FOUNDATION · CHAPTER 1 OF 2

Nonlinear elements: operating points and linearisation

How to solve a diode circuit, when a straight-line diode model will do, and why a small signal on a bias sees the diode as a resistor set by its bias current. The diode and amplifier circuits that follow rest on these ideas.

Updated 13 Oct 2026

What you will be able to do

You will solve a one-diode circuit by a table and by an iteration. You will judge when a straight-line diode model is good enough, and measure its error. You will get the diode's small-signal resistance from the slope of its law. You will say how small a signal must be to add it to a bias.

Before you start

Exponentials and natural logarithms: $e^{a + b} = e^{a} e^{b}$ , $ln (ab) = ln a + ln b$
Differentiation of $e^{x / a}$ and of $ln x$ , with the chain rule
Linear approximation near a point: $f (x) \approx f (a) + f^{'} (a) (x - a)$
The series $e^{x} = 1 + x + x^{2} /2 + \dots$
The identity $sin^{2} θ = (1 - cos 2 θ) /2$
Ohm's law and Kirchhoff's voltage law

One diode and one resistor: two ways to solve

A diode conducts easily from anode to cathode; its current is an exponential function of its voltage (taken as given). To solve a diode circuit, find the voltage where its current equals what the rest of the circuit allows. $V_{T} = k T / q$ is the thermal voltage ( $k$ : Boltzmann's constant, J/K; $T$ : temperature, K; $q$ : electron charge, C), taken as 25 mV throughout (at 300 K it is about 25.9 mV). $n = 1$ unless stated.

DIODE LAW

I_{D} = I_{S} (e^{V_{D} / (n V_{T})} - 1)

$V_{D}$ : anode minus cathode voltage (V). $I_{D}$ : anode-to-cathode current (A). $I_{S}$ : saturation current (A). $n$ : ideality factor (no unit, usually 1 to 2).

DIODE VOLTAGE FROM ITS CURRENT

V_{D} = n V_{T} ln \frac{I _{D}}{I _{S}}

Holds when $I_{D} \geq 100 I_{S}$ : dropping the $- 1$ then changes the current by at most 1 %. Take the natural logarithm of $I_{D} \approx I_{S} e^{V_{D} / (n V_{T})}$ . Result in V.

WORKED EXAMPLE

A diode has $I_{S} = 1 0^{- 14}$ A and $n = 1$ . What voltage does it need to carry 1 mA?

Solution

$I_{D} / I_{S} = 1 0^{- 3} /1 0^{- 14} = 1 0^{11}$ , far above 100, so the $- 1$ may be dropped.
$ln 1 0^{11} = 11 \times 2.3026 = 25.33$ , so $V_{D} = 25 mV \times 25.33 = 0.633$ V.

Answer: $V_{D} \approx 0.633$ V.

The circuit used from here, as a netlist (each element and the two nodes it joins). VCC is the supply node, out the output node; DC means constant.

Element	From	To	Value
DC source $V_{S}$	VCC	ground	5 V
Resistor $R$	VCC	out	1 kΩ
Diode (anode at out)	out	ground	$I_{S} = 1 0^{- 14}$ A, $n = 1$

LOAD LINE

I_{D} = \frac{V _{S} - V _{D}}{R}

Kirchhoff's voltage law around the loop. $V_{S}$ in V, $R$ in Ω, $I_{D}$ in A. $I_{D}$ must also satisfy the diode law.

$V_{D}$ (V)	Law current (mA)	Resistor current (mA)	Larger
0.60	0.265	4.400	resistor
0.66	2.920	4.340	resistor
0.67	4.356	4.330	law

Try voltages and see which current is larger. The law's current rises with $V_{D}$ and the resistor's falls, so they cross once: here between 0.66 V and 0.67 V. More trials close the gap slowly; an iteration is faster. Repeat a rule $V_{k + 1} = g (V_{k})$ from a guess until $V$ stops changing, at a fixed point.

FIXED-POINT ITERATION

V_{k + 1} = g (V_{k}) = n V_{T} ln \frac{V _{S} - V _{k}}{R I _{S}}

Valid while $V_{k} < V_{S}$ and each step's current is at least $100 I_{S}$ . The load line gives the current at $V_{k}$ ; the law turns it into the next voltage.

Step $k$	$V_{k}$ (V)	Resistor current (mA)	$V_{k + 1}$ (V)
0	0.7000	4.3000	0.6697
1	0.6697	4.3303	0.6699
2	0.6699	4.3301	0.6699

Why it settles fast: near the answer $V^{*}$ , where $g (V^{*}) = V^{*}$ , the linear approximation gives $V_{k + 1} - V^{*} \approx g^{'} (V^{*}) (V_{k} - V^{*})$ . With $g^{'} (V) = - n V_{T} / (V_{S} - V)$ and $V_{S} - V^{*} = R I_{D}$ , each step multiplies the error by $- n V_{T} / (R I_{D})$ : the guesses alternate about the answer and close in when the factor's size is below 1. Here it is $25 mV / (1 kΩ \times 4.33 mA) = 0.0058$ . So $V_{D} = 0.670$ V, $I_{D} = 4.33$ mA.

WATCH OUT FOR

Dropping the $- 1$ unchecked. Test: after solving, confirm $I_{D} \geq 100 I_{S}$ ; if not, solve with the full law.

Straight-line models and the slope of the law

A curve can be replaced by straight pieces. Lowering $V_{D}$ by $n V_{T} ln 10 \approx 58$ mV divides the law's current by 10. In the circuit above, 0.60 V gives 0.265 mA, 6 % of the 4.4 mA the resistor allows. 58 mV lower, at about 0.542 V, the law gives about 0.026 mA, under 1 % of the 4.46 mA the resistor would then allow. So below a cut-in voltage $V_{γ}$ the diode is taken as OFF, above it ON, on a straight line. ON needs $I_{D} \geq 0$ ; OFF needs $V_{D} < V_{γ}$ . $V_{γ}$ is where the ON line meets $I_{D} = 0$ :

Ideal ( $V_{γ} = 0$ ): ON means $V_{D} = 0$ .
$V_{γ}$ battery: ON means $V_{D} = V_{γ}$ , often 0.7 V.
$V_{γ}$ plus $r_{f}$ : ON means $V_{D} = V_{γ} + r_{f} I_{D}$ , with $r_{f}$ a small series resistance (Ω).
In every model, OFF means $I_{D} = 0$ .

DYNAMIC RESISTANCE

\frac{1}{r _{d}} r_{d} = \frac{d I _{D}}{d V _{D}} = \frac{I _{S}}{n V _{T}} e^{V_{D} / (n V_{T})} = \frac{I _{D} + I _{S}}{n V _{T}} \approx \frac{I _{D}}{n V _{T}} = \frac{n V _{T}}{I _{D}}

Valid where $I_{D} ≫ I_{S}$ . The law's slope at the bias point: the circuit's solved steady current and voltage. $r_{d}$ in Ω, $n V_{T}$ in V, $I_{D}$ in A. At 4.33 mA, $r_{d} = 5.77$ Ω.

TANGENT-LINE MODEL

V_{D} = V_{γ} + r_{f} I_{D}, r_{f} = r_{d}, V_{γ} = V_{D 0} - n V_{T}

ON only, near the bias point $I_{D 0}$ (A), $V_{D 0}$ (V): here 4.33 mA, 0.670 V. The tangent there is $V_{D} = V_{D 0} + r_{d} (I_{D} - I_{D 0})$ ; as $r_{d} I_{D 0} = n V_{T}$ , it meets $I_{D} = 0$ at $V_{D 0} - n V_{T} = 0.645$ V. Here $r_{f} = 5.8$ Ω.

Each model against the law, $R = 1$ kΩ. The ideal model is within 5 % only at 20 V. The 0.7 V battery is within 1 % at 5 V and 20 V but 23 % low at 1 V. The tangent model is exact at its own point, 5 V, and drifts away from it.

Model	$V_{S} = 1$ V	$V_{S} = 5$ V	$V_{S} = 20$ V
Law (by iteration)	0.390 mA	4.330 mA	19.293 mA
Ideal	1.000 mA (+156 %)	5.000 mA (+15.5 %)	20.000 mA (+3.7 %)
$V_{γ} = 0.7$ V	0.300 mA (-23.1 %)	4.300 mA (-0.7 %)	19.300 mA (+0.04 %)
$V_{γ} = 0.645$ V, $r_{f} = 5.8$ Ω	0.353 mA (-9.6 %)	4.330 mA (0.0 %)	19.243 mA (-0.3 %)

WATCH OUT FOR

Reading $r_{d}$ as $V_{D} / I_{D}$ . Test: at 0.67 V and 4.33 mA, $V_{D} / I_{D} = 155$ Ω, but the slope gives $r_{d} = 5.8$ Ω.

Small signals on a bias

In amplifiers a diode's voltage is often a bias, a steady DC voltage $V$ with current $I$ , plus a small time-varying signal $v$ . The usual method solves each alone, with the diode as $r_{d}$ for the signal, and adds the answers. The expansion below measures its error.

EXPANSION ABOUT THE BIAS POINT

i_{D} \approx I e^{x} = I (1 + x + \frac{x ^{2}}{2} + \dots), x = \frac{v}{n V _{T}}

Holds with $I ≫ I_{S}$ , the $- 1$ dropped: $i_{D} = I_{S} e^{(V + v) / (n V_{T})} = I e^{x}$ , then the series. $I = I_{S} e^{V / (n V_{T})}$ : bias current (A); $v$ : signal across the diode (V). Terms: bias current, $I x = v / r_{d}$ , then the error.

SIZE OF THE SECOND-ORDER TERM

\frac{second-order term}{first-order term} = \frac{x}{2} = \frac{v}{2 n V _{T}}

Divide the third term by the second; the $r_{d}$ model holds when this is much less than 1 in size. For $n = 1$ , a 1 % error needs $∣ v ∣ \leq 0.02 \times 25 mV = 0.5$ mV, at any bias current.

Why adding fails for large signals: with $v = v_{1} + v_{2}$ , the term $x^{2} /2$ holds $v_{1} v_{2} / (n V_{T})^{2}$ . So the response to the sum misses the sum of the separate responses by about $I v_{1} v_{2} / (n V_{T})^{2}$ , unless the squared term is negligible.

WORKED EXAMPLE

A diode with $n = 1$ has a 1 mA bias, and a 5 mV peak sine is added to its voltage. Find $r_{d}$ , the signal current and the size of the second-order term.

Solution

$r_{d} = n V_{T} / I = 25 mV /1 mA = 25$ Ω, set by the bias current.
The signal current peaks at $v / r_{d} = 5 mV /25 Ω = 0.2$ mA.
The peak of $x = v / (n V_{T})$ is $x_{m} = 5/25 = 0.2$ , so the second-order term is $x_{m} /2 = 10$ % of the first there.
With $v = v_{m} sin ω t$ and $sin^{2} ω t = (1 - cos 2 ω t) /2$ , the term $I x^{2} /2$ becomes $I x_{m}^{2} (1 - cos 2 ω t) /4$ .
Its amplitude is $1 mA \times 0.04/4 = 10 μ$ A at twice the signal frequency, plus a 10 µA DC shift: 5 % of the signal current.

Answer: $r_{d} = 25$ Ω; 0.2 mA peak; 10 % at the peak: a 10 µA second harmonic and a 10 µA DC shift.

WATCH OUT FOR

Testing the source, not the diode. Test: through a series $R$ , a source $v_{s}$ puts $v_{s} r_{d} / (R + r_{d})$ across the diode. Apply the condition to that.

Check yourself

CHECK YOURSELF 1

In the fixed-point iteration, each step multiplies the error by about $n V_{T} / (R I_{D})$ in size. In terms of the dynamic resistance $r_{d}$ , this factor is:

(A)
$r_{d} / R$
(B)
$R / r_{d}$
(C)
$r_{d} R$
(D)
$I_{S} / I_{D}$

Answer and solution

Answer: (A)

The dynamic resistance is $r_{d} = n V_{T} / I_{D}$ .
So $n V_{T} / (R I_{D}) = r_{d} / R$ .

CHECK YOURSELF 2

A diode with $n = 1.5$ carries 2.5 mA DC. Take $V_{T} = 25$ mV. Find its dynamic resistance in ohms.

Answer and solution

Answer: $15$ (accepted range $\pm 0.1$ )

$n V_{T} = 1.5 \times 25 mV = 37.5$ mV.
$r_{d} = n V_{T} / I_{D} = 37.5 mV /2.5 mA = 15$ Ω.

CHECK YOURSELF 3

A diode with $n = 1$ carries a DC bias far above $I_{S}$ . Take $V_{T} = 25$ mV. What is the largest peak signal voltage across it, in mV, for which the second-order term is at most 4 % of the first-order term?

Answer and solution

Answer: $2$ (accepted range $\pm 0.02$ )

The ratio of the second-order to the first-order term is $v / (2 n V_{T})$ .
Require $v / (2 \times 25 mV) \leq 0.04$ .
So $v \leq 0.08 \times 25 mV = 2$ mV.

CHECK YOURSELF 4One or more options are correct

A diode is biased with $I_{D} ≫ I_{S}$ and a small voltage $v$ is added. Which statements are correct?

(A)
The error of the $r_{d}$ model is measured by $v / (2 n V_{T})$ .
(B)
The condition on $v$ gets easier to meet as the bias current rises.
(C)
Doubling the bias current halves $r_{d}$ .
(D)
Adding the bias and signal results is exact for any $v$ , as Kirchhoff's law is linear.

Answer and solution

Answer: (A), (C)

A is true: it is the ratio of the second-order to the first-order term.
B is false: the ratio does not depend on the bias current.
C is true: $r_{d} = n V_{T} / I_{D}$ .
D is false: the diode law is not linear, so the squared term is present.

Nonlinear elements: operating points and linearisation

What you will be able to do

Before you start

One diode and one resistor: two ways to solve

Straight-line models and the slope of the law

Small signals on a bias

Check yourself

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