ELET 354 Electronics I
Fall 2007
MOS Circuits at DC
Triode Region: vGS ≥ Vt (induced channel) and vDS ≤ vGS – vt (continuous channel)
iD = kn’ (W/L) [ ( vGS – Vt )vDS – ½ vDS2 ]
iD ≈ kn’ (W/L) ( vGS – Vt ) vDS
rDS = ( ΔvDS / Δi D ) for vDS
small
(linear)
= [ kn’(W/L) ( VGS – Vt )] –1
Saturation Region: vGS ≥ Vt (induced channel) and vDS ≥ vGS – vt
( pinched-off channel)
iD = ½ kn’ (W/L) ( vGS – Vt )2
Boundary between triode and saturation regions:
Cut off Region:
vDS = vGS – vt
vGS < Vt
kn’ = μn Cox
μn is the electron mobility in Si = 1360 cm2/Vs
Cox is the capacitance per unit gate area of the oxide layer = εox / tox
tox is the thickness of the oxide layer
εox = 3.9 εo = 3.9 x 8.85 x 10 –12 = 3.45 x 10 –11 F/m
εox is the dielectric or permittivity of the silicon dioxide
εo is the vacuum permittivity (or dielectric constant)
Example 1.
Design the following NMOS circuit so that the transistor operate at ID = 0.3 mA and VD = +0.4V. Consider that
the NMOS transistor has Vt = 1.0 V, μn Cox = 60 μA/V2, L = 3 μm, W = 120 μm, and VDD = – VSS = 2.5 V
Ans: Rs = 3.3 kΩ, RD = 7 kΩ
R. Alba-Flores
ELET 354 Electronics I
Fall 2007
Example 2.
Design the following NMOS circuit so that the transistor operate at ID = 0.16 mA. Consider that the NMOS
transistor has Vt = 0.6 V, μn Cox = 200 μA/V2, L = 0.8 μm, W = 8 μm, and VDD = 3 V
Notice that VDS = VGS
Therefore:
VDS >> VGS– Vt
is true, and
the transistor is in the saturation region
thus
2
ID = ½ [μn Cox (W/L)] ( VGS – Vt )
solving for VGS – Vt
( VGS – Vt )2 = ( 2 ID ) / [ μn Cox (W/L) ]
2
2
Substituting values, we obtain: ( VGS – Vt ) = 0.16 V ;
VGS – Vt = 0.4 V ; VGS = 0.4 + Vt = 1 V
VGS = 1 V, VDS = VD = 1 V
RD = (VDD – VD ) / ID = (3 V – 1V) / 0.16 mA = 12.5 kΩ
Example 3.
Assume that a second NMOS transistor is connected to the previous example, as shown below. Consider that
RD2 = 10 kΩ, VDD = 3 V, and Q1 = Q2. Find ID2 and VD2
From example 2 we have:
ID1 = 0. 16 mA, RD1 = 12.5k Ω, VD1 = 1 V
VGS2 = VGS1 =VDS1 =1 V,
Assuming that:
VDS2 > ( VGS2 – Vt2 ) ( = 1V– 0.6V =0.4V) (saturation)
Then we use the equation
ID2 = ½ [μn Cox (W/L)] ( VGS2 – Vt 2 )
2
substituting values:
ID2 = 160 μΑ = 0.16 mΑ
VD2 = VDD – ID2 RD2 = 3 V – (0.16 mA)(10k Ω ) = 1.4 V
R. Alba-Flores
ELET 354 Electronics I
Fall 2007
Example 4.
Design the following NMOS circuit so that the transistor operate at VD = 0.05 V. Consider that the NMOS
transistor has Vt = 1 V, k'n ( W / L) = 1 mΑ/V2 , and VDD = 5 V. What is the effective resistance between drain
and source at this operating point ?
Notice that:
VDS = 0.05 V < VGS – Vt = 5V – 1V = 4 V
Therefore that transistor is in the Triode Region
ID = kn’ (W/L) [ ( VGS – Vt )VDS – ½ VDS2 ]
substituting values we obtain:
ID = 0.2 mA
RD = ( VDD – VD ) / ID = ( 5 V – 0.05V ) / 0.2 mA = 24.75 kΩ
The effective Drain to Source resistance is:
rDS = VDS/ ID = 0.05V/0.2mA = 250 Ω
Example 5.
For the circuit shown below, assume that the NMOS transistor has Vt = 1 V, k'n ( W / L) = 1 mΑ/V2 , and
RG1 = R G2 = 10MΩ, RD = RS = 6kΩ, and VDD = 10 V.
a) Determine the voltages at all nodes and the currents through all branches
b) What is the largest value that RD can have while the transistor remains in the saturation mode
Because IG = 0, we can apply voltage divider to the gate:
VG= RG2/(RG1+RG2) VDD = 5 V
Because VG > Vt the transistor is on
Assume saturation region, solve the problem, and then verify
Notice that: VS = (ID )(RS ) = 6 ID
Then:
(ID in mA)
VGS = VG – VS = 5 V – 6 ID
ID = ( ½ kn’)(W/L)( VGS – Vt )2 = ( 1/2 mA/V2)(5 V – 6 ID – 1V )2
18 ID2 - 25 ID + 8 = 0
Solving for ID we obtain: ID = 0.89 mA and ID = 0.5 mA
Notice that if we take ID = 0.89 mA then VS = ID RS = 0.89 mA ( 6 kΩ) = 5.34 V,
Then VS > VG and the transistor is OFF (this does not make sense). Therefore, take the ID = 0.5 mA
VS = 0.5 mA (6 kΩ ) = 3V ; VGS = VG – VS = 5 V – 3 V = 2V
VD = VDD – ID RD = 10 V – (0.5mA)(6 kΩ) = 7 V
Because VD > ( VG - Vt ) the transistor is in saturation, as assumed
b) In the limit of saturation we have :
VDS = VD – VS
then
VDS = VGS – Vt = 2 – 1 = 1 V
VD = VDS + VS = 1 V + 3 V = 4 V
RD = ( VDD – VD ) / ( ID) = (10 V – 4 V) / 0.5 mA = 12 kΩ
R. Alba-Flores