n-type
Schottky contact / ohmic contact
Schottky contact
EF,s
EF,m
metal
EF,m
metal
EF,s
Ohmic contact:
linear resistance
specific contact resistance:
⎛ ∂J ⎞
Rc = ⎜
⎟
⎝ ∂V ⎠
−1
Ω-cm 2
Tunnel contacts
For high doping, the Schottky barrier is so thin that electrons can
tunnel through it.
metal
p+
p
Degenerate doping at a tunnel contact
metal
n+
n
Tunnel contacts have a linear resistance.
Contacts
Transport mechanisms
Drift
Diffusion
Thermionic emission
Tunneling
All mechanisms are always present.
One or two transport mechanisms can dominate depending on the
device and the bias conditions.
In a forward biased pn-junction, diffusion dominates.
In a tunnel contact, tunneling dominates.
In a Schottky diode, thermionic emission dominates.
Student projects
Calculate the diffusion current for a Schottky diode
Calculate the thermionic emission current for a pn diode
Institute of Solid State Physics
Technische Universität Graz
JFETs - MESFETs - MODFETs
Junction Field Effect Transistors (JFET)
Metal-Semiconductor Field Effect Transistors (MESFET)
Modulation Doped Field Effect Transistors (MODFET)
n
JFET
n-channel JFET
n
For NA >> ND
2ε (Vbi − V )
xn =
eN D
Depletion mode
Enhancement mode
h > xn =
2ε Vbi
eN D
2ε Vbi
h < xn =
eN D
conducting at Vg = 0
nonconducting at Vg = 0
n-channel JFET
drain
depletion region
p
n
p
gate
JFETs are often discrete devices
source
MESFET
Metal-Semiconductor Field Effect Transistors
n
Depletion layer created by Schottky barrier
xn =
2ε (Vbi − V )
eN D
Fast transistors can be realized in n-channel GaAs, however GaAs has
a low hole mobility making p-channel devices slower.
JFET
n
D
n-channel JFET
G
2ε (Vbi − V )
xn =
eN D
S
n-channel JFET
Pinch-off at h = xn
eN D h 2
Vp =
2ε
At Pinch-off, Vp = Vbi - V.
D
G
S
p-channel JFET
Vp = pinch-off voltage
JFET
The drain is the side of
the transistor that gets
pinched off.
JFET
xL
h
xn
h
xR
dV.
There is a long derivation to determine how the current depends on VG and VD.
We will find a relatively simple formula (probably familiar to electrical engineers).
Understanding the derivation is important for knowing when this formula is valid.
JFET
xL
h
xn
xR
h
dV.
dV = I D dR = I D
ρ dy
Z (h − xn ( y ))
1
1
1
ρ= =
=
σ neμn N D eμn
dV = I D
ID is constant throughout the transistor
dy
eμn N D Z (h − xn ( y ))
I D dy = eμn N D Z (h − xn ( y ))dV
JFET
I D dy = eμn N D Z (h − xn ( y ))dV
xL
h
xn
xR
h
VG is a forward bias
V(y) is a reverse bias.
dV.
depletion width is a function of position xn ( y ) =
differentiate
2ε (Vbi + V ( y ) − VG )
eN D
dxn ( y ) 1 ⎛ 2ε (Vbi + V ( y ) − VG ) ⎞
= 2⎜
⎟
dV
eN
D
⎝
⎠
dV =
eN D xn
ε
dxn
−1/ 2
2ε
ε
=
eN D eN D xn ( y )
JFET
xL
h
xn
h
xR
from last slides
I D dy = eμn N D Z (h − xn ( y ))dV
dV =
eN D xn
ε
dxn
I D dy = eμn N D Z (h − xn ( y ))
L
I D ∫ dy = eμn N D Z
0
eN D
ε
eN D
ε
xn dxn
xR
∫ (h − x ( y)) x dx
n
n
n
xL
μn N D2 Ze 2 ⎡
2
2
3
3
2
⎤
ID =
h
x
x
x
x
−
−
−
(
)
(
)
R
L
R
L
3
⎣
⎦
2 Lε
JFET
xn
xL
h
xR
h
μn N D2 Ze 2 ⎡
2
2
3
3
2
⎤
ID =
h
x
x
x
x
−
−
−
(
)
(
R
L
R
L )⎦
3
⎣
2 Lε
h=
2ε V p
eN D
2ε (Vbi − VG )
xL =
eN D
xR =
2ε (Vbi − VG + VD )
eN D
JFET - drain current
⎡V
2 ⎛ Vbi + VD − VG
D
⎢
ID = I p
− ⎜
Vp
⎢ V p 3 ⎜⎝
⎣
μn N D2 Ze 2 h3
Ip =
2 Lε
valid in the linear regime
(until pinch-off)
⎞
⎟⎟
⎠
3/ 2
2 ⎛ Vbi − VG
+ ⎜
3 ⎜⎝ V p
⎞
⎟⎟
⎠
3/ 2
⎤
⎥
⎥
⎦
Junction FET simulation
http://www-g.eng.cam.ac.uk/mmg/teaching/linearcircuits/jfet.html
JFET - Linear regime
⎡V
2 ⎛ Vbi + VD − VG
D
⎢
ID = I p
− ⎜
Vp
⎢ V p 3 ⎜⎝
⎣
⎞
⎟⎟
⎠
3/ 2
2 ⎛ Vbi − VG
+ ⎜
3 ⎜⎝ V p
In the linear regime VD << Vsat.
⎡1
dI D
1 ⎛ Vbi + VD − VG
⎢
= Ip
− ⎜
dVD
Vp
⎢V p V p ⎝⎜
⎣
dI
ID ≈ D
dVD
1/ 2
⎞
⎟⎟
⎠
Ip ⎡
V −V
VD =
⎢1 − bi G
V p ⎢⎣
Vp
VD = 0
variable resistor
⎤
⎥
⎥
⎦
⎤
⎥ VD
⎥⎦
⎞
⎟⎟
⎠
3/ 2
⎤
⎥
⎥
⎦
JFET - Saturation regime
⎡V
2 ⎛ Vbi + VD − VG
D
⎢
ID = I p
− ⎜
Vp
⎢ V p 3 ⎜⎝
⎣
⎡1
dI D
1 ⎛ Vbi + VD − VG
⎢
= Ip
− ⎜
dVD
Vp
⎢V p V p ⎜⎝
⎣
1/ 2
⎞
⎟⎟
⎠
⎞
⎟⎟
⎠
3/ 2
2 ⎛ Vbi − VG
+ ⎜
3 ⎜⎝ V p
⎤
⎥=0
⎥
⎦
set dID/dVD = 0 to find Vsat
will be 0 when Vbi+ VD - VG = VP
Vsat = V p − Vbi + VG
I sat
⎡1 V −V
2 ⎛ Vbi − VG
bi
G
⎢
= Ip −
+ ⎜
3 ⎜⎝ V p
Vp
⎢3
⎣
⎞
⎟⎟
⎠
3/ 2
⎤
⎥
⎥
⎦
No VD dependence
Voltage controlled current source
⎞
⎟⎟
⎠
3/ 2
⎤
⎥
⎥
⎦
JFET - transconductance
In the saturation regime,
I sat
⎡1 V −V
⎛ Vbi − VG
2
bi
G
= Ip ⎢ −
+ ⎜
3
3 ⎜⎝ V p
V
⎢
p
⎣
transconductance (describes how good
the voltage controlled current source is)
dI sat I p ⎛
Vbi − VG
gm =
= ⎜1 −
dVG V p ⎜⎝
Vp
⎞
⎟
⎟
⎠
dI sat 2 Z μn eN D h ⎛
Vbi − VG
gm =
1
=
−
⎜
⎜
dVG
L
Vp
⎝
large μ small L increases gm
⎞
⎟
⎟
⎠
⎞
⎟⎟
⎠
3/ 2
⎤
⎥
⎥
⎦
JFET
High frequencies
iin = 2π fCG vG
iout = g m vG
iin < iout
for gain:
f <
average capacitance:
gm
= fT
2π CG
CG = ZL
ε
xn
μn eN D h 2
fT =
2πε L2
fT is the frequency
where the gain drops
below 1
For velocity saturation, the approximation dV = I D
Ohm's law assumes vd = μE
fT ≈
vs
L
ρ dy
Z (h − xn ( y ))
is not valid
JFET/MESFET
JFET: small gate current (reverse leakage of the gate-to-channel junction)
More gate leakage than MOSFET, less than bipolar.
JFET has higher transconductance than the MOSFET.
Used in low-noise, high input-impedance op-amps and sometimes used in
switching applications.
MESFET: usually constructed in compound semiconductor technologies
lacking high quality surface passivation such as GaAs, InP, or SiC, and are
faster but more expensive than silicon-based JFETs or MOSFETs.
Production MESFETs are operated up to approximately 30 GHz, and are
commonly used for microwave frequency communications and radar.
Majority carrier device (like Schottky diode).
MODFET (HEMT)
Modulation doped field effect transistor (MODFET)
High electron mobility transistor (HEMT)
Modulation doped field effect transistor
High electron mobility transistor
VT = Threshold voltage = voltage where charge is depleted
Heterostructure
pn junction formed from two semiconductors with different band gaps
MODFET/HEMT
undoped channel
HEMT: HEMT devices are found in cell phones, electronic warfare systems,
microwave and millimeter wave communications, radar, and radio astronomy.
PhD Thesis Sergey Smirnov
http://www.iue.tuwien.ac.at/phd/smirnov/node71.html
MODFET (HEMT)
j = nevd = neμn E y
I = jZt = Zeμn ns E y
ns
n=
t
t is the thickness of the 2DEG
ns is the sheet charge at the interface in C/cm2.
VG − V ( y ) is the voltage between the gate and the 2DEG
ns = 0 when VG − V ( y ) = VT
gate
y
2DEG
q = CV
−ens ( y ) = −Cg (VG − V ( y ) − VT )
MODFET (HEMT)
−ens ( y ) = Cg (VG − V ( y ) − VT ) is the charge on the 2DEG at point y
The charge is zero when
VG − V ( y ) = VT
solve for ns
ns ( y ) =
−Cg (VG − V ( y ) − VT )
e
Substitute this in Ohm's law:
I = jZt = Zeμn ns E y
MODFET (HEMT)
I = jZt = Zeμn ns E y
ns ( y ) =
−Cg (VG − V ( y ) − VT )
e
substitute ns in the top equation and substitute
dV ( y )
I = Z μnC (VG − VT − V ( y ) )
dy
integrate along the length of the channel
L
VD
∫ Idy = ∫ Z μ C (V
n
0
G
− VT − V ( y ) ) dV
0
⎡
VD2 ⎤
Z
I D = μnC ⎢(VG − VT ) VD − ⎥
2 ⎦
L
⎣
− dV ( y )
Ey =
dy
MODFET (HEMT)
⎡
VD2 ⎤
Z
I = μnC ⎢(VG − VT ) VD − ⎥
2 ⎦
L
⎣
dI
Z
= μnC ⎡⎣(VG − VT ) − VD ⎤⎦ = 0
dVD L
Set the derivative = 0 to find
saturation voltage
Vsat = (VG − VT )
Substitute the saturation voltage into the formula at the top to find
saturation current
Z
2
I sat =
2L
μnC (VG − VT )