MESFET Operation [ edit ]
Assume an N channel MESFET with uniform doping and sharp depletion
region shown in figure 1.
The depletion region
W
n
{\displaystyle W_{n}}
is given by the depletion width for a
diode. Where the voltage is the voltage from the gate to the
channel, where the channel voltage is given for a position x along
the channel as
V
g
c
(
x
)
{\displaystyle V_{gc}(x)}
.
W
n
(
x
)
=
2
ε
0
ε
r
(
Ψ
−
V
g
c
(
x
)
)
q
N
d
{\displaystyle W_{n}(x)={\sqrt {\frac {2\varepsilon _{0}\varepsilon _{r}(\Psi -V_{gc}(x))}{qN_{d}}}}}
W
n
(
x
)
2
=
2
ε
0
ε
r
(
Ψ
−
V
g
c
(
x
)
)
q
N
d
{\displaystyle W_{n}(x)^{2}={\frac {2\varepsilon _{0}\varepsilon _{r}(\Psi -V_{gc}(x))}{qN_{d}}}}
W
n
(
x
)
2
q
N
d
2
ε
0
ε
r
=
Ψ
−
V
g
c
(
x
)
{\displaystyle {\frac {W_{n}(x)^{2}qN_{d}}{2\varepsilon _{0}\varepsilon _{r}}}=\Psi -V_{gc}(x)}
V
g
c
(
x
)
=
Ψ
−
W
n
(
x
)
2
q
N
d
2
ε
0
ε
r
{\displaystyle V_{gc}(x)=\Psi -{\frac {W_{n}(x)^{2}qN_{d}}{2\varepsilon _{0}\varepsilon _{r}}}}
d
V
g
c
(
x
)
d
W
n
(
x
)
=
−
2
W
n
(
x
)
q
N
d
2
ε
0
ε
r
{\displaystyle {\frac {dV_{gc}(x)}{dW_{n}(x)}}=-{\frac {2W_{n}(x)qN_{d}}{2\varepsilon _{0}\varepsilon _{r}}}}
(1)
The current density in the channel is given by:
J
n
=
σ
ξ
{\displaystyle J_{n}=\sigma \xi }
I
n
(
x
)
=
σ
ξ
⋅
W
⋅
b
(
x
)
{\displaystyle I_{n}(x)=\sigma \xi \cdot W\cdot b(x)}
I
n
(
x
)
=
−
σ
d
V
g
c
(
x
)
d
x
W
(
a
−
W
n
(
x
)
)
{\displaystyle I_{n}(x)=-\sigma {\frac {dV_{gc}(x)}{dx}}W(a-W_{n}(x))}
where:
ξ
=
−
d
V
g
c
(
x
)
d
x
{\displaystyle \xi =-{\frac {dV_{gc}(x)}{dx}}}
Therefore,
I
n
(
x
)
=
−
σ
a
W
(
1
−
W
n
(
x
)
a
)
d
V
g
c
(
x
)
d
W
n
(
x
)
d
W
n
(
x
)
d
x
{\displaystyle I_{n}(x)=-\sigma aW{\bigg (}1-{\frac {W_{n}(x)}{a}}{\bigg )}{\frac {dV_{gc}(x)}{dWn(x)}}{\frac {dWn(x)}{dx}}}
∫
0
L
I
n
(
x
)
d
x
=
∫
0
L
−
σ
a
W
(
1
−
W
n
(
x
)
a
)
d
V
g
c
(
x
)
d
W
n
(
x
)
d
W
n
(
x
)
d
x
d
x
{\displaystyle \int _{0}^{L}I_{n}(x)\,dx=\int _{0}^{L}-\sigma aW{\bigg (}1-{\frac {W_{n}(x)}{a}}{\bigg )}{\frac {dV_{gc}(x)}{dW_{n}(x)}}{\frac {dW_{n}(x)}{dx}}\,dx}
I
n
⋅
L
=
−
σ
a
W
∫
W
n
(
0
)
W
n
(
L
)
(
1
−
W
n
(
x
)
a
)
d
V
g
c
(
x
)
d
W
n
(
x
)
d
W
n
(
x
)
{\displaystyle I_{n}\cdot L=-\sigma aW\int _{Wn(0)}^{W_{n}(L)}{\bigg (}1-{\frac {W_{n}(x)}{a}}{\bigg )}{\frac {dV_{gc}(x)}{dW_{n}(x)}}\,dW_{n}(x)}
Substituting from equation 1:
I
n
=
−
σ
a
W
L
∫
W
n
(
0
)
W
n
(
L
)
(
1
−
W
n
(
x
)
a
)
(
−
2
W
n
(
x
)
q
N
d
2
ε
0
ε
r
)
d
W
n
(
x
)
{\displaystyle I_{n}={\frac {-\sigma aW}{L}}\int _{W_{n}(0)}^{W_{n}(L)}{\bigg (}1-{\frac {W_{n}(x)}{a}}{\bigg )}{\bigg (}-{\frac {2W_{n}(x)qN_{d}}{2\varepsilon _{0}\varepsilon _{r}}}{\bigg )}\,dWn(x)}
I
n
=
σ
a
W
2
q
N
d
2
ε
0
ε
r
L
∫
W
n
(
0
)
W
n
(
L
)
(
W
n
(
x
)
−
W
n
(
x
)
2
a
)
d
W
n
(
x
)
{\displaystyle I_{n}={\frac {\sigma aW2qN_{d}}{2\varepsilon _{0}\varepsilon _{r}L}}\int _{W_{n}(0)}^{W_{n}(L)}{\bigg (}W_{n}(x)-{\frac {W_{n}(x)^{2}}{a}}{\bigg )}\,dWn(x)}
I
n
=
2
σ
a
W
q
N
d
2
ε
0
ε
r
L
[
W
n
2
(
x
)
2
−
W
n
3
(
x
)
3
a
]
W
n
(
0
)
W
n
(
L
)
{\displaystyle I_{n}={\frac {2\sigma aWqN_{d}}{2\varepsilon _{0}\varepsilon _{r}L}}{\bigg [}{\frac {W_{n}^{2}(x)}{2}}-{\frac {W_{n}^{3}(x)}{3a}}{\bigg ]}_{W_{n}(0)}^{W_{n}(L)}}
I
n
=
2
σ
a
W
q
N
d
2
ε
0
ε
r
L
[
W
n
2
(
L
)
−
W
n
2
(
0
)
2
−
W
n
3
(
L
)
−
W
n
3
(
0
)
3
a
]
{\displaystyle I_{n}={\frac {2\sigma aWqN_{d}}{2\varepsilon _{0}\varepsilon _{r}L}}{\bigg [}{\frac {W_{n}^{2}(L)-W_{n}^{2}(0)}{2}}-{\frac {W_{n}^{3}(L)-W_{n}^{3}(0)}{3a}}{\bigg ]}}
I
n
=
2
σ
a
W
q
N
d
a
2
6
L
⋅
2
ε
0
ε
r
[
3
(
W
n
2
(
L
)
−
W
n
2
(
0
)
)
a
2
−
2
(
W
n
3
(
L
)
−
W
n
3
(
0
)
)
a
3
]
{\displaystyle I_{n}={\frac {2\sigma aWqN_{d}a^{2}}{6L\cdot 2\varepsilon _{0}\varepsilon _{r}}}{\bigg [}{\frac {3(W_{n}^{2}(L)-W_{n}^{2}(0))}{a^{2}}}-{\frac {2(W_{n}^{3}(L)-W_{n}^{3}(0))}{a^{3}}}{\bigg ]}}
One defines constant Β as the channel conductance with no
depletion. And the work function to deplete the channel
W00 [1]:
W
00
=
Ψ
−
V
t
o
=
q
N
d
a
2
2
ε
0
ε
r
{\displaystyle W_{00}=\Psi -V_{to}={\frac {qN_{d}a^{2}}{2\varepsilon _{0}\varepsilon _{r}}}}
β
=
σ
a
3
L
W
00
{\displaystyle \beta ={\frac {\sigma a}{3LW_{00}}}}
We now define Vto , the voltage such that the channel is pinched off. d is the ratio of channel depletion to maximum depletion for the drain. s the ratio of channel depletion to
maximum depletion for the source.
d
=
W
n
(
L
)
a
=
2
ε
0
ε
r
(
Ψ
−
V
g
d
)
q
N
d
2
ε
0
ε
r
(
Ψ
−
V
t
o
)
q
N
d
=
Ψ
−
V
g
d
W
00
{\displaystyle d={\frac {W_{n}(L)}{a}}={\frac {\sqrt {\frac {2\varepsilon _{0}\varepsilon _{r}(\Psi -V_{gd})}{qN_{d}}}}{\sqrt {\frac {2\varepsilon _{0}\varepsilon _{r}(\Psi -V_{to})}{qN_{d}}}}}={\sqrt {\frac {\Psi -V_{gd}}{W_{00}}}}}
s
=
W
n
(
0
)
a
=
2
ε
0
ε
r
(
Ψ
−
V
g
s
)
q
N
d
2
ε
0
ε
r
(
Ψ
−
V
t
o
)
q
N
d
=
Ψ
−
V
g
s
W
00
{\displaystyle s={\frac {W_{n}(0)}{a}}={\frac {\sqrt {\frac {2\varepsilon _{0}\varepsilon _{r}(\Psi -V_{gs})}{qN_{d}}}}{\sqrt {\frac {2\varepsilon _{0}\varepsilon _{r}(\Psi -V_{to})}{qN_{d}}}}}={\sqrt {\frac {\Psi -V_{gs}}{W_{00}}}}}
Substituting:
I
n
=
W
⋅
σ
a
⋅
W
00
3
L
[
3
(
d
2
−
s
2
)
−
2
(
d
3
−
s
3
)
]
{\displaystyle I_{n}=W\cdot {\frac {\sigma a\cdot W_{00}}{3L}}{\big [}3(d^{2}-s^{2})-2(d^{3}-s^{3}){\big ]}}
I
n
=
W
⋅
β
W
00
2
[
3
(
d
2
−
s
2
)
−
2
(
d
3
−
s
3
)
]
{\displaystyle I_{n}=W\cdot \beta W_{00}^{2}{\big [}3(d^{2}-s^{2})-2(d^{3}-s^{3}){\big ]}}
(2)
Equation 2 is Shockley's expression [2] for drain current in the linear region. When the device enters saturation, one end is pinched off(normally the drain). Thus $d=1$ and one may derive the equation for the saturation region:
I
s
a
t
=
β
W
00
2
(
1
−
3
s
2
+
2
s
3
)
{\displaystyle I_{sat}=\beta W_{00}^{2}(1-3s^{2}+2s^{3})}
g
m
=
3
β
W
00
(
s
−
1
)
{\displaystyle g_{m}=3\beta W_{00}(s-1)}
G
D
S
=
3
β
W
00
(
1
−
d
)
{\displaystyle G_{DS}=3\beta W_{00}(1-d)}
Simpler Model [ edit ]
I
d
s
=
3
2
β
W
00
2
[
(
V
g
s
−
v
t
o
)
2
W
00
2
−
(
V
g
d
−
v
t
o
)
2
W
00
2
]
{\displaystyle I_{ds}={\frac {3}{2}}\beta W_{00}^{2}{\bigg [}{\frac {(V_{gs}-v_{to})^{2}}{W_{00}^{2}}}-{\frac {(V_{gd}-v_{to})^{2}}{W_{00}^{2}}}{\bigg ]}}
g
m
=
3
β
W
00
(
V
g
s
−
V
t
o
)
{\displaystyle g_{m}=3\beta W_{00}(V_{gs}-V_{to})}
G
d
s
=
3
β
W
00
(
V
g
d
−
V
t
o
)
{\displaystyle G_{ds}=3\beta W_{00}(V_{gd}-V_{to})}
General power law: [ edit ]
It was found that a general power law provided a better fit for real devices [3].
I
d
s
=
β
[
(
V
g
s
−
V
t
o
)
Q
−
(
V
g
d
−
V
t
o
)
Q
]
{\displaystyle I_{ds}=\beta {\big [}(V_{gs}-V_{to})^{Q}-(V_{gd}-V_{to})^{Q}{\big ]}}
Where Q is dependent on the doping profile and a good fit is usually obtained for Q between 1.5 and 3. A general power law is approximately equal to Shockley's equation for Q = 2.4. Β is also empirically chosen and is proportion to the previous Β
β
proportial to
σ
a
W
3
L
W
00
{\displaystyle \beta {\mbox{ proportial to }}{\frac {\sigma aW}{3LW_{00}}}}
Modelling the various regions is done though model binning. This however infers that a sharp transition exists from one region to another, which may not be accurate.
I
d
s
=
{
0
V
g
s
<
V
t
o
β
[
(
V
g
s
−
V
t
o
)
Q
−
(
V
g
d
−
V
t
o
)
Q
]
V
g
s
≤
V
g
d
β
(
V
g
s
−
V
t
o
)
Q
V
g
s
>
V
g
d
{\displaystyle I_{ds}=\left\{{\begin{matrix}0&V_{gs}<V_{to}\\\beta {\big [}(V_{gs}-V_{to})^{Q}-(V_{gd}-V_{to})^{Q}{\big ]}&V_{gs}\leq V_{gd}\\\beta (V_{gs}-V_{to})^{Q}&V_{gs}>V_{gd}\end{matrix}}\right.}
References [ edit ]
[1] A. E. Parker. Design System for Locally Fabricated Gallium Arsenide Digital
Integrated Circuits. PhD thesis, Sydney University, 1990.
[2] W. Shockley. A unipolar field-effect transistor. IEEE Trans/ Electron Devices, 20(11):1365–1376, November 1952.
[3] I. Richer and R.D. Middlebrook. Power-law nature of field-effect transistor experimental characteristics. Proc. IEEE, 51(8):1145–1146, August 1963.