pn Junction under Reverse Bias
■
First, we must understand the complete structure of the pn junction-- starting in
thermal equilibrium:
,,,,
,,,,
−
φpm
+
−Wp
metal contact to
p side
p
metal interconnection
−xpo
−
φB
+
xno
,,,,
,,,,
n
Wn
■
−
φmn
+
metal contact
to n side
How can VD = 0 and the built-in potential barrier be φB = 1 V (approx.)?
Answer: look at the complete circuit ... including the potential barriers at the
p-type silicon-to-metal (φpm) and the metal-to-n-type silicon (φmn) junctions.
■
KirchhoffÕs Voltage Law:
0 = φ pm + φ B + φ mn
therefore, the built-in voltage is given by:
φ B = Ð φ pm Ð φ mn
EE 105 Spring 1997
Lecture 8
Potential Plot through pn Junction
■
The potential in the metal is the same on both ends of the pn junction in thermal
equilibrium, with the metal-semiconductor contact potentials (ÒbatteriesÓ)
cancelling out the built-in potential
φo(x)
φn
−xpo
−Wp
−
φpm
+
xno
φB
φp
−
+
φmn
Wn
+ x
−
(b)
Note: we show potential changes at metal-silicon contacts as vertical, which is
not correct. The details are left for an advanced device physics course.
■
Now we apply a battery VD ... with VD < 0 (reverse bias)
−Wp
ID ≈ 0 A
VD (< 0 V)
,,,,
,,,,
−
φpm
+
ohmic contact to
p side
p
+
−xp
−
xn
x
Wn
−
φj
+
,,,,
,,,,
n
−
φmn
+
ohmic contact
to n side
EE 105 Spring 1997
Lecture 8
pn Junction under Reverse Bias (cont.)
■
Potential plot under reverse bias: contact potentials donÕt change ...
they are ohmic contacts. Only place for change is at the pn junction
(a)
φ(x)
−xp
−Wp
φn
−
+ VD < 0 V
−
φpm
+
φj
φp
−
+
xn
φmn
Wn
+x
−
(b)
■
The new potential barrier is called φj
KVL:
Ð V D Ð φ pm Ð φ j Ð φ mn = 0
φ j = ( Ð φ pm Ð φ mn ) Ð V D = φ B Ð V D
■
The potential barrier is increased over the built-in barrier by the reverse bias ...
which widens the depletion region (xn > xno, xp > xpo)
EE 105 Spring 1997
Lecture 8
−xp(−6.4)
−xp(−2.4)
ρ(x)
−xp(0)
+qNd
n-side
x
p-side
−qNa
xn(0) xn(−2.4) x (−6.4)
n
(a)
E(x)
p-side
−xp
n-side
xn
x
thermal equilibrium (VD = 0 V)
VD = − 2.4 V
VD = − 6.4 V
qNd xn(VD)
E(x = 0) = −
εs
(b)
φ(x)
thermal equilibrium (VD = 0 V,
n-side
φ = φ = 0.8 V)
p-side
B
j
x
VD = − 2.4 V, φj = 3.2 V
−2.4 V
−2.5 V
−6.4 V
−5V
VD = − 6.4 V, φj = 7.2 V
− 7.5 V
(c)
EE 105 Spring 1997
Lecture 8
Quantitative Results
■
■
Substitute φj for φB in the equilibrium depletion width and we Þnd the depletion
width under reverse bias (the math is the same):
x p(V D) =
 2ε s ( φ B Ð V D )  N d 
 ----------------------------------  -------------------- = x po 1 Ð ( V D ⁄ φ B )
qN a

  N d + N a
x n(V D) =
 2ε s ( φ B Ð V D )  N a 
 ----------------------------------  -------------------- = x no 1 Ð ( V D ⁄ φ B )
qN d

  N d + N a
X d(V D) =
2ε s ( φ B Ð V D ) 1
1
 ---------------------------------  ------- + ------- = X do 1 Ð ( V D ⁄ φ B )

N
q
N 
a
d
Note xpo, xno, and Xdo are the widths in thermal equilibrium
EE 105 Spring 1997
Lecture 8
Capacitance
■
Basic circuits: C = Q / V ... linear capacitor
■
Semiconductor devices: C = dq / dv ... nonlinear charge-storage -->
capacitance is deÞned as a small signal quantity
vd +
_
VD < 0 V
+
_
,,,,,
p
qJ = QJ + qj
− −
−
− −− − − − − −
++++++++++
,,,,,
n
(a)
+qj
vd +
_
Cj = Cj (VD)
−qj
(b)
Symbols:
* break up the total applied voltage (symbol: vD) into two parts:
total voltage = DC voltage + small-signal voltage
vD = VD + vd
* break up the charge qJ on the p-side of the junction similarly:
total charge in the depletion region = DC charge + small-signal charge
qJ = QJ + qj
EE 105 Spring 1997
Lecture 8
Depletion Capacitance
■
Find the function qJ = qJ(vD) from xp(vD):
q J(v D) = Ð qN a x p(v D) = Ð qN a x po 1 Ð ( v D ⁄ φ B )
■
Normalized plot:
−14
−12
−10
−8
−6
−4
−2
qJ
|qJo|
vD / φB
−1
−2
−3
−4
−5
■
To Þnd the depletion capacitance Cj we simply take the derivative and evaluate it
at the particular DC voltage
dq J
C j = C j(V D) = ---------dv D
VD
■
Math --> no insight into the concept of capacitance!
EE 105 Spring 1997
Lecture 8
Graphical Interpretation
■
Derivative is the slope of the plot of qJ(vD):
−14
−12
−10
−8
−6
−4
−2
qJ
|qJo|
vD / φB
−1
−2
−3
−4
−5
■
The small-signal charge is related to the small-signal voltage by the slope at
point (QJ, VJ):
 dq J
q j =  --------- dv D

 ⋅ v d = C j(V D) ⋅ v d

VD
EE 105 Spring 1997
Lecture 8
Physical Interpretation
■
Small-signal voltage changes the depletion width (vd > 0 --> reverse bias is
reduced --> depletion width is slightly narrower)
ρ(x)
p-side
(a)
−xp
+qNd
n-side
x
xn
vD = VD
−qNa
QJ = −qNaxp
ρ(x)
+qNd
p-side −x −x
p
p
(b)
n-side
x
xn xn
vD = VD + vd > VD-->
xp < xp,| qJ | < | QJ |
−qNa
qJ = −qNaxp
qj = qNa∆xp
∆ρ(x) = ρ(x) − ρ(x)
+ qNd
Xd
p-side
(c)
xn xn
−xp −xp
qj = qj − Qj > 0
= −qNaxp − (−qNaxp)
= qNa (xp −xp)
= qNa∆xp
n-side
x
− qNa
−qj = −qNd ∆xn
EE 105 Spring 1997
Lecture 8
Depletion Capacitance Equation
■
Derivative can be evaluated (see Chapter 3), but the incremental charge is two
sheets separated by a distance Xd(VD) --> use parallel plate capacitor formula:
C jo
εs
qj
εs
C j = ----- = ------------------- = ------------------------------------------ = ------------------------------vd
X d(V D)
X do 1 Ð V D ⁄ φ B
1 Ð V D ⁄ φB
■
Plot of depletion capacitance (normalized to Cjo):
Cj /Cjo
1
0.75
0.5
0.25
−14
−12
−10
−8
−6
−4
−2
0
VD /φB
Typical numbers: Xdo = 0.4 µm --> Cjo = 2.6 x 10-8 F/cm2 = 0.26 fF/µm2
φB = 0.8 V --> VD = - 6.4 V = - 8 φB -->
(1 - VD / φB)1/2 = 3 --> Cj = Cjo / 3 = 86 aF/µm2
EE 105 Spring 1997
Lecture 8