Problem Set # 10
ECEN 3320 Fall 2013
Semiconductor Devices
November 18, 2013 – Due December 2, 2013
1. An n-channel MOSFET has W/L = 10 and oxide thickness xox =30 nm. Assume the carrier
mobility in the channel is constant at µn =500 cm2 /Vs.
(a) Calculate the required inversion layer carrier concentration, Qn /q, in order to obtain
channel resistance of 500 Ω.
Solution: The drain current, ID is given by
ID = µn Cox
W
[VGS − VT ]VDS
L
in the linear regime and
"
W
V2
ID = µn Cox
[VGS − VT ]VDS − DS
L
2
#
in the quadratic regime. The inversion layer charge, Qn (= Qinv ) is given by
Qn = Cox [VGS − VT ].
The channel conductance is given by
gd =
∂ID
|V
∂VDS GS
and evidently is a linear conductance when ID is expressed in the linear approximation such that the linear conductance is given by
gd = µn Cox
W
W
[VGS − VT ] = µn Qn
L
L
The channel resistance , rd = 1/gd , can be expressed as
rd =
L
µn Qn W
Qn =
L
.
µn rd W
and the charge, Qn as
1
For a 500 Ω channel, the Qn then is
Qn =
1
= 4 × 10−7 C/cm2 = 2.5 × 1012 charges/cm2 .
10 × 500 × 500
(b) Calculate the gate voltage above the threshold, VG −VT , required to produce the inversion
layer charge, Qn /q, found in (a).
Solution: We can write that
VGS − VT =
Qn
Qn xox
4 × 10−7 3 × 10−6
=
=
= 3.48 V
Cox
ox
3.9 × 8.85 × 10−14
2. An n-channel enhancement-mode MOSFET is biased as shown below. The threshold voltage
was found to be VT =1V. Using the long-channel MOSFET theory, calculate and plot ID vs
VDS for
Figure 1: An n-channel enhancement mode MOSFET.
(a) VGD = 0,
Solution: Here we have that VGS = VDS . We cannot have quadratic behavior (VDS
cannot be less than VGS − VT ). We have
(
ID =
ID0
VDS ≤ VT
qVDS
ID0 exp kT
VDS > VT
(b) VGD = VT /2 and
Solution: Here we have VGS = VDS + VT /2. Again, we can have no quadratic behavior.
We have
(
ID =
ID0
VDS ≤ VT /2
qVDS
ID0 exp kT
VDS > VT /2
(c) VGD = 2VT .
2
Solution: Here we have that VGS = VDS + 2VT . We will have quadratic behavior. VD S
is always less than VGS − VT . We still have cut-off when VGS ≤ VT , that is, when
VDS ≤ −VT . Knowing that
"
W
V2
ID = µn Cox
(VGS − VT )VDS − DS
L
2
#
when VDS ≥ −VT
and plugging in for VGS in terms of VDS and VT , we find
(
ID =
ID0
W
µn Cox L [(VT VDS
VDS ≤ −VT
+
2
VDS
2 ]
VDS > −VT
Figure 2: An n-channel enhancement mode MOSFET I-V curves. VT = 1 was assumed for all of the
numerical calculations.
3. Suppose that the I − V curve for an n-channel MOSFET has saturation drain current IDsat =
2 × 10−4 A, the drain voltage at the onset of saturation VDSsat = 4 V and threshold voltage
VT = 0.8 V.
(a) Find the gate voltage VGS at the onset of saturation
Solution: We have that
ID = µn Cox
i
W h
2
(VGS − VT )VDS − VDS
/2 when VDS ≤ VGS − VT .
L
We then know that
VDS,sat = VGS,sat − VT
so that
VGS,sat = VDS,sat + VT = 4.8 V
(b) Find the value of the conduction parameter,
3
W µn Cox
.
2L
Solution: Above saturation, we can write that
W 2
ID,sat = µn Cox VDS,sat
2L
or that
W
ID,sat
µn Cox
= 1.25 × 10−5 mhos/V
= 2
2L
VDS,sat
(c) Determine ID when VGS = 2V and VDS = 2V.
Solution: This is just plugging in to the expression for saturated (VDS ≥ VGS − VT )
current
i
W h
2
ID = µn Cox
(VGS − VT )VDS − VDS
/2
L
W
= µn Cox (VGS − VT )2 /2
L
= 1.25 × 10−5 × 1.22 = 1.8 × 10−5 amps
4. Start with the inversion layer charge density equation, Qn = −Cox [VG − VT − V (y)].
Solution: We have that
Jn = Dn q
∂n
+ qnµn E = Jdif f + Jdrif t .
∂y
The Qn (y) we are using is just qn(y) × A × L where A is the channel cross sectional area
and L is the channel length in the above notation. We can then write that
∂Qn
J n × A × L = Dn
+ Qn µn E.
∂y
(a) Derive an expression for diffusion current in the channel.
Solution: Starting from the definition of the diffusion current above, we find
Jdif f = Dn
∂Qn
∂E(y)
= −Dn Cox
∂y
∂y
(b) Derive an expression for drift current in the channel.
Solution: Starting from the definition of the drift current above, we find
Jdrif t = Qn µn E = −µn Cox [VG − VT − V (y)]
∂E(y)
∂y
(c) Derive an expression for the ratio of diffusion and drift currents.
Solution: Taking the ratio of drift to diffusion currents, we find
Jdrif t
µn
=
[VG − VT − V (y)].
Jdif f
Dn
Using the Einstein relation that Dn =
kT
q µn ,
we find that
Jdrif t
(VG − VT − V (y))
=
.
Jdif f
kT /q
The drift current is greater than the drift by essentially the amount that the driving
voltage is greater than the thermal noise voltage kT /q.
4