# Unstable points in a basic memoryless circuit

I have been told that there can be more than a single solution to a circuit, and that some of those solutions might be unstable. As a simple example, a resistor with a constant power source, connected to a diac (which, I've also been told, has the "S"-like I vs V characteristic.

The big blue dot I drew was the point that supposedly is unstable. I can't understand why clearly from the explanation attempts I got from the teachers (I don't think the ones I asked understand it clearly either). I wonder if anyone in here can help out ## 1 Answer

In the vicinity of the big blue dot, the diac exhibits negative dynamic resistance which means that an increase of current through the diac results in a decrease of voltage across the diac.

One way to see that the solution given by the big blue dot is unstable is to consider the effect of perturbing the voltage across the diac; perturb the voltage across the diac $\epsilon$ more positive.

Because of the negative dynamic resistance of the diac in the vicinity of the big blue dot solution, there will be less series current. But with less current through the resistor, the voltage across the diac would be larger still with corresponding less series current etc. A small perturbation from the solution "runs away"; a disturbance in one direction implies a further disturbance in the same direction.

However, for the two solutions shown where the diac has positive dynamic resistance, the opposite occurs.

A positive diac voltage perturbation implies more series current which implies less voltage across the diac. A small perturbation from the solution "runs back"; a disturbance in one direction implies a correction in the opposite direction.

• Thanks +1. Let's see, increase Vd by e. That decreases Id, and that means less Vr than what we had at the beginning, which means higher Vd than what we had at the beginning. But does that mean it's higher than Vd+e? I think looking at the plot that it's lower – Rojo Aug 31 '12 at 14:16
• @Rojo, peruse this and see if it helps. – Alfred Centauri Aug 31 '12 at 18:24