Resistors

[eblc1388]

Your turn.

That's impressive work towards finding the definition of Ohm's Law.

But, the cream of the cake, which is yet to be presented, is something that is published which clearly stated that Ohm's Law also applies to cases where the resistance is not constant and changes with current. Also it will be a nail in the coffin for all the above argument if you can find any literature which state that diode and active junction obey ohm's law.

JimB has given one on diode and junction from AOE that stated otherwise.

Also, the equation I=V/Z in the case of AC was taught to me as "equivalent of ohm's law" and not ohm's law. One must be careful to note under what situation a law is being discovered and stated. If you extends the scope of the variables, I think it will cease to be called the same law.

 

[eng1]

It is my belief that Ohm's law applies only to phisical resistances (resistors, wires..). Instead the resistance of a diode (assuming that it is biased in the active region and you're applying a signal) is an incremental (or small-signal) quantity that has the dimensions of resistance.

 

[mechie]

Georg Ohm might have been right ...

As a counter to RadioRon I offer the following from The Art of Electronics by Horowitz and Hill, Second edition, page 44:

"Before jumpimg into some circuits with diodes, we should point out two things:
(a) A diode doesn't actually have a resistance (it doesn't obey Ohms law).
(b) If you put some diodes in a circuit, it wont have a Thevenin equivalent."

JimB

I disagree with (a) ...
If I have a diode and resistor in series across a 10v power supply ...
and if I adjust the resistor to allow the power supply to supply exactly 1mA (minimising any heating effect) ...
if I measure 0.7v dropped across the diode - - - will there not be 9.3v across the resistor ?

If the resistor is dropping 9.3v then its resistance will be R=V/I = 9.3/0.001 = 9300 ohms --- simple Ohm's law

If the diode is dropping 0.7v then its resistance will be R=V/I = 0.7/0.001 = 700 ohms ### IF Ohm's law is correct ###

If the whole series circuit (diode and resistor) has a resistance of 9300+700 = 10000 ohms and passes 1mA then Ohm's law gives ...
V=I*R = 0.001 * 10000 = 10v (the power supply's output)

So why does Ohm's law not apply ?

I agree that this is at one instant in time with controlled conditions (I stated heating - which can usually be ignored in small power circuits) but Ohm's law IS an instantaneous equation - there is no time term in it !

Books like The Art of Electronics always seem to wade far too deep to be real to me - especially as this started out as a simple enquiry about the application of a simple formula.

 

[JimB]

OK
So you have 1mA flowing through the diode.
What happens when the current is increased to 2mA?
The volt drop across the diode is 0.7volts (plus a little bit), certainly not 1.4volts as would be predicted by Ohms law.

Just to turn this debate around a bit, how many data sheets have you seen which show the "resistance" if a diode and how it varies with current (and temperature)?

JimB

 

[dknguyen]

A common filament lamp bulb does not obey ohm's law between the range from zero to normal working current because its resistance has changed many times. But it will still obey ohm's law if it's element is immersed in liquid nitrogen and being kept at a constant temperature.

It's a non-linear equation, so it's easier just to use a graph an I-V graph and specify the forward voltage drop since it changes so little with current. I still wouldn't exactly call it resistance though, but nothing says the R term in V=IR does not have to be a constant. My post was referring specifically to the argument that light bulb filaments do not follow Ohm's law (not diodes) since it heats up a lot and therefore resistance changes a lot during operation (I am arguing that it does and is meant for instaneous points in time before any more heating increases the temperature and changes the resistance of the material).

 

[Marks256]

Doesn't anyone begin to wonder; "Maybe there are anomalies in our life"?
Mabey ohm's law is defective, or maybe it isn't?

 

[eng1]

Yes, I hope to get off to sleep tonite

 

[mechie]

OK
So you have 1mA flowing through the diode.
What happens when the current is increased to 2mA?
The volt drop across the diode is 0.7volts (plus a little bit), certainly not 1.4volts as would be predicted by Ohms law.
JimB

...10v power supply ...
...exactly 2mA

total resistance will be R = V/I = 10/2e-3 = 5000 ohms (so the 9300 ohm resistor can no longer exist - or the current would still be 1mA)

... diode or no diode, linear or not, I am still using Ohm's law ?

REMEMBER I said instantaneous ?

 

[RadioRon]

As a counter to RadioRon I offer the following from The Art of Electronics by Horowitz and Hill, Second edition, page 44:

"Before jumpimg into some circuits with diodes, we should point out two things:
(a) A diode doesn't actually have a resistance (it doesn't obey Ohms law).
(b) If you put some diodes in a circuit, it wont have a Thevenin equivalent."

JimB

Well, okay. Even though I think these statements in Art of Electronics are nonsense, I will concede that you may be able to counter every reference I can provide and stalemate is the potential outcome. And since I will accept only text references (or perhaps the statement of someone with supreme credibility, like the head electrical professor of some universities) there is nothing more I can offer to bend you to my will. So while I do not agree, I will argue no longer.

Thanks, that was fun.

PS: to MARKS256. No, Ohm's law is not defective. I've used it for every kind of device for 33 years and it has never failed me. So don't worry, I does indeed equal E/R.

 

[Marks256]

I never said it was. I just said there is that possablity. It could also be human error....

Edit: Tried to fix some spelling, but it, once again, didin't help....

 

[_nox_]

I never said it was. I just said there is that possablity. It could also be human error....

Edit: Tried to fix some spelling, but it, once again, didin't help....

1.possibility
2.didn't

 

[JimB]

Just to drag the debate a little further, a couple of extracts from:

https://en.wikipedia.org/wiki/Ohm's_law

The relation V = IR can also be applied to non-ohmic devices, but it then ceases to represent Ohm's Law. In non-ohmic cases, R depends on V and is no longer a constant of proportionality but a variable called differential resistance. To check whether a given device is ohmic or not, one plots V versus I and checks that the curve is a straight line.

Ohm's law applies to conductors whose resistance is (substantially) independent of the applied voltage (or equivalently the injected current). That is, Ohm's law only applies to the linear portion of the I vs. V curve centered around the origin. The equation is just too simple to encompass devices described by a more complicated I vs. V relationship.

These infer that Ohms law should only be applied to devices with linear V/I characteristics.

JimB

 

[RadioRon]

Must have the last word, eh? Well, despite the fact that I don't accept anything posted on the internet (i.e. Wikipedia doesn't count) I read the article anyway.

I can't help but think that this difference of opinion is like arguing about the most correct form of a word or phrase in English. Words and grammar evolve and whether we like it or not, common usage of a thing often, after a time, dictates what the thing is no matter what the original inventor intended. We see every day that new words are invented and old words are perverted by the democratic masses and while some of these things happen by dragging linguists kicking and screaming to the alter of practicality, at some point in history the new form becomes "correct". This is how we get saddled with words like "parenting" for example. This is true of any language, and I think this is true of poor old Georg Ohm's old Law. It may originally have been meant to apply only to circuits with elements who's V/I behavior was constant for any I, but that was then and this is now. In any case, it doesn't matter because the relationship I=V/R is universally taught and applies to differential resistance just as well as ohmic resistance. In my text books and my experience, this relationship is commonly called Ohm's Law and that is the modern fact of life. If you want to be a purist and a Luddite, like our Wikipedia authors seem to be, then OK, Ohm's Law is the special case where I=V/R but only when R is not a function of I. But the rest of us who speak modern English, not the English common in 1840, accept the modern version, I=V/R (period) and we feel no compunction to calling this Ohm's Law. Why shouldn't we accept it, as it is the more general case and works no matter what the material causing the resistance. In other words, who really cares if the original Ohm's law only applied to constant R when in fact I=V/R works differentially too. I think Georg would have been proud that the general case is attributed to him, even though this was not his original claim.

glad to be here
RR

 

[tanmanknex]

i don't get it. sorry but it just isn't clicking. okay, so i have to take 5V down to ~2.14V. what resistor would i use and why? would i have to use a whole different power supply or what?

 

[JCLrd]

There are two sides to this question.

Does ohms law only apply to a material with fixed resistance. R(constant)=V/I
OR
Does ohms law apply to an instant in time where resistance will change according to voltage and current. R(instataneous)i=V/I.

No matter how you look at it when needing to calculate a resistance be it linear or nonlinear, you can use the equation R=V/I for that particular set of values measured od voltage and current. So some say that ohms law applies only if R is constant and some say not.

Bottom line to your (OP) post, if you emperically measure a current and a voltage in a circuit than you can solve for the resistance that the component in question is by using the equation V/I=R.

Wether this Resistance is a physical resistor or a nonlinear device OFFERING resistance, it is still a resistance to the flow of current and therefore can be used in the equation V/I=R...

All conductors OFFER resistance, to current flow, yes (OP) you can even calculate the resistance of a straight piece of wire if you try to connect a motor to a supply voltage using long thin wires you will have a voltage drop across that wire that would make your motor run with less voltage. If you measured the current through the wires and took the difference in voltage across the wires you could use V/I=R to solve for resistance the wires are OFFERING to the circuit.
The motor itself OFFERS a resistance to current flow, thats why it APPEARS as a low resistance at start up, due to low resiistance windings (physical constant resistance)
but then current drops as motor reaches its max speed, physical resistance hasn't changed the windings still offer same resistance, but now due to CEMF and all the magnetic changes and the like of the motor, causes the current to see a OFFERING of higher resistance, even though that resistance is not physical in nature but a reactance of inductance and such, HOWEVER if you can measure the current through the circuit in that branch and the voltage drop across the motor, you can determine the amount of resistance (not physical resistor) that the motor is OFFERONG to the flow of current in that branch, by using R=V/I...

 

[Sceadwian]

JCLRD this thread is over 4 years old, you're arguing with ghosts. We've already done an Ohm's law thread recently. It hit 200 posts.

https://www.electro-tech-online.com/threads/im-confused-about-ohms-law.92334/

 

[JCLrd]

Thanks

I just got so caught up in reading them that I never looked at the post dates.

Thanks again...

 

[Roff]

i don't get it. sorry but it just isn't clicking. okay, so i have to take 5V down to ~2.14V. what resistor would i use and why? would i have to use a whole different power supply or what?

That all depends on what the 2.14V is going to be used for. What are you really trying to do?