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Beginning - Microelectronic Devices and Circuits - Solved Exam, Exams of Microelectronic Circuits

Bharat Ratna Dr. B. R. Ambedkar UniversityMicroelectronic Circuits

Main points of this past exam are: Beginning, Common Parameters, Values, Amplifier, Small-Signal Parameters, Transistor, Minimum Number

Typology: Exams

2012/2013

Uploaded on 03/22/2013

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EE 105 Midterm-2 Solution
Prof. Ming C. Wu Spring 2007
q 1.6 10 19−C⋅:= ni 1010cm 3−
:= Vth 0.026V:=
ε0 8.854 10 14−
⋅F
cm
:= εs 11.7 ε0⋅:= εox 3.9 ε0⋅:=
mS 0.001S:= fF 10 15−F⋅:=
(1) g_m 0.1mS:= ro 100kΩ:= Cgd 10fF:= Cgs 100fF:= Cdb 10fF:=
Rs 1kΩ:= R_L 100kΩ:=
(a) Rout ro R_L⋅
ro R_L+
:= Rout 5 104
×Ω=
Av g_m−Rout⋅:= Av 5−=
(c) There are 2 poles, one at input and another at output
Cin Cgs Cgd 1 Av−()⋅+:= Cin 1.6 10 13−
×F=
Cout Cdb Cgd 1 1
Av
−
⎛
⎜
⎝
⎞
⎟
⎠
⋅+:= Cout 2.2 10 14−
×F=
ωp_in 1
Rs Cin⋅
:= ωp_in 6.25 109
×1
s
=
ωp_out 1
Rout Cout⋅
:= ωp_out 9.091 108
×1
s
=
(d) Output pole is the dominant pole. The 3-dB frequency is
f_3dB 1
2π⋅ ωp_out⋅:= f_3dB 144.686 MHz=
(2) μnCox 100 μA
V2
:= μpCox 50 μA
V2
:= V_TH 0.4V:= Vdd 2V:=
W_L1 10:= W_L2 20:= W_L3 40:= W_L4 20:= V_GS1 1V:=
(a) Id1 = Id2 (magnitude)
V_overdrive1 V_GS1 V_TH−:= V_overdrive1 0.6V=
V_overdrive2 V_overdrive1 μnCox W_L1⋅
μpCox W_L2⋅
⋅:= V_overdrive2 0.6 V=
Vx Vdd V_TH−V_overdrive2−:= Vx 1 V=
V_overdrive3 Vdd Vx−V_TH−:= V_overdrive3 0.6V=
V_overdrive4 V_overdrive3 μpCox W_L3⋅
μnCox W_L4⋅
⋅:= V_overdrive4 0.6 V=
Vout V_overdrive4 V_TH+:= Vout 1 V=
pf3

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Download Beginning - Microelectronic Devices and Circuits - Solved Exam and more Exams Microelectronic Circuits in PDF only on Docsity!

EE 105 Midterm-2 Solution Prof. Ming C. Wu Spring 2007

q := 1.6 10⋅ −^19 C ni := 10 10 cm −^3 Vth :=0.026V ε 0 8.854 10⋅ −^14 F cm := εs := 11.7 ⋅ε 0 εox :=3.9 ⋅ε 0 mS := 0.001S fF := 10 −^15 ⋅F

(1) (^) g_m := 0.1mS ro := 100kΩ Cgd := 10fF Cgs := 100fF Cdb :=10fF

Rs := 1kΩ R_L :=100kΩ

(a) (^) Rout ro R_L⋅ ro +R_L := Rout = 5 × 10 4 Ω

Av := −g_m ⋅Rout Av =− 5 (c) There are 2 poles, one at input and another at output

Cin := Cgs +Cgd ⋅( 1 −Av) Cin =1.6 × 10 −^13 F Cout Cdb Cgd 1 1 Av

:= + ⋅ Cout =2.2 × 10 −^14 F ωp_in 1 Rs Cin⋅ := ωp_in 6.25 × 1091 s

ωp_out 1 Rout Cout⋅ := ωp_out 9.091 × 1081 s

(d) (^) Output pole is the dominant pole. The 3-dB frequency is

f_3dB 1 2 ⋅ π

:= ⋅ωp_out f_3dB =144.686 MHz

(2) (^) μnCox 100 μA V^2

:= μpCox 50 μA V^2

:= V_TH := 0.4V Vdd :=2V W_L1 := 10 W_L2 := 20 W_L3 := 40 W_L4 := 20 V_GS1 :=1V (a) Id1 = Id2 (magnitude) V_overdrive1 := V_GS1 −V_TH V_overdrive1 =0.6 V V_overdrive2 V_overdrive1 μnCox W_L1⋅ μpCox W_L2⋅

:= ⋅ V_overdrive2 =0.6 V

Vx := Vdd − V_TH−V_overdrive2 Vx =1 V V_overdrive3 := Vdd − Vx−V_TH V_overdrive3 =0.6 V V_overdrive4 V_overdrive3 μpCox W_L3⋅ μnCox W_L4⋅

:= ⋅ V_overdrive4 =0.6 V

Vout := V_overdrive4 +V_TH Vout =1 V

1 3 1 3 2 4 2 4

1 3 2 4

1 3 2 4

m m m m V m m m m

n ox p ox p ox n ox

g g g g

A

g g g g

C W L C W L

C W L C W L

W L W L

W L W L

(b)

Av W_L1 W_L3⋅ W_L2 W_L4⋅

:= Av = 1

(3) (^) μnCox 100 μA V^2

:= μpCox 50 μA V^2

:= λn 0.1^1 V

:= ⋅ λp 0.1^1 V

:= ⋅ Rs := 100 Ω

(a) This is a common gate amplifier. M1 is the main amplifying transitor

(b) The current going through M4-M5 can be found from the current mirror W_L6 := 10 W_L5 := 10 W_L2 := 10 Iref := 100 μA Id5 Iref W_L W_L

:= ⋅ Id5 = 100 μA

Id2 Iref W_L W_L

:= ⋅ Id2 = 100 μA

(c) The output current is equal to the input current, with opposite sign. So Ai = -

(d) (^) W_L1 := 10

Id1 := Id2 gm1 := 2 ⋅μ nCox⋅W_L1 ⋅Id gm1 =4.472 × 10 −^4 S

2 1 1 2 1 1

o m m s in in S o S m m

r

v v g^ v g

R r R

g g

( ) e

vs_vin_ratio

1 gm Rs 1 gm

:= vs_vin_ratio =0.