How Does a Diode's Small-Signal Resistance Change With Bias Current?
It falls in inverse proportion to the bias current: rd = nVT ÷ ID, so ten times the current gives one tenth the resistance. For an ideal diode (n = 1) at 300 K, where VT = 25.85 mV, that is 258.5 Ω at 0.1 mA, 25.85 Ω at 1 mA and 2.585 Ω at 10 mA.
What you'll learn
- Where rd = nVT ÷ ID comes from, in two lines of calculus
- Which thermal voltage to use, and what the ideality factor n changes
- The small-signal resistance at 0.1 mA, 1 mA and 10 mA, read off the diode curve
- A worked example: a 5 V supply, 4.3 kΩ and a 10 mV signal
- How small the signal has to be before the model's error matters
Where does rd = nVT ÷ ID come from?
Start from the Shockley equation, ID = Is(e^(VD ÷ nVT) − 1), with ID the forward current from anode to cathode, Is the saturation current, n the ideality factor and VT = kT ÷ q the thermal voltage. The small-signal resistance is the slope of the voltage against the current at the bias point, rd = dVD ÷ dID, which is the inverse of dID ÷ dVD. Differentiating gives dID ÷ dVD = (Is ÷ nVT)e^(VD ÷ nVT) = (ID + Is) ÷ nVT. Under forward bias ID is many orders of magnitude larger than Is, so rd = nVT ÷ ID. Nothing about the diode's material or size survives except n: in this model, which ignores the small series resistance of the diode's bulk and contacts, two quite different silicon diodes biased at the same current have the same rd if their n is the same.
Which value of VT should you use?
With the exact SI values of Boltzmann's constant and the electron charge, VT = kT ÷ q is 25.85 mV at 300 K, 25.87 mV at 27 °C (300.15 K) and 24.99 mV at 290 K. Textbooks round this to 25 mV or 26 mV, and GATE questions usually state the value to use. If a question gives VT, use it; if it gives only the temperature, compute kT ÷ q. This post uses 25.85 mV throughout.
What does the ideality factor change?
The ideality factor n is 1 for an ideal junction where diffusion current dominates, and closer to 2 when recombination in the depletion region dominates. Since rd = nVT ÷ ID, it scales rd directly: the same diode model with n = 2 has twice the small-signal resistance at every bias current. Check whether a question states n; if it says nothing, n = 1 is the usual assumption.
Small-signal resistance at three bias points
| Bias current ID | rd with n = 1 | rd with n = 2 |
|---|---|---|
| 0.1 mA | 258.5 Ω | 517.0 Ω |
| 1 mA | 25.85 Ω | 51.70 Ω |
| 10 mA | 2.585 Ω | 5.170 Ω |
Read the picture as two separate facts. The tangent at each point is steeper than the one before, because the slope ID ÷ nVT grows with the current. Meanwhile the voltage hardly moves: each tenfold rise in current adds only nVT × ln 10 to VD, which is 59.5 mV for n = 1. That is why a fixed 0.7 V drop works well for the DC operating point, while the small-signal resistance still changes by a factor of 100 across the same range.
Worked example: a 5 V supply, 4.3 kΩ and a 10 mV signal
- DC analysis first, with the signal set to zero and the constant-drop model VD = 0.7 V: ID = (5 − 0.7) ÷ 4.3 kΩ = 1.00 mA.
- Small-signal resistance at that point, with n = 1 and VT = 25.85 mV: rd = 25.85 mV ÷ 1 mA = 25.85 Ω.
- Small-signal circuit: the DC supply becomes a short and the diode becomes rd, leaving the 10 mV source driving 4.3 kΩ in series with 25.85 Ω.
- Voltage divider: vd = 10 mV × 25.85 ÷ (4300 + 25.85) = 0.0598 mV peak.
- Signal current: id = 10 mV ÷ 4325.85 Ω = 2.31 µA peak, riding on the 1 mA DC current.
How small does the signal have to be?
The model replaces an exponential with a straight line, so it is only accurate when the signal voltage across the diode is small compared with nVT. At 1 mA bias (n = 1) the exact exponential and the straight-line estimate compare like this. A +1 mV swing in VD gives an exact current change of 0.0394 mA against 0.0387 mA from rd, about 2% low. A +5 mV swing gives 0.2134 mA against 0.1934 mA, about 9% low, and −5 mV gives −0.1759 mA against −0.1934 mA, about 10% too large. At +10 mV the straight line underestimates by 18%. The positive and negative halves come out unequal, which is the start of distortion. A common rule is to keep the diode's signal voltage below about a fifth of VT, around 5 mV; in the worked example it is only 0.06 mV, so the model is excellent there.
Common mistakes
- Using rd = VD ÷ ID. That is the DC (static) resistance, 700 Ω at 0.7 V and 1 mA, not the slope.
- Taking the bias current from the wrong circuit: rd needs the DC current, found with the signal switched off.
- Forgetting n when the question gives it. With n = 2 every rd doubles.
- Mixing 25 mV, 25.85 mV and 26 mV in one answer. Use the value the question gives, or kT ÷ q at its temperature.
- Applying the model to a large signal. Once the diode's signal voltage approaches VT the exponential takes over.
Check your understanding
- Q1. What is rd for an ideal diode (n = 1) biased at 2.5 mA, with VT = 25.85 mV? Answer: 25.85 ÷ 2.5 = 10.34 Ω.
- Q2. A diode with n = 2 must present rd = 50 Ω at 300 K. What bias current does it need? Answer: ID = 2 × 25.85 mV ÷ 50 Ω = 1.034 mA.
- Q3. The current through an ideal diode rises from 1 mA to 10 mA. By how much does VD rise? Answer: VT × ln 10 = 25.85 mV × 2.303 = 59.5 mV.
Related
Official source
Last verified
The GATE 2027 EC syllabus lists the P-N junction under Section 3, Electronic Devices, and diode circuits and small-signal analysis under Section 4, Analog Circuits (checked against the official PDF from IIT Madras on 10 October 2026). Every number here was recomputed in code from the exact SI constants.