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```p-n junction photodetectors (a.k.a. photodiodes)
Asymmetrical p+ - n or n+ - p junction can be used
as a fast photodetector
•
Reverse bias applied to the p-n junction creates a
depletion region with high electric field.
•
Photons absorbed in the depletion regions create
electron-hole pairs, which are separated by the
electric field and contribute to the photocurrent.
Equilibrium conditions
(in the dark, zero bias applied)
– the diode current is zero
The potential barrier BLOCKS the
electrons in n-type material from diffusing
into p-region
The potential barrier BLOCKS the
holes in p-type material from diffusing
into n-region
Junction under illumination
Light is
generating the
e-h pairs
--
+
The light creates the carriers that move in the same
directions as the minority carriers in the reverse biased
junction (in the darkness).
Therefore, under illumination, there is a photocurrent,
which direction corresponds to the reverse current of
the junction.
The origin of this photo-current is the DRIFT of
photo-generated carriers.
Junction photocurrent
Pinc
φ0 =
(1 − Θ R
A hν
Incident photon flux
(the number of photons per unit area, per second):
ΘR is the reflection coefficient of the top surface, A is the device area, Pinc is the
incident optical power.
The drift current density, Jdr assuming that all the carriers
are swept out by the electric field in the depletion region:
(
J dr = q φ0 1 − e−αW
)
In case the depletion region is thick enough, i.e. αW >> 1,
Jdr max = q ϕ0
The total drift current:
(
I dr = Aq φ0 1 − e−αW
)
)
The external quantum efficiency of the photodiode:
The responsivity of the photodiode:
R=
η q η λ ( µm)
= ext = ext
( A /W )
Pinc
hν
1.24
I ph
Photodiode response time
The diode response time has two components:
1) transit time
2) RC-limited time constant
1) Photo-carrier transit time
If the electric field in the depletion region is strong enough, both
electrons and holes move with the saturation velocity,
vS ≈ 107 cm/s.
Transit time ttr ≅ W/vS
Note that W depends on the
applied voltage:
For N a >> N d , (V0 − V ) =
1q
Nd W 2
2ε
2) RC-time constant
P-n junction capacitance
E(x)
V=0
V<0
-xp
- xn
xn
V=0
xn
V<0
E0
W
VV0bi-- V
V
Depletion region capacitance
Since the p-n junction has a space charge region there has to be a capacitance
associated with it.
The static and differential capacitances.
First, consider a simple plain capacitor:
The electric field between the plates is related to the charge per unit area:
Q1
σ
E=
=
ε ε0 ε ε0
++++++
d
------
Q1 is the charge per unit area.
The voltage drop across the capacitor, V = E × d;
V=
Q1
ε ε0
×d
Q=
ε ε0 V
d
×A
, A is the cross section area
of the junction
The static capacitance
C0 = Q/V
For the plain capacitor
Q=
ε ε0 V
d
C0 =
×A
ε ε0 A
d
The differential capacitance
which is the most important for the speed of response
Cd =
∂Q
∂V
For the plain capacitor,
Cd =
ε ε0 A
d
∂ Q εε 0
=
A
∂V
d
= C0
Static capacitance of the p-n junction
For the depletion region of a p-n junction, the
exact separation between positive and negative
charges is not obvious: the charges are distributed
within the depletion width W.
We use the definition C0 = Q/V to find the capacitance.
The total (positive) charge,
(n-side of the junction)
QD = q×ND×Xn × A
The total negative charge
|-QA| = q×NA×Xp × A
(p-side of the junction),
Static capacitance of the p-n junction (cont.)
The voltage across the junction
1 q N AN D
(Vbi −V ) =
W2
2 εε 0 N A + N D
where W = xn + xp;
1 q
For N A >> N D , (Vbi −V ) =
ND W 2
2 εε 0
and W ≈ xn;
The static capacitance (for |-V| >> Vbi, NA >> ND)
Q = q×ND×W × A
C0 = Q/V =
q ND W A
q ND W 2
2ε ε0 =
2ε ε0 A ε ε0 A
=
W
(W / 2)
This capacitance is equivalent to that of plain capacitor with
d = W/2;
Differential capacitance of the p-n junction
Cd =
∂ Q1
∂V
Consider asymmetric p-n junction, NA >> ND
(Vbi −V ) =
1 q
ND W 2
2 εε 0
Q = q×ND×W × A
∂Q ∂Q ∂W
=
×
∂V ∂W ∂V
∂Q
= q×N D× A
∂W
∂W
1
=
∂V ∂ V ∂W
∂V
q
ND W
=
∂W ε ε 0
⎛ qN DW ⎞ εε 0 A
∂Q
Cd =
= q×N D×A ⎜
⎟=
∂V
⎝ εε 0 ⎠ W
(omitting the voltage negative sign)
Cd =
εε 0 A
W
P-n junction as a variable capacitor
(Vbi −V ) =
p+
region
W1 @ |V| =V1
W2 @ |V2| > |V1|
1 q
ND W 2
2 εε 0
Cd =
εε 0 A
W
W(V)
C(V)
Vbi - V
The RC component of the photodiode speed of response:
An intrinsic RC – time constant of the photoresponse:
C=
εε 0 A
p+ region
W1 @ |V| =V1
d
W2 @ |V2| > |V1 |
n- region
RS = ρ
W
(d − W )
A
τ RCi = C×Rs
(A is the diode area)
When the photodetector is connected to the external load (i.e. the amplifier),
τRC = C (RS + RL),
where RL is the load resistance
Photodiode overall response time
RC – time constant of the photoresponse:
Transit time
Total response time:
τ RCi = C×Rs
ttr ≅ W/vS
τT ≅ ttr + τRCi
p+ region
W(V), ttr(V)
W1 @ |V| =V1
d
W2 @ |V2| > |V1 |
RS(V)
n- region
C(V)
Vbi - V
As W increases, ttr increases but τRCi decreases because C decreases;
hence there is an optimal W value for the fastest photoresponse.
```
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