted the current until couldn't detect a difference between PS on and off in a dark area (under the bench, not exactly a lightproof room).

The result was 11μA.

[EDIT: typos and punctuation fixed.]

Last edited:

#### MrSalts

Joined Apr 2, 2020
2,767
Second, LEDs are non-linear with respect to current. That is, as oyu apply voltage the current rises much faster than the voltage does, following a non-linear curve.

People read that "LEDs are non-linear" and keep repeating it without looking at the graph they post. I'm not sure if the complicated or convoluted point that someone is attempting to make but, I don't think anybody's eye could sense the nonlinearity of this LED. I think it is better described as "essentially linear".

#### MrChips

Joined Oct 2, 2009
28,127
People read that "LEDs are non-linear" and keep repeating it without looking at the graph they post. I'm not sure if the complicated or convoluted point that someone is attempting to make but, I don't think anybody's eye could sense the nonlinearity of this LED. I think it is better described as "essentially linear".
Don’t confuse dynamic resistance and static resistance.

#### BobTPH

Joined Jun 5, 2013
6,293
People read that "LEDs are non-linear" and keep repeating it without looking at the graph they post. I'm not sure if the complicated or convoluted point that someone is attempting to make but, I don't think anybody's eye could sense the nonlinearity of this LED. I think it is better described as "essentially linear".
Notice that the “linear” part if the curve starts at twice the operating current.

Bob

#### SamR

Joined Mar 19, 2019
4,547

#### MrSalts

Joined Apr 2, 2020
2,767
Notice that the “linear” part if the curve starts at twice the operating current.

Bob
Who announced the specified operating current? A white LED with a 2.6v Vf is unlikely to be a 20mA device.

#### MrChips

Joined Oct 2, 2009
28,127

The black lines show the DC or static resistance of the device. The slope represents 1/R. A gentle slope represents high resistance. As can be seen, the resistance is very high at low forward voltages and decreases as voltage increases. This makes the device very non-linear.

The slope of the blue line is the reciprocal of the dynamic or AC resistance. A steep slope reflects a low resistance. This represents the inherent series resistance to be included in the model of the device.

#### DickCappels

Joined Aug 21, 2008
9,615
I/V curve slope is often also affected by heating of the die which reduces the current (increases the die's resistance) and also reduces the light output per unit of current.