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1 . Lecture #9
OUTLINE
– Metal-semiconductor contacts (cont.)
» I-V characteristics
» practical ohmic contacts
» small-signal capacitance
Reading: Finish Chapter 14
Spring 2003 EE130 Lecture 9, Slide 1
Review: Schottky Diode (n-type Si)
metal n-type Si
ΦM > ΦS
Eo
χSi
ΦM
qVbi = ΦBn – (Ec – EF)FB
ΦBn
Ec Equilibrium (VA = 0)
EF -> EF continuous,
constant
ΦBn = ΦM – χ
Ev
W
Spring 2003 EE130 Lecture 9, Slide 2
1
2 . Schottky Diode (p-type Si)
metal p-type Si
Eo ΦM < ΦS
χSi
Ec
ΦM
Equilibrium (VA = 0)
-> EF continuous,
EF constant
Ev
ΦBp qVbi = ΦBp– (EF – Ev)FB
ΦBp = χ + EG - ΦM
W
Spring 2003 EE130 Lecture 9, Slide 3
Depleted Layer Width, W
Last time, we found that
− qN D
V (x ) = (W − x )2
2K Sε 0
At x = 0, V = - (Vbi - VA)
2ε s (Vbi − VA )
⇒ W=
qN D
• W increases with increasing –VA
• W decreases with increasing ND
- (Vbi - VA)
Spring 2003 EE130 Lecture 9, Slide 4
2
3 . W for p-type Semiconductor
qN A
V (x ) = (W − x )2
2 K Sε 0
At x = 0, V = Vbi + VA
2ε s (VA + Vbi )
⇒ W=
qN A
• W increases with increasing VA
• W decreases with increasing NA
Spring 2003 EE130 Lecture 9, Slide 5
Current Flow in a Schottky Diode
• Diode current is determined by majority-
carrier flow across the MS junction
– Under forward bias, majority-carrier
diffusion from the semiconductor into the
metal dominates the current
– Under reverse bias, majority-carrier
diffusion from the metal into the
semiconductor dominates the current
Spring 2003 EE130 Lecture 9, Slide 6
3
4 . Thermionic Emission Theory
• Electrons can cross the junction if
1
mv x ≥ q(Vbi − VA )
2
KE x =
2
2q
v x ≥ vmin ≡ (Vbi − VA )
mn*
• The current for electrons at a given velocity is:
I s • → M , v x = − qAv x n(vx )
• So, the total current over the barrier is:
− v min
I s • → M = −qA ∫ v n(v )dv
−∞
x x x
Spring 2003 EE130 Lecture 9, Slide 7
Schottky Diode I - V
• Given that
4πkTmn* 2 ( E − E ) / kT −(m* / 2 kT )v 2
n (v x ) = 3 e F c e n x
h
• We obtain
4πqmn k 2
I S •→ M = AT 2e − qΦ B / kT e qVA / kT
h3
= AT 2 J S e qVA / kT , where J S ≈ 120e −qΦ B / kT A/cm 2
• In the other direction, we always see the same barrier
ΦBn: I M • → S = − I S • → M (VA = 0)
• Therefore I = I S ( e − 1) where I S = AT 2 J S
qV A / kT
Spring 2003 EE130 Lecture 9, Slide 8
4
5 . Applications of Schottky Diodes
• IS of a Schottky diode is 103 to 108 times larger than a pn
junction diode, depending on ΦB .
⇒ Schottky diodes are preferred rectifiers for low voltage,
high current applications.
Spring 2003 EE130 Lecture 9, Slide 9
Practical Ohmic Contact
• In practice, most M-S contacts are rectifying
• To achieve a contact which conducts easily in
both directions, we dope the semiconductor
very heavily
Æ W is so narrow that carriers can tunnel directly
through the barrier
Spring 2003 EE130 Lecture 9, Slide 10
5
6 . ΦBn – VA
2ε s Φ Bn - -
- - Ec , EFS
W ≅ Ec , EF EFM V
qN D
Ev
Ev
− H ( Φ Bn −V A ) ND
tunneling probabilit y P = e
H = 4π ε s mn / h = 5.4 × 109 mn / mo cm −3/2 V −1
− H ( Φ Bn −V A ) / N D
J S → M ≈ qN D vthx P = qN D kT / 2πmn e
Spring 2003 EE130 Lecture 9, Slide 11
Specific Contact Resistance
Spring 2003 EE130 Lecture 9, Slide 12
6
7 .Voltage Drop across an Ohmic Contact
Spring 2003 EE130 Lecture 9, Slide 13
Review: MS-Contact Charge Distribution
• In a Schottky contact, charge is stored on either side
of the MS junction
• This charge is modulated by the applied voltage
Spring 2003 EE130 Lecture 9, Slide 14
7
8 .Schottky Diode: Small-Signal Capacitance
• If an A.C. voltage is applied in series with the D.C.
bias VA, the charge stored in the Schottky contact will
be modulated
→ displacement current will flow
εs
C= A
W
Spring 2003 EE130 Lecture 9, Slide 15
Using C-V Data to Determine ΦB
1 2(Vbi − VA )
=
C 2
qN Dε s A2
Once Vbi is known, ΦΒn can be determined:
Nc
qVbi = qΦ Bn − ( Ec − EF ) FB = qΦ Bn − kT ln
ND
Spring 2003 EE130 Lecture 9, Slide 16
8
9 . Summary
Spring 2003 EE130 Lecture 9, Slide 17
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