# Diode small signal model and exponential model comparison

I'm working my way through the chapter on diodes in Sedra and Smith Fifth Edition

Consider a diode with $n=2$ biased at $1ma$. Find the change in current as a result of changing the voltage by $(a) -20mV (b) -10mv (c)...$ etc. In each case , do the calculations $(i)$ using the small-signal model and $(ii)$ using the exponential model.

so for $(i)$ we find the diode small-signal resistance with: (page 161.)

$r_d = \frac{nV_t}{I_d}$ with $V_t = 25mv$

$r_d = \frac{50mV}{1mA}$

$r_d = 50\Omega$

so then it's simply

$\Delta I = \frac{\Delta V}{50\Omega}=\frac{-20mV}{50\Omega} = -0.40mA$

This agrees with the answer given.

For $(ii)$, the exponential model:

$I = I_S e^{\frac{V}{nV_t}}$

and as per the book, this can be used to get: (page 150.)

$\frac{I_2}{I_1} = e^{(V_2-V_1)/nV_t}$

substituting in the values:

$\frac{I_2}{1mA} = e^{-20mV/50mV}$

$I_2 = 0.67mA$

this does NOT agree with the book:

$0.33mA$

which can be gotten by

$0.67mA - 1mA$

so I think I have made a small mistake, but I have been stuck on this for last two nights so I thought it was time for some help.