# Maximum Power Transfer with Negative Thevenin Resistance?

Hello everyone! I have a general question about how to find out the maximum power transfer to a load resistor with a negative Thevenin Resistance (Rth < 0). I know for positive cases, P = R*(Vth/(R + Rth))^2, in which R is the load resistor connected to the Thevenin equivalence circuit, and Pmax = (Vth)^2/(4*Rth). But what if Rth < 0? I don't think that the Pmax equation still holds true in negative cases for Rth. Can someone help me explain this phenomenon please? I very much appreciate your help! Thank you.

Sincerely,
gta100

## Answers and Replies

berkeman
Mentor
Hello everyone! I have a general question about how to find out the maximum power transfer to a load resistor with a negative Thevenin Resistance (Rth < 0). I know for positive cases, P = R*(Vth/(R + Rth))^2, in which R is the load resistor connected to the Thevenin equivalence circuit, and Pmax = (Vth)^2/(4*Rth). But what if Rth < 0? I don't think that the Pmax equation still holds true in negative cases for Rth. Can someone help me explain this phenomenon please? I very much appreciate your help! Thank you.

Sincerely,
gta100

Welcome to the PF.

What is a negative Thevenin resistance? Is it physical? Or is it related to some complex impedance (in which case there is a good answer)?

Hello berkeman! Thank you very much for your reply!

By negative Thevenin resistance, I mean Rth < 0 in the Thevenin equivalent circuit. According to my textbook, negative Rth may appear when there are ONLY DEPENDENT sources inside "the circuit". By "the circuit", I mean the circuitry with CONSTANT PHYSICAL QUANTITIES FOR ALL ELEMENTS and with two terminals connected to a load resistor with a VARYING RESISTANCE (rheostat). I hope that is clear? Thank you!

gta100

berkeman
Mentor
Hello berkeman! Thank you very much for your reply!

By negative Thevenin resistance, I mean Rth < 0 in the Thevenin equivalent circuit. According to my textbook, negative Rth may appear when there are ONLY DEPENDENT sources inside "the circuit". By "the circuit", I mean the circuitry with CONSTANT PHYSICAL QUANTITIES FOR ALL ELEMENTS and with two terminals connected to a load resistor with a VARYING RESISTANCE (rheostat). I hope that is clear? Thank you!

gta100

So to figure this out, we should take a typical circuit that exhibits negative output resistance, and figure out how to get maximum power to a load impedance (by varying the load impedance in this case).

Can you post a typical circuit that exhibits negative resistance at its two output terminals? And then see if you might be able to answer your own question by trying some different resistances connected to those terminals?

So to figure this out, we should take a typical circuit that exhibits negative output resistance, and figure out how to get maximum power to a load impedance (by varying the load impedance in this case).

Can you post a typical circuit that exhibits negative resistance at its two output terminals? And then see if you might be able to answer your own question by trying some different resistances connected to those terminals?

Hi Berkeman! Sure! Let me go ahead and post a sample circuit for you (Please see attachment).

So in this circuit, I find Rth = -15000/11 ohms = - 1363.636364 ohms , and Vth = - 2680/11 V (Also negative). I need to find the maximum power transfer to the rheostat R as indicated.

Thank you for your help!

gta100

#### Attachments

• Negative Thevenin Resistance.jpg
34.3 KB · Views: 1,301
rude man
Homework Helper
Gold Member
Forgetting Thevenin equivalent reistance for a moment, have you determined i(R) for arbitrary R?

Hello rude man! Thank you for your reply.

I have determined i(R) for sure. It is Vth/(R + Rth), right? I actually included it in the formula for the power transferred to the load resistor, namely, P = R*(Vth/(R + Rth))^2.

I even plotted the equation in MATLAB, and it turned out that P will reach infinitely high when R = - Rth (Recall that Rth < 0 now, so R = -Rth > 0). So it seems like there is no max power. Could you help me investigate if I calculated the wrong Rth? I attached the sample problem in my last post. Thank you very much!

gta100

uart
Science Advisor
Hello rude man! Thank you for your reply.

I have determined i(R) for sure. It is Vth/(R + Rth), right? I actually included it in the formula for the power transferred to the load resistor, namely, P = R*(Vth/(R + Rth))^2.

I even plotted the equation in MATLAB, and it turned out that P will reach infinitely high when R = - Rth (Recall that Rth < 0 now, so R = -Rth > 0). So it seems like there is no max power. Could you help me investigate if I calculated the wrong Rth? I attached the sample problem in my last post. Thank you very much!

gta100

Yes, the correct value of $R_{thv}=$ is $-\frac{15}{11}k \Omega$. (Hand calculated the good old fashioned way. )

Yes the maximum power is unlimited (infinity if you like), as R_L approaches -R_thv.

The direction that Berkeman was trying to point you toward was to consider what type of real circuit components can produce a negative resistance. When you do this you'll find that in all cases the negative resistance is a result of some active component that is only valid over a limited operating range. There is no such thing as a true (or intrinsic) negative resistance.

Common examples of negative resistances are.

1. Small signal linearized examples like the tunnel diode, where both V and I are actually in the same direction, but $dV/dI$ is negative over some limited range of its V/I characteristics.

2. Active circuits using opamps and transistors. (remember that these always have a limited linear operating range).

3. Electro-mechanical systems like motors and generators. Again these will always be subject to a finite operating range.

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Hello uart! Thank you very much for your reply! That is very inspirational and reassuring! Thank you for quoting some real-world examples which would produce negative Thevenin resistances!

gta100

uart,

Yes, the correct value of Rthv= is −1511kΩ. (Hand calculated the good old fashioned way. )

Yes the maximum power is unlimited (infinity if you like), as R_L approaches -R_thv.

Did you check your negative resistance value by finding the power dissipated by R at that negative value? If you did, you would find that the power is a small negative value, which means that mathematically, R is generating power at that negative value. However, at R=1363.636364, the power dissipated is 2.962740001E21, not infinity. See the power plot I posted above.

Ratch

uart
Science Advisor
uart,

Did you check your negative resistance value by finding the power dissipated by R at that negative value? If you did, you would find that the power is a small negative value, which means that mathematically, R is generating power at that negative value. However, at R=1363.636364, the power dissipated is 2.962740001E21, not infinity. See the power plot I posted above.

Ratch

Hi Ratch. Your results are simply an artifact of the simulation software only evaluating the response at a finite number of discrete values of R.

For any circuit that has a negative Thevenins resistance (and a non zero Thevenins voltage) it is easy to see that if you make the external resistance equal to the absolute value of the Thevenins resistance that the total resistance becomes zero. The current, and therefore the load power, must be infinite.

As I explained to gta100 however, in reality this can never happen, as the active components forming the negative resistance will limit or saturate at some finite level.

uart,

Your results are simply an artifact of the simulation software only evaluating the response at a finite number of discrete values of R.

I did not use simulation software. As shown in post #8 of this thread, I found the voltage across R by using node analysis. Then using the well known formula V^2/R, I plotted and calculated the R value at the highest power dissipation. That occurs when R is positive, not negative. Anytime R is negative, the resistor is not dissipating power, because V^2/R will always be negative. I stand by my answer of R=1363.636364, and again invite you to calculate the power the resistor dissipates at your value and mine.

Ratch

uart
Science Advisor
uart,
I did not use simulation software.
Ok your plotting software then.

That occurs when R is positive, not negative. Anytime R is negative, the resistor is not dissipating power, because V^2/R will always be negative. I stand by my answer of R=1363.636364, and again invite you to calculate the power the resistor dissipates at your value and mine.
What are you talking about, they're basically the same value, you just had a decimal approximation. I said to use R = -R_thv, which is positive 15/11 k-ohm. As a decimal this is $1363.\dot{6}\dot{3}\, \Omega$ (that's the .63 repeated).

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uart,

Ok your plotting software then.

My plotting software does "evalute the response at a finite number of discrete values of R". However, that is not what I used to determine the value of R.

I said to use R = -R_thv, which is positive 15/11 k-ohm.

I don't know about the Thevenin value, but if you agree that R is positive, then the matter is resolved.

Ratch

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uart
Science Advisor
I don't know about the Thevenin value, but if you agree that R is positive, then the matter is resolved.
I do know about the Thevenin value, and I also know that -(-15/11) is a positive number. So yes this is resolved.

uart
Science Advisor
My plotting software does "evalute the response at a finite number of discrete values of R". However, that is not what I used to determine the value of R.

Ratch. Are you honestly saying that you cannot see that the maximum value of,
$$\frac{2.173 R}{(0.00605R - 8.25)^2}$$
increases without bound as R approaches $\frac{8.25}{0.00605}$?

* Equation taken from your own working (with factor of $10^{24}$ removed top and bottom).

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uart,

Ratch. Are you honestly saying that you cannot see that the maximum value of,

increases without bound as R approaches 8.250/.00605?

Sure I can. As the denominator goes to zero, the term becomes infinite. I plugged the decimal value of R into V^2/R and got a very high value of dissipated power. Had I used the exact value of R, then I would have gotten infinity.

Ratch

uart
Science Advisor
I plugged the decimal value of R into V^2/R and got a very high value of dissipated power. Had I used the exact value of R, then I would have gotten infinity.

OK, I'm glad that you've finally realized that. Now we're done.

uart,

OK, I'm glad that you've finally realized that. Now we're done.

Except for one thing. I did not "finally" realize that, I always knew that I did not have the exact value of R. Therefore I knew that the calculated dissipated power by the approximate R was finite, and not infinite.

Ratch

uart
Science Advisor
uart,
Except for one thing. I did not "finally" realize that, I always knew that I did not have the exact value of R. Therefore I knew that the calculated dissipated power by the approximate R was finite, and not infinite.

Ok I'm happy to concede that. It was really just your reply #11 that confused me into thinking otherwise. It seems that most of the discussion after reply #10 in this thread is therefore pointless.