# I'm confused about ohms law

Discussion in 'General Electronics Chat' started by suhasm, Apr 27, 2009.

1. ### MikebitsWell-Known Member

Joined:
May 24, 2008
Messages:
6,217
Likes:
175
Location:
San Diego, Ca
I thought my graph explanation was good, but alas, I failed .

2. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,049
Likes:
961
Location:
NJ

Yes, R=V/I at every instant, that is true, but that doesnt mean it
is Ohm's Law. Ohm's Law is a methodology used to calculate something
*new*, not something we already know.
Can we even call this R in R=V/I you talk about 'ohms', or do we have to
resort to the more generic, "volts per ampere" ?

you and others will read carefully.

3. ### suhasmNew Member

Joined:
Sep 12, 2008
Messages:
47
Likes:
0
Location:
Bangalore
Suppose we had a function that consolidated all the different variables that affect the resistance of a conductor , and we plugged that in to yield V=IR(v).

Then would we not be able to predict V or I at any given time?
Anyway, i doubt that would be called the ohms law , but still i think it would work.

4. ### DaveNew Member

Joined:
Jan 12, 1997
Messages:
-
Likes:
0

Joined:
Oct 27, 2006
Messages:
14,047
Likes:
141
Location:
Rochester, US

The ratio works flawlessly to any degree excluding quantum effect. It's just like Newtonian physics before Einstein.

6. ### unclejed613Well-Known Member

Joined:
Apr 3, 2009
Messages:
1,828
Likes:
99
because of the nonlinear nature of semiconductors, most books teach Kirchoff's law as a way to analyze transistor circuits, because to try to apply Ohm's law right away would be a bit more complex than most people would want to learn.

for all intents and purposes, Ohm's law applies.... always.... semiconductors have a nonlinear resistance curve, where either the voltage or current can maintain an almost constant value, while the resistance changes when the other quantity changes. take for instance a diode junction. at 0.6V might have 160 microamps flowing through it, at 0.65V 1.6 milliamps, and at 0.7V it will be 16 milliamps. what changed with the voltage?, the resistance did. if you changed the current in the same increments as seen in the test where the voltage was changed (160uA, 1.6mA, 16mA), you will see the voltage developed aross the junction follows the fixed voltage values seen in the first test. the same action takes place in the operation of a transistor. simplified, the action is along these lines..... if a transistor with a beta (Hfe) of 100 is biased for normal operation, a B-E current of 1mA will cause a collector current flow of 100mA. . the C-E resistance is controlled by the B-E current. the C-E resistance is inversely proportional to the B-E current. Ohm's law still applies, but it's applied to some devices with nonlinear resistances.

a constant current source (or sink) appears to defy Ohm's law' as you change the applied voltage. whether it's 1 volt or 50 volts, you get 1mA (for example). if you have a 100 ohm resistor in series with this current source, you will have 0.1V across itn no matter what the applied voltage is across the series combination of the current source and resistor. so what's changing?, the resistance of the current source. it does this by monitoring the voltage across an internal resistor, and comparing it to a fixed voltage reference, and maintaining that voltage across the fixed resistance no matter what the applied voltage is. anything placed in series with the circuit sees only a 1mA current. if you change the load resistor, the current remains the same, and the voltage across the load changes in accordance with Ohm's law. this is why digital ohmmeters use a constant current source. all you have to do is measure the voltage, and you then know the resistance. analog ohmmeters used the opposite relationship, by measuring the current through the unknown resistor from a fixed voltage source...

want to see Ohm's law really turned on it's head? do a search for NIC (Negative Impedance Converter). this is a device using an op amp that exhibits a negative relationship between voltage and current (E/-I=-R?). if you analyze the circuit carefully, you will find that all of the devices in it actually obey Ohm's law, but use amplification to invert the relationship. the circuit is useful for counteracting the losses from too low a load impedance in audio systems, but the circuit works for DC as well. you can actually build one of these devices to act as a -1 ohm resistor (or any value you want)..... this makes for an interesting classroom "black box" experiment "apply a variable voltage source across the two terminals, measure the current as you change the voltage, and tell me what's in the box....." ...."the current goes down as the voltage goes up????? what???????"

do they even do "black box" analysis in school anymore?

Last edited: May 5, 2009
7. ### LeftyretroNew Member

Joined:
Mar 11, 2007
Messages:
1,420
Likes:
13
Location:
Hercules, California
I seem to recall that tunnel diodes can exhibit 'negative resistance'? I would think that could add some spice to our topic?

Lefty

8. ### suhasmNew Member

Joined:
Sep 12, 2008
Messages:
47
Likes:
0
Location:
Bangalore
^^ I was just thinking about that

Can anyone explain how tunnel diodes exactly work at the electron level? I cant find much documentation of them on the net....

Joined:
Oct 27, 2006
Messages:
14,047
Likes:
141
Location:
Rochester, US

Joined:
Oct 27, 2006
Messages:
14,047
Likes:
141
Location:
Rochester, US
By the way unclejed, you're doing the same thing that me and everyone else in this thread is doing. The V=IR equation is not Ohm's law. Please read back a few posts for the quote I found from Maxwell in a translation of Ohm's findings. I think it's going to be pointless to try to correct people every time it's mis-quoted though, it's become so ingrained in common usage the definition of "Ohms law" has basically changed to become the ratio in common usage.

11. ### suhasmNew Member

Joined:
Sep 12, 2008
Messages:
47
Likes:
0
Location:
Bangalore
I have one more question :

This is may seem like a bit dumb , but still :

When an ohmic resistor is being driven by a semiconductor device (say , a BJT), will the VI of the resistor be equal to its R in all cases?

What i mean is that , suppose we have a BJT in CE config ,and a resistor of 1K is connected from the collector to the +V. Will the Voltage across the resistor times the current in it be always equal to 1K?

Or does the interaction with the BJT make the resistor disobey ohms law?

Joined:
Oct 27, 2006
Messages:
14,047
Likes:
141
Location:
Rochester, US
Nope, that would require bending reality in some fasion =) A 1K resistor is just a 1k resistor. Mind you real world resistors aren't perfectly linear. If you had a high resolution scope and well calibrated voltage/current sources you'd see the linearity change very slightly over it's functional range, this is however purely academic, in the real world just consider it 1k.

13. ### unclejed613Well-Known Member

Joined:
Apr 3, 2009
Messages:
1,828
Likes:
99
actually, carbon composition resistors have a readily measurable nonlinearity (around 0.1-5%).... seems to be one of the distortion mechanisms in vintage guitar amps

tunnel diodes.... seems to me i remember seeing some energy state diagrams that explained their behavior pretty well, because the math is beyond most engineers as well. a physicist want's to know the spin of the electron as it jumps from one energy state to another. all an engineer needs to know is the bias point needed for negative resistance, and the limits of that -R portion of the curve

Last edited: May 6, 2009
14. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,049
Likes:
961
Location:
NJ
Hello again,

For a resistance that follows Ohm's Law over a given range of operation and not
for other commonly operated range is said to be ohmic, but only within that
region, and the word 'ohmic' still does not characterize the device adequately.

I think i have narrowed down the way that devices are considered either
ohmic (which follow Ohm's Law) and those which are non ohmic (those that dont).

The three equations normally associated with Ohm's Law are these:
v=i*R
i=v/R
R=v/i

and note that R is constant.

The point is, a device is said to be ohmic if *all three* of these equations can be
used for the device, and that *all three* are actually useful for that device.

Having said that, note that for a resistor all three of those equations can provide
new useful information, while for a diode only R=v/i can be useful because we always
have to know v and i in order to know what R is...R is not constant.

Note that for a thermistor this also holds, so that a thermistor is basically ohmic.
That's because for any given R held constant with temperature and measured
by measuring v and i (like a diode) we can then use the equations to calculate
new v and i (unlike a diode).

It appears to me now that if some people are still interested in debating these
issues it might be more productive to discuss individual devices and why or
why not they follow Ohm's Law.

In the mean time, i'll present here an experiment which shows what i was talking

One little thing i forgot to comment on...
E=IZ is a phasor concept. In an electrical circuit analysis handbook at one time used for
a college course in electrical engineering, it states that the rule E=IZ is, and i quote,
an "Ohm-like" rule, and later it goes on to say that "we use a rule similar to Ohm's when
we calculate E=IZ". It's interesting that the professors who wrote that book took the
time to explain that, as well as v=Ri with those exact cases when explaining Ohm's Law.

Anyway, my main reason for posting today was to show something interesting about Ohm's
Law when doing some maybe every day normal calculations involving both an ordinary
resistor and an ordinary si diode. I promise that if you follow this experiment carefully
that you will find this interesting too.

THE EXPERIMENT:

We are given two devices, one is an unmarked resistor and one is an unmarked diode except
for the band on the diode. We are also given a small power supply that is not a voltage
supply, but is a DC current supply. That is, when we adjust the dial the current output
goes up or down so that we can adjust the current from zero to 1 amp. We are also given
one digital voltmeter and one digital current meter.

The point of the experiment is that we want to find the voltage across the two devices
for various current levels.

WE START THE EXPERIMENT:

First we take the resistor and connect it to the generator. We adjust the generator to
output 0.001 amp using the current meter. Now we measure the voltage with the voltmeter,
and we measure 1 volt. We now calculate the resistance with R=V/I and find that 1/0.001
equals 1000 ohms.

Next, we connect the diode to the current generator and again set the current for 0.001 amp
and again using the voltmeter we measure the voltage across the diode. It reads 0.567 volts
this time. We calculate the resistance again using R=V/I and we find that 0.567/0.001 equals
567 ohms.

Next a funny thing happens...we drop the dang voltmeter and it breaks into pieces.
Now we are stuck with one current meter and a current generator, but no voltmeter.

But we are not done yet, because we want to know the voltage across the elements with various
current levels, so we try to continue...

We again connect the resistor, and now set the current to 0.002 amps. We know that the
current is 0.002 amps and that the resistor we have is 1000 ohms, so we can use the formula
V=I*R
to get the new voltage. We do this and come up with the new voltage and it equals 2 volts.
We repeat with 0.003 amps, and we come up with a calculation of 3 volts.
We now have three readings for the resistor:
1. 0.001 amps gave us 1 volt
2. 0.002 amps gave us 2 volts
3. 0.003 amps gave us 3 volts

Now we need the new data for the diode, so we again connect the diode and set the current to
0.002 amps. Now we know the current is 0.002 amps, and the resistance was 567 ohms when the
current was 0.001 amps, but gee, how do we calculate the voltage now since we dont know
what the resistance is with a current of 0.002 amps either? We end up with two unknowns
rather than one so we can not use V=I*R.

Isnt that interesting? We cant even apply a derivative of Ohm's Law to the diode here to
calculate anything new, but we had no problem at all with the resistor.

That's just something to think about, and now here is something else:

We do not have to call what we calculate as V/I actually "Ohm's". We can go back to
the more generic terminology: "Volts per ampere=V/I". In other words, the more
generic term for Ohms is "Volts per Ampere". The question can then be asked, are
we calculating volts per ampere or are we calculating ohms when we form V/I ?

15. ### suhasmNew Member

Joined:
Sep 12, 2008
Messages:
47
Likes:
0
Location:
Bangalore
That was a brilliant explanation. I finally understood it !!!
Thank you so much!!

16. ### ericgibbsWell-Known MemberMost Helpful Member

Joined:
Jan 4, 2007
Messages:
21,233
Likes:
645
Location:
Ex Yorks' Hants UK
hi Al,

For starters, any equation of 3 terms, where 2 terms are unknown, cannot be calculated to give a numerical value.

Consider the resistor in your example, measured using the known constant current and applied voltage. ie: 1Volt at 1mA = 1000R

You claim that when you apply a constant test current of 2mA, you 'know' that resistor is still 1000R, that is an assumption.!

The resistor could have changed value due to self heating as a result of the increase in current.
Therefore to be confident that the resistor hasnt changed value, you MUST measure the voltage across the resistor at the current of 2mA.

Consider this experiment:
I give you an enclosed box, with just wires coming fom the box, I say that within the box the two wires are connected to an electronic component.

The component is either a resistor, thermistor or a diode.

On the test bench is a constant current source that can be switched to give known currents, at this stage you dont have a voltmeter.

How would you determine the resistance value of the unknown component.?

Obviously you cannot, as you need for the equation, Volts and Amps, R= V/I

Say you now have a voltmeter, you would measure the voltage across the two leads, at say 10 different known currents.

If you now plotted these values [ V versus I ] you would produce a graph showing the resistance of the component over the 10 current values.

From the shape of the graph you could only make a informed guess of what the component is.

Summary:
To measure the resistance of any material, if apply a known voltage and measure the current thru the material,
I can solve the R = V/I equation to give me the resistance of that material at these two known values of Voltage and Current.

This all that Dr Ohms Law claims....
ie; That the current thru a conductor is directly proportional to the applied voltage and inversely proportional to the resistance

amps = voltage / resistance

ohmic and non ohmic dosn't enter the equation.

17. ### unclejed613Well-Known Member

Joined:
Apr 3, 2009
Messages:
1,828
Likes:
99
that's a great explanation of the differences between "ohmic" devices and nonlinear devices. if however, you did drop the voltmeter as in the above experiment, you wouldn't know what's in either black box. you only had one known data point for each box when the voltmeter got broken. you "peeked" into the box, because you already knew what was in the box. the whole idea of "black box" testing methods is that you create a set of data points and determine from your data what the box contains. one of my favorite "stumpers" for black boxes is the following:

a black box is connected to a 10V power supply with a 1 ohm output resistance with the following instruments attached, a voltmeter, an ammeter and a wattmeter. the voltmeter reads 5V, the ammeter reads 5A, and the wattmeter reads 0. what's in the box?

18. ### LeftyretroNew Member

Joined:
Mar 11, 2007
Messages:
1,420
Likes:
13
Location:
Hercules, California

That sounds like an interesting puzzle, however without showing where the various measurements are being taken there is too much room for misrepresentation. A picture (or in this case a schematic drawing) is worth a thousand words in cases like this. Also are we to assume that you are specifiying a DC power supply and not a AC at some frequency?

Lefty

19. ### MrAlWell-Known MemberMost Helpful Member

Joined:
Sep 7, 2008
Messages:
11,049
Likes:
961
Location:
NJ

Ohmic, as you can see, has the name "Ohm" right in it, and this term
is used to indicate whether or not the device follows Ohm's Law.
If you dont like that, then re read my post and substitute "Ohm's Law"
when necessary.

The point was, not that we can not measure the parameters of objects,
but that once we do measure one parameter we know a lot about that
object if it follows Ohm's Law, yet not as much if it doesnt.
If you dont follow this reasoning, then you tell me what Ohm's Law *is*
used for. Tell me a few experiments where Ohm's Law actually does
something.

If you are worried that the resistor will heat up and change resistance,
then measure the temperature of the resistor when you do the experiment
outlined in my previous post, and oh yeah, measure the temperature of
the diode too just in case we over current that and it burns up and either
shorts out or goes open circuit.
If you dont like this either there are several other things we can do to
eliminate the possibility that the resistor value changes during the
experiment. There seems to be nothing we can do about the diode
changing resistance however, another interesting point.

unclejed613:
Once you know the resistance of the resistor, you no longer need the
help of the voltmeter as long as the resistor does not heat up.

Joined:
Oct 27, 2006
Messages:
14,047
Likes:
141
Location:
Rochester, US
MrAL, I think the point of contention here is that Ohm's experiments where this 'law' came from used conductors with a fixed R to get the results that made sense. You have to remember that these experiments were done over 150 years ago, the basic reasons for these forces were almost completely unknown, in order for his experiments to work he early on realized that he had to restrict the variables in his experiments to fixed conductors and fixed temperatures, because he knew that he couldn't explain flows in ionic fluids or complex materials. So the root of Ohms law is in fact based on these highly limited set of original test conditions, however the root applicability of the equation extends to non-ohmic materials as well. Ohm's law CAN be applied to non-ohmic materials you have as yet to disprove this. A PN junction or any of the junctions in a transistor will have an equivalent resistance and voltage across it and that current will balance out perfectly so that there is no imbalance in the circuit at any given point in time. If using the V=IR equation on non-ohmic materials isn't Ohm's Law then why do the numbers ALWAYS work out? R is in fact a complex thing, even in metallic conductors, Ohm just picked a set of experiments that kept R almost completely static to prove the general proportional response of the V and I which at the time were better understood than resistance.

Again, I'll repeat myself one more time, just because his experiments used fixed resistances does NOT mean that Ohm's Law does not apply to variable resistance. At any given moment in time an electric circuit will have a set V=IR values that will ALWAYS work out

One caveat, I'm a little iffy on saying always, simply because ohms law like most macro laws break (or at least become insanely complex) on a quantum level.

21. ### The ElectricianActive Member

Joined:
Jan 13, 2009
Messages:
551
Likes:
70
Assuming the voltmeter and ammeter are average responding instruments, a switch opening and closing with a 50% duty cycle would do it.