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

Exercise 3.15 asks:

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:

the answer given is:


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.

All advice is appreciated!!!


1 Answer 1


They say find the CHANGE in current, not just I2. So once you have I2, then you figure out what the change is: 0.67 mA - 1 mA = 0.33 mA. So you have done all the right math, you just need to pick out which number is the answer the book is looking for.

  • \$\begingroup\$ oh god, I can beleive I did that.... \$\endgroup\$ Commented Dec 12, 2013 at 11:10
  • \$\begingroup\$ Eh, I think we've all been there, especially late at night... \$\endgroup\$ Commented Dec 12, 2013 at 11:14

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