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Solutions for HW8: Determining Rpsuch when I1 = 2IREF, Exercises of Electronics

The University of Texas at AustinElectronics

The solutions for problem 8 in a circuit analysis homework. It covers the calculation of Rp, the output resistance of a MOSFET transistor, when I1 = 2IREF. The document also includes the derivation of the equation for Rp and its independence of VT H. Additionally, it discusses the relationship between ID1, ID2, and Gm, the transconductance of the MOSFET.

Typology: Exercises

2019/2020

Uploaded on 01/17/2020

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Solutions for HW8
Divija Gogineni
November 7, 2019
(9.50)
Determine Rpsuch that I1= 2IREF
First calculate VGS
VGS1=q2I1
µnCox(W
L)+VT H = 2qIREF
µnCox(W
L)+VT H Eqn 1
Assuming I1is in saturation:
IREF =1
2µnCox W
L(VGS,REF −VT H )2=1
2µnCox W
L(VGS1−IREF RP−VT H )2
Substitute Eqn 1 into IREF :
IREF =1
2µnCox W
L2qIREF
µnCox(W
L)−IREF Rp2
Eqn 2
Solve for Rp=(2−√2)
pIREF µnCox(W
L)
From Eqn 2, we find that Rpis independent of any change in VTH ,∆V
(9.53)
VGS,REF =VGS,1=VGS
1
pf3
pf4
pf5

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Download Solutions for HW8: Determining Rpsuch when I1 = 2IREF and more Exercises Electronics in PDF only on Docsity!

Solutions for HW

Divija Gogineni

November 7, 2019

Determine Rpsuch that I 1 = 2IREF

First calculate VGS

VGS 1 =

2 I 1

μnCox( WL )

+ VT H = 2

IREF

μnCox( WL )

+ VT H Eqn 1

Assuming I 1 is in saturation:

IREF =

2 μnCox

W

L

(VGS,REF − VT H )

2 μnCox

W

L

(VGS 1 − IREF RP − VT H )

Substitute Eqn 1 into IREF :

IREF =

2 μnCox

W

L

[

IREF

μnCox( WL )

− IREF Rp

] 2

Eqn 2

Solve for Rp =

(^2 −

IREF μnCox( WL )

From Eqn 2, we find that Rpis independent of any change in VT H , ∆V

VGS,REF = VGS, 1 = VGS

(a)

IREF =

2 μnCox

W

L

(VGS − VT H )

(1 + λVGS )

Icopy =

2 μnCox

W

L

(VGS − VT H )

(1 + λVDS 1 )

For IREF = Icopy => VDS 1 = VGS

(b)

IREF

Icopy =^

1+λVGS 1+λ(VGS −VT H )

Icopy = IREF

λVT H 1+λVGS

ID 1 − ID 2 =

μnCox

W

L

(Vin 1 − Vin 2 )

4 ISS

μnCox WL

− (Vin 1 − Vin 2 )

Vov = (VGS − VT H )equil =

ISS

μnCox WL

=> μnCox

W

L

ISS

V (^) ov^2

ID 1 − ID 2 =

ISS

(Vin 1 −Vin 2 ) V (^) ov^2

4 V ov^2 − (Vin 1 − Vin 2 )

Gm =

∂(ID 1 −ID 2 )

∂(Vin 1 −Vin 2 ) =^

ISS

2 V (^) ov^2

[√

4 V ov^2 − (Vin 1 − Vin 2 )

(Vin 1 −Vin 2 )^2

4 V (^) ov^2 −(Vin 1 −Vin 2 )^2

]

ISS

2 V (^) ov^2

4 V (^) ov^2 −2(Vin 1 −Vin 2 )^2

4 V (^) ov^2 −(Vin 1 −Vin 2 )^2

μnCox

W

L

4 ISS μnCox WL

−2(Vin 1 −Vin 2 )^2

4 ISS μnCox WL

−(Vin 1 −Vin 2 )^2

Vin 1 − Vin 2 = 0 => Gm,mox =

μnCox

W

L

ISS

From problem 46,

Gm,mox =

μnCox

W

L ISS^ =

ISS

V ov^2 ISS^ =^

ISS

Vov

if Gm =

ISS

Vov we have

(b)

Rout = ro 3 ||

gm 1 ro 1

RSS

+ ro 1 +

RSS

To calculate Gm 1

vs ro 1 ||

RSS 2

+ gm 1 vs = gm 1 vin => vs =

gm 1 vin gm 1 + (^) R^1 SS 2 ||ro^1

Gm =

io vin

vs RSS 2

vin

2 gm 1 RSS

gm 1 + (^) R^1 SS 2 ||ro^1

Av = −GmRout

(c)

Rout = (gm 3 ro 3 RS + ro 3 + RS ) ||

gm 1 ro 1

RSS

2 +^ ro^1 +^

RSS

Gmfor this circuit is equal to the one for part (b) so:

Gm = +

2 gm 1 RSS

gm 1 + (^) R^1 SS 2 ||ro^1

=> Av = −GmRout