# How can the maximum average power transfer be negative in a power delivery circuit?

It is possible that the Thevenin resistance $$\R_{Th}\$$ of a circuit is negative. This occurs in non-linear circuits. Likewise, in AC circuits, the real part of $$\Z_{Th}\$$ can be negative.

For AC power transfer, $$\P = \frac{1}{2}|\mathbf{I}|^2 R_L\$$, or $$P = \frac{1}{2}\frac{|\mathbf{V}_{Th} |^2 R_L}{(R_{Th} +R_L)^2 + (X_{Th} + X_L)^2 }$$

It is well known[1] that the power is at a maximum when $$\X_{Th} =-X_L\$$ and when $$\R_L=R_{Th}\$$. This leads to a generally well known equations for maximum average power transfer:

$$P_{\text{max}} =\frac{|\mathbf{V}_{Th} |^2}{8 R_{Th}}$$

Unfortunately, if $$\R_{Th}<0\$$, this yields negative power. It also implies that $$\R_L<0\$$, which also does not make sense.

How to find the maximum power for such a circuit?

Edit: The maximum is found by taking two derivatives, $$\\frac{\delta P}{\delta X_L}\$$ and $$\\frac{\delta P}{\delta R_L}\$$, and setting them equal to zero. For $$\\frac{\delta P}{\delta X_L}\$$, it is easy to see that $$\X_{Th}=-X_L\$$. For $$\\frac{\delta P}{\delta R_L}\$$, the derivative is:

$$\frac{\delta P}{\delta R_L}\ = \frac{|\mathbf{V}_{Th} |^2 R_L[(R_{Th}+R_L)^2-2R_L(R_{Th}+R_L)]}{2(R_{Th}+R_L)^4}$$

This to find the maximum, we set $$\\frac{\delta P}{\delta R_L}=0\$$. The resulting equation simplifies to:

$$R_L = \sqrt{R_{Th}^2}$$

Mathematically, this has two roots, $$\R_L = R_{Th}\$$ and $$\R_L = -R_{Th}\$$. Both seem physically unrealistic to me; the positive root give a negative power; which doesn't make sense for a passive element. The negative root gives an infinite power transfer, from:

$$P = \frac{1}{2}\frac{|\mathbf{V}_{Th} |^2 R_L}{(R_{Th} - R_{Th})^2 } =\frac{|\mathbf{V}_{Th} |^2 R_L}{0}$$

This implies an infinite power transfer, which would be great, right? But I'm not sure how realistic that is either.

[1] For example, section 11.3 in Fundamentals of Electrical Circuits by Alexander

• Negative power simply means that your circuit is delivering power. What is the problem with this? Isn’t that what a negative resistance is supposed to do? Commented Mar 19, 2020 at 6:36
• Actually you are wrong!...If Rth is negative Rl = -Rth (a positive Rl) gives maximum power. In fact maximum power is infinite using above formula. Commented Mar 19, 2020 at 8:44
• @BenFM If we could find a way for passive elements to deliver power, that would be a fantastic energy source. Unfortunately, I don't think that is realistic. Commented Mar 19, 2020 at 12:23
• @sarthak If $R_L = -R_th$, the power give 1/0, or infinity. Infinity power deliver is great! Why don't we do this in real life? Commented Mar 19, 2020 at 12:25
• @axsvl77 You are missing something. If you have a source impedance with a negative real part, it delivers power yes. But remember: no passive component is negative! To build that negative resistance, you need active circuitry whom consume dc power. The very fact you mentioned (infinite power loop) is the concept behind designing oscillators. They consume dc power to be able to maintain a constant oscillation amplitude. Maximum power transfer is meaningless unless it is defined for a certain scenario. I don’t think there exists any power delivery scenario whom source resistance is negative! Commented Mar 19, 2020 at 13:29