Solutions for HW7: Electronic Circuits Analysis and Design, Exercises of Electronics

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Solutions for HW

Divija Gogineni

October 31, 2019

(a)

First compute RX IX = gm 2 vx + (^) rvox 2 => RX = v ixx = (^) gm 21 + (^) r 1 02 Therefore Rout = gm 1 ro 1 RX = (^) ggmm 21 +r (^) ro^11 o 2 (b)

Equivalent circuit: Therefore, Rout = gm 1 ro 1 ro 2

(c)

By KCL, Vy = ixro 2 Eqn 1 ix = gm 1 (−vy ) + gm 3 (−vy ) + (vx − vy )

ro 1 +^ r^1 o 3

Eqn 2 Substitute Eqn 1 into Eqn 2 and re-arrange Rout = v ixx = (ro 1 ||ro 3 ) + ro 2 (ro 1 ||ro 3 )(gm 1 + gm 2 ) + ro 2 ≈ ro 2 (ro 1 ||ro 3 )(gm 1 + gm 3 ) Intuitively this makes sense because we have 2 NMOSs in parallel gmvgsadds up, and ros are splitting total current, ix. This is as if an equivalent NMOS replacing M 1 and M 3 with gm = gm 1 + gm 3 and ro = (ro 1 ||ro 3 )

(d)

Rout = 40kΩ = (gm 1 ro 2 + 1)ro 1 + ro 2 gm 1 =

2 μpCox

( W

L

1 IBIAS^ =

( R

out r−ro 2 o 1 −^1

ro 2 ( (^) W L

[( R

out r−ro 2 o 1 −^1

ro 2

] 2

2 μpCoxIBIAS

{[ (^) (40kΩ)−[(0.2)(0. 5 m)]− 1 [(0.2)(0. 5 m)]−^1 −^1

]

[0. 2 ∗ 0. 5 m]

(^2) ( 50 μAV 2 )(0. 5 mA) ≈ 0. 8

(a)

RX is the input impedance of a common-gate configuration: “Looking into” the source of M 2 Rx = (^) gm^12 ||ro 2 Therefore, Rout = gm 2 ro 1 Rx = gm 1 ro 1 ( (^) gm^12 ||ro 2 )

(b)

From observation, R 3 = ro 3 (SinceVSG = 0 in AC) R 2 = ro 2 (SinceVSG = 0 in AC) Therefore, Rout = gm 1 ro 1 (R 2 ||R 3 ) = gm 1 ro 1 (ro 2 ||ro 3 )

(c)

By observation, R 2 = ro 2 (VS = VG =AC GND) R 3 = ro 3 (VS = VG =AC GND) Therefore, Rout = gm 1 ro 1 (R 2 ||R 3 ) = gm 1 ro 1 (ro 2 ||ro 3 )

(d)

Rx = gm 2 ro 2 ro 3 Rout = gm 1 ro 1 Rx = gm 1 gm 2 ro 1 ro 2 ro 3

(a)

Gm = (^) viino ≈ gm 1 Since gmro >> 1 Rp = gm 3 ro 3 ro 4 Rn = gm 2 (ro 2 ||Rp)ro 1 Therefore AV = −GmRout = −gm 1 [gm 3 ro 3 ro 4 ||gm 2 (ro 2 ||Rp)ro 1 ]

Gm = (^) viino ≈ gm 5 Since gmro >> 1 Rp = gm 3 ro 3 (ro 4 ||ro 5 ) Rn = gm 2 ro 1 ro 2 Therefore AV = −GmRout = gm 5 [gm 3 ro 3 (ro 4 ||ro 5 )||gm 2 ro 1 ro 2 ]