## Chapter 2

## Q. 2.SP.26

Using ideal diodes, resistors, and batteries, synthesize a function-generator circuit that will yield the i-v characteristic of Fig. 2-41(*a*).

## Step-by-Step

## Verified Solution

Since the i-v characteristic has two breakpoints, two diodes are required. Both diodes must be oriented so that no current flows for v < -5 \text{V}. Further, one diode must move into forward bias at the first breakpoint, v = -5 \text{V}, and the second diode must begin conduction at v = +10 \text{V}. Note also that the slope of the i-v plot is the reciprocal of the Thévenin equivalent resistance of the active portion of the network.

The circuit of Fig. 2-41(*b*) will produce the given i-v plot if R_1 = 6 kΩ, R_2 = 3 kΩ, V_1 = 5 \text{V}, and V_2 = 10 \text{V}. These values are arrived at as follows:

1. If v < -5 \text{V}, both v_{D1} \text{ and }v_{D2} are negative, both diodes block, and no current flows.

2. If -5 ≤ v < 10 \text{V}, D_1 is forward-biased and acts as a short circuit, whereas v_{D2} is negative, causing D_2 to act as an open circuit. R_1 is found as the reciprocal of the slope in that range:

R_1 = \frac{10 – (-5)}{0.0025} = 6 kΩ

3. If v ≥ 10 \text{V}, both diodes are forward-biased,

R_{Th} = \frac{R_1R_2}{R_1 + R_2} = \frac{Δv}{Δi} = \frac{20 – 10}{(7.5 – 2.5) × 10^{-3}} = 2 kΩ

and R_{2} = \frac{R_1R_{Th}}{R_1 – R_{Th}} = \frac{( 6 × 10^3)(2 × 10^3)}{4 × 10^{3}} = 3 kΩ