Diode small-signal resistance

The diode small-signal resistance $r_{d}$ is the effective linear resistance a forward-biased Diode presents to a small time-varying signal riding on top of a DC bias. It’s the inverse of the slope of the Diode equation evaluated at the Operating point: the fourth diode model, used once the DC bias is known and you want to know what the signal does.

Derivation

The diode is non-linear, so it has no single “resistance”. But over a small enough voltage swing the exponential $i$ – $v$ curve looks locally straight, and a straight $i$ – $v$ line is a resistor. The slope we want is the derivative of the Diode equation with respect to $v_{D}$ :

$\frac{d i _{D}}{d v _{D}} = \frac{d}{d v _{D}} (I_{S} e^{v_{D} / V_{T}}) = \frac{I _{S} e ^{v_{D} / V_{T}}}{V _{T}} = \frac{i _{D}}{V _{T}}$

The last step uses $i_{D} = I_{S} e^{v_{D} / V_{T}}$ itself, which is the trick: the slope is just the current divided by the Thermal voltage. Evaluate this at the DC operating point, where $i_{D} = I_{D}$ . The small-signal resistance is the inverse of that slope (resistance is $d v / d i$ , conductance is $d i / d v$ ):

$r_{d} = \frac{V _{T}}{I _{D}}$

$V_{T} \approx 25$ mV is the Thermal voltage and $I_{D}$ is the DC bias current set by the Operating point. Notice $r_{d}$ depends only on the bias current and temperature, not on $I_{S}$ or the device area. (Strictly $r_{d} = n V_{T} / I_{D}$ with the diode ideality factor $n$ ; the $n = 1$ convention used throughout this course reduces it to $V_{T} / I_{D}$ .) This is the diode analogue of the BJT’s $r_{e} = V_{T} / I_{E}$ .

$r_{d} = V_{T} / I_{D}$ depends only on DC bias current and thermal voltage.

Numbers and intuition

At $I_{D} = 1$ mA: $r_{d} = 25 mV /1 mA = 25 Ω$ . At $I_{D} = 100$ µA: $r_{d} = 25 mV /100 μ A = 250 Ω$ .

The smaller the bias current, the larger $r_{d}$ . Intuition: at high $I_{D}$ the exponential is rising steeply, so a tiny voltage wiggle causes a big current wiggle, low resistance. At low $I_{D}$ the curve is flatter, so the same voltage wiggle barely moves the current, high resistance. The validity condition is that the signal swing $v_{d}$ must be much smaller than $V_{T}$ (~25 mV), because the linear approximation comes from truncating the Taylor expansion of the exponential. See Linearisation around an operating point.

Worked example: signal across a single diode

A small AC source $v_{s} (t) = V_{s} sin (ω t)$ is in series with a 10 k $Ω$ resistor and a forward-biased diode powered by a $V^{+} = 10$ V supply. Find the AC voltage across the diode.

Step 1 — DC (find the bias). Zero the AC source. KVL: $V^{+} = I_{D} R + V_{D}$ . Using the Constant-voltage-drop model for the DC part, $V_{D} \approx 0.7$ V, so

$I_{D} = \frac{10 - 0.7}{10 k Ω} \approx 0.93 mA, r_{d} = \frac{V _{T}}{I _{D}} = \frac{25 mV}{0.93 mA} \approx 26.9 Ω$

Step 2 — AC (use $r_{d}$ ). Zero the DC supply (short it). The diode becomes the resistor $r_{d}$ . The signal source now sees a simple voltage divider made of $R$ in series with $r_{d}$ :

$v_{d} = v_{s} \cdot \frac{r _{d}}{R + r _{d}} \approx v_{s} \cdot \frac{26.9}{10000 + 26.9} \approx v_{s} \times 0.00268$

So a small input signal produces a tiny AC voltage across the diode, attenuated by $r_{d} / (R + r_{d})$ . If $v_{s}$ has, say, a 1 V amplitude, then $v_{d} \approx 2.68$ mV, comfortably less than $V_{T} = 25$ mV, so the small-signal approximation was justified.

With $V^{+} = 10$ V, $R = 10$ kΩ, $I_{D} \approx 0.93$ mA ⇒ $r_{d} \approx 26.9 Ω$ ; a divider sets $v_{d} \approx 2.68$ mV $≪ V_{T}$ , so the small-signal approximation holds.

DC bias first, then AC with the device replaced by its small-signal element: that’s the general Small-signal analysis procedure, here on its simplest device.

Idriss Rami — Notes

Explorer

Diode small-signal resistance

Derivation

Numbers and intuition

Worked example: signal across a single diode

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