Ohm's law violation?

[easilyconfused]

I'm perfectly capable of making embarrassing mistakes. So maybe that's what I've done. But here's what I see: A circuit consisting of one LED, a pot adjusted to 4.19k, ammeter reading of 1.49mA, voltmeter reading of 7.92. When I crunch the numbers with Ohm's law (for current) I get 1.89mA. I was building the circuit to show my grandson how Ohm's law would coincide with the meters. I didn't factor any resistance for the LED. My 17B Fluke, fresh from China, shows OL for it in either direction.

[DrGeoff]

You must account for Vf (forward voltage) across the LED.
You will have (1.49 x 4.19) = 6.24V across the pot, and 1.68V across the LED and ammeter. There will be a small voltage drop across the ammeter, the rest is the Vf of the LED.

If you want to demonstrate ohm's law, get rid of the LED.

[EEVblog]

The LED is a semiconductor and doesn't really have a "resistance" as such. It has a voltage drop at a particular current. That varies on the LED colour and type.

Dave.

[Zero999]

Yes.
I = (Vs-Vf)/R
Vs = supply voltage
Vf = LED forward voltage

I disagree, it's still a good example of Ohm's law - it's how we calculated the voltage across the LED and resistor.

[EEVblog]

If you want to demonstrate ohm's law, get rid of the LED.

Because it's a series circuit you can simply measure the voltage across the resistor.

Or you can do it the hard way and measure Vf and then subtract it from the measured battery voltage.

Dave.

[armandas]

You will have (1.49 x 7.92) = 6.24V across the pot, and 1.68V across the LED and ammeter.

That should read something like this Vf = Vin - (I * R):

Code: [Select]

>>> 7.92 - (0.00149 * 4190)
1.6768999999999998


[easilyconfused]

Thanks guys. By the time I got back to check your responses I had found my problem on a site which shows that my voltage (for crunching) is not the battery voltage but the voltage across the pot. And, best I can tell, that's what you guys are telling me. And that makes it come out juuuuust right.
http://www.hawestv.com/ohm/ohm_exps1.htm

[easilyconfused]

If you want to demonstrate ohm's law, get rid of the LED.

That's true. But... I didn't know that. I thought you could take any circuit, use the supply voltage (for V), the total resistance (for R), and calculate the current. Today I purchased Make: Electronics, by Charles Platt. The circuit I used was from the book with the exception of the pot values. I didn't have the one he prescribed. He asks you to measure the voltages across the pot and the LED. I observed that they were different of course. He doesn't ask you to check the current with the LED. He gets to current a few pages later AFTER asking you to remove the LED. I felt that he was doing that simply to make it less cumbersome physically. I decided to go ahead and measure the existing circuit complete with LED. I had three voltages to choose from: the battery, the voltage across the LED, the voltage across the pot. I was dead sure you would use the battery voltage. Now I know better. I'm going to recommend (to the publisher) that they discuss the issue in the next printing. Hobby electronics is LED crazy and I think it would be smart to do the current-flow exercise WITH the LED. Kills two birds with one stone: You learn Ohm's Law and which voltage to use. All this is of little interest unless you're a stone beginner like me and my grandson. Blind leading the blind.

[DrGeoff]

The reason I said this is that a simple ohm's law example demonstrates the relationship between voltage, current and real resistance. Adding non-linear devices can cause confusion for beginners, and is not necessary in a simple demonstration.

[easilyconfused]

"...Adding non-linear devices can cause confusion for beginners, and is not necessary in a simple demonstration."

Good point. Thanks

[NiHaoMike]

The LED has resistance and follows Ohm's law, it's just not a constant resistance.
To illustrate, if I measured the voltage across the LED and the current flowing through it, then replaced it with a resistor having the same calculated resistance as the LED, the circuit voltages and currents will remain the same under the same conditions.

An incandescent lamp will not appear to follow Ohm's law either, because its resistance changes with temperature.

[EEVblog]

If you want to demonstrate ohm's law, get rid of the LED.

That's true. But... I didn't know that. I thought you could take any circuit, use the supply voltage (for V), the total resistance (for R), and calculate the current. Today I purchased Make: Electronics, by Charles Platt. The circuit I used was from the book with the exception of the pot values. I didn't have the one he prescribed. He asks you to measure the voltages across the pot and the LED. I observed that they were different of course. He doesn't ask you to check the current with the LED. He gets to current a few pages later AFTER asking you to remove the LED. I felt that he was doing that simply to make it less cumbersome physically. I decided to go ahead and measure the existing circuit complete with LED. I had three voltages to choose from: the battery, the voltage across the LED, the voltage across the pot. I was dead sure you would use the battery voltage. Now I know better. I'm going to recommend (to the publisher) that they discuss the issue in the next printing. Hobby electronics is LED crazy and I think it would be smart to do the current-flow exercise WITH the LED. Kills two birds with one stone: You learn Ohm's Law and which voltage to use. All this is of little interest unless you're a stone beginner like me and my grandson. Blind leading the blind.

As you've found, there are two issues here. One is ohms law, and the other is knowing how to apply it correctly in the circuit you are using.
The circuit you have involves ohms law of course, but because it has more than one element in series, you have to know the principle of Kirchhoff's Voltage Law as well and how it applies.

Resistors in a circuit are useful, because you know you can just simply measure the voltage across it and get the current flowing through it using ohms law. The rest of the circuit won't matter.

Dave.

[Simon]

The LED has resistance and follows Ohm's law, it's just not a constant resistance.
To illustrate, if I measured the voltage across the LED and the current flowing through it, then replaced it with a resistor having the same calculated resistance as the LED, the circuit voltages and currents will remain the same under the same conditions.

An incandescent lamp will not appear to follow Ohm's law either, because its resistance changes with temperature.

well essentially no because ohms law was formulated for a fixed resistance and aims at showing the fixed relationship between V and I which is determined by R. the led does not have a fixed resistance, if anything you are using ohms law to calculate it apparent resistance under those conditions only you cannot use it to predict what the V or I will be at different values of I and V

[Zero999]

well essentially no because ohms law was formulated for a fixed resistance and aims at showing the fixed relationship between V and I which is determined by I. the led does not have a fixed resistance, if anything you are using ohms law to calculate it apparent resistance under those conditions only you cannot use it to predict what the V or I will be at different values of V and I

True, lots of loads are non-Ohmic, meaning their resistance change as the current/voltage changes and you normally can calculate the current/voltage but it's more complicated.

An LED follows the diode equation, the voltage has a logarithmic response to a change in current.
http://en.wikipedia.org/wiki/Diode#Shockley_diode_equation

An incandescent will have a logarithmic current vs voltage characteristic, the current will increase a little bit as the voltage is increased.

This is all too confusing for someone who doesn't know Ohm's law, so it's best to stick with resistors for now.

[easilyconfused]

"...you have to know the principle of Kirchhoff's Voltage Law as well and how it applies. Dave."


I will study up on KVL to get a better understanding. I'm not turning many pages in my new book but I'd rather have a thorough understanding of one or two pages as to half-ass understand the whole book.

[djsb]

http://ibiblio.org/kuphaldt/electricCircuits/index.htm

David

[easilyconfused]

http://ibiblio.org/kuphaldt/electricCircuits/index.htm

Thanks. That helps.