Unraveling the Mystery of Ohms Law: Answers to Your Burning Questions

In summary: Not really. If you're talking about materials that don't follow the traditional model of Ohm's Law, then the equation is not valid to calculate a third variable. But the equation is still valid to calculate current, voltage, and power.
  • #1
suhasm
11
0
After seeing contradictory posts on many forums , and getting differing opinions from my teachers , I'm completely lost with the ohms law...

I have a number of questions. Can someone please answer them for me?

1) Do semiconductors obey Ohms law? and why do they obey/not obey?
if they obey , then why is the drop across a ideal diode constant even when the current in it changes?

2) I heard someone say in another thread "semiconductors obey ohms law , but semiconductor junctions dont". why is it so?

3) A thread in another forum said that liquids don't obey ohms law as ions are the current carriers... true or false?

4) Is ohm's law a special case of V=IR or is V=IR a special case of ohms law?

5) Is Ohms law a law at all?

6) Is there any theory in physics which explains why ohms law works the way it does?
 
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  • #2
suhasm said:
After seeing contradictory posts on many forums , and getting differing opinions from my teachers , I'm completely lost with the ohms law...

Believe it or not, you're asking a fairly deep and interesting question. It's a great question, but getting a coherent answer that makes sense to you (and is correct!) might take a lot of tries. :)

I'd love to be the one who finally answers it right... we'll see. Someone will probably stop me if I say something wrong.

1) Do semiconductors obey Ohms law?

Normally, yes. If you have nothing more complicated than a lump of silicon, it obeys the law pretty well, I believe.

and why do they obey/not obey?

Well, uhh... err... they just do. Okay, there are reasons, but they are complicated. Basically, the idea (warning: classical approximation, not really valid in quantum physics) is that the electrons just move in the direction that the applied electric field tells them to, until they hit an atom, and then they bounce off in some random direction, until the field starts pulling them back around again to the direction it wants them to travel in, etc. After lots of math (I can elaborate if you're interested) you get Ohm's Law (under certain conditions).

The thing you need to know is that there are "ohmic" metrials and "non-ohmic" materials. Ohmic materials obey Ohm's Law. Non-ohmic materials do not. Of course, really, for high enough values of V and/or I, pretty much everything becomes non-ohmic eventually. So Ohm's Law is an approximation that holds in certain regions of voltage and current, for certain materials.

if they obey , then why is the drop across a ideal diode constant even when the current in it changes?

Because a diode is not a semiconductor any more than a house is a brick. It is made out of semiconductor materials, but a diode is a device. It is a machine constructed from different materials such that the junction between them has special properties (because of quantum physics). Quick way to think about it: N-type and P-type silicon are both fairly ohmic by themselves, but when you put them together, crazy quantum junk happens at the boundary between them and all the normal rules fly out the window. At that point, right at the junction, you can kiss Ohm's Law (and a few other "laws" you may be used to!) goodbye.

2) I heard someone say in another thread "semiconductors obey ohms law , but semiconductor junctions dont". why is it so?

Well, the answer requires you to be able to think of a doped semiconductor as having more "holes" than "free electrons" or vice-versa. Start with http://en.wikipedia.org/wiki/P-n_junction" . But be warned: these things are sometimes hard to get a firm grasp on without a quantum mechanics background.

3) A thread in another forum said that liquids don't obey ohms law as ions are the current carriers... true or false?

Hmmm, I think that sometimes they obey Ohm's Law.

4) Is ohm's law a special case of V=IR or is V=IR a special case of ohms law?

V=IR is a special case of Ohm's Law, I suppose, but not very special. Basically, that is Ohm's Law exactly.

5) Is Ohms law a law at all?

No, not if the word "law" means something that is always true for anything. It's a rule that some materials follow sometimes. Another (better) way to look at it is that it is just the definition of "resistance". Anything not following that law doesn't have a well-defined "resistance".

6) Is there any theory in physics which explains why ohms law works the way it does?

Heh. I think electrical conduction is pretty well understood in normal cases.
 
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  • #3
I would like to add a question.

When a material doesn't 'obey Ohm's law', does that mean U/I isn't constant but the equation is still valid to calculate a third variable if you know two? E.g. V=5V, R=10 Ohm for a non-Ohmic material, is: I= 5 / 10 = 0,5A anyway ?
 
  • #4
ImAnEngineer said:
I would like to add a question.

When a material doesn't 'obey Ohm's law', does that mean U/I isn't constant but the equation is still valid to calculate a third variable if you know two? E.g. V=5V, R=10 Ohm for a non-Ohmic material, is: I= 5 / 10 = 0,5A anyway ?

Huh? Non-ohmic means that the resistance isn't defined.
 
  • #5
Xezlec said:
Huh? Non-ohmic means that the resistance isn't defined.
So if there's a current of 1A, and a voltage of 2V, the resistance isn't 2/1=2Ohm if it's a non-ohmic material? So then what is the resistance? Is there any way to figure that out?
 
  • #6
ImAnEngineer said:
So if there's a current of 1A, and a voltage of 2V, the resistance isn't 2/1=2Ohm if it's a non-ohmic material? So then what is the resistance? Is there any way to figure that out?

Not sure what you're trying to figure out. Sure, if you want you can just define a resistance over a smaller range, and say something like, "oh, this material is ohmic over the range of 1.99 to 2.01 volts, and in that range it has a resistance of 2 ohms."

Here, tell me what you would consider to be the resistance is of the non-ohmic device with this I-V relationship:

http://www.utc.edu/Faculty/Tatiana-Allen/IVfig6.gif
 
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  • #7
1) Do semiconductors obey Ohms law? and why do they obey/not obey?
if they obey , then why is the drop across a ideal diode constant even when the current in it changes?

At certain values (of current or V) they do, and at certain they don't.

Not obeying ohms law is a property of the material, usually metals (i.e kernels in a sea of electrons) obey ohms law.

Here is a section of my notes, might be of some help -

"Ohm's law -
V = IR.
This relation is experimentally derived.
Explain in terms of K.E of electrons -
The resistance offered by a conductor is a final function of the no. of kernels it encounters through the conductor.
If a circuit is in a constant voltage source, and the length of the is resistance suppose x, and if the length is reduced to half; the resistance is also reduced to half and so each electron will have to dissipate more energy per unit length so as to give out all the energy it possesses.
So finally; what determines the energy generated throughout the circuit per electron? The only thing can be is current, that's the only reason; if energy dissipation is seen by the resistance, it should decrease on decrement of the resistance since the no. of kernels will reduce resulting in lesser kernels on which the electron can dissipate energy on, but this is not so; it remains a constant.
Answer -- As the current increases the velocity of each electron also increases, and so the momentum, thereby reduction of the kernels is compensated by increase in momentum of the electrons and if the length or resistance is made half, the momentum will increase 2 folds maintaining energy dissipation per electron.
In the case of a constant current source, the potential decreases 2 folds cause the momentum of the election is maintained at a constant and so the energy released per kernel changes; since the total no. of kernels each election encounter has been reduced to half (the resistance has been reduced to half) the energy dissipated will too be half."

2) I heard someone say in another thread "semiconductors obey ohms law , but semiconductor junctions dont". why is it so?

na na na...many people are wrong.

3) A thread in another forum said that liquids don't obey ohms law as ions are the current carriers... true or false?

We have nothing called kernels in fluids, considering that, it can't be compared.

4) Is ohm's law a special case of V=IR or is V=IR a special case of ohms law?

Same thing.

5) Is Ohms law a law at all?

:biggrin:

6) Is there any theory in physics which explains why ohms law works the way it does?

As said before, this is practically derived, but we can prove it using drift velocity and all.
 
  • #8
dE_logics said:
"3) A thread in another forum said that liquids don't obey ohms law as ions are the current carriers... true or false?"

We have nothing called kernels in fluids, considering that, it can't be compared.

What do you mean by "kernels"? Ohm's Law can be useful whenever charge carriers exist and respond linearly to an electric field. Electrolytes (liquids containing ions) certainly are modeled with Ohm's Law.
 
  • #9
i always thought that ohm's law is the actual definition of resistance. Like Ohm did a number of experiments and found out V/I=constant and called it resistance.

I also heard that Ohm didn't give the law as V=IR, but in some terms of current density and electric field. like E=pJ something where p is resistivity

From all i know, ohm's law can be derived as a special case. One way to do it is by assuming something called as drift velocity of electrons

My textbook has some explanations but it uses statistics assuming a large number of electron collide but on average they move with constant velocity, but couldn't grasp it well
 
  • #10
Actually, in the case of non-ohmic materials We can speak about resistance, but not as a quotient of voltage and current, but the derivative U'(I). This is a so called differential resistance. It was widely used for radio lamps - very old radioelemnts based on electron emissions from heated cathode.
 
  • #11
All electrical currents always obey Ohm's law. Always. Always always.

The trouble is that the resistance is rarely a simple number, fixed, for all time.

Much more often, the resistance is a complex function of voltage, distance, time, capacitance, doping ratios, the Lotto numbers, etc.

- Warren
 
  • #12
chroot said:
Much more often, the resistance is a complex function of voltage, distance, time, capacitance, doping ratios, the Lotto numbers, etc.
Which is practically the definition of not obeying Ohm's law...:uhh:
 
  • #13
diazona said:
Which is practically the definition of not obeying Ohm's law...:uhh:

According to whom?

- Warren
 
  • #14
diazona said:
Which is practically the definition of not obeying Ohm's law...:uhh:

Well, I got confused with this 'definition'. Never heard about it. :wink:
 
  • #15
chroot said:
All electrical currents always obey Ohm's law. Always. Always always.

The trouble is that the resistance is rarely a simple number, fixed, for all time.

Much more often, the resistance is a complex function of voltage, distance, time, capacitance, doping ratios, the Lotto numbers, etc.

- Warren

Maybe we should first agree what is "Ohm's law".
If you allow for "resistance" to vary with current, voltage and other things, what would you say Ohm's law means?
Definitely the voltage won't be proportional with the current (if "resistance" depends on the current).
 
  • #17
Uh, atyy, your book reference supports me.
Sometimes relationship (9.2) [Ohm's law] is used to define a resistance even for non-ideal resistors, having a non-linear V-I relationship.

-Warren
 
  • #18
chroot said:
Uh, atyy, your book reference supports me.

Well, the reference first says that Ohm's law is RI=V where R is constant. Then it says RI=V can be used to define resistance even if R is not constant.
 
  • #19
The more fundamental form of Ohm's Law is:
[tex]\mathbf{J} = \sigma \mathbf{E}[/tex]

You can get the circuit form by taking the spatial integral across the cross-section and path of the wire/device. Conductivity/resistivity do not have to be constants for this relation to hold. There is nothing in Ohm's Law that requires a linear relationship between voltage and current. Another way of putting it, in the circuit model, would be
[tex] R = \frac{\partial V}{\partial I}[/tex]

While these forms are not as useful when the device is non-linear, like a diode or transistor, you can still use the derivative form to find the resistance to calculate the ohmic loss in power or to find the effective resistance.

Taylor series expansions of the voltage-current relationship in non-linear devices can result in estimates of effective linear resistances that are used to create a small-signal model for easier circuit analysis. For example, the turn on state of the diode can be estimated as a voltage source in series with a resistor in the small-signal realm. Taking more terms of the Taylor's series would allow you to expand the range of voltage signals the model would be valid across.
 
  • #20
Born2bwire said:
The more fundamental form of Ohm's Law is:
[tex]\mathbf{J} = \sigma \mathbf{E}[/tex]

.

This is just an approximation. In general the current is some function of E.
j=f(E)

You can expand the function in Taylor series in powers of E. If all the powers higher than 1 are negligible then we have j=sigma*E .
I think that this is what is called an ohmic case and this is Ohm law.
You may say that j=f(E) in general is Ohm's law but then is just a matter of definition.

One reason the formula j=sigma*E with sigma=constant is that sigma contains the relaxation time. This is considered constant and it is a very good assumption at low fields. But higher field may change the relaxation time and then sigma is not a constant, it depend on the field.
 
  • #21
Born2bwire said:
The more fundamental form of Ohm's Law is:
[tex]\mathbf{J} = \sigma \mathbf{E}[/tex]

For isotropic conductors that is a good approximation. But in non-isotropic media, sigma is not a scalar, but rather, a dyadic (a scalar is a 0th order tensor whereas a dyadic is a 2nd order tensor). Since J and E are vectors, a tensor of rank 2 for sigma, describes the relation in 3 dimensions including interaction between spatial axes. The conductor has a y component of sigma that influences current in the x , y, and z directions. Hence tensor form is needed.

Claude
 
  • #22
nasu said:
This is just an approximation. In general the current is some function of E.
j=f(E)

You can expand the function in Taylor series in powers of E. If all the powers higher than 1 are negligible then we have j=sigma*E .
I think that this is what is called an ohmic case and this is Ohm law.
You may say that j=f(E) in general is Ohm's law but then is just a matter of definition.

One reason the formula j=sigma*E with sigma=constant is that sigma contains the relaxation time. This is considered constant and it is a very good assumption at low fields. But higher field may change the relaxation time and then sigma is not a constant, it depend on the field.
I am talking about this in the confines of circuit theory though. You don't take the Taylor's expansion of the electric field, this is in relation to the small signal models in which you take the Taylor's expansion of the I-V curve. The I-V curve of a diode in the forward bias is approximately
[tex]I = I_s\left(e^{V_d/(nV_T)}-1\right)[/tex]
[tex]V = nV_T\ln\left(\frac{I}{I_s}+1\right) \approx nV_T\left[\frac{I}{I_s}-... \right][/tex]
So the resistance here is approximately [tex]\frac{nV_T}{I_s}[/tex] for small currents. There is also the fact that the diode has a turn-on voltage usually at around 0.7 V and so the small signal model is a voltage source of 0.7V and a resistor in series.

cabraham said:
For isotropic conductors that is a good approximation. But in non-isotropic media, sigma is not a scalar, but rather, a dyadic (a scalar is a 0th order tensor whereas a dyadic is a 2nd order tensor). Since J and E are vectors, a tensor of rank 2 for sigma, describes the relation in 3 dimensions including interaction between spatial axes. The conductor has a y component of sigma that influences current in the x , y, and z directions. Hence tensor form is needed.

Claude

How far are we meaning to take this though? It seems to me that the OP is talking about in the confines of general circuit theory with linear and non-linear devices. We could talk about anisotropic medium and high-order effects but isn't it a little beyond what is needed or has the topic diverged since I first checked on it when it was originally made?
 
  • #23
atyy said:
Well, the reference first says that Ohm's law is RI=V where R is constant. Then it says RI=V can be used to define resistance even if R is not constant.

My understanding was always that R should be independent of I and V in order that an object obeys Ohm's Law. I.e., Ohm's law says that V and I are directly proportional to each other, and R is the proportionality constant.

But I could be wrong.
 
  • #24
Redbelly98 said:
My understanding was always that R should be independent of I and V in order that an object obeys Ohm's Law. I.e., Ohm's law says that V and I are directly proportional to each other, and R is the proportionality constant.

But I could be wrong.

Yes, I agree, and that's what I thought the reference said too.

Of course naming things is not a big deal really, just might get confusing if there are multiple conventions.
 
  • #25
actually i want to know what is the relationship between ohm's law and resistance thermometry bridge or precision thermometry bridge. i know it's bcause of the V I and R but can i know the detail of it.
really hopes someone could answer my problem..
 
  • #26
suhasm said:
1) Do semiconductors obey Ohms law? and why do they obey/not obey?
if they obey , then why is the drop across a ideal diode constant even when the current in it changes?

2) I heard someone say in another thread "semiconductors obey ohms law , but semiconductor junctions dont". why is it so?
Semiconductors do not obey Ohms Law. The voltage-current dependence is well known and is logarithmic. See:
http://en.wikipedia.org/wiki/Diode
From the equation I = Isat[exp(eV/kT) -1]
we get
V=(kT/e) Ln[(I/Isat) + 1]
where Isat is about 10-10 to 10-8 amps, and kT/e is about 26 millivolts at room temperature.
 
  • #27
Bob S said:
Semiconductors do not obey Ohms Law. The voltage-current dependence is well known and is logarithmic. See:
http://en.wikipedia.org/wiki/Diode
From the equation I = Isat[exp(eV/kT) -1]
we get
V=(kT/e) Ln[(I/Isat) + 1]
where Isat is about 10-10 to 10-8 amps, and kT/e is about 26 millivolts at room temperature.

This is the property of a semiconductor junction.
 
  • #28
can someone answer my question..
 
  • #29
I think this is true only in certain points within the operation of the "semiconductive" region of operation. I believe that this Ohms law is more of a static, first order approximation equation. When you bring a semi-conductor and its small signal properties into play, you are no longer in a first order approximation simple environment. If you look at the small signal approximation equations for the semiconductor active elements, you will find that simple equations will not apply unless you are running "saturated" or off type conditions for the logic states of the silicon. Once you get into non-linear operation, do you expect linear equations to all apply? I think not - just simple approximations at certain time stamps along the operation of the currents driving the "semiconductor".

Liquids, in general would not have any applicability to ohms law as their concentrations and transmissions of electrons between nucleic mechanics do not replicate anything found in a transmission line impedance line matched characteristic of a wire or line of copper in a circuit board. Fluids have their own fundamental properties of current flow depending on the chemical mixtures of the fluid. Short answer - No.

Am i wrong on this, or ok? thanks!
 
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  • #30
aninamie said:
can someone answer my question..

aninamie, welcome to Physics Forums.

It's best for you to start a new thread with your question, rather than posting it in an already ongoing discussion like this one.

Click the "New Topic" button at the following web page to start a new thread:
https://www.physicsforums.com/forumdisplay.php?f=111
(The "New Topic" button appears just below the words "Physics Archives")
 

1. What is Ohm's Law?

Ohm's Law is a fundamental principle in physics that describes the relationship between voltage, current, and resistance in an electrical circuit. It states that the current flowing through a conductor is directly proportional to the voltage applied, and inversely proportional to the resistance of the conductor.

2. Why is Ohm's Law important?

Ohm's Law is important because it allows us to calculate and predict the behavior of electrical circuits. It is the foundation of many electronic devices and is essential for designing and troubleshooting circuits.

3. How do you use Ohm's Law to calculate voltage, current, and resistance?

To use Ohm's Law, you need to know two of the three variables (voltage, current, or resistance) and use the appropriate formula. For example, to calculate current, you can use the formula I = V/R, where I is current, V is voltage, and R is resistance. To calculate voltage, you can use V = IR, and to calculate resistance, you can use R = V/I.

4. What are some real-life applications of Ohm's Law?

Ohm's Law has many real-life applications, including in household appliances, lighting systems, electronic devices, and power distribution networks. It is also used in the design of electrical circuits for cars, airplanes, and other vehicles.

5. Are there any limitations to Ohm's Law?

Yes, there are some limitations to Ohm's Law. It assumes that the conductor is a perfect, linear material, which is not always the case in real-life circuits. It also does not take into account factors such as temperature and frequency, which can affect the behavior of a circuit. Additionally, Ohm's Law is only applicable to direct current (DC) circuits and not alternating current (AC) circuits.

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