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