Q.The V-I characteristic of a silicon diode is shown in Fig. 14.17. Calculate the resistance of the diode at

Concept: Diode Resistance Calculation — The resistance depends on the operating region: dynamic (ac) resistance in forward bias from the slope of the V-I curve, and static (dc) resistance in reverse bias from Ohm's law.

Reasoning:

1. Forward bias (a): The diode is in the steep, nearly linear region. Use the dynamic resistance formula $r_d = \frac{\Delta V}{\Delta I}$ between two points on the straight-line portion. From the graph: at $I=10\ \text{mA}$, $V=0.7\ \text{V}$; at $I=20\ \text{mA}$, $V=0.8\ \text{V}$.

2. Compute $\Delta V = 0.8 - 0.7 = 0.1\ \text{V}$, $\Delta I = 20 - 10 = 10\ \text{mA} = 0.01\ \text{A}$.

   $r_d = \frac{0.1}{0.01} = 10\ \Omega$. …

The key idea is that a diode does not have a single, fixed resistance. Its resistance depends on where you are on the V-I curve. In forward bias, the diode conducts heavily and the resistance is small. In reverse bias, almost no current flows, so the resistance is enormous.

We calculate the resistance differently depending on whether we want the static (DC) resistance or the dynamic (AC) resistance. The problem asks for "the resistance of the diode" at a given current or voltage. From the graph, we can extract the necessary values.

1. Forward bias case: $I_D = 15\ \text{mA}$

The forward characteristic is steep and approximately linear between the marked points. This means we can treat it as a straight line in that region.

The two given points on the forward branch are:

• Point A: $V = 0.7\ \text{V}$, $I = 10\ \text{mA}$
• Point B: $V = 0.8\ \text{V}$, $I = 20\ \text{mA}$

Since the curve is nearly a straight line through these points, the dynamic resistance (the slope resistance at any point between them) is constant. This is the resistance that matters for small signal changes around the operating point.

Using the two points:

Therefore:

2. Reverse bias case: $V_D = -10\ \text{V}$ …

Method: Reading Resistance from a V-I Characteristic Graph

A diode's resistance is NOT a single fixed number -- it depends on where you are on the V-I curve, and on whether you want the static (DC) resistance or the dynamic (AC/slope) resistance.

Step 1 -- Identify what kind of resistance each part asks for.

Part (a) gives a current in the steep, nearly-linear forward-bias region -- here the natural quantity is the dynamic resistance, the slope of the curve. Part (b) gives a voltage in the flat reverse-bias region -- here the static resistance (plain Ohm's law, $R = V/I$) is the meaningful one, since the dynamic (slope) resistance would be enormous and not very informative.

Step 2 -- Forward bias, $I_D = 15\ \text{mA}$: use the slope.

Read two points that bracket 15 mA on the graph's linear forward segment: $(V=0.7\ \text{V}, I=10\ \text{mA})$ and $(V=0.8\ \text{V}, I=20\ \text{mA})$. …

Mistake 1: Using the static resistance formula in forward bias

A student might compute $R = V/I$ at $I=15\ \text{mA}$ instead of the slope. Reading $V\approx0.75\ \text{V}$ from the graph gives $R = 0.75/0.015 = 50\ \Omega$ -- a very different number from the correct $10\ \Omega$. The two questions ("what is V/I here?" vs "what is the slope here?") measure different things, and the forward-bias part of a diode curve is exactly where they diverge most.

How to avoid: When the curve is steep and nearly straight (the forward branch), the question is almost always asking for the dynamic (slope) resistance $r_d=\Delta V/\Delta I$, found from two nearby points -- not the static $V/I$ at a single point.

Mistake 2: Using the slope formula in reverse bias

In reverse bias the curve is nearly flat, so the slope $\Delta V/\Delta I$ is enormous and not a useful number to report from a graph reading. Here the meaningful quantity is the plain static resistance $R=V/I$ at the one given point. …