We owe a debt to Bell Labs with Shockley, Bardeen and Brattain for inventing the semiconductor. All have now passed on. I remember seeing my first transistor radio in the early 60's. Shockley developed the very important diode equation. I was reading through one of my good old engineering books where the dynamic resistance of a diode is derived. I'm trying to understand this because it's basic to many applications for transistors, current mirrors, long tailed pairs, etc. The dynamic resistance is used to get the volt drop over the small signal ac resistance. They give the equation $$I =I_s \left( e^{eV/kT}-1 \right) \tag{1}$$ where $I$ is the diode current, $V$ is the diode voltage, $k$ is Boltzmann's constant, $T$ is the temperature in Kelvin, and $I_s$ is the reverse leakage current.

They then derive the dynamic resistance as $dV/dI$ and get an answer as roughly $1/40I$.

I can't get the same answer that they do. Instead, I have to use this equation from Wikipedia to get the same answer: $$I=I_s \left( e^{V/V_t}-1 \right) \tag{2}$$ where $V_t$ is the thermal volts, which is 25mV at room temperature. Are they using the wrong equation in (1), or have I worked it out wrong?

  • 2
    $\begingroup$ They're the same thing.... en.wikipedia.org/wiki/… $\endgroup$ Jun 10, 2012 at 8:56
  • 2
    $\begingroup$ As Chris says, if you calculate kT/e at 298K it's 25mV i.e. the same as $V_t$ in your second equation, so your two equations are exactly the same. If you get different answers with the two equations it must be a slip of the pen. Can you expand your post to show your working - we might be able to spot the problem. $\endgroup$ Jun 10, 2012 at 19:10
  • $\begingroup$ Hi John, Vt=kT/q =1.38x10^-23x298/1.6x10^-19volts =25mV $\endgroup$
    – johann
    Jun 11, 2012 at 23:53
  • 2
    $\begingroup$ e here is not 2.718..., but the electron charge: 1.602E-19 coulomb. $\endgroup$
    – Art Brown
    Jun 16, 2012 at 6:46
  • 2
    $\begingroup$ OK just to be clear, there are two e's: the one in the exponent is the magnitude of the electron charge (1.602E-19 coulombs) and the one being raised to a power is indeed the familiar 2.718... I can see that could be confusing... Often the charge is written as q instead. $\endgroup$
    – Art Brown
    Jun 16, 2012 at 17:32

1 Answer 1


The dynamic resistance is defined as $r_d = \frac{du}{di}$

from the Shockley equation $I_F = I_R(e^\frac{U_F}{m U_{Th}} -1)$


  • $I_F$ : foreward current
  • $U_F$ : foreward voltage
  • $U_{Th}$ : the thermal voltage $U_{th} = \frac{kT} {e}$
  • m : an adjusting factor to real diodes (m = 1..2) for ideal diodes m = 1

$$\frac{1} {r_d} = \frac{dI_F} {dU_F} = \frac{dI_R(e^\frac{U_F}{m U_{Th}} -1)}{dU_F}=I_R e^\frac{U_F}{m U_{Th}}\frac{1}{m U_{Th}}$$ \If the forward current $I_F$ is much greater than the reverse saturation current $I_R$, then it does not mater whether we add or substract $I_{R}$ from $I_{F}$ therefore $$I_R e^\frac{U_F}{m U_{Th}}\approx I_R e^\frac{U_F}{m U_{Th}}-I_R = I_R (e^\frac{U_F}{m U_{Th}}-1)=I_F$$ $$\frac{1}{r_d} = \frac{I_F} {m U_{Th}}$$ and we finaly get $$r_d = \frac{m U_{Th}} {I_F}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.