My objective in this circuit is to find the value of RL that achieves maximum power transfer.


simulate this circuit – Schematic created using CircuitLab

To do this, find Voc and Isc, as shown below.


simulate this circuit

Upon performing the calculations, it is determined that \$V_ \text{OC} = 4~V\$ and \$I_\text{SC} = -2~mA\$, indicating that \$R_\text{TH} = -8~k\Omega\$.

Well, assuming that the calculations and simulations are correct, I agree that it doesn't make sense to talk about negative resistance. The direction of the short-circuit current and the polarity of the open-circuit voltage seem to be consistent. Now, what would be the theoretical explanation for this result or where could the error be?


1 Answer 1


It does make sense to have a negative Thevenin resistance. This doesn't mean anything more than what it could be literally interpreted: the higher voltage it sees, the higher (in absolute value) but reversed sided will be the current. This situation only can happen when you have active dependent sources in your circuit—not in pure resistive networks, because the resistance of passive elements is always positive. In your case, it is VCCS1 the active and dependent source that makes the resistence to be negative. Thou still can't believe it? Think it simpler, consider this circuit:

enter image description here

$$ V_{AB} = (1A + k V_{AB}) \cdot 1\Omega$$

$$ R_{AB} = \frac{V_{AB}}{I_{AB}} = \frac{V_{AB}}{V_{AB}/1\Omega - k V_{AB}} = \frac{1}{1 \Omega^{-1} - k} $$

If k > 0, then the relationship between the voltage and the current in your terminals will be negative. I don't think, however, that makes sense to talk about maximum transfer of power with negative resistances. In those cases when you try to maximize the power it is considered that the Thevenin resistance cannot be smaller, so you try to put the load impedance that suites best. On the other hand, the purpose of a negative resistance is usually to cancel the positive resistance so your circuit doesn't have any ‘internal dissipation’ of energy. This is just an active element that supplies an extra amount of energy per time unit so the energy inside your system doesn't go to zero when you don't have any excitation (for this see how a linear sinusoidal oscillator, like quadrature oscillator, works).

But if this is just an excersise, follow the definition of maximum power transfer and put on the load an impedance equal to the conjugate of the Thevenin impedance (in this case is -8kOhms)


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