# negative resistance

#### dujo

Joined Jan 19, 2006
3
1) Combined resistance of R1 and R2 in parallel is found from
a) 1/Rc = (1/R1) + (1/R2) or B) Rc = (R1 x R2)/ (R1 + R2)

2) Given R1=R ohms(a real physical resistor) and R2 is a negative resistor =-R ohms
Then from a) 1/Rc = (1/R) + (1/-R) = (1/R) - (1/R) = 0 and Rc = 1/0
Or from B) Rc = (R x -R)/ (R-R) = - (R squared)/0
The results for Rc are different so what is wrong?

#### beenthere

Joined Apr 20, 2004
15,819
Hi,

A resistor can't have a negative resistance. You can go down to 0 ohms, and that's it.

#### CrktMan

Joined Nov 29, 2005
34
Originally posted by beenthere@Jan 19 2006, 03:36 PM
Hi,

A resistor can't have a negative resistance. You can go down to 0 ohms, and that's it.
[post=13255]Quoted post[/post]​
If you think practically what a resistance is, then you would find the answer without math.
For example: resistance means there is some sort of obstacle in flowing current in a conduction path. Negative idea does not make any sense here. Because two thins are only possible: either some obstacle or no obstacle.

You might be thinking of negative direction of current that has something to do with sign (+/-) convention.

#### n9xv

Joined Jan 18, 2005
329
There's no such beast as negative resistance. If you encounter a "negative" resistance mathematically then your indicating an opposed direction for current flow. Thats like saying you have a negative short circuit.

#### CoulombMagician

Joined Jan 10, 2006
37
You might do a search on frequency dependent negative resistors or FDNRs. They are used in synthesizing active filters.

The mathematical answer to your question is that infinity is not a number and you can't put it in an equality. You need a limiting process( calculus) to deal with the infinite.
Electrically(ideal circuit elements) it means that you can have arbitrary current through that branch with zero voltage across it. Impedance or admittance is a constraint which relates the current and voltage of a two terminal element. Instead of thinking of the resistance as going off to positive or negative infinity think of the admittance as going to zero from either the positive or negative side.
How's that?

#### n9xv

Joined Jan 18, 2005
329
dujo is talking about a resistance being negative. not infinite. A negative number is less than 0. Infinite is some unknown/undefined point much much greater than 0. An FDNR is a manipulation of the transfer function of a filter. It means the individual components (inductors, capacitors, resistors) "act" as other components than what they are. It does not mean the use negative resistors.

#### dujo

Joined Jan 19, 2006
3
Originally posted by n9xv@Jan 20 2006, 07:14 AM
dujo is talking about a resistance being negative. not infinite. A negative number is less than 0. Infinite is some unknown/undefined point much much greater than 0. An FDNR is a manipulation of the transfer function of a filter. It means the individual components (inductors, capacitors, resistors) "act" as other components than what they are. It does not mean the use negative resistors.
[post=13284]Quoted post[/post]​
I accept that no real physical resistor can have negative R but real negative resistance can be created by active circuits (e.g. using op amps - see application note 1868 at www.maxim-ic.com)
My question is how such neg resistors combine with the common positive resistors e.g.
In series, Rc = R1 + R2, e.g. +1 ohm and -1 ohms give Rc = 0 ohm No problems.
But in parallel case, ref.my original posting, neg and pos resistors combine give different results according to how Rc is calculated . Where is my error?

#### n9xv

Joined Jan 18, 2005
329
Lets take a look at this again.

1) Combined resistance of R1 and R2 in parallel is found from

a) 1/Rc = (1/R1) + (1/R2) (For any number of resistance in parallel)

or

Rc = (R1 x R2)/ (R1 + R2) (For only two resistances in parallel)

2) Given R1=R ohms(a real physical resistor) and R2 is a negative resistor =-R ohms

Then

from a) 1/Rc = (1/R) + (1/-R) = (1/R) - (1/R) = 0 and Rc = 1/0 (Only true if R = R)

Or from Rc = (R x -R)/ (R-R) = - (R squared)/0 (Only true if R1 = R2)

The results for Rc are different so what is wrong?

When you "introduce" a negative sign into the formulas you change the formulas from unconditional to conditional. Your mathematically right given the above conditions, but beyond that there is no meaning in the formulas.

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by dujo@Jan 19 2006, 12:45 PM
1) Combined resistance of R1 and R2 in parallel is found from
a) 1/Rc = (1/R1) + (1/R2) or B) Rc = (R1 x R2)/ (R1 + R2)

2) Given R1=R ohms(a real physical resistor) and R2 is a negative resistor =-R ohms
Then from a) 1/Rc = (1/R) + (1/-R) = (1/R) - (1/R) = 0 and Rc = 1/0
Or from B) Rc = (R x -R)/ (R-R) = - (R squared)/0
The results for Rc are different so what is wrong?

[post=13248]Quoted post[/post]
First you have to differentiate between absolute negative resistance and negative differential resistance. Please refer to this page for more explanations.

Both types of negative resistances do not have any meaning outside the particular circuits in which they are referred to. If we analyse absolute negative resistance, then it has the property of increasing current with decreasing voltage across the defined terminals. Putting a resistor across these terminals would actually result in different circuit action or modify the response of the circuit and would not be considered simply as two parallel resistances regardless of the values. Therefore, the parallel resistance formula is not actually valid for this purpose.

Negative differential resistance would behave similarly with the above. However, the response might be the opposite depending on where on the response function the resulting circuit ended up. It might bring the circuit out of negative region.

As n9xv mentioned, mathematical treatments of the above do not have any meaning and correlation to the actual physical occurences. Most mathematical models were afterall derived with certain underlying assumptions. Let us see how the formula for the parallel resistance was derived:

From Ohm's law

I = V/R (1)

Consider two resistors in parallel with voltage V across, current across each resistor is given by:

I1 = V/R1 (2)
I2 = V/R2 (3)

If we want to find an equivalent resistance that can replace the two resistor then:

Iequiv = I1 + I2 (4)

Equivalent resistance would therefore be given as the resistance value that passes equal amount of current:

Requiv = V/Iequiv (5)

Substituting (4) into (5):

Requiv = V/(I1 + I2)
V/Requiv = I1 + I2 (6)

Substituting (2) and (3) into (6):

V/Requiv = V/R1 + V/R2

1/Requiv = 1/R1 + 1/R2

As you can see, there are assumptions during the derivation that are violated if we consider the resistance is negative, hence the meaningless and inconsistent results when you apply the formula to a negative resistance.

#### dujo

Joined Jan 19, 2006
3
Originally posted by n9352527@Jan 20 2006, 10:18 AM
First you have to differentiate between absolute negative resistance and negative differential resistance. Please refer to this page for more explanations.

Both types of negative resistances do not have any meaning outside the particular circuits in which they are referred to. If we analyse absolute negative resistance, then it has the property of increasing current with decreasing voltage across the defined terminals. Putting a resistor across these terminals would actually result in different circuit action or modify the response of the circuit and would not be considered simply as two parallel resistances regardless of the values. Therefore, the parallel resistance formula is not actually valid for this purpose.

Negative differential resistance would behave similarly with the above. However, the response might be the opposite depending on where on the response function the resulting circuit ended up. It might bring the circuit out of negative region.

As n9xv mentioned, mathematical treatments of the above do not have any meaning and correlation to the actual physical occurences. Most mathematical models were afterall derived with certain underlying assumptions. Let us see how the formula for the parallel resistance was derived:

From Ohm's law

I = V/R (1)

Consider two resistors in parallel with voltage V across, current across each resistor is given by:

I1 = V/R1 (2)
I2 = V/R2 (3)

If we want to find an equivalent resistance that can replace the two resistor then:

Iequiv = I1 + I2 (4)

Equivalent resistance would therefore be given as the resistance value that passes equal amount of current:

Requiv = V/Iequiv (5)

Substituting (4) into (5):

Requiv = V/(I1 + I2)
V/Requiv = I1 + I2 (6)

Substituting (2) and (3) into (6):

V/Requiv = V/R1 + V/R2

1/Requiv = 1/R1 + 1/R2

As you can see, there are assumptions during the derivation that are violated if we consider the resistance is negative, hence the meaningless and inconsistent results when you apply the formula to a negative resistance.
[post=13290]Quoted post[/post]​

#### CoulombMagician

Joined Jan 10, 2006
37
As n9xv mentioned, mathematical treatments of the above do not have any meaning and correlation to the actual physical occurences. Most mathematical models were afterall derived with certain underlying assumptions. Let us see how the formula for the parallel resistance was derived:

From Ohm's law

I = V/R (1)

Consider two resistors in parallel with voltage V across, current across each resistor is given by:

I1 = V/R1 (2)
I2 = V/R2 (3)

If we want to find an equivalent resistance that can replace the two resistor then:

Iequiv = I1 + I2 (4)

Equivalent resistance would therefore be given as the resistance value that passes equal amount of current:

Requiv = V/Iequiv (5)

Substituting (4) into (5):

Requiv = V/(I1 + I2)
V/Requiv = I1 + I2 (6)

Substituting (2) and (3) into (6):

V/Requiv = V/R1 + V/R2

1/Requiv = 1/R1 + 1/R2

As you can see, there are assumptions during the derivation that are violated if we consider the resistance is negative, hence the meaningless and inconsistent results when you apply the formula to a negative resistance.
Exactly what are the assumptions here and where are they violated?

The formula for equivalent parallel resistance works just fine with negative resistance(s) in the equation. Try it with 100 Ohm and -10 Ohm, you get 90 Ohm series resistance and -11.11... Ohm parallel resistance. In both cases the voltage and current are correct.
In the original post the poster was confused in the case where the two resistances are R and -R.
2) Given R1=R ohms(a real physical resistor) and R2 is a negative resistor =-R ohms
Then from a) 1/Rc = (1/R) + (1/-R) = (1/R) - (1/R) = 0 and Rc = 1/0
Or from Rc = (R x -R)/ (R-R) = - (R squared)/0
The results for Rc are different so what is wrong?
What's wrong with both is that you cannot divide by zero.
1/Rc = 0 does not mean Rc = 1/0!
and
Rc = (R x -R)/ (R-R) <> - (R squared)/0 !

If we take
I1 = V/R1 (2)
I2 = V/R2 (3)

At face value then V is the independent variable and I1,I2 are the dependent variables related by the parameters R1,R2 whereas by convention the impedance z of a port is defined with the current as the independent parameter and the voltage as the dependent parameter. When is z resistive? When
1) The magnitude of z, |z| is a constant independent of frequency and
2) The phase of z is a constant value of 0 radians.

In the case of the series combination of R and -R the ESR is really 0 because you will always have zero port voltage regardless of the port current. In the case of the parallel combination you will always have zero port current for any port voltage and the z parameter does not exist but the y parameter, the admittance does exist and is zero.

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by CoulombMagician+Jan 22 2006, 11:27 AM--><div class='quotetop'>QUOTE(CoulombMagician @ Jan 22 2006, 11:27 AM)</div><div class='quotemain'>Exactly what are the assumptions here and where are they violated?
[post=13321]Quoted post[/post]​
[/b]

The assumptions are:
1. The parallel equivalent resistance is the resistance value that gives equal current as with the original resistances across the given nodes.
2. That equal current is established at equal voltage across the nodes (i.e. the equivalent resistance is established on an equivalent voltage condition).

Putting a resistor across two nodes that exhibit negative resistance property (in parallel) would not produce the equal current as the mathematically treated equivalent parallel resistance. And more often than not it'd alter the operation of the circuit and change the voltage across the nodes.

Originally posted by CoulombMagician@Jan 22 2006, 11:27 AM
The formula for equivalent parallel resistance works just fine with negative resistance(s) in the equation. Try it with 100 Ohm and -10 Ohm, you get 90 Ohm series resistance and -11.11... Ohm parallel resistance. In both cases the voltage and current are correct.
[post=13321]Quoted post[/post]​
Of course it works. Mathematically. However, physically it doesn't mean a thing. If you have experience working with circuits that exhibit negative resistance properties then you'd know that putting a resistor across the concerned nodes would not produce the equivalent effects as observed by putting the calculated parallel resistance values instead. What I am saying is, if we have a circuit that exhibit say -10 ohm resistance across two nodes, putting a 100 ohm resistor across the nodes would not result in an observed negative resistance of -11.11 ohm (per your example).

An example of this is if you have some current flowing between two nodes with a -10 ohm resistance across them. Putting a 10 ohm resistor would therefore mathematically suggest that the total parallel resistance to be infinity, or the current to be zero. This is just pure impossible physically. If the voltage is still there, how would it possible for the current not flowing by just putting a 10 ohm resistor (which is a simple passive device) across the nodes? Where does the current that previously flow across the nodes gone? I don't think the current would now be looping back through the 10 ohm resistor to create a null current effect. And if the voltage is not there anymore, that would be a violation of the second assumption above.

Originally posted by CoulombMagician@Jan 22 2006, 11:27 AM
In the original post the poster was confused in the case where the two resistances are R and -R.
What's wrong with both is that you cannot divide by zero.
1/Rc = 0 does not mean Rc = 1/0!
and
Rc = (R x -R)/ (R-R) <> - (R squared)/0 !
[post=13321]Quoted post[/post]​
True. I totally agree with you on that division by zero thing. But I think it is wise to point out to him that although the formula mathematically appears to work with unequal value of resistances (where no division by zero occurs), it really doesn't correspond to the physical occurences. In fact, mathematical treatments involving negative resistance using the above formula are not valid.

<!--QuoteBegin-CoulombMagician
@Jan 22 2006, 11:27 AM
If we take
I1 = V/R1 (2)
I2 = V/R2 (3)

At face value then V is the independent variable and I1,I2 are the dependent variables related by the parameters R1,R2 whereas by convention the impedance z of a port is defined with the current as the independent parameter and the voltage as the dependent parameter. When is z resistive? When
1) The magnitude of z, |z| is a constant independent of frequency and
2) The phase of z is a constant value of 0 radians.

In the case of the series combination of R and -R the ESR is really 0 because you will always have zero port voltage regardless of the port current. In the case of the parallel combination you will always have zero port current for any port voltage and the z parameter does not exist but the y parameter, the admittance does exist and is zero.
[post=13321]Quoted post[/post]​
[/quote]

You lost me here. I am sorry. I am not brainy enough to understand these paragraphs.

#### chesart1

Joined Jan 23, 2006
269
Originally posted by n9352527@Jan 23 2006, 06:51 AM
The assumptions are:
1. The parallel equivalent resistance is the resistance value that gives equal current as with the original resistances across the given nodes.
2. That equal current is established at equal voltage across the nodes (i.e. the equivalent resistance is established on an equivalent voltage condition).

Putting a resistor across two nodes that exhibit negative resistance property (in parallel) would not produce the equal current as the mathematically treated equivalent parallel resistance. And more often than not it'd alter the operation of the circuit and change the voltage across the nodes.
Of course it works. Mathematically. However, physically it doesn't mean a thing. If you have experience working with circuits that exhibit negative resistance properties then you'd know that putting a resistor across the concerned nodes would not produce the equivalent effects as observed by putting the calculated parallel resistance values instead. What I am saying is, if we have a circuit that exhibit say -10 ohm resistance across two nodes, putting a 100 ohm resistor across the nodes would not result in an observed negative resistance of -11.11 ohm (per your example).

An example of this is if you have some current flowing between two nodes with a -10 ohm resistance across them. Putting a 10 ohm resistor would therefore mathematically suggest that the total parallel resistance to be infinity, or the current to be zero. This is just pure impossible physically. If the voltage is still there, how would it possible for the current not flowing by just putting a 10 ohm resistor (which is a simple passive device) across the nodes? Where does the current that previously flow across the nodes gone? I don't think the current would now be looping back through the 10 ohm resistor to create a null current effect. And if the voltage is not there anymore, that would be a violation of the second assumption above.
True. I totally agree with you on that division by zero thing. But I think it is wise to point out to him that although the formula mathematically appears to work with unequal value of resistances (where no division by zero occurs), it really doesn't correspond to the physical occurences. In fact, mathematical treatments involving negative resistance using the above formula are not valid.
You lost me here. I am sorry. I am not brainy enough to understand these paragraphs.
[post=13357]Quoted post[/post]​

#### chesart1

Joined Jan 23, 2006
269
Hi,

I think some confusion exists. The term negative resistance does not actually refer to a negative resistor as is indicated by the problem proposed by the student. The negative resistance region refers to the voltage-current characteristic of a tunnel diode where the forward bias voltage increases while the current decreases.

Therefore, I cannot define the circuit as actually having a negative resistance. The actual resistance is still +10 ohms but increases with increasing voltage.

An actual negative resistor value cannot exist.

Therefore a clarification is needed. What was the exact problem presented and what was the discussion in class related to the problem?

John.

#### aac

Joined Jun 13, 2005
35
Originally posted by dujo@Jan 19 2006, 06:45 AM
1) Combined resistance of R1 and R2 in parallel is found from
a) 1/Rc = (1/R1) + (1/R2) or B) Rc = (R1 x R2)/ (R1 + R2)

2) Given R1=R ohms(a real physical resistor) and R2 is a negative resistor =-R ohms
Then from a) 1/Rc = (1/R) + (1/-R) = (1/R) - (1/R) = 0 and Rc = 1/0
Or from B) Rc = (R x -R)/ (R-R) = - (R squared)/0
The results for Rc are different so what is wrong?
[post=13248]Quoted post[/post]​

First this is a math problem and the electronics is not the issue. The formula is:

1/Rc = 1/R + 1/-R. You essentially changed the formula when you did this:

1/Rc = 1/R - 1/R. If you had solved this like this:

1/Rc = -R/(R*(-R)) + R/(R*(-R)) creating a common denominator to add the fractions. This becomes:

1/Rc = 0/-(R**2) which is consistant with your other method. What is wrong is that you can't change the formula. You used two positive resistances in a new formula you made up.

For everyone who doesn't think there is a negitive resistor, look up what a tunnel diode is and how it is used to make oscillators. This is in addition to using amplifiers as a negitave impedance converters.

#### chesart1

Joined Jan 23, 2006
269
Hi,

Again, negative resistance does not exist. The proper way to describe the negative resistance effect is show the curves as they are shown in the data sheet for any tunnel diode.

The pure mathematical exercise has no value in the real world.
And even is the problem is about negative resistance, the point is still missed.

negative resistance definition:
A resistance such that when the current through it increases the voltage drop across the resistance decreases.

That has nothing to do with the actual resistance which is still a positive value. Negative resistance does not mean that the actual resistance is negative.

John

#### n9xv

Joined Jan 18, 2005
329
A tunnel diode has a "negative resistance" region. The current through the diode increases up to a certain voltage, than with increasing voltage the current decreases until a voltage point is reached and the current begins to increase again.

The term negative refers to the V/I characteristics of the device. Its characteristic curves as viewed in a data book or on a curve tracer etc. "Negative" here, means that at a certain point on the transfer curve an increase in voltage will cause a decrease in current. Which is not what you would "normally" expect to see.

The above does not define negative ohms. Its "refered to" as negative resistance because of the way it responds to V/I.

Infinity ohms is X > 1-ohms^ (infinity power).

Does negative ohms imply something less than a dead short? If so then how can you define it or give it meaning?

You can have negative voltage.
You can have negative current (current flowing in the "opposite" direction).
You can have numbers < 0 or negative, but you can not have negative ohms of resistance.

Again, we're mixing terms and/or atempting to apply that which has no meaning purely from a circuit analysis standpoint.

#### aac

Joined Jun 13, 2005
35
Originally posted by n9xv@Jan 23 2006, 03:31 PM
A tunnel diode has a "negative resistance" region. The current through the diode increases up to a certain voltage, than with increasing voltage the current decreases until a voltage point is reached and the current begins to increase again.

The term negative refers to the V/I characteristics of the device. Its characteristic curves as viewed in a data book or on a curve tracer etc. "Negative" here, means that at a certain point on the transfer curve an increase in voltage will cause a decrease in current. Which is not what you would "normally" expect to see.

The above does not define negative ohms. Its "refered to" as negative resistance because of the way it responds to V/I.

Infinity ohms is X > 1-ohms^ (infinity power).

Does negative ohms imply something less than a dead short? If so then how can you define it or give it meaning?

You can have negative voltage.
You can have negative current (current flowing in the "opposite" direction).
You can have numbers < 0 or negative, but you can not have negative ohms of resistance.

Again, we're mixing terms and/or atempting to apply that which has no meaning purely from a circuit analysis standpoint.
[post=13381]Quoted post[/post]​
OK, lets look at is purely from a circuit analysis standpoint. One of the things you can do with a tunnel diode is create and amplifier. To do this you use a voltage divider with a negative resistor. Let's say the load is some normal linear resistor ® and the series arm is a tunnel diode (Rd). We will take the output voltage (Vo) between the diode and load. Then the circuit is easy to anlayze.

Vo = V*R/(R+(-Rd)) => Vo/V = R/(R-Rd)

Now this is a circuit that can be built, and works, simplified of course. It has gain because Rd is negative. How would you evaluate (R+Rd) if (Rd) was not in Ohms? We can't just add things with different units. And both the numerator and denometer must be in Ohms for voltage gain to be unitless.

The original question for a negative resistor in parallel with a resistor is the way to use a tunnel diode as a current amplifier by the way.

If you agree that resistance is taking energy away form electrons by collisions with the lattice of the material, then negiatve resistance would be imparting energy to the electrons. This is exactly what happens to some of the electrons in a tunnel diode.

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by aac@Jan 25 2006, 01:56 PM
OK, lets look at is purely from a circuit analysis standpoint. One of the things you can do with a tunnel diode is create and amplifier. To do this you use a voltage divider with a negative resistor. Let's say the load is some normal linear resistor ® and the series arm is a tunnel diode (Rd). We will take the output voltage (Vo) between the diode and load. Then the circuit is easy to anlayze.

Vo = V*R/(R+(-Rd)) => Vo/V = R/(R-Rd)

Now this is a circuit that can be built, and works, simplified of course. It has gain because Rd is negative. How would you evaluate (R+Rd) if (Rd) was not in Ohms? We can't just add things with different units. And both the numerator and denometer must be in Ohms for voltage gain to be unitless.
[post=13461]Quoted post[/post]​
Pardon my ignorance, but I thought tunnel diodes do not have absolute negative resistance. Instead these have negative differential resistance region in which the diodes exhibit 'negative resistance' property, i.e. current decreases with voltage increase. This region, and the relationship of I and V in this region, is not linear and not absolute and therefore simple I=V/-R and Vo=V*R/(Rtot) formulaes do not really apply in this case. Combining and analysing a circuit with a positive resistance and a not-absolute and non-linear negative resistance with a simple linear absolute Ohm's law equation, in my opinion, is an erroneous exercise.

As I pointed out in my previous post, we need to distinguish between absolute negative resistance (does this really exist in real life?) and negative differential resistance.

Just a simple thought, might be worth something or probably not #### n9xv

Joined Jan 18, 2005
329
I for one am not dissagreeing that a tunnel has the characteristiscs it does. But your simply describing just that - a characteristic - or the way a device response to voltage & current. Engineers develope & apply the mathematics as necessary in order to calculate or "controll" the phenomena so as to have a means of producing a desired result. Its only logical that they would "name" this phenomena negative resistance and apply the term ohms to it.

Thats no different than "imaginary" numbers or the "J-operator" as used in the field of electronics for example. A given inductor or capacitor exhibits X amount of ohms of reactance at a specified frequency. The reactance of the device being the way the device "responds" to voltage/current at the specified frequency. This responce "acts" in such a way to impead the flow of current. Impeading the flow of current sounds like something a resistor would do. So, guess what units reactance is in - right - ohms. In reality those "ohms" of reactance dont exist without the frequency variable. In that same sense, "ohms" of negative resistance in a tunnel diode (or anywhere else) dont exist either without the voltage variable. I'am not arguing about the way things are engineered, I'am just looking at it from a "George Simon Ohm" point of view.