Chapter 4 - Diodes

Ideal Diode

The diode has an anode (positive terminal) and a cathode (negative terminal).

The ideal diode may be considered the most fundamental nonlinear circuit element.

• If a negative voltage is applied to the diode, no current flows and the diode behaves as an open circuit.
• If a positive voltage is applied, zero voltage drop appears and the diode behaves as a short circuit.

The positive terminal of the diode is called the anode and the negative terminal the cathode.

i-v characteristic of the Ideal Diode

The ideal diode’s i–v curve is strongly nonlinear, but can be represented by simple piecewise segments. In the idealized picture there are two regions:

• Reverse/zero bias ($V_D \lt 0$): the diode blocks current and behaves like an open circuit.
• Forward conduction ($V_D \gt 0$): the diode conducts and, in the ideal model, behaves like a short circuit.

Each segment is linear by itself, so when the terminal voltage stays within a single segment the diode can be treated as a linear element for the duration of that operating condition. However, if the signal crosses a segment boundary (a “break point”) the operating region changes and a single linear model no longer applies.

For real diodes the break is not perfectly sharp. Practical points to remember:

• A silicon diode typically begins to conduct noticeably near $~0.6–0.8 V$ (the “threshold” or “knee”), and a Schottky diode has a lower threshold.

• Around a chosen DC operating point (bias), small AC perturbations can be analysed with a linear small‑signal model. The diode is represented by its dynamic (differential) resistance $r_d = V_T / I_D$ where $V_T ≈ 25–26 mV$ at room temperature and $I_D$ is the DC bias current.

• The small‑signal linear approximation is valid only while the AC excursion remains small compared to the distances to the nearest breakpoints (e.g., does not drive the diode from blocking to strongly forward conduction).

Terminal Characteristics of Junction Diodes

The most common implementation of the diode involves a pn junction. Its characteristic curve consists of three regions:

1. Forward-bias: $v\gt 0$
2. Reverse-bias: $v\lt 0$
3. Breakdown: $v\lt -V_{BR}$

Forward-bias Region

In the forward-bias region, the i-v relationship is closely approximated by

\[i = I_S(e^{v/V_T}-1)\]

where the saturation current $I_S$ is a constant for a given diode at a given temperature; we recall that $I_S$ is directly proportional to the cross-sectional area of the diode.

We note that $I_S$ is a strong function of temperature; further, that $I_S$ approximately doubles for each $5\degree C$ rise in temperature.

The thermal voltage $V_T = \frac{kT}{q}$, where $k$ is Boltzmann’s constant ($8.62\times 10^{-5} J/K$), $T$ is the absolute temperature in Kelvins ($273 + \degree C$), and $q$ is the magnitude of electronic charge ($1.60\times 10^{-9} \text{ coulombs}$).

In rapid approximate circuit analysis, we shall assume $V_T \approx 25 mV$ at room temperature.

When $i \gg I_S$, we approximate

\[i \approx I_Se^{v/V_T}\]

or alternatively

\[v = V_T \ln \frac{i}{I_S}\]

And we observe:

\[\frac{I_2}{I_1} = e^{(V_2-V_1)/V_T}\]

or

\[V_2 - V_1 = V_T \ln \frac{I_2}{I_1}\]

Reverse-bias Region

In the reverse-bias region, the i-v relationship is closely approximated by

\[i \approx -I_S\]

and while $I_S$ is observed to roughly double for every $5\degree C$ rise in temperature, $-I_S$ is observed to approximately double for every $10\degree C$ rise in temperature.

Breakdown Region

The diode enters the breakdown region when $v \lt -V_{BR}$. $-I_S$ increases rapidly, although the increase in voltage drop is very small.

Diode Models

There are a few models for the diode, among which are the:

• Exponential model
• Constant-Voltage-Drop model
• Ideal-diode model
• Small-signal model

Diode Models

Exponential Model

The exponential model is the most accurate representation of the diode’s behavior and is based on the fundamental physics of the pn junction.

Mathematical Description

The complete exponential model is given by:

\[I_D = I_Se^{V_D/V_T}\]

and

\[I_D = \frac{V_{DD}-V_D}{R}\]

How to Apply the Exponential Model

1. Set up the circuit equation using Kirchhoff’s laws
2. Substitute the exponential relationship for the diode
3. Solve the resulting nonlinear equation (often requires iterative methods or graphical solutions)

Iterative Analysis for Exponential Diode Model

Iterative analysis is used to solve the nonlinear equation that results from combining the exponential diode equation with the circuit equation.

Step-by-Step Iterative Process

1. Set Up the Problem

Given a circuit with:

• Supply voltage: $V_{DD}$
• Series resistance: $R$
• Diode with saturation current: $I_S$

The equations are:

• Circuit equation: $I_D = \frac{V_{DD} - V_D}{R}$
• Diode equation: $I_D = I_S e^{V_D/V_T}$

2. Choose an Iterative Method

Method A: Direct Substitution

Start with an initial guess for $V_D$ and iterate:

1. Initial guess: $V_{D0} = 0.7V$ (typical forward voltage)

2. Calculate current: $I_{D1} = I_S e^{V_{D0}/V_T}$

3. Find new voltage: $V_{D1} = V_{DD} - I_{D1} \cdot R$

4. Repeat: Use $V_{D1}$ as the new guess and continue until convergence

Method B: Newton-Raphson Method

Define the function: $f(V_D) = I_S e^{V_D/V_T} - \frac{V_{DD} - V_D}{R} = 0$

1. Calculate derivative: $f’(V_D) = \frac{I_S}{V_T} e^{V_D/V_T} + \frac{1}{R}$

2. Iterative formula: $V_{D(n+1)} = V_{D(n)} - \frac{f(V_{D(n)})}{f’(V_{D(n)})}$

3. Example Calculation

Given:

• $V_{DD} = 5V$
• $R = 1k\Omega$
• $I_S = 10^{-12}A$
• $V_T = 26mV$ (at room temperature)

Iteration using Method A:

Iteration	$V_D$ (V)	$I_D$ (mA)	$V_{D,new}$ (V)	Error
0	0.700	4.32	0.568	-
1	0.568	0.31	0.691	0.123
2	0.691	3.45	0.576	0.115
3	0.576	0.39	0.688	0.112
4	0.688	3.17	0.579	0.109
…	…	…	…	…

4. Convergence Criteria

Stop iterations when:

• Voltage criterion: $

  V_{D(n+1)} - V_{D(n)}

  < \epsilon_V$ (e.g., $\epsilon_V = 1mV$)

• Current criterion: $

  I_{D(n+1)} - I_{D(n)}

  < \epsilon_I$ (e.g., $\epsilon_I = 0.1mA$)

5. Algorithm Flowchart

START
  ↓
Set initial guess V_D0
  ↓
Calculate I_D = I_S * exp(V_D/V_T)
  ↓
Calculate V_D_new = V_DD - I_D * R
  ↓
Check: |V_D_new - V_D| < tolerance?
  ↓ NO        ↓ YES
V_D = V_D_new  SOLUTION FOUND
  ↓             ↓
Go back to     END
calculate I_D

I-V Curve & Q Point

Load line from $i=\frac{V_{DD}}{R}$ to $v=V_{DD}$. Q point at intersection with diode characteristic, at coordinates $(I_D, V_D)$.

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Constant-Voltage-Drop Model

The constant-voltage-drop model assumes that a conducting diode maintains a constant voltage drop across its terminals, typically 0.7 V for silicon diodes and 0.3 V for Schottky diodes.

Mathematical Description

\[v_D = \begin{cases} V_{on} & \text{if } i > 0 \text{ (forward conducting)} \\ \text{undefined} & \text{if } i = 0 \text{ and } v_D \leq V_{on} \text{ (cutoff)} \end{cases}\]

where $V_{on} = 0.7$ V for silicon diodes.

How to Apply the Constant-Voltage-Drop Model

1. Assume the diode is conducting and set $v_D = 0.7$ V
2. Analyze the circuit using this assumption
3. Check if the resulting current is positive
   • If $i > 0$: assumption is correct
   • If $i \leq 0$: diode is off, set $i = 0$ and solve for $v_D$

Example: Voltage Divider with Diode

Consider a voltage divider with $V_S = 10$ V, $R_1 = 2$ kΩ, $R_2 = 3$ kΩ, and a diode in parallel with $R_2$.

Step 1: Assume diode is ON ($v_D = 0.7$ V)

Step 2: Current through $R_1$: \(i_{R1} = \frac{V_S - v_D}{R_1} = \frac{10 - 0.7}{2000} = 4.65 \text{ mA}\)

Step 3: Current through $R_2$: \(i_{R2} = \frac{v_D}{R_2} = \frac{0.7}{3000} = 0.233 \text{ mA}\)

Step 4: Diode current: \(i_D = i_{R1} - i_{R2} = 4.65 - 0.233 = 4.42 \text{ mA} > 0\)

Since $i_D > 0$, our assumption is correct.

Multiple Diode Circuits

For circuits with multiple diodes, test all possible combinations of ON/OFF states:

Example: Two diodes with different voltage sources

• Test: D1 ON, D2 OFF
• Test: D1 OFF, D2 ON
• Test: D1 ON, D2 ON
• Test: D1 OFF, D2 OFF

Choose the solution where all assumed states are consistent with calculated currents.

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Ideal-Diode Model

The ideal diode model treats the diode as a perfect switch: a short circuit when forward-biased and an open circuit when reverse-biased.

Mathematical Description

\[\begin{cases} v_D = 0, & i \geq 0 \text{ (forward conducting)} \\ i = 0, & v_D < 0 \text{ (reverse blocking)} \end{cases}\]

How to Apply the Ideal-Diode Model

1. Assume diode state (ON or OFF)
2. Analyze the linear circuit with assumption
3. Verify consistency of assumption with results
4. If inconsistent, try opposite state

Example: Peak Detector Circuit

Consider a peak detector with an AC source $v_s = 10\sin(\omega t)$ V, diode, capacitor $C$, and load resistor $R_L$.

Analysis:

• When $v_s > v_C$: diode conducts ($v_D = 0$), capacitor charges to $v_s$
• When $v_s < v_C$: diode blocks ($i_D = 0$), capacitor discharges through $R_L$

Result: Output voltage $v_o$ follows the peak of the input signal.

Example: Rectifier Circuit

Half-wave rectifier with $v_s = V_m\sin(\omega t)$:

• Positive half-cycle ($v_s > 0$): Diode ON, $v_o = v_s$
• Negative half-cycle ($v_s < 0$): Diode OFF, $v_o = 0$

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Small-Signal Model

The small-signal model linearizes the diode around a DC operating point, useful for analyzing small AC variations superimposed on a DC bias.

Mathematical Description

The small-signal resistance (dynamic resistance) is:

\[r_d = \frac{dv}{di}\bigg|_{Q-point} = \frac{V_T}{I_D}\]

where $I_D$ is the DC bias current at the Q-point.

At room temperature: $r_d = \frac{25 \text{ mV}}{I_D}$

How to Apply the Small-Signal Model

1. Find the DC operating point using one of the previous models
2. Calculate the small-signal resistance $r_d = V_T/I_D$
3. Replace the diode with $r_d$ for AC analysis
4. Analyze the linear AC circuit

Example: Diode with AC Source

Circuit: DC source $V_{DC} = 5$ V, AC source $v_{ac} = 0.1\sin(\omega t)$ V, resistor $R = 1$ kΩ, and diode in series.

Step 1: Find DC operating point (using constant-voltage-drop model): \(I_D = \frac{V_{DC} - 0.7}{R} = \frac{5 - 0.7}{1000} = 4.3 \text{ mA}\)

Step 2: Calculate small-signal resistance: \(r_d = \frac{25 \text{ mV}}{4.3 \text{ mA}} = 5.81 \text{ Ω}\)

Step 3: AC analysis - total AC resistance: \(R_{total} = R + r_d = 1000 + 5.81 ≈ 1006 \text{ Ω}\)

Step 4: AC voltage across diode: \(v_{d,ac} = v_{ac} \cdot \frac{r_d}{R + r_d} = 0.1 \cdot \frac{5.81}{1006} ≈ 0.58 \text{ mV}\)

Validity Conditions

The small-signal model is valid when:

• AC variations are small compared to $V_T$ (typically < 5-10 mV)
• The diode remains forward-biased throughout the AC cycle
• Frequency is not too high (parasitic capacitances become significant)

Applications

• Amplitude modulation circuits
• Voltage regulation analysis
• High-frequency small-signal amplifiers
• Mixing circuits in RF applications