Lec5
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View Lec5 as PDF for free.
More details
- Words: 1,396
- Pages: 19
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-1
Lecture 5 - PN Junction and MOS Electrostatics (II) pn Junction in Thermal Equilibrium September 22, 2005 Contents: 1. Introduction to pn junction 2. Electrostatics of pn junction in thermal equilibrium 3. The depletion approximation 4. Contact potentials Reading assignment: Howe and Sodini, Ch. 3, §§3.3-3.4
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-2
Key questions • What happens if the doping distribution in a semiconductor abruptly changes from n-type to p-type? • Is there a simple description of the electrostatics of a pn junction?
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-3
1. Introduction to pn junction • pn junction: p-region and n-region in intimate contact • Why is the p-n junction worth studying? It is present in virtually every semiconductor device! Example: CMOS cross section PMOS
n+
p+
n
NMOS
p+
n+
n+
p+
p
n
Understanding p-n junction is essential to understanding transistor operation.
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-4
2. Electrostatics of p-n junction in equilibrium Focus on intrinsic region:
p type n type x p type (a)
(Na)
metal contact to p side
p
x=0 x n
(Nd)
metal contact to n side (b)
Doping distribution of abrupt p-n junction: N p-region
n-region
Na
Nd
0
x
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-5
What is the carrier concentration distribution in thermal equilibrium? First think of two sides separately: p-region majority carrier
n-region log po, no
Na
po
majority carrier no Nd
minority carrier
po ni2 Na
no
minority carrier ni2 Nd x
Now bring them together. What happens? Diffusion of electrons and holes from majority carrier side to minority carrier side until drift balances diffusion.
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-6
Resulting carrier profile in thermal equilibrium: log po, no Na
Nd
po
no
ni2 Nd
ni2 Na 0
x
• Far away from metallurgical junction: nothing happens – two quasi-neutral regions • Around metallurgical junction: carrier drift must cancel diffusion – space-charge region
6.012 - Microelectronic Devices and Circuits - Fall 2005
In a linear scale:
Lecture 5-7
po, no
Na
Nd
po
+
-
no
0
x
E
Thermal equilibrium: balance between drift and diffusion Jpdiff Jpdrift Jndiff Jndrift
Can divide semiconductor in three regions: • two quasi-neutral n- and p-regions (QNR’s) • one space charge region (SCR) Now, want to know no (x), po (x), ρ(x), E(x), and φ(x). Solve electrostatics using simple, powerful approximation.
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-8
3. The depletion approximation • Assume QNR’s perfectly charge neutral • assume SCR depleted of carriers (depletion region) • transition between SCR and QNR’s sharp (must calculate where to place −xpo and xno) depletion approximation
po, no Na
Nd exact po
-
-xpo 0
+
no
• x < −xpo
po (x) = Na, no (x) =
• − xpo < x < 0
po (x), no (x) Na
• 0 < x < xno
no (x), po (x) Nd
• xno < x
x
xno
no (x) = Nd, po(x) =
n2i Na
n2i Nd
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-9
• Space charge density ρ
depletion approximation exact
qNd 0
-xpo
0
xno
x
-qNa
ρ(x) = = = =
0 −qNa qNd 0
x < −xpo − xpo < x < 0 0 < x < xno xno < x
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-10
• Electric field Integrate Gauss’ equation: 1 Z x2 E(x2 ) − E(x1 ) = ρ(x)dx s x1 ρ qNd 0
-xpo
0
xno
x
-qNa E 0
-xpo
xno 0
x
Eo
• x < −xpo
E(x) = 0
• − xpo < x < 0
x E(x) − E(−xpo ) = 1s −x −qNadx po −qNa x a x| = (x + xpo) = −qN −x s s po
R
qNd (x s
• 0 < x < xno
E(x) =
• xno < x
E(x) = 0
− xno)
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-11
• Electrostatic potential (with φ = 0 @ no = po = ni): φ=
kT no ln q ni
φ=−
kT po ln q ni
In QNR’s, no , po known ⇒ can determine φ: in p-QNR: po = Na ⇒ φp = − kTq ln Nnia in n-QNR: no = Nd ⇒ φn =
kT q
ln Nnid
φ φn 0
φB -xpo
0
xno
x
φp
Built-in potential: kT NaNd ln 2 φB = φn − φp = q ni General expression: did not use depletion approximation.
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-12
To get φ(x) in between, integrate E(x): φ(x2) − φ(x1 ) = −
Z
x2 x1
E(x)dx
ρ qNd 0
-xpo
0
xno
x
-qNa E 0
-xpo
xno 0
x
Eo φ φn 0 φp
φB -xpo
0
xno
x
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-13
φ φn 0
φB -xpo
0
xno
x
φp
• x < −xpo
φ(x) = φp
• − xpo < x < 0
φ(x) − φ(−xpo ) Rx qNa − (x + xpo )dx = − −x s po 2 a = qN (x + x ) po 2s 2 a φ(x) = φp + qN (x + x ) po 2s
• 0 < x < xno
d (x − x )2 φ(x) = φn − qN no 2s
• xno < x
φ(x) = φn
Almost done...
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-14
Still don’t know xno and xpo ⇒ need two more equations 1. Require overall charge neutrality: qNaxpo = qNd xno 2. Require φ(x) continuous at x = 0: qNa 2 qNd 2 φp + xpo = φn − xno 2s 2s Two equations with two unknowns. Solution: xno
v u u u u u t
2sφB Na = q(Na + Nd)Nd
v u u u u u t
2sφB Nd xpo = q(Na + Nd)Na
Now problem completely solved.
6.012 - Microelectronic Devices and Circuits - Fall 2005
Other results: Total width of space charge region: xdo = xno + xpo =
v u u u u u t
2sφB (Na + Nd) qNaNd
Field at metallurgical junction: v u u u u u t
2qφB NaNd |Eo | = s(Na + Nd)
Lecture 5-15
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-16
Three cases: • Symmetric junction: Na = Nd ⇒ xpo = xno • Asymmetric junction: Na > Nd ⇒ xpo < xno • Strongly asymmetric junction: i.e. p+n junction: Na Nd xpo xno
v u u u u u t
2sφB 1 √ ' xdo ' ∝ qNd Nd v u u u u u t
2qφB Nd √ |Eo| ' ∝ Nd s The lowly-doped side controls the electrostatics of the pn junction. ρ qNd
ρ
ρ
qNd
qNd
x
x
x
-qNa -qNa
-qNa
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-17
4. Contact potentials Potential distribution in thermal equilibrium so far:
p
-
p-QNR
+
n
SCR
n-QNR
φ
0
φB -xpo
0
xno
x
Question 1: If I apply a voltmeter across diode, do I measure φB ? 2 yes
2 no
2 it depends
Question 2: If I short diode terminals, does current flow on outside circuit? 2 yes
2 no
2 sometimes
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-18
We are missing contact potential at metal-semiconductor contacts: p
-
p-QNR
n
+
SCR
n-QNR
φ φmn φmp
φB -xpo
0
xno
x
Metal-semiconductor contacts: junctions of dissimilar materials ⇒ built-in potentials: φmn, φmp Potential difference across structure must be zero ⇒ cannot measure φB ! φB = φmn + φmp
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-19
Key conclusions • Electrostatics of pn junction in equilibrium: – a space-charge region – surrounded by two quasi-neutral regions ⇒ built-in potential across p-n junction • To first order, carrier concentrations in space-charge region are much smaller than doping level ⇒ depletion approximation. • Contact potential at metal-semiconductor junctions: ⇒ from contact to contact, there is no potential buildup across pn junction
Lecture 5-1
Lecture 5 - PN Junction and MOS Electrostatics (II) pn Junction in Thermal Equilibrium September 22, 2005 Contents: 1. Introduction to pn junction 2. Electrostatics of pn junction in thermal equilibrium 3. The depletion approximation 4. Contact potentials Reading assignment: Howe and Sodini, Ch. 3, §§3.3-3.4
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-2
Key questions • What happens if the doping distribution in a semiconductor abruptly changes from n-type to p-type? • Is there a simple description of the electrostatics of a pn junction?
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-3
1. Introduction to pn junction • pn junction: p-region and n-region in intimate contact • Why is the p-n junction worth studying? It is present in virtually every semiconductor device! Example: CMOS cross section PMOS
n+
p+
n
NMOS
p+
n+
n+
p+
p
n
Understanding p-n junction is essential to understanding transistor operation.
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-4
2. Electrostatics of p-n junction in equilibrium Focus on intrinsic region:
p type n type x p type (a)
(Na)
metal contact to p side
p
x=0 x n
(Nd)
metal contact to n side (b)
Doping distribution of abrupt p-n junction: N p-region
n-region
Na
Nd
0
x
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-5
What is the carrier concentration distribution in thermal equilibrium? First think of two sides separately: p-region majority carrier
n-region log po, no
Na
po
majority carrier no Nd
minority carrier
po ni2 Na
no
minority carrier ni2 Nd x
Now bring them together. What happens? Diffusion of electrons and holes from majority carrier side to minority carrier side until drift balances diffusion.
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-6
Resulting carrier profile in thermal equilibrium: log po, no Na
Nd
po
no
ni2 Nd
ni2 Na 0
x
• Far away from metallurgical junction: nothing happens – two quasi-neutral regions • Around metallurgical junction: carrier drift must cancel diffusion – space-charge region
6.012 - Microelectronic Devices and Circuits - Fall 2005
In a linear scale:
Lecture 5-7
po, no
Na
Nd
po
+
-
no
0
x
E
Thermal equilibrium: balance between drift and diffusion Jpdiff Jpdrift Jndiff Jndrift
Can divide semiconductor in three regions: • two quasi-neutral n- and p-regions (QNR’s) • one space charge region (SCR) Now, want to know no (x), po (x), ρ(x), E(x), and φ(x). Solve electrostatics using simple, powerful approximation.
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-8
3. The depletion approximation • Assume QNR’s perfectly charge neutral • assume SCR depleted of carriers (depletion region) • transition between SCR and QNR’s sharp (must calculate where to place −xpo and xno) depletion approximation
po, no Na
Nd exact po
-
-xpo 0
+
no
• x < −xpo
po (x) = Na, no (x) =
• − xpo < x < 0
po (x), no (x) Na
• 0 < x < xno
no (x), po (x) Nd
• xno < x
x
xno
no (x) = Nd, po(x) =
n2i Na
n2i Nd
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-9
• Space charge density ρ
depletion approximation exact
qNd 0
-xpo
0
xno
x
-qNa
ρ(x) = = = =
0 −qNa qNd 0
x < −xpo − xpo < x < 0 0 < x < xno xno < x
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-10
• Electric field Integrate Gauss’ equation: 1 Z x2 E(x2 ) − E(x1 ) = ρ(x)dx s x1 ρ qNd 0
-xpo
0
xno
x
-qNa E 0
-xpo
xno 0
x
Eo
• x < −xpo
E(x) = 0
• − xpo < x < 0
x E(x) − E(−xpo ) = 1s −x −qNadx po −qNa x a x| = (x + xpo) = −qN −x s s po
R
qNd (x s
• 0 < x < xno
E(x) =
• xno < x
E(x) = 0
− xno)
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-11
• Electrostatic potential (with φ = 0 @ no = po = ni): φ=
kT no ln q ni
φ=−
kT po ln q ni
In QNR’s, no , po known ⇒ can determine φ: in p-QNR: po = Na ⇒ φp = − kTq ln Nnia in n-QNR: no = Nd ⇒ φn =
kT q
ln Nnid
φ φn 0
φB -xpo
0
xno
x
φp
Built-in potential: kT NaNd ln 2 φB = φn − φp = q ni General expression: did not use depletion approximation.
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-12
To get φ(x) in between, integrate E(x): φ(x2) − φ(x1 ) = −
Z
x2 x1
E(x)dx
ρ qNd 0
-xpo
0
xno
x
-qNa E 0
-xpo
xno 0
x
Eo φ φn 0 φp
φB -xpo
0
xno
x
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-13
φ φn 0
φB -xpo
0
xno
x
φp
• x < −xpo
φ(x) = φp
• − xpo < x < 0
φ(x) − φ(−xpo ) Rx qNa − (x + xpo )dx = − −x s po 2 a = qN (x + x ) po 2s 2 a φ(x) = φp + qN (x + x ) po 2s
• 0 < x < xno
d (x − x )2 φ(x) = φn − qN no 2s
• xno < x
φ(x) = φn
Almost done...
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-14
Still don’t know xno and xpo ⇒ need two more equations 1. Require overall charge neutrality: qNaxpo = qNd xno 2. Require φ(x) continuous at x = 0: qNa 2 qNd 2 φp + xpo = φn − xno 2s 2s Two equations with two unknowns. Solution: xno
v u u u u u t
2sφB Na = q(Na + Nd)Nd
v u u u u u t
2sφB Nd xpo = q(Na + Nd)Na
Now problem completely solved.
6.012 - Microelectronic Devices and Circuits - Fall 2005
Other results: Total width of space charge region: xdo = xno + xpo =
v u u u u u t
2sφB (Na + Nd) qNaNd
Field at metallurgical junction: v u u u u u t
2qφB NaNd |Eo | = s(Na + Nd)
Lecture 5-15
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-16
Three cases: • Symmetric junction: Na = Nd ⇒ xpo = xno • Asymmetric junction: Na > Nd ⇒ xpo < xno • Strongly asymmetric junction: i.e. p+n junction: Na Nd xpo xno
v u u u u u t
2sφB 1 √ ' xdo ' ∝ qNd Nd v u u u u u t
2qφB Nd √ |Eo| ' ∝ Nd s The lowly-doped side controls the electrostatics of the pn junction. ρ qNd
ρ
ρ
qNd
qNd
x
x
x
-qNa -qNa
-qNa
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-17
4. Contact potentials Potential distribution in thermal equilibrium so far:
p
-
p-QNR
+
n
SCR
n-QNR
φ
0
φB -xpo
0
xno
x
Question 1: If I apply a voltmeter across diode, do I measure φB ? 2 yes
2 no
2 it depends
Question 2: If I short diode terminals, does current flow on outside circuit? 2 yes
2 no
2 sometimes
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-18
We are missing contact potential at metal-semiconductor contacts: p
-
p-QNR
n
+
SCR
n-QNR
φ φmn φmp
φB -xpo
0
xno
x
Metal-semiconductor contacts: junctions of dissimilar materials ⇒ built-in potentials: φmn, φmp Potential difference across structure must be zero ⇒ cannot measure φB ! φB = φmn + φmp
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-19
Key conclusions • Electrostatics of pn junction in equilibrium: – a space-charge region – surrounded by two quasi-neutral regions ⇒ built-in potential across p-n junction • To first order, carrier concentrations in space-charge region are much smaller than doping level ⇒ depletion approximation. • Contact potential at metal-semiconductor junctions: ⇒ from contact to contact, there is no potential buildup across pn junction
