7.11. The current source shown in Fig 7.44 are to be designed for Ix = Iy = 0.6 mA. If VB1 = 1.1 V, VB2 = 1.0 V, λ =
0.1 V–1 and L1 = L2 = 0.25 μm, calculate W1 and W2. Calculate output resistances of these current sources.
Ix 
Solution
n
0.6 mA 
2
(VGS  Vtn ) 2
n C0  W1
2  L1
W1
200  106

 (1.1  0.4) 2  196W1
6
2
0.25  10
 W1  3  m.
(Vin  Vtn ) 2 
Similarly
Iy 
n
2
W2
200  106

 (1  0.4)2  144W2
2
0.25  106
0.6 mA
 W2 
 4  m.
144
(VGS  Vt )2 
Output resistances
Rout1 = Rout2 
1
1

 16.67 k.
 I D 0.1  0.6  103
7.15. For the circuit shown in Fig 7.48, (W/L)1 = 4/0.15, (W/L)2 = 10/0.2, λ1 = 0.1 V–1 and λ2 = 0.12 V–1, determine
the inversion point Vin (= VB) for VX = 0.8 V.
Solution At inversion point, IDM1 = IDM2 and both are in saturation.
So,
W 
W 
 n Cox    (Vgs  Vtn ) 2   p Cox   (VDD  Vin  Vtp ) 2
 L n
 L p
200  10 6 
4
10
 (VB  0.4)2  100  10 6 
 (1.8  VB  0.4) 2
0.15
0.2
 VB  1.301V.
19. For the circuit shown in Fig. 7.52, calculate Rout for given data.
I D = 1mA(W / L)2 = 5 / 1 (W / L )1 = 10 / 1, l2 = 0.1V-1 , l1 = 0.1 V-1.
Solution Diode-connected load M2
rO 2 =
1
lID
=
1
0.1 ´ 10-3
= 10 kW.
Similarly
rO1 =
1
= 10 kW.
l ID
5
g m2 = 2 ´ 200 ´ 10-6 ´ ´ 1 ´ 10-3 = 0.00141S.
1
and
So,
Rout =
1
rO 2 rO1 = 709 10 kΩ 10 kΩ ; 709 W.
gm 2
7.22. For the circuit of Fig.7.55, find voltage gain (λ ≠ 0).
I D  1.5mA
(W / L) 1  10/1 (W / L)
1  2  0.1V
Solution
rO1  rO 2
2
 15 / 1
1
1
1


 6.67 k
 I D 0.1  1.5  103
g m2  2  100  106 
15
 1.5  103  0.00212S
1
10
 1.5  103  0.00173S
1
So, we know from theory of CS stage with diode-connected MOS device,
 1

Av   g m 2 
rO1 rO 2 
g
 m1

1


 0.00212 
667k 6.67k 
 0.00173

 0.00212(578)
g m1  2  100  106 
 1.22 V/V.
7.23. The CS stage shown in Fig 7.56 must achieve a gain of 7. If (W/L)2 = 2/0.18, compute required value of
(W/L)1.
Solution We know that
I D2
æ 1
ö
AV = - gm1 ç
r01 r02 ÷
ç gm
÷
è 2
ø
æ 1 ö
; - gm1 ç
÷.
ç gm ÷
è 2ø
m C æW ö
m C æW ö
= n 0 ´ ç ÷ (VGS - VT )2 = n 0 ´ ç ÷ (VGS - VT ) 2
2
2
è L ø2
è L ø2
200 ´ 10 -6
2
´ (1.8 - 0.4) 2
´
0.18
2
= 2.17 mA.
=
I D2
I D2 = I D1 .
So,
2
æW ö
´ 2.17 ´ 10-3
gm2 = 2 ´ mn C0 ´ ç ÷ ´ I D2 = 2 ´ 200 ´ 10-6 ´
0.18
è L ø2
= 0.00311S.
æW ö
g m1 = 2 ´ 200 ´ 10 -6 ´ ç ÷ ´ 2.17 ´ 10-3
è L ø1
æW ö
Þ g m1 = 868 ´ 10 -9 ´ ç ÷ .
è L ø1
æ 1 ö
æW ö é 1 ù
AV = - g m1 ç
÷ Þ 7 = 868 ´ 10 -9 ç ÷ ´ ê
ú.
ç gm ÷
L
0.00311
è
ø
ë
û
1
è 2ø
W
æ
ö
Squaring = 49 = 868 ´ 10-9 ´ ç ÷ ´ [103.3K]
è L ø1
æW ö
Þ ç ÷ = 546.
è L ø1
7.25. For the circuit shown in Fig.7.58, determine the gate voltage at which M1 operates at the edge of saturation.
Solution We know at the edge of saturation of M1
(VGS  Vt )  VDS .
Here,
(Vin  0)  VGS .
So, (Vin  Vt )  VDS . Therefore,
VDS  VDD  I D RD
 1.8  I D RD .
As
(Vin  Vt )  1.8  I D RD
 Gate voltage Vin  1.8  I D RD  Vt
7.26. For the circuit shown in Fig. 7.59, which should give gain of 5 with a bias current of 0.5 mA. Assume a drop
of 250 mV across RS and λ = 0.
(a) If RD = 2 kΩ, determine the required value of (W/L).
(b) If (W/L) = 40/0.18, determine the value of RD.
Solution
(a) RD = 2 kΩ
We know, AV   g m RD  5  g m  (2k )  g m  2.5 mS.
In solution
W
 ID
L
W
 2  200  106   0.5  103
L
g m  2n C0 
2.5  103
Vout  1.8  2  103  0.5  103  1.8  1  0.8 V.
 VDS  Vout  250 mV  0.8 V  0.25 V  0.55 V.
We have VT = 0.4 V  VDS  VT  M 1 in saturation.
(b) W/L = 40/0.18, RD = ?
ID 
n C0
2

W
200  106
40
(VGS  Vt )2  0.5  103 

 (VDS )2
L
2
0.18
 VDS  0.15 V.
KVL drain source  1.8  0.5  103  RD  VDS  250 mV  0.
 RD  2.8 kΩ.
29. The circuit in fig 7.62 has bias current of 0.8 mA. If RD = 1.5 kW, l = 0.1V -1 , compute required value of
W/L for gate voltage of 1 V. What is voltage gain of circuit?
Solution
I D = 0.8 mA, RD = 1.5 kW, l =0.1V -1.
KVL from drain-source is given by
VDD - I D RD - Vout = 0
That is,
1.8 - 0.1 ´ 10-3 ´ 1.5 ´ 103 - Vout = 0.
Þ Vout = 1.8 - 1.2 = 0.6 V = VDS .
ID =
200 ´ 10-6 W
× ´ (0.6)2 = 0.8mA
L
L
2
2
0.8 ´ 10-3 ´ 2
Þ W /L =
= 22.22 ; 23.
(0.6) 2 ´ 200 ´ 10-6
mn C0 W
´
(VGS - Vt )2 =
gm = 2 ´ 200 ´ 10-6 ´
23.0
´ 0.8 ´ 10-3 = 7360 ´ 10-9 = 0.00085
1
= 0.85 ms.
Gain = g m RD = 0.85 ´ 10 -3 ´ 1.5 kW
=1.275 V/V.