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Solutions for HW
Divija Gogineni
October 20 2019
Given, AV = 5 => gmRD = 5
Power= IDS ∗ VDD
Since P ≤ 1 mW
IDS ∗ 1. 8 ≤ 1 mW => IDS ≤ 0. 556 mA
gm =
2 βIDS =
2 ∗ 200 ∗ 10 −^6 ∗ ( (^020). 18 ) ∗ 0. 556 mA = 0. 00497 Ω−^1
Therefore RD = (^0). 004975 ≈ 1006Ω
(7.20)
(a)
|AV | = gm 1 (ro 1 ||r 02 ) = 10
ro 1 = (^) λ 11 I 1 = (^0). 1 ∗ 0.^15 ∗ 10 − 3 = 20kΩ ro 2 = (^) λ 21 I 1 = (^0). 15 ∗ 01. 5 ∗ 10 − 3 = 13. 3 kΩ Therefore,gm 1 = (^20) k||^1013. 3 k = 0.00138Ω−^1
gm 1 =
2 β 1 IDS => β 1 = 0. 00192 200 ∗ 10 −^6 ∗ ( WL ) 1 = 0. 00192 W L ≈^9.^6
(b) Given, ( WL ) 2 = (^020). 18
ID = 12 μpCox WL (VGS − VT H )^2
- 5 ∗ 10 −^3 = 12 (100 ∗ 10 −^6 )(1. 8 − VB − 0 .4)^2 020. 18 VB = 1. 1 V
(7.22)
(a) AV = gm 1 (ro 1 ||ro 2 )
When length of M 1 and M 2 doubles, rodoubles as it is proportional to L
ro 1 ||ro 2 = (^) rroo 11 +rro^2 o 2 ∝ L
2 L i.e (ro 1 ||ro 2 ) ∝ L gm 1 is constant because both ( WL ) 1 , 2 and IDS remain the same Therefore, Voltage gain is doubled
(b) When both length and bias current double, roremains the same
Sincegm 1 ∝
IDS
Voltage gain increases by
(a)
To find z 2 ,the impedance presented by M 2 ,apply a test voltage VT at the drain of M 2 .From teh small signal model,
Av = −gm 1 (ro 1 ||z 2 ) where z 2 is the impedance depicted on the right. The equivalent small signal is
it = gm 2 vt + (^) rvox 2 = gm 2 vt + vt− rRo 2 D^ it z 2 = v itt = (^) grmo^22 +roR 2 +1D Av = −gm 1 (ro 2 || (^) grmo^22 +roR 2 +1D )
(7.32)
Eqn 1: vo = −iiRD i 1 = gmv 1 + vo r−ov 1 = (gmro− r1)ov^1 +vo i 1 ≈ gmv 1 + v roo Eqn 2: − (^) RvoD = gmv 1 + v roo vin = v 1 + i 1 RS Eqn 3: v 1 = vin + (^) RvoD RS Eqn 2 combines with Eqn 3
− (^) RvoD = gmvin + gmvo RRDS + v roo
−vo
[
1 RD +^ gm
RS RD +^
1 ro
]
= gmvin
Therefore, Av = (^) vvino = −
[
gm ro+gmRS ro+RD
]
(roRD )
