9.5. MOSFET OPERATION 461
The saturation current is now, from equation 9.5.11,
ID(sat)=
(25× 10 −^4 )(600)
2(1. 5 × 10 −^4 )
(3.9)(8. 85 × 10 −^14 )
500 × 10 −^8
[5. 0 − 0 .04]^2
=8.5mA
Example 9.8Consider a silicon NMOS device at 300 K characterized by
φms=0,Na=4× 1014 cm−^3 ,dox= 200A ̊,L=1. 0 μm,Z=10μm. Calculate the
drain current for a gate voltage ofVGS= 5 V and drain voltage of 4 V. The electron
mobility in the channel is 700 cm^2 /V·s.
We start by calculating the threshold voltage. The potentialφFis given by
φF=(0.026) ln
(
4 × 1014
1. 5 × 1010
)
=0.264 V
The threshold voltage is, from equation 9.3.12,
VT =0.528 +
[
4(1. 6 × 10 −^19 )(11. 9 × 8. 85 × 10 −^14 )(4× 1014 )(0.264)
] 1 / 2
3. 9 × 8. 85 × 10 −^14
·
(
2 × 10 −^6
)
=0.58 V
The saturation voltage for a gate bias of 5 V is, from equation 9.5.7,
VDS(sat)=4.42 V
The saturation current is now, from equation 9.5.11,
ID(sat)=11.8mA
Example 9.9Consider ann-channel MOSFET with gate widthZ=10μm, gate lengthL
=2μm and oxide capacitanceCox=10−^7 F/cm^2. In the linear region, the drain current is
found to have the following values atVDS=0.1V:
VGS =1. 5 V, ID=50μA
=2. 5 V, ID=80μA
The intercept of theID−VGScurve is at− 0. 16 V, which is the threshold voltage.
Example 9.10Consider ann-channel MOSFET with a substrate doping of
Na=2× 1016
cm−^3 at 300 K. The SiO 2 thickness is 500A and a source-body bias of 1.0 V is applied. ̊
Calculate the shift in the threshold voltage arising from the body bias.
The potentialφFis given by
φF=0.026 ln
(
Na
ni
)
=0.026 ln
(
2 × 1016
1. 5 × 1010
)
=0.367 V
