Assume that the \$C-V\$ curve of a Silicon diode is the following

enter image description here

The equivlant \$\dfrac{1}{C^2}=f(V)\$ is shown next

enter image description here

The slope will give me \$slope=\dfrac{2}{\epsilon q N_d}\Rightarrow N_d=2.50\times 10^{10}\$ while the interception will give me the built-in pottential \$\Phi_i:\Phi_i=-\dfrac{\text{intercept}}{\text{slope}}\Rightarrow \Phi_i=0.8V\$.

Are those values reasonable? I am asking because I don't know anything about the diode(\$N_a, N_d\$)...

The fact that in higher voltages, the capacitance is constant means that the depletion region has reached its maximum length-correct me if I am wrong.

How can I find this length from the aforementioned diagram? In Sze's, Physics of Semiconductor Devices, p.\$85\$, relation \$24\$ it is stated that


but \$C_D\$ is the capacitance per unit area, \$W_D\$ the depletion region's length and \$\epsilon_s=\epsilon_0 \epsilon_r^{(Si)}\$

If I use this, \$W_D\approx 2.35m\$. This is way too much... To be honest i would expected it to be near the diode's thickness, which is \$325\mu m\$.

How to calculate the depletion region's length from the \$C-V\$ curve?

  • \$\begingroup\$ Are you sure that the y-axis in the plot is capacitance per unit area and not just capacitance? \$\endgroup\$
    – Justin
    Jun 17, 2013 at 12:54
  • \$\begingroup\$ @Justin: Thank you very much for your comment! The y axis is just capacitance, so-you are right- I cannot use the formula \$C_D=\frac{\epsilon_s}{W_D}\$... I've just mentioned it because it's my starting point! \$\endgroup\$
    – Thanos
    Jun 17, 2013 at 12:58
  • \$\begingroup\$ Is the capacitance really constant at higher voltages, or is it just decreasing at a rate too small to see at this scale? I know very little about the particular physics of this situation, but I bet if you make the voltage scale logarithmic, you will see that the capacitance continues to decrease in a straight line, forever. This would make sense: moving from 200V to 205V is not the same as 5V to 10V. \$\endgroup\$
    – Phil Frost
    Jun 17, 2013 at 14:06
  • \$\begingroup\$ @PhilFrost: It may be decreasing, but somehow I am suppossed to use this behaviour in order to find the region's length. On a second thought, if the voltage increases too much, the depletion region will reach the thickness of the diode. This means that the capacitance will remain constant. \$\endgroup\$
    – Thanos
    Jun 17, 2013 at 14:16
  • \$\begingroup\$ Since this appears to be a homework problem I suggest you look a little further into Sze. The equation you need deals with the built in voltage. and yes, the voltage should almost go flat. While possible you should not be expecting the junction to deplete the whole thickness of Si, that would require material that is doped so little that it would be impractical. \$\endgroup\$ Jun 17, 2013 at 14:22

1 Answer 1


In Britney Spears' Guide To Semiconductor Physics, on the page about pn-junctions, you can find this equation:

enter image description here

For your purposes, ignore the final version and just use the first and third parts of the equality

\$C = \dfrac{A e N_0 w_n}{V}\$

This equation comes from an assumption that the p- and n-sides of the device are equally and uniformly doped (\$N_{a} = N_{d}\$), and wn is half the total depletion width.

If you don't have equal doping, you'll need to work out the depletion width for the n-side and p-side separately using equations (16) and (17) from Brtiney's page. This is not so bad in the relatively common case that one side is extremely highly (degenerately) doped --- which makes the depletion width on that side negligibly small.

  • \$\begingroup\$ Thank you very much for your answer! The diode has \$N_a\approx 10^{19} m^{-3}\$ and \$N_d\approx 10^{10}\$(this comes from the slope of the diagram \$\frac{1}{C^2}=f(V)\$). In this case I used eq. 16 but the result is \$0.7m\$ which doesn't make any sense at all! \$\endgroup\$
    – Thanos
    Jun 18, 2013 at 7:17

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