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 is the inverse of the slope of the Diode equation evaluated at the Operating point, and it is the fourth diode model — the one 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 relevant slope is the derivative of the Diode equation with respect to $v_D$:

$\frac{d{i}_D}{d{v}_D}=\frac{d}{d{v}_D}\left(I_S{e}^{v_D/V_T}\right)=\frac{I_S{e}^{v_D/V_T}}{V_T}=\frac{i_D}{V_T}$

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 $dv/di$, conductance is $di/dv$):

$\boxed{r_d=\frac{V_T}{I_D}}$

$V_T\approx 25$ 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 $r_e=V_T/I_E$ and lives in the same family as the MOSFET small-signal parameters.

$r_d=V_T/I_D$ depends only on DC bias current and thermal voltage.

Numbers and intuition

At $I_D=1$ mA: $r_d=25\text{mV}/1\text{mA}=25\Omega$. At $I_D=100$ µA: $r_d=25\text{mV}/100\mu \text{A}=250\Omega$.

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 and Small-signal analysis.

Worked example: signal across a single diode

A small AC source $v_s(t)=V_s\sin (\omega t)$ is in series with a 10 k$\Omega$ 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

$I_D=\frac{10-0.7}{10\text{k}\Omega }\approx 0.93\text{mA},r_d=\frac{V_T}{I_D}=\frac{25\text{mV}}{0.93\text{mA}}\approx 26.9\Omega$

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 ⇒ $r_d\approx 26.9\Omega$; a divider sets $v_d\approx 2.68$ mV $\ll V_T$, so the small-signal approximation holds.

This three-step pattern — DC bias, then AC with the device replaced by its small-signal element — is the universal technique of Small-signal analysis, applied here to its simplest possible device.