UNIVERSITY OF CALIFORNIA AT BERKELEYCollege of Engineering

Department of Electrical Engineering and Computer Sciences

Final Exam Solutions

EE 105 Spring 2007Prof. Wu

1. (a) (B) holes

(b) (A) electrons

(c) (A) electrons

(d) (B) holes

(e) VFB = φp − φp+ = (60mV) log Na

ni+ 550mV = −300mV + 550mV = 250mV

(f)

Vox =1

Cox

√2qεsNa (−2φp) (1)

=tox

εox

√2qεsNa (−2φp) (2)

=10nm3.9ε0

√2qεs1015 (600mV) (3)

= 40.88mV (4)

(g) R = VT

10mA = 2.6Ω

(h) Recall the equation for the junction capacitance of a PN junction:

Cj =Cj0√1− VD

φbi

(5)

So the largest capacitance will occur with the smallest reverse bias VD. Thus, the ratio is asfollows:

Cj,max

Cj,min=

Cj(−1)Cj(−10)

(6)

=Cj0√1 + 1

φbi

√1 + 10

φbi

Cj0(7)

=

√1 + 10

φbi√1 + 1

φbi

(8)

=

√1 + 10

720mV

1 + 1720mV

(9)

= 2.497 (10)

1

2. (a)

VX = VCC − VEB,3 (11)

= 3− VT lnIREF

IS,3(12)

= 2.341V (13)

(b) Since the emitter area of Q2 is 10 times more than that of Q3, the collector current of Q2 is also10 times larger. This means the collector current of Q1 is also 10 times larger, so IC,1 = 1mA.

gm,1 =IC

VT= 38.46mS (14)

rπ,1 =β

gm,1= 2.6kΩ (15)

ro,1 =VA

IC= 50kΩ (16)

(c) Noting that ro2 = ro1, we have:

Av = − gm,1 (ro,2||RL)

1 + RE

(gm,1 + 1

rπ,1

)+ RS

rπ,1

(17)

= − gm,1 (ro,1||RL)

1 + RE

(gm,1 + 1

rπ,1

)+ RS

rπ,1

(18)

= −31.87 (19)

(d) Rin = RS + rπ,1 + (1 + β) RE = 104.6kΩ

(e)

IREF = IC,3 + IB,3 + IB,2 (20)

=(

1 +1β

)IC,3 +

1β

IC,2 (21)

=(

1 +1β

+10β

)IC,3 (22)

IC,3 = 90.09µA (23)

IC,1 = IC,2 = 10IC,3 = 900.9µA (24)

g′m,1 = 34.65mS (25)

r′π,1 = 2.886kΩ (26)

r′o,1 = 55.5kΩ (27)

A′v = −34.03 (28)

Error =A′

v −Av

A′v

(29)

= 0.06351 (30)

= 6.351% (31)

2

3. (a)

VY = VT lnIREF

IS,3= 0.6585V (32)

VX = VY + VGS,7 (33)

= VY +

√2IREF

(W/L)7 µnCox+ VTn (34)

= 1.512V (35)

(b)

AE,2

AE,3= 4 (36)

AE,2 = 4AE,3 (37)

= 16µm2 (38)

(c)

Rout = Rup||Rdown (39)Rup = gm,4ro,4ro,1 + ro,1 + ro,4 (40)

Rdown = gm,5ro,5ro,2 + ro,2 + ro,5 (41)

gm,4 =√

2 (W/L)4 µpCoxID (42)

=√

2 (W/L)4 µpCox4IREF (43)

= 3.2mS (44)

gm,5 =√

2 (W/L)5 µnCox4IREF (45)

= 2.263mS (46)

ro,4 = ro,5 =1

λID(47)

=1

λ4IREF(48)

= 50kΩ (49)

ro,1 = ro,2 =VA

IC(50)

=VA

4IREF(51)

= 25kΩ (52)

Rout = 1.695MΩ (53)

3

4. (a)

12

(W

L

)1

µpCox (VSG,1 + VTp)2 (1 + λVSD,1) =

12

(W

L

)2

µnCox (VGS,2 − VTn)2 (1 + λVDS,2)

(54)

VGS,2 = VB (55)VSG,1 = VDD − VB (56)VDS,2 = Vout = 1V (57)VSD,1 = VDD − Vout = 1V (58)

(1.5− VB)2 (1 + λ) = 2 (VB − 0.5)2 (1 + λ) (59)

(1.5− VB)2 = 2 (VB − 0.5)2 (60)

1.5− VB = ±√

2 (VB − 0.5) (61)

1.5− VB = ±√

2VB ∓ 0.5√

2 (62)

1.5± 0.5√

2 = x(1±

√2)

(63)

VB =1.5± 0.5

√2

1±√

2(64)

= 0.9142V or − 1.914V (65)

VB = 0.9142V (66)

(b) This is was solved in the homework, but even if you didn’t remember you could easily re-deriveit from the small-signal model.

Av = − (gm,1 + gm,2) (ro,1||ro,2||RL) (67)gm,1 = (W/L)1 µpCox (VSG,1 + VTp) (1 + λVSD,2) = 3.075mS (68)gm,2 = (W/L)2 µnCox (VGS,1 − VTn) (1 + λVDS,1) = 4.349mS (69)

ID =12

(W

L

)2

µnCox (VGS,1 − VTn)2 (1 + λVDS,1) = 900.8µA (70)

ro,1 =1

λID= 22.20kΩ (71)

ro,2 = 22.20kΩ (72)

Av = −6.811 (73)

(c)

ωp,in =1

RS [2CGS + 2CGD (1−Av)](74)

= 8.467Grad/s (75)

ωp,out =1

(ro,1||ro,2||RL) [2CDB + 2CGD (1− 1/Av)](76)

= 66.19Grad/s (77)

ω3dB = min (ωp,in, ωp,out) (78)= 8.467Grad/s (79)

= 1.348GHz (80)

4

5. (a) We ignore channel length modulation for parts of this problem not involving small-signal param-eters.

ID,1 = 2IBIAS = 200µA (81)

ID,3 = 10IBIAS = 1mA (82)

(b)

Av = [−gm,1 (ro,1||ro,2)] [−gm,3 (ro,3||ro,4)] (83)= gm,1gm,3 (ro,1||ro,2) (ro,3||ro,4) (84)

gm,1 =√

2 (W/L)1 µpCoxID,1 = 632.5µS (85)

gm,3 =√

2 (W/L)3 µnCoxID,3 = 4.472mS (86)

ro,1 =1

λID,1= 100kΩ (87)

ro,2 = ro,1 = 100kΩ (88)ro,3 = 20kΩ (89)ro,4 = ro,3 = 20kΩ (90)

Av = 1, 414 (91)

Rout = ro,3||ro,4 (92)

= 10kΩ (93)

(c) For large RL (ten times ro,3 and ro,4), a negligible amount of current will flow through the load.We can do our normal analysis using Vdsat:

Vout,min = Vdsat,3 =

√2ID,3

(W/L)3 µnCox= 447.2mV (94)

Vout,max = VDD − Vdsat,4 = 3−

√2ID,4

(W/L)4 µpCox= 2.368V (95)

(d) For small RL (100 times smaller than ro,3 and ro,4), we have to worry about being able to supplyenough current to the load if we swing too high or too low. We need to calculate the maximumand minimum currents we can supply. This is fixed by VGS,3, since VSG,4 is already fixed by thecurrent mirror.

5

ID3,max =12

(W

L

)3

µnCox (VGS3,max − VTn)2 (96)

VGS3,max = VDD − Vdsat,1 = 3−

√2ID,1

(W/L)1 µpCox= 2.368V (97)

ID3,max = 17.44mA (98)

ID3,min =12

(W

L

)3

µnCox (VGS3,min − VTn)2 (99)

VGS3,min = Vdsat,2 =

√2ID,2

(W/L)2 µnCox= 447.2mV < VTn ⇒ VGS3,min = VTn = 0.5V (100)

VGS3,min = VTn = 0.5V (101)ID3,min = 0 (102)Iout,max = ID,4 − ID3,min = 1mA (103)Iout,min = ID,4 − ID3,max = 1mA− 17.44mA = −16.44mA (104)

Vout,max = Iout,maxRL = 0.1V (105)

Vout,min = Iout,minRL = −1.644V (106)

Note that these are AC values.

6