# What does negative impedance mean in the context of miller's theorem

Miller's theorem states that for a linear circuit of two nodes with voltages $$V_1,V_2$$ which are connected by an impedance, $$Z$$in series, circuit could be represented by an equivalent one with two grounded impedance. The first impedance written as $$\frac{Z}{(K-1)}$$ and the second as $$\frac{ZK}{(K-1)}$$ where $$K=\frac{V_2}{V_1}$$ Suppose V1=2V and V2=3V. The impedance of the linear equivalent circuit according to miller's theorem would be negative for the first impedance since 1<3/2 and positive for the second part since 3/2>. Similarly, suppose V2=2 and V1=3, the first impedance would be positive and the second impedance would be negative in the resulting linear circuit. Blow is schematic

simulate this circuit – Schematic created using CircuitLab

My question is, what would the negative impedance mean in both cases?

The circuit you have is not entirely accurate. The voltage sources must be connected to ground as the node voltages must be defined with respect to ground for Miller's theorem to work. Also, there is usually some well-defined relationship between the voltages. For example, the nodes might be connected with an amplifier with gain K that sets the voltage of one node based on the voltage of the other node. It would be more accurage to replace V2 with a dependent source that is 3/2 V1. If K is not constant, then the impedances will depend on the ratio of the actual voltages, which likely won't help simplify a circuit very much.

As for the impedances, they aren't representative of physical impedances. It's not possible to make a resistor with negative resistance. It is, however, possible to make a circuit that acts like a negative resistance in a limited capacity. In this case, since the 2V node is connected to a node at a higher voltage via a resistor, current will be flowing into the node through the resistor. When that connection is redefined as an impednace to ground, it must be a negative impedance otherwise the current would be flowing the wrong way. Check this by calculating what the current would be through the original impedance and for the transformed impedances.

These sorts of theorems are mathematical tools to help simplify larger circuits so that they are easier to characterize mathematically. Unfortunately, that means that the implications can be rather abstract as sometimes you end up with seemingly nonphysical results like negative impedances.

A very interesting question... Congratulations, @pyler!

I am closely connected with the Miller theorem and, a few years ago, I spent a lot of efforts to reveal the idea on which it is based. In 2011, I created and fully wrote the Wikipedia page about the theorem. More precisely speaking, this page was dedicated not so much to the very theorem, but rather to the extremely useful applications of this Miller arrangement (a voltage source connected in series to an impedance element) in analog circuitry.

Besides its use as a "mathematical tool" (as alex.forencich said), it is an extremely powerful tool for creating virtual elements; I have dedicated a special question about it in ResearchGate - How do we create virtual electrical elements in electronics? The recipy is extremely simple - insert an additional voltage source in series with the impedance and vary its voltage proportionally to the input voltage. Thus you can decrease up to zero and even invert... or increase up to infinty and invert again the impedance... Now about the concrete question; here are my considerations.

My note to alex.forencich is that it is not mandatory the circuit to be grounded; it is enough just to close the loop. But if we ground the sources, we will more easily see that the second source, compared with the first, has an opposite polarity (+) with respect to ground. Тhis requires the coefficient K in the expression to be negative and then the denominator is K = 2.5 instead of -0.5; just to correct that the theorem states that Z1 = Z/(1 - K).

So, the new resistance is Z1 = 100/2.5 = 40 ohm... it is not negative... it is just the (2.5 times) decreased old resistance Z. Now, the most interesting part of my comment - I assert that we cannot obtain a negative resistance with this simple Miller arrangement... and I will try to explain why.

In this arrangement, the two sources are connected in series and in the same direction when travelling the loop (+ -, + -). So, the second source V2 increases the overall voltage (2 + 3 = 5 V vs the initial 2 V) and thus increases 2.5 times the common current (I = 5/100 = 50 mA vs the initial 2/100 - 20 mA). As a result, the new input resistance is 2/50 = 40 ohm; it is simply decreased. We can decrease it even more by increasing further V2 but we can reach only zero... we cannot go beyond this point... we cannot invert this resistance to obtain a negative resistance.

To invert the resistance, we have to invert the overal voltage across the network Z - V2 (i.e., the voltage across the new virtual "resistor" Z1). But we cannot invert this voltage since this network is connected to an ideal voltage source that defines the voltage. To do it, we have to connect another resistor in series to the input voltage source. Then the current will keep its old direction but the voltage across the network will change its polarity to the opposite... and the ratio V/I will become negative. This kind of circuits are named voltage inversion negative impedance converters (VNIC). See my Wikibooks stories about the S-shaped negative resistance and the operation of VNIC.