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Solutions for HW9: Common-Mode and Differential-Mode Analysis of Amplifiers, Exercises of Electronics

The University of Texas at AustinElectronics

Microelectronics Homework and solutions

Typology: Exercises

2019/2020

Uploaded on 01/17/2020

delawer
delawer 🇺🇸

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Solutions for HW9
Divija Gogineni
November 14, 2019
(10.67)
The same value for the inputs common-mode leads to the following
circuit:
gm1=gm2=2ISS /2
(VGS−VT h )eq =ISS
(VGS−VT H )eq
4Vout,CM
4Vin,CM =−RD/2
1
2gm1+ro3=−RD
1
gm+2ro3=−RD
(VGS−VT H )eq
ISS +2
λISS
ACM =−RDISS
2
λ+(VGS−VT H )eq
(10.68)
1
pf3
pf4

Partial preview of the text

Download Solutions for HW9: Common-Mode and Differential-Mode Analysis of Amplifiers and more Exercises Electronics in PDF only on Docsity!

Solutions for HW

Divija Gogineni

November 14, 2019

The same value for the inputs common-mode leads to the following

circuit:

gm 1 = gm 2 =

2 ISS / 2

(VGS −VT h)eq =^

ISS

(VGS −VT H )eq 4 Vout,CM 4 Vin,CM

RD / 2

1 2 gm 1 +ro^3

−RD

1 gm +2ro^3

−RD

(VGS −VT H )eq ISS +^

2 λISS

ACM = −

RD ISS

2 λ +(VGS^ −VT H^ )eq

λ > 0 , gmro >> 1

ro 3 = ro 4 ; ro 1 = ro 2 ; gm 1 = gm 2

For the common-mode input we have:

RP = ro 3 ||ro 4 =

ro 3 2

ro 4 2

RN = ro 5 +

ro 1 2

+ 2gm 1

ro 1 2

ro 5 = gm 1 ro 1 ro 5 + ro 5 +

ro 1 2 io vin,CM

= Gm =

2 gm 1 vgs 1 vin 1 ,CM

= 2gm 1

1 2 gm 1 1 2 gm 1 +ro^5

Gm =

2 gm 1 1+2gm 1 ro 5

ro 5

ACM = −GmRo

Ro = RP ||RN =

ro 4 2

gm 1 ro 1 ro 5 + ro 5 +

ro 1 2

ro 4 2

||gm 1 ro 2 ro 5 ≈

ro 4 2

ACM = −

ro 5

ro 4 2

ro 4 2 ro 5

(a)

To calculate ADM , using the half circuit, we have

ADM =

vout vin 1

= −gm 1 RD

Similar to part (a) we have:

ACM −DM = −

4 RD

1 gm 1 +2[gm^3 ro^3 ro^4 +ro^3 +ro^4 ]

CM M R =

ADM

ACM −DM = (1 + 2gm^1 [gm^3 ro^3 ro^4 +^ ro^3 +^ ro^4 ])^

RD

∆RD

Notice that CMMR of part(b) is much higher than the one for part (b)