Diode models and diode circuit analysis

A diode lets current flow easily in one direction and blocks it in the other. That one-way behaviour turns AC into DC, protects circuits from reversed batteries and voltage spikes, steers signals and forms the basis of every transistor. But a real diode is not a perfect switch. It needs a forward voltage of a few tenths of a volt before it conducts much, its current depends exponentially on that voltage, it leaks a little in reverse, it changes with temperature and it takes time to turn off. A designer has to decide, for each job, how much of that behaviour matters, and choose a model accordingly. This lesson builds the four models used in practice and shows how to analyze circuits with each.

At Cedarline, Amara's first task on the pedestrian controller is the 24 V DC input from the cabinet supply. The board must survive a technician connecting the supply backwards, and the tachometer input, which comes from a sensor on a long cable beside traffic, must survive spikes. Both jobs use diodes, and each needs a different level of modelling.

The diode's terminal characteristic

The diode symbol is an arrow pointing in the direction of conventional forward current, from the anode (the p side) to the cathode (the n side), with a bar marking the cathode. On a real part the cathode is marked by a band. The current–voltage curve has three regions. In forward bias, current is negligible until v_D reaches roughly 0.5 V for a small silicon diode, then rises so steeply that over a wide range of currents the voltage stays between about 0.6 and 0.8 V. In reverse bias, the current is a tiny leakage, nanoamps in a small diode at room temperature. Beyond the breakdown voltage the reverse current rises sharply.

In the forward region the Shockley equation from lesson 1 reduces to i_D ≈ I_S e^{v_D/(nV_T)}, because the exponential is huge compared with 1. Taking logarithms gives a useful rule. For two currents I₁ and I₂ at voltages V₁ and V₂,

With n = 1 and V_T = 25 mV, 2.3 n V_T ≈ 60 mV: the forward voltage rises by about 60 mV for every tenfold increase in current. With n = 2 it is about 120 mV per decade. A diode carrying 1 mA at 0.70 V will therefore need about 0.76 V at 10 mA and 0.82 V at 100 mA, ignoring the extra drop across the resistance of the silicon itself, which becomes important at high current.

Temperature shifts the curve. At constant current the forward voltage falls by about 2 mV for each degree Celsius rise. This is large enough to matter: a diode at 0.70 V at 25 °C will sit near 0.58 V at 85 °C. It is also useful, since a forward-biased junction is a cheap temperature sensor, and many chips measure their own die temperature this way.

Model 1: the ideal diode

The simplest model treats the diode as a perfect switch: a short circuit when forward current flows and an open circuit when reverse voltage is applied. It is good enough whenever the circuit voltages are much larger than 0.7 V, for example when checking which way current flows in a 120 V rectifier. To analyze a circuit with ideal diodes, assume a state for each diode, solve the linear circuit, and check the assumption: an on diode must carry current in the forward direction, and an off diode must have zero or reverse voltage across it. If a check fails, change that assumption and solve again.

Model 2: constant voltage drop

The workhorse model replaces a conducting diode with a fixed voltage source V_D, usually 0.7 V for small silicon diodes, 0.3 V for Schottky diodes and about 2 V for a red LED, in the forward direction. An off diode is still an open circuit. This model gives answers within a few percent whenever the circuit voltages are a few volts or more, and it is what experienced designers use for almost all first-pass calculations.

Model 3: piecewise linear

When currents vary widely, the constant-drop model can be refined by adding a series resistance r_D: a conducting diode becomes a source V_D0 in series with r_D. The values are chosen so the straight line fits the real curve over the current range of interest. For a small diode between 1 and 10 mA, V_D0 ≈ 0.65 V and r_D ≈ 10 Ω is typical. For a power rectifier, the datasheet often gives a threshold voltage and slope resistance directly, for example 0.75 V and 25 mΩ. This model is what power-electronics designers use to estimate conduction losses, P = V_D0 I_avg + r_D I_rms².

Model 4: the exponential model and iteration

For accurate answers, especially at low currents or when only a small voltage is available, use the exponential equation itself. A diode in series with a resistor R and a source V_DD obeys two equations at once: the diode law i_D = I_S e^{v_D/(nV_T)} and Kirchhoff's voltage law i_D = (V_DD − v_D)/R. These cannot be solved in closed form, so we iterate. Guess v_D, compute i_D from the resistor, compute a new v_D from the diode law, and repeat. Two or three passes are usually enough. Newton's method converges even faster and is what circuit simulators use internally.

The same two equations can be solved graphically. The resistor equation is a straight line on the i_D–v_D plane, the load line, crossing the voltage axis at V_DD and the current axis at V_DD/R. The operating point, or quiescent point (Q-point), is where the load line meets the diode curve. The load line idea returns in lesson 6, where it shows how a transistor amplifier swings.

import math
Is, n, VT = 2.0e-14, 1.0, 0.02585     # small-signal silicon diode
VDD, R = 5.0, 1000.0

# Newton's method on f(v) = Is*(exp(v/nVT) - 1) - (VDD - v)/R
v = 0.6
for step in range(6):
    f = Is * (math.exp(v / (n * VT)) - 1) - (VDD - v) / R
    df = Is / (n * VT) * math.exp(v / (n * VT)) + 1 / R
    v -= f / df
    print(f"step {step}: vD = {v:.5f} V")
i = (VDD - v) / R
print(f"exponential model: vD = {v:.3f} V, iD = {i*1000:.3f} mA")
print(f"constant drop 0.7 V: iD = {(VDD - 0.7)/R*1000:.3f} mA")
print(f"ideal diode:        iD = {VDD/R*1000:.3f} mA")

step 0: vD = 1.00347 V
step 1: vD = 0.97762 V
step 2: vD = 0.95177 V
step 3: vD = 0.92592 V
step 4: vD = 0.90007 V
step 5: vD = 0.87422 V
exponential model: vD = 0.874 V, iD = 4.126 mA
constant drop 0.7 V: iD = 4.300 mA
ideal diode:        iD = 5.000 mA

The comparison is the lesson. With a 5 V source, the constant-drop model is almost perfect and the ideal model is noticeably off. With a 1 V source, the constant-drop model would be badly wrong, because 0.7 V is most of the supply, and only the exponential model gives a sensible answer.

Small-signal resistance

Often a diode carries a steady DC current with a small AC signal on top, for example in a radio detector or a bias network. For signals small enough that the curve looks straight, the diode behaves like a resistor whose value is the inverse slope of the curve at the Q-point:

At 1 mA with n = 1, r_d ≈ 25 Ω; at 10 µA it is 2.5 kΩ. The resistance depends on the DC current, which means a diode can act as a current-controlled resistor, the idea behind some attenuators and automatic gain controls. Lesson 10 uses the same reasoning for transistors, where it is the foundation of all amplifier design. The rule for small-signal analysis is always the same: solve the DC circuit first to find the Q-point, then replace each nonlinear device with its small-signal model and analyze the AC circuit with DC sources set to zero.

Clamping a sensor input

The tachometer's shaft sensor delivers pulses of 0 to 12 V on a long cable, but the counter logic runs from 3.3 V and its input must stay between −0.3 V and 3.6 V. Daniel's standard input circuit uses a series resistor and two clamp diodes: one from the input node to the 3.3 V rail and one from ground to the input node. When the input tries to rise above 3.3 V plus one diode drop, the upper diode conducts and dumps the excess current into the rail; when it tries to go below ground by a diode drop, the lower diode conducts. The series resistor limits the current to a value the diodes and the rail can absorb.

That last check is a classic trap. Clamp current flowing into a rail has to go somewhere; if the board draws less current than is being injected, the rail can be pushed up. Some designs clamp to a Zener diode or a dedicated protection device instead of the logic supply for exactly this reason.

Switching behaviour and datasheet ratings

A conducting junction diode stores minority charge, and before it can block reverse voltage that charge must be removed. When the voltage reverses, a reverse current flows briefly, the reverse recovery current, until the charge is gone. Ordinary rectifiers take several microseconds; fast-recovery types take tens to hundreds of nanoseconds. Schottky diodes, which use a metal–semiconductor junction and carry current by majority carriers only, have almost no stored charge and switch very fast, at the cost of higher leakage and lower reverse voltage ratings.

The rule in the last row of the table deserves repeating, because it applies to every component in this course. Typical values describe the middle of production; a circuit that works only with typical parts will fail in the field. Cedarline's design reviews ask for every calculation to be done with worst-case limits at the coldest and hottest operating temperatures, which for traffic equipment is commonly −40 °C to +74 °C inside the cabinet.