Diffusion Capacitance: Forward Bias

Please see attached image. I'm struggling to understand how the book went from line 2 to line 3; it doesnt seem obvious how they did the differential.

Thanks

• You've skipped out the part where Q_p is defined, – placeholder Jun 2 '16 at 14:02
• The end of these UofI lecture notes looks very close to your final equation. Actually, there are several university lecture notes on the subject. But the terms in the solutions don't match exactly. I suspect, given some time, you could find the steps in these lectures you need to prove the above to your self. – st2000 Jun 2 '16 at 14:06
• Attached Qp (Excess Hole charges in N-region) – Arsenal123 Jun 2 '16 at 14:39

Well, there are several ways to look at this question, you have some simple relationships that are developed probably much earlier in a chapter on P-N junctions. From the equations you have we can look at the parameters in the final equation and try and make sense of what is going on. Using the relation that $I = \frac{dQ}{dt}$, looking at the final expression arrived at in your question: \begin{equation} C_{diff} = \frac{dQ_p}{dV} = \frac{dQ_p}{dt}\frac{dt}{dV} = I\frac{e\tau}{kT}. \end{equation} Using the relationship $V_T = \frac{kT}{e}$, where $V_T$ is the thermal voltage for a given temperature we see that: \begin{equation} \frac{dV}{dt} = \frac{kT}{e\tau} = \frac{V_T}{\tau}. \end{equation} Where in $\tau$ is generally the forward diffusion time constant for carriers in the junction region. So this derived diffusion capacitance is really just telling you how the current and voltage of the junction change with time.