6.44 (a) 6. Q(x) = W C ox ox (V GS − V (x) − V T H )
= W C ox ox (V GS − V T H ) − W C ox ox V (x) )
x
(
Q
W C ox (V GS − V T H )
Increasing V DS
L
x
The curve that intersects the axis at x = L (i.e., the curve for which the channel begins to pinch off) corresponds to V DS = V GS − V T H . (b) RLocal (x) ∝
1 µQ(x)
)
x
(
l a c o L
R
Increasing V DS
L
x
Note that R Local diverges at x = L when V DS = V GS − V T H .
6.15 I D
Increasing V DS
V T H
V GS
Initially, when V GS is small, the transistor is in cutoff and no current flows. Once V GS increases beyond V T H , the curves start following the square-law characteristic as the transistor enters saturation. However, once V GS increases past V DS + V T H (i.e., when V DS < V GS − V T H ), the transistor goes into triode and the curves become linear. As we increase V DS , the transistor stays in saturation up to larger values of V GS , as expected.
6.17 I D = gm
1 W µn C ox (V GS − V T H )α , α < 2 2 L
∂I D ∂V GS α W 1 = µn C ox (V GS − V T H )α− L 2
=
αI D V GS − V T H
6.21 Since they’re being used as current sources, assume M 1 and M 2 are in saturation for this problem. To find the maximum allowable value of λ , we should evaluate λ when 0.99I D2 = I D1 and 1.01I D2 = I D1 , i.e., at the limits of the allowable values for the currents. However, note that for any valid λ (remember, λ should be non-negative), we know that I D2 > I D1 (since V DS 2 > V DS 1 ), so the case where 1.01I D2 = I D1 (which implies I D2 < I D1 ) will produce an invalid value for λ (you can check this yourself). Thus, we need only consider the case when 0 .99I D2 = I D1 . 1 W 2 0.99I D2 = 0.99 µn C ox (V B − V T H ) (1 + λV DS 2 ) L 2 = I D1 1 W 2 = µn C ox (V B − V T H ) (1 + λV DS 1 ) 2 L 0.99 (1 + λV DS 2 ) = 1 + λV DS 1 λ = 0.02 V −1
5.27 V DD
−
I D RD = V GS = V T H +
2I D µn C ox W L
= ( V DD
−
2I D
µn C ox W L
V T H − I D RD )
W 1 (V DD I D = µn C ox L 2
−
2
V T H )
2
−
2I D RD (V DD
−
2
2
V T H ) + I D RD
We can rearrange this to the standard quadratic form as follows:
W 2 W 1 2 µn C ox R D I D − µn C ox R D (V DD 2 L L
−
V T H ) + 1
I D +
W 1 µn C ox (V DD 2 L
−
2
V T H ) = 0
Applying the quadratic formula, we have: I D =
=
=
W ox L
µn C
RD (V DD
µn C ox W RD (V DD L
−
−
µn C ox W R D (V DD L
V T H ) + 1
±
V T H ) + 1 ±
−
V T H ) + 1 ±
µn C ox W R D (V DD L
2
1 2
−
2 µn C ox W RD L
µn C ox W R D (V DD L
−
V T H ) + 1
µn C ox W R 2D L
RD (V DD 1 + 2 µn C ox W L
2
V T H ) + 1
−
2
−
−
4
1 2
µn C ox W R D (V DD L
µn C ox W RD (V DD L
−
−
V T H )
V T H )
2 µn C ox W R D L
Note that mathematically, there are two possible solutions for I D . However, since M 1 is diodeconnected, we know it will either be in saturation or cutoff. Thus, we must reject the value of I D that does not match these conditions (for example, a negative value of I D would not match cutoff or saturation, so it would be rejected in favor of a positive value).
V T H )
2
2
6.33 (a) Assume M 1 is operating in saturation. V GS = 1 V
1 W 2 µn C ox (V GS − V T H ) (1 + λV DS ) RD 2 L V DS = 1 .35 V > V GS − V T H , which verifies our assumption I D = 4 .54 mA V DS = V DD − I D RD = V DD −
gm = µ nC ox ro =
1 λI D
W (V GS − V T H ) = 13.333 mS L
= 2.203 kΩ
+ vgs
gm vgs
ro
RD
−
(b) Since M 1 is diode-connected, we know it is operating in saturation. V GS = V DS = V DD − I D RD = V DD −
W 1 2 (V GS − V T H ) (1 + λV GS ) RD µnC ox L 2
V GS = V DS = 0 .546 V I D = 251 µ A gm = µ n C ox ro =
1 λI D
W (V GS − V T H ) = 3.251 mS L
= 39.881 kΩ
+ vgs
gm vgs
ro
RD
−
(c) Since M 1 is diode-connected, we know it is operating in saturation. I D = 1 mA gm = ro =
2µn C ox
1 λI D
W I D = 6.667 mS L
= 10 kΩ
+ vgs
gm vgs
ro
−
(d) Since M 1 is diode-connected, we know it is operating in saturation. V GS = V DS W 1 (V GS − V T H )2 (1 + λV GS ) (2 kΩ) µn C ox L 2 V GS = V DS = 0 .623 V I D = 588 µ A
V DD − V GS = I D (2 kΩ) =
gm = µ n C ox ro =
1 λI D
W (V GS − V T H ) = 4.961 mS L
= 16.996 kΩ
+ vgs
gmvgs
2 kΩ
ro
−
(e) Since M 1 is diode-connected, we know it is operating in saturation. I D = 0 .5 mA gm = ro =
2µn C ox
1 λI D
W I D = 4.714 mS L
= 20 kΩ
+ vgs
−
gmvgs
ro
6.38 (a) vout
+ vgs2
gm2 vgs2
ro2
gm1 vgs1
ro1
RD
−
vin
+ vgs1
−
(b) vin
vout
+ vgs1
gm1 vgs1
ro1
RD
−
+ gm2 vgs2
ro2
vgs2
−
(c) vin
vout
+ vgs1
gm1 vgs1
ro1
gm2 vgs2
ro2
−
+ vgs2
−
(d)
RD
vin
+ vgs1
gm1 vgs1
ro1
− vout
+ vgs2
gm2 vgs2
ro2
−
(e) vout
+ vgs1
gm1 vgs1
ro1
RD
− vin
+ gm2 vgs2
ro2
vgs2
−
6.43 (a) Assume M 1 is operating in triode (since | V GS | = 1.8 V is large). |V GS | = 1.8 V W 1 2 2 (|V GS | − |V T H |) |V DS | − |V DS | (500 Ω) V DD − | V DS | = | I D | (500 Ω) = µ p C ox L 2 |V DS | = 0.418 V < |V GS | − |V T H | , which verifies our assumption
|I D | = 2.764 mA (b) Since M 1 is diode-connected, we know it is operating in saturation. |V GS | = | V DS | W 1 (|V GS | − | V T H |)2 (1 kΩ) µ p C ox L 2 |V GS | = | V DS | = 0.952 V
V DD − | V GS | = | I D |(1 kΩ) =
|I D | = 848 µ A (c) Since M 1 is diode-connected, we know it is operating in saturation. |V GS | = |V DS | |V GS | = V DD − | I D |(1 kΩ) = V DD − | I D |(1 kΩ) = |V GS | = |V GS | = 0.952 V |I D | = 848 µ A
W 1 2 µ p C ox (|V GS | − |V T H |) (1 kΩ) L 2
6.44 (a) I X
Saturation
Cutoff
V DD − V T H
V DD
V X
V DD
V X
M 1 goes from saturation to cutoff when V X = V DD − V T H = 1 .4 V.
(b) I X
1 + V T H
Saturation
M 1 goes from saturation to triode when V X = 1 + V T H = 1 .4 V.
(c)
Triode
I X V DD − V T H
Saturation
V DD
V X
V DD
V X
Cutoff
M 1 goes from saturation to cutoff when V X = V DD − V T H = 1 .4 V.
(d) I X
Cutoff
Saturation
V T H
M 1 goes from cutoff to saturation when V X = V T H = 0 .4 V.
