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Home Assignment - VI (EEL732)Adersh Miglani
Assignment: Find the model for the diffusion capacitanceof forward bias PN junction.
Solution: When p-type and n-type semiconductor materialsare brought together, majority charge carriers diffuse acrossthe metallurgical junction. The diffused carriers are termedas minority carriers or injected carriers in the other region.These charge carriers leave uncovered impurity atoms whichcreate depletion region. Under thermal equilibrium, no mobilecharge carriers are present in the depletion region and diffusionof majority carriers is opposed by build-in potential barriercreated due to the induced electric field in the depletion region.
A. Diffusion Capacitance in term of IncrementalResistance
These injection carriers, stored at the edges of depletionregion, cause diffusion capacitance which dominates underforward bias condition. The diffusion capacitance due toelectrons in the p-side is given by the following expression
Cdn =dQndV
where dQn is a change in the negative charge due of dV , achange in the forward bias.
Say γn is the minority carrier electron lifetime and rdn isthe incremental resistance.
rdn =dV
dIdn
The expression for Cdn is reduced to
Cdn = γndQnγndV
= γndIdndV
=γnrdn
Similar expression can be developed for the holes available inn-side
Cdp =γprdp
Total diffusion capacitance under forward bias is
C = Cdn + Cdp =γnrdn
+γprdp
Answer
The diffusion capacitance increases with the increase in for-ward bias current. Larger forward bias current results in thenet increase in the injected charge carriers outside of depletionregion.
B. Diffusion Capacitance (Small Signal Model)
In small signal model, a sinusoidal voltage is superimposedon the dc voltage. The minority carrier currents are used
to derive small signal impedance. The total voltage underforward bias is Va = V0 + v1(t), where V0 is dc voltage andv1(t) = V1 sin ωt ac voltage. In small signal model, theminority carrier electron and hole concentration at the edgesof space charge region becomes a function of time becauseapplied voltage is also a function of time. For convenience,let us assume origin at the edge of space charge region on then-side.
So, minority carrier hole concentration at x = 0 is given by
pn(0) = pn0 exp(e VakT
)= pn0 exp
(e(V0 + v1(t))
kT
)(1)
pn(0) = pdc exp(e v1(t)
kT
)(2)
where pdc = pn0 exp(e V0
kT
)If we assume magnitude of
v1(t) � kTe , the exponential term can be expanded using
Taylor’s expansion (higher order terms can be ignored) andexpression can be simplified to
pn(0, t) ≈ Pdc(1 +
v1(t)
Vt
)= Pdc
(1 +
V1 ejωt)
Vt
)(3)
We assume that electric field if zero for x > 0. So, the timedependent diffusion equation for holes can be written as
Dp∂2(δpn)
∂x2− δpnτp0
=∂(δpn)
∂t(4)
If ac signal is sinusoidal and is superimposed over dc volt-age, the solution for δpn is also constant component with asinusoidal superimposed component as given by
δpn(x, t) = δp0(x) + p1(x)ejωt (5)
Use this result in second order differential equation
Dp∂2(δp0(x) + p1(x)e
jωt)
∂x2
−δp0(x) + p1(x)ejωt
τp0=∂(δp0(x) + p1(x)e
jωt)
∂t(6)
Dp
[∂2(δp0(x))
∂x2+∂2(p1(x))
∂x2ejωt
]− δp0(x) + p1(x)e
jωt
τp0
= jωp1(x)ejωt (7)[
Dp∂2(δp0(x))
∂x2− δp0(x)
τp0
]+
[Dp
∂2(p1(x))
∂x2− p1(x)
τp0− jωp1(x)
]ejωt = 0 (8)
Let us consider the time dependent component
Dpd2(p1(x))
dx2−(1 + jωτp0
τp0
)p1(x) = 0 (9)
Say, L2p = Dpτp0 and C2
p =(
1+jωτp0L2
p
)d2(p1(x))
dx2− C2
p p1(x) = 0 (10)
The general solution to this equation is
p1(x) = K1e−Cpx +K2e
+Cpx (11)
The coefficient K2 = 0 because of p1(x → +∞) = 0boundary condition. So, we get
p1(x) = K1e−Cpx (12)
At x = 0,
pn(0, t) = Pdc
(1 +
V1 ejωt)
Vt
)(13)
Using this boundary condition, we get
K1 = pdcV1Vt
(14)
The hole diffusion current density at x = 0 is given by
Jp = −eDp∂pn∂x
(15)
If we consider a homogeneous semiconductor, the derivative ofthe hole concentration will be just the derivative of the excesshole concentration. The Jp would have dc and time dependentcomponent as given by
Jp = Jp0 + Jp(t) (16)
where
Jp0 =eDppn0Lp
[exp
(eV0kT
)− 1
](17)
and
Jp(t) = Jpejωt = −eDp
∂p1(x)
∂xejωt (18)
where
Ip = eDp(−Cp)
[pdc
V1Vt
]e−Cpx (19)
Total hole current phasor
Ip = AJp = eADpCp pdcV1Vt
(20)
Ip =eADp pdc
Lp
√1 + jωτp0
V1Vt
(21)
Ip = Ip0√1 + jωτp0
V1Vt
(22)
where Ip0 =eADp pdc
Lp.
Similarly, the electron current can be derived to the follow-ing expression
In = In0√
1 + jωτn0V1Vt
(23)
where In0 = eADn pdcLn
.
The PN junction admittance is Y =Ip+In
V1. It is simplified
to
Y =
(Ip0√1 + jωτp0 + In0
√1 + jωτn0
Vt
)(24)
Let us approximate ωτp0 � 1 and ωτn0 � 1 and frequencyof ac signal is small. So,
√1 + jωτp0 ≈ 1 +
jωτp02 and√
1 + jωτn0 ≈ 1+ jωτn0
2 . The final expression for admittanceis
Y =Ip0 + In0
Vt+ jω
(Ip0τp0 + In0τn0
2Vt
)(25)
There, the small signal diffusion capacitance (under abovestated assumptions) is given by the following expression
Cd =
(Ip0τp0 + In0τn0
2Vt
)Answer (26)
