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```EXAMPLE 7.1
BJECTIVE
Determine the total bias current on an IC due to subthreshold current.
Assume there are 107 n-channel transistors on a single chip, all biased at VGS = 0 and
VDS =2 V. Assume Isub = 10-10 A for each transistor for this bias condition and for a
threshold voltage of VT = 0.5 V. What happens to the total bias current on the IC if the
threshold voltage is reduced to VT = 0.25 V, all other parameters remaining the same.
 Solution
The total bias current is the bias current of each transistor times the number of
transistors, or
IT = Isub(107) = (10-10)(107)  1 mA
We can write
so
10 10
 VGS  VT
I sub  I 0 exp 
 Vt
 0  0.5 
 I 0 exp 
  I0
 0.0259 



 0.0242
Now, if the threshold voltage changes to VT = 0.25, then the subthreshold current at
VGS = 0 becomes
I sub
or
 VGS  VT
 I 0 exp 
 Vt

0  0.25 

  0.0242 exp 

 0.0259 

Isub = 1.56  106 A
Now, the total bias current for this IC chip would be
IT = (1.56  10-6)(107) = 15.6 A
 Comment
This example is intended to show that, taking into account subthreshold currents, the
threshold voltage must be designed to be a “reasonable” value such that the zero-bias
gate currents are not excessive.
EXAMPLE 7.2
OBJECTIVE
To determine the effect of channel length modulation on the value of drain current.
Consider a n-channel MOSFET with a substrate impurity doping concentration of Na = 2 
1016 cm-3, a threshold voltage of VT = 0.4 V, and a channel length of L = 1 m. The device is
biased at VGS = 1 V and VDS = 2.5 V.
 Solution
We find that
Fp
and
 Na
 Vt ln 
 ni

 2 1016 
  0.0259 ln 
  0.365V
10 
 1.5 10 

VDS(sat) = VGS  VT = 1  0.4 = 0.6 V
Now



 211.7  8.85  10 14 
L  

19
2  1016 
 1.6  10

or
We can write
or
 Comment

1/ 2


0.365  0.6  2.5  0.6  
0.365  0.6
L = 0.181 m

ID
L
1


ID
L  L 1  0.181

ID
 1.22
ID
Due to channel length modulation, the drain current is 22 percent larger than the ideal
long channel value.

EXAMPLE 7.3
OBJECTIVE
To calculate the effective electric field at threshold for a given semiconductor doping.
Consider a p-type silicon substrate at T = 300 K and doped Na = 3  1016 cm-3.
 Solution
From Equation (6.8b) in Chapter 6, we can calculat
and
 Na 
 3 1016 

  0.376V
Fp  Vt ln 
 0.0259 ln 
10 


 1.5 10 
 ni 
1/ 2
1/ 2
 4 s  Fp 
 411.7 8.85  10 14 0.376  
xdT  




eN
1.6  10 19 3  1016 




a


which is xdT = 0.18 m. Then
 max 
QSD
8.64 108

 8.34 104 V/cm
14
11.7  8.85 10


At the threshold inversion point, we may assume that Qn = 0, so the effective electric
field from Equation (7.10) is found as
 eff
1
8.64 10 8
4
 max  

QSD

8
.
34

10
V/cm
14
11.7 8.85 10 
s
 Comment
We can see, from Figure 7.10, that this value of effective transverse electric field at the
surface is sufficient for the effective inversion charge mobility to be significantly less
than the bulk semiconductor value.
EXAMPLE 7.4
OBJECTIVE
To determine the ratio of drain current under the velocity saturation condition to the ideal longchannel value.
Assume an n-channel MOSFET with a channel length L = 0.8 m, a threshold voltage of VT =
0.5 V, an electron mobility of n = 700 cm2/V-s, and vsat = 5  106 cm/s. Assume that the
transistor is biased at (a) VGS = 2 V and (b) VGS = 3 V.
 Solution
We can write
ID
v ,sat
ID
ideal


vsat
2 0.8  10 4 5  10 6


 n VGS  VT
700
VGS  0.5
2L
For (a) VGS = 2 V, we find
and for (b) VGS = 3 V, we obtain
 Comment
ID
v ,sat
ID
ideal
ID
v ,sat
ID
ideal
 0.762
 0.457
We see that as the applied gate-to-source voltage increases, the ratio decreases. This
effect is a result of the velocity saturation current being a linear function of VGS  VT ,
whereas the ideal long-channel current is a quadratic function of VGS  VT .
EXAMPLE 7.5
BJECTIVE
Calculate the threshold voltage shift due to short-channel effects.
Consider an n-channel MOSFET with Na = 5  1016 cm-3 and tox = 200
Å. Let L = 0.8 m and assume that rj = 0.4 m.
 Solution
We can determine the oxide capacitance to be
Cox
ox 3.98.85 10 14 
7
2



1
.
73

10
F/cm
tox
200 10 8
and can calculate the potential as
Fp
 Na 
 5 1016 
  0.0259 ln 
  0.389V
 Vt ln 
10 
 1.5 10 
 ni 
the maximum space charge width is found as
1/ 2
xdT
 4 s Fp N a 




eN a




 411.7  8.85 10 14 0.389 

  0.142m
19
16
1.6 10
5 10





Finally, the threshold voltage shift, from Equation (7.22), is
VT
or

1.6 10 5 10 0.142 10   0.4 

 
19
4
16
1.73 10
7
20.142 

1


1


0
.
8
0
.
4
 

VT = 0.101 V
 Comment
If the threshold voltage of this n-channel MOSFET is to be VT = 0.40 V,
for example, a shift of VT = 0.101 V due to short-channel effects is
significant and needs to be taken into account in the design of this device.
EXAMPLE 7.6
OBJECTIVE
Design the channel width that will limit the threshold shift because of narrow channel effects to a
specified value..
Consider a n-channel MOSFET with Na = 5  1016 cm-3 and tox = 200 Å. Let  =  / 2.
Assume that we want to limit the threshold shift to VT = 0.1 V.
 Solution
From Example 7.5, we have
Cox  1.73 107 F/cm 2
xdT  0.142m
and
From Equation (7.28), we can express the channel width as
W 
or

eN a x
Cox VT
2
dT
 
5 10   0.142 10  4
2
1.73  10 7 0.1

1.6 10 



19
16



2

W = 1.46 m
 Comment
We can note that the threshold shift of VT = 0.1 V occurs at a channel width of W = 1.46 m,
which is approximately 10 times larger than the induced space charge width xdT .
EXAMPLE 7.7
BJECTIVE
Calculate the theoretical punch-through voltage assuming the abrupt
junction approximation.
Consider an n-channel MOSFET with source and drain doping
concentrations of Nd = 1019 cm-3 and a channel region doping of Na =
1016 cm-3. Assume a channel length of L = 1.2 m, and assume the
source and body are at ground potential.
 Solution
The pn junction built-in potential barrier is given by
 Na Nd
Vbi  Vt ln 
2
n
i






 1016 1019 

  0.0259 ln 
  0.874V
10 2
 1.5 10


The zero-biased source-substrate pn junction width is
1/ 2
xd 0
 2 s Vbi 

 
 eN a 


 211.7  8.85 10 14 0.874  

  0.336 m
19
16
1.6  10
10





The reverse-biased drain-substrate pn junction width is given by
 2 s Vbi  VDS  
xd  

eN a


1/ 2
At punch-through, we will have
xd 0  xd  L
or
0.336  xd  1.2
Which fives xd = 0.864 m at the punch-through condition. We can then
find
2
4 2
19
16
Vbi  VDS



   5.77V

xd eN a
0.864 10
1.6 10 10


14
2 s
211.7  8.85 10
The punch-through voltage is then found as
VDS = 5.77  0.874 = 4.9 V
 Comment
As the two space charge regions approach punch-through, the abrupt
junction approximation is no longer a good assumption.
EXAMPLE 7.8
BJECTIVE
Design the ion implant dose required to adjust the threshold voltage to a
specified value.
Consider an n-channel MOSFET with a doping of Na = 5  1015 cm-3,
and oxide thickness of tox = 500 Å, and an initial flat-band voltage of
VFBO = 1.25 V. Determine the ion implantation dose such that a
threshold voltage of VT = +0.70 V is obtained.
 Solution
We may calculate the necessary parameters as
Fp0
 Na
 Vt ln 
 ni

 5 1015
  0.0259  ln 
10
1
.
5

10



  0.329V

1/ 2
14
 4 s  Fp 0 
   411.7 8.85 10 0.329    0.413m
xdT 0  


19
15





eN a
1
.
6

10
5

10





ox
3.9 8.85 10 14 
8
2
Cox 


6
.
9

10
F/cm
tox
500  10 8
1/ 2
The initial pre-implant threshold voltage is
VT 0  VFBO  2 Fp0
eN a xdT 0

Cox

1.6 10 5 10 0.413 10 
 1.25  20.329 
19
4
15
6.9 10 8
 0.113V
The threshold votage after implant, from Equation (7.31), is
so that
eDI
VT  VT 0 
Cox

1.6 10 D
 0.70  0.113 
19
I
6.9 108
Which gives
DI = 3.51  1011 cm2
If the uniform step implant extends to a depth of xI = 0.15 m, for
example, then the equivalent acceptor concentration at the surface is
DI 3.511011
16
-3
Ns  Na 


2
.
34

10
cm
xI 0.15 10 4
or
Ns = 2.84  1016 cm3
 Comment
The required implant dose to achieve the desired threshold voltage is DI
= 3.51  1011 cm-2. This calculation has assumed that the induced space
charge width in the channel region is greater than the ion implant depth
xI. We can show that this requirement is indeed satisfied in this example.
```
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