The given circuit has two ideal diodes connected as shown in figure below. Calculate the current flowing through the resistance $R_1$. ![Printed circuit (page 9): the 10 V battery drives current through R1 = 2 Ω to a top node; from the top node two parallel branches return to — Class 12 Physics question](/figures/tn-phy-semiconductor-electronics/printed-circuit-page-battery-drives-current-top-fig-pyq-i4k0y5.svg)

The given circuit has two ideal diodes connected as shown in figure below. Calculate the current flowing throu

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Printed circuit (page 9): the 10 V battery drives current through R1 = 2 Ω to a top node; from the top node two parallel branches return to — Class 12 Physics question — Figure

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Concept understanding — Diode Resistance Calculation

Diode Resistance: What Does It Even Mean?

Think of a diode as a one-way valve for electricity. When you push current through it in the forward direction, the diode doesn't just let everything through freely — it resists the flow, just like any other component. But here's the twist: that resistance isn't a fixed number like a resistor's 100 Ω. It changes depending on how much voltage you apply.

Why? Because a diode is a semiconductor device. Its current-voltage relationship follows the Shockley equation:

I = I_S (e^{V / \eta V_T} - 1)

where $I_S$ is the reverse saturation current, $V$ is the applied voltage, $\eta$ is the ideality factor (usually 1 for silicon), and $V_T \approx 26 \text{ mV}$ at room temperature.

This exponential curve means that a tiny change in voltage can cause a huge change in current. So the "resistance" you measure depends entirely on where you are on that curve.

Two Kinds of Diode Resistance

Because the I-V curve is nonlinear, we define two different resistances — each useful in different situations.

1. Static (DC) Resistance

This is the simplest idea: just apply Ohm's law using the total voltage and total current at a given operating point.

R_{DC} = \frac{V}{I}

For example, if a diode has 0.7 V across it and 10 mA flowing through it, its DC resistance is:

R_{DC} = \frac{0.7}{0.01} = 70 \ \Omega

Note

Static resistance tells you the average opposition to current at that specific point. It's useful for power calculations ( $P = I^2 R_{DC}$ ) but not for small signal analysis.

2. Dynamic (AC) Resistance

This is the more important one for circuit design. It tells you how the diode responds to small changes in voltage around a fixed operating point.

Mathematically, dynamic resistance is the slope of the I-V curve at that point:

r_d = \frac{dV}{dI}

For a forward-biased diode obeying the Shockley equation, we can derive a clean formula. Starting from:

I = I_S e^{V / \eta V_T}

(ignoring the -1, which is negligible in forward bias)

Differentiate:

\frac{dI}{dV} = \frac{I}{\eta V_T}

Therefore:

r_d = \frac{dV}{dI} = \frac{\eta V_T}{I}

r_d = \frac{\eta V_T}{I}

At room temperature with $\eta = 1$ and $V_T = 26 \text{ mV}$ :

r_d = \frac{26 \text{ mV}}{I}

So if the diode current is 10 mA, $r_d = 2.6 \ \Omega$ — much smaller than the 70 Ω DC resistance.

Watch out

Dynamic resistance is not a physical resistor inside the diode. It's a small-signal model parameter. You cannot use it with DC voltages or large signals — only for tiny variations around the operating point.

When Do You Use Each?

Situation	Use
Finding DC power dissipation	$R_{DC}$
Designing a biasing circuit	$R_{DC}$
Analyzing small-signal amplifier response	$r_d$
Calculating voltage regulation in a Zener diode	$r_d$ (called Zener impedance)

A Quick Example to Tie It Together

A silicon diode ( $\eta = 1$ ) is forward biased with $V = 0.7 \text{ V}$ and carries $I = 20 \text{ mA}$ .

Static resistance: …

Why this formula?

Diode Resistance: Why It Changes with Operating Point

A diode is not a linear resistor. Its current-voltage relationship follows the Shockley equation:

I = I_S (e^{V / \eta V_T} - 1)

where $I_S$ is the reverse saturation current, $\eta$ is the ideality factor (typically 1–2), and $V_T = kT/q \approx 26\,\text{mV}$ at room temperature.

Because the $I$ – $V$ curve is exponential, the diode's resistance depends entirely on where you are on that curve. There are two distinct resistances we care about: DC resistance (static) and AC resistance (dynamic).

DC Resistance (Static Resistance)

Definition: The ratio of the DC voltage across the diode to the DC current through it at a given operating point.

R_{DC} = \frac{V}{I}

Why this formula? It's simply Ohm's law applied to the DC values. If you put 0.7 V across a diode and get 10 mA through it, the DC resistance is $0.7 / 0.01 = 70\,\Omega$ . But this number is misleading — it doesn't tell you how the diode responds to a small change in voltage.

Watch out

DC resistance is rarely useful in circuit analysis because diodes are never operated as fixed resistors. It's just a snapshot at one point.

AC Resistance (Dynamic Resistance)

Definition: The slope of the $I$ – $V$ curve at a given operating point — i.e., the ratio of a small change in voltage to the resulting small change in current.

r_d = \frac{dV}{dI}

Why this formula? For small signals (like an AC voltage superimposed on a DC bias), the diode behaves approximately linearly around that bias point. The dynamic resistance is the local slope of the $I$ – $V$ curve.

Now let's derive the actual expression.

Derivation of $r_d = \frac{\eta V_T}{I}$

Start from the Shockley equation. For forward bias where $V \gg V_T$ , the $-1$ term is negligible:

I \approx I_S e^{V / \eta V_T}

Take the derivative with respect to $V$ :

\frac{dI}{dV} = I_S \cdot \frac{1}{\eta V_T} \cdot e^{V / \eta V_T} = \frac{I}{\eta V_T}

The dynamic resistance is the reciprocal:

r_d = \frac{dV}{dI} = \frac{\eta V_T}{I}

r_d = \frac{\eta V_T}{I}

Key insight: The dynamic resistance is inversely proportional to the DC current $I$ . At higher currents, the diode's $I$ – $V$ curve is steeper, so a small voltage change produces a larger current change — meaning lower resistance.

Why This Matters …

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The two ideal diodes are drawn with OPPOSITE polarities: $D_1$ is reverse biased and blocks its branch, while $D_2$ is forward biased and conducts. Only the $R_3$ branch carries current, so the circuit is simply $R_1$ in series with $R_3$ , giving $I = 10/(2+2) = 2.5\,A$ through $R_1$ .

1. Circuit reading. A $10\,V$ battery drives current through $R_1 = 2\,\Omega$ in series. After $R_1$ the circuit splits into two parallel branches: one branch has diode $D_1$ in series with $R_2 = 3\,\Omega$ , the other has diode $D_2$ in series with $R_3 = 2\,\Omega$ . As printed, the two diodes point in OPPOSITE directions relative to the current the battery drives: $D_1$ 's cathode faces the top node, so $D_1$ is reverse biased and blocks its branch, while $D_2$ is oriented to conduct and is forward biased.

2. Which branch conducts. An ideal diode behaves as a perfect conductor (zero resistance) when forward biased and as an open circuit (infinite resistance) when reverse biased. Hence the $D_1$ - $R_2$ branch carries NO current, and the whole current flows through the $D_2$ - $R_3$ branch.

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Q.The given circuit has two ideal diodes connected as shown in figure below. Calculate the current flowing through the resistance R1R_1R1​.

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Q.The given circuit has two ideal diodes connected as shown in figure below. Calculate the current flowing through the resistance $R_1$ .

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