Diode Small-Signal Resistance Calculator
Find the dynamic (AC) resistance of a diode at its DC operating point: rd = n·VT / ID.
Inputs
Drag the handle (or focus it and press ↑/↓) to reorder. Scientific notation such as 1.2e-3 is accepted. Pick any unit; conversion is automatic.
Results
Enter values and press Calculate. Your answer will appear here.
rd vs. diode current (log–log)
On log–log axes the small-signal resistance follows a straight line of slope −1 while the junction dominates: ten times the current gives one tenth of rd. Series resistance bends the curve flat at high current.
What is being calculated
Common examples
| Diode current | r_d, n = 1 | r_d, n = 2 |
|---|
At 25 °C with no series resistance. Illustrative values, not manufacturer data.
Typical diode parameters (approximate)
| Device | Ideality factor n | r_d at 1 mA, 25 °C |
|---|---|---|
| 1N4148 silicon signal diode | 1.8 – 2.0 | ≈ 46 – 51 Ω |
| 1N4007 silicon rectifier | 1.7 – 2.0 | ≈ 44 – 51 Ω |
| 1N5817 Schottky rectifier | 1.05 – 1.2 | ≈ 27 – 31 Ω |
| Red LED, 5 mm | 1.8 – 2.2 | ≈ 46 – 57 Ω |
| Blue / white LED | 2.5 – 4.0 | ≈ 64 – 103 Ω |
n and IS vary strongly with temperature, current level and process. Extract them from a measured I–V curve for precision.
Key relations and unit conversions
| Quantity | Relation |
|---|---|
| Thermal voltage | V_T = kT/q = 25.69 mV at 25 °C · 25.85 mV at 27 °C · 26.99 mV at 40 °C |
| Small-signal resistance | r_d = n·V_T / I_D · r_d ≈ 25.7 mV / I_D for n = 1 at 25 °C |
| Small-signal conductance | g_d = 1 / r_d = (I_D + I_S) / (n·V_T) |
| With series resistance | r_d,total = r_s + n·V_T / (I_D + I_S) |
| DC resistance | R_DC = V_D / I_D (always larger than r_d) |
| Shockley equation | I_D = I_S · (exp(V_D / (n·V_T)) − 1) |
| Current units | 1 A = 10³ mA = 10⁶ µA = 10⁹ nA = 10¹² pA |
Formula
- rd: small-signal (dynamic) resistance, Ω
- n: ideality factor, dimensionless (1 ideal, 1.5–2 silicon, 2–3 LEDs)
- VT: thermal voltage, ≈ 25.69 mV at 25 °C
- ID: DC bias current through the diode, A
- k = 1.380649 × 10⁻²³ J/K, q = 1.602176634 × 10⁻¹⁹ C
Example: at 25 °C, VT = 25.69 mV. With n = 1 and ID = 1 mA, rd = 25.69 mV / 1 mA ≈ 25.7 Ω. Raising the current to 10 mA drops rd to ≈ 2.57 Ω; lowering it to 100 µA raises rd to ≈ 257 Ω.
Advanced mode also solves the Shockley equation for ID from a given VD, adds a series resistance rs, and reports the junction-only resistance, the DC resistance VD/ID, the conductance and the power dissipation. Because rd is a derivative, it describes the diode only for signals small enough that the operating point does not move appreciably.
References
- Wikipedia, Shockley diode equation: ID = IS(e^(V_D/nV_T) − 1)
- Wikipedia, Diode modelling: small-signal and large-signal models
- Wikipedia, Thermal voltage: VT = kT/q and its value at room temperature
- NIST SP 811, Guide for the Use of the SI: unit conversion factors
How to Use This Tool
- Pick a preset, or type the DC bias current ID and choose its unit.
- Set the ideality factor n and the junction temperature T (these set VT).
- In Advanced, switch to voltage input, add IS and rs, or change the resistance unit.
- Click Calculate.
- Read rd in the highlighted result and check the log–log chart.
What This Tool Offers
- Basic and Advanced modes on one calculation engine
- Current-driven and voltage-driven (full Shockley) operating points
- Ideality factor, junction temperature and thermal voltage
- Optional reverse saturation current and series resistance
- Junction resistance, DC resistance, conductance and power dissipation
- Log–log rd vs. ID chart and labelled I–V tangent diagram
- Ready-made presets and a mobile-friendly layout
What it means
The small‑signal resistance of a forward‑biased diode is r_d = V_T / I_D, where V_T is the thermal voltage (≈25mV at room temperature) and I_D is the DC bias current. This is the dynamic resistance for AC signals superimposed on a DC bias. It is the slope of the diode I‑V curve at the operating point. The small‑signal resistance is used in AC analysis of diode circuits, such as rectifiers, clippers, and detectors. For a given bias current, the resistance is inversely proportional to the current; higher current gives lower resistance. Example: At I_D = 1mA, r_d = 25mV/1mA = 25Ω. At I_D = 10mA, r_d = 25mV/10mA = 2.5Ω. This small resistance affects the voltage gain of diode‑based circuits.
Worked example
Diode Small‑Signal Resistance – Practical Example
Real‑World| Parameter | Value |
|---|---|
| VT | 26 mV |
| ID | 1 mA = 1×10⁻³ A |
| Formula | rd = VT / ID |
Common mistakes
- Diode current I_D: The DC bias current through the diode.
- V_T: Thermal voltage (~25 mV).
- Small‑signal: This is the AC resistance at the operating point.
- Dynamic resistance: r_d = dV/dI – valid for small signal variations.
- Reverse bias: For reverse bias, r_d is very large (leakage).
Applications
Diode small‑signal resistance r_d = V_T / I_D is the AC resistance of a forward‑biased diode at a given DC current. It is used to model the diode's dynamic behaviour in amplifiers and rectifiers. Engineers use it to calculate the voltage gain of diode‑loaded stages, to design clipper and clamper circuits, and to analyse rectifier circuits. The resistance decreases with increasing current, which is important for linearity in mixers and detectors. By understanding this concept, professionals can accurately predict the AC response of diode circuits. This formula is essential for analog and RF circuit design.
- Rectifier and clipper circuit analysis
- Mixer and detector circuit design
- Voltage‑controlled resistance applications (PIN diodes)
- Temperature stabilisation in bias circuits
- Educational understanding of diode dynamic resistance
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