Piecewise-linear approximation
(extracted from Chua, Desoer, Kuh, Linear and Nonlinear Circuits, McGraw-Hill, 1987)

Piecewise-linear approximation is useful in dealing with both simple and general circuits made up of nonlinear resistors. We can use the piecewise-linear method and concave and/or convex resistors to help us in understanding qualitatively the nonlinear behavior of circuits. Let us see an example of using concave resistors and piecewise-linear analysis to approximate the characteristic of a tunnel-diode.

The piecewise-linear approximation of a tunnel-diode characteristic is shown in Fig. 1 below.

tunnel-diode PWL characteristic
Fig.1 - PWL approximation of a tunnel-diode characteristic
The three linear segments have slopes

Region 1:   G = Ga    for   v <= E1   (1)
Region 2:   G = Gb    for   E1 < v <= E2   (2)
Region 3:   G = Gc    for   v > E2    (3)

These define the three regions as shown in Fig. 1. Beginning from left to right, we can decompose the piecewise-linear characteristic into the sum of three components as shown in Fig. 2(a): a straight line passing through the origin with slope G0, a concave resistor characteristic which starts at E1 with a negative slope G1, and a concave resistor characteristic which starts at E2 with a positive slope G2. The corresponding circuit is shown in Fig. 2(b).

pwl circuit Fig. 2 - (a) Decomposition of the piecewise-linear characteristic and (b) the corresponding circuit realization.
Adding the three branch currents, we have the driving-point characteristic of the one port:

i = i0 + i1 + i2 = i(v)     (4)

To make it identical with the piecewise-linear characteristic of the tunnel-diode, the three parameters G0, G1, and G2 in Fig. 2a must satisfy the following:

Region 1:   G0 = Ga  (5)
Region 2:   G0 + G1 = Gb   (6)
Region 3:   G0 + G1 + G2 = Gc   (7)

Thus    G0 = Ga ,   G1 = – Ga + Gb ,   and   G2 = – Gb + Gc

We may generalize the above by using the function representation of the concave resistor. Combining

i0 = G0 v ;    i1 = ½ G1 [|v – E1| + (v – E1)] ;    i2 = ½ G2 [|v – E2| + (v – E2)]

with Eq. (4), we obtain

i = – ½ (G1 E1 + G2 E2) + (G0 + ½ G1 + ½G2) v + ½ G1|v – E1| + ½ G2|v – E2|
This may be written in the following general form for future use;
i = a0 + a1 v + b1|v - E1| + b2|v - E2|