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

Question

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

Answer 1

If you have an impedance with a negative real part, it delivers power, yes. But remember: no actual passive component has a negative value! To build that negative resistance, you need active circuitry that do 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. Most discussion about this theorem is made in dc where no negative resistances are possible. In amplifiers, maximum power transfer results in only 50% efficiency and is not always of interest. In Radio Frequency circuits, this theorem translates to impedance matching and reflection reduction. To wrap up, having a negative Thevenin equivalent resistance does result in a theoretical infinite power delivery but you have to keep in mind where that resistance is resulted from and what the circuit aims for.