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ECSE-2210 Microelectronics Technology
Class Activity 16 – Solution
1. What is the physical origin of the junction capacitance (or depletion layer
capacitance)?
The in and out movement of the majority carriers about the steady state depletion
width in response to the applied ac signal gives rise to the junction capacitance.
Effectively, it looks like the plus and the minus charges are alternatively being added
and subtracted from two planes inside the diode with a width W, similar to what
happens in a parallel plate capacitor.
2. What is the physical origin of the diffusion capacitance?
Changes in the stored minority carriers in the quasi-neutral region in response to the
applied ac signal gives rise to the diffusion capacitance. In response to an ac signal,
the minority carrier distributions oscillate about their dc values resulting in an
additional capacitance.
3. Why is the diffusion capacitance negligible under reverse bias?
The concentration of minority carriers stored in reverse-bias is very small. So, any
change will be even smaller.
4. What is the value of the reverse-bias conductance for an ideal diode?
The reverse bias conductance is zero for an ideal diode.
5. Write the equation for the reverse bias conductance for a real p+-n diode if IR-G is
dominating in reverse bias.
IR-G = – q A ni W / 2
τ
o with W = [2
ε
/ q ND × (Vbi – VA)]1/2
So, dI / dV = q A ni W / 2
τ
o × [1 / (Vbi – VA)]
6. Given a planar p+-n Si step junction diode with an n-side doping of ND = 1015 cm-3 at
T = 300 K, determine the junction capacitance at –1 V reverse bias. Assume that the
Fermi-level is at the valence band edge in the p+-region. Junction area A = 1 cm2. At
zero bias, will the junction capacitance be higher or lower than the one calculated
above? (Hint: Calculate Vbi, then W and then the junction capacitance).
First find Vbi.
Vbi = 0.55 V+ 0.0259 V × ln (1015/1010)
= 0.848 V
W = [2
ε
Si/qND × (Vbi – VA)]1/2
W = [(2
ε
Si/q) × (1/ND)(0.848 V + 1 V)]1/2
= 1.52 × 10-4 cm.
So, Cj =
ε
Si A/W
= 10-12x 1/1.52 × 10-4
= 6.57 nF.
At zero bias, the junction capacitance will be larger than the above value since the
depletion layer width will be smaller.
