ECE102 (Fall 2012), Solution of the Final

Problem 1. Find the mid-band gain and the lower cut-off frequency of the amplifier

below (µnCox(W/L) = 20 mA/V2, Vt = 0.6 V, λ = 0, and Cc1 = Cc2 = 1 µF).

C c1

C c2 vovi

vsig

1k 10k

1.8 V

50k

1k

Bias: (VG = 1.8 V because IG = 0)

1.8 = VOV + Vt + 103ID

ID = 0.5µnCox(W/L)V 2

OV

1.8 = VOV + 0.6 + 103 × (0.5× 20× 10−3V 2

OV )

10V 2

OV + VOV − 1.2 = 0 → VOV = 0.3 V

ID = 0.5× 20× 10−3V 2

OV = 0.9 mA

gm =2IDVOV

= 6× 10−3 A/V & ro = ∞

Signal: Source Follower

vovi

=gm(ro ‖ R′

L)

1 + gm(ro ‖ R′

L)

ro ‖ R′

L = R′

L = 1 k ‖ 10 k = 909 Ω

vovi

=6× 10−3 × 909

1 + 6× 10−3 × 909= 0.845

vivsig

=50 k

50 k + 1 k= 0.980

vovsig

=vovi

×vivsig

= 0.828

C c1vsig

1k

50k

C c2 vo

1/gm 1k 10k

Cc1 : Rcc1 = 50 k + 1 k = 51 k

fp1 =1

2π(51× 103 × 10−6)= 3.12 Hz

Cc2 : Rcc2 = 10 k + (1 k ‖ 167) = 10.1 k

fp2 =1

2π(10× 103 × 10−6)= 15.7 Hz

fp = fp1 + fp2 = 3.12 + 15.7 = 18.8 Hz

Problem 2. The amplifier below should achieve a gain of 10 at a bias current of 0.5 mA.

In addition to vi, a DC voltage of VG1 is applied to Q1 (not shown). Find the required

W1, VG1, and VG2 (µnCox = 200 µA/V2, µpCox = 100 µA/V2, Vtn = |Vtp| = 0.4 V,

λn = 0.10 /V, λp = 0.15 /V, L1 = L2 = 0.18 µm, and W2 = 20 µm).

vi

vo

VG2 Q2

Q1

3.6 VCS Amp:

ID1 = ID2 = 0.5 mA

gm1 =2IDVOV 1

ro1 =1

λnID, ro2 =

1

λpID∣

∣

∣

∣

vovi

∣

∣

∣

∣

= gm1(ro1 ‖ ro2)

∣

∣

∣

∣

vovi

∣

∣

∣

∣

=2IDVOV 1

×

(

1

λnID‖

1

λpID

)

=2IDVOV 1

×1

(λn + λp)ID=

2

(λn + λp)VOV 1

10 =2

(0.1 + 0.15)VOV 1

→ VOV 1 = 0.8 V

ID1 = 0.5µnCox(W/L)1V2

OV 1

0.5× 10−3 = 0.5× 200× 10−6(W/L)1(0.8)2

(W/L)1 = 7.81 → W1 = 7.81× L1 = 1.41 µm

Bias:

Q1 : VGS1 = VOV 1 + Vtn = 0.8 + 0.4 = 1.2

VG1 = VGS1 − VS1 = 1.2− 0 = 1.2 V

Q2 : ID2 = 0.5µpCox(W/L)2V2

OV 2

0.5× 10−3 = 0.5× 100× 10−6(20/0.18)V 2

OV 2→ VOV 2 = 0.30 V

VSG2 = VOV 2 + |Vtp| = 0.3 + 0.4 = 0.7

VG2 = VS2 − VSG2 = 3.6− 0.7 = 2.9 V

Problem 3. Transistors in the differential amplifier circuit below all have the same VOV

and the same λ (Q1 and Q2 are identical. Q3 and Q4 are also identical). Find vo/vi in

terms of gm1 and ro1.

−VSS

VDD VDD

VG3

VG5

vi

v2v1

vo

Q2Q1

Q4Q3

Q5

+−

First, we find all gm and ro in terms of gm1 and ro1:

ID1 = ID2 = ID3 = ID4 = 0.5ID5

VOV 1 = VOV 2 = VOV 3 = VOV 4 = VOV 5

gm =2IDVOV

→ gm1 = gm2 = gm3 = gm4

ro =1

λID→ ro1 = ro2 = ro3 = ro4 = 2ro5

Decompose v1 = vi and v2 = 0 into differential and common-mode signals and note that

Ac = 0 because of the circuit symmetry:

vd = v2 − v1 = −vi

vc = 0.5(v2 + v1) = 0.5vi

vo = vo2 − vo1 = Advd + Acvc = Advd = −Advi

vo/vi = −Ad

VDD

0.5v d−

ro3

vo1,d

Q1

Differential half-circuit is shown:

Ad =vo1,d

−0.5vd= −gm1(ro1 ‖ ro2) = −0.5gm1ro1

vo/vi = 0.5gm1ro1

Problem 4. The amplifier below has a power dissipation of 4 mW at its bias point (ignore

power dissipation in the 100 k load). Find its mid-band gain. (µnCox = 200 µA/V2,

µpCox = 50 µA/V2, λn = 0.1 /V, λp = 0.2 /V, W/L = 20/0.18 for all transistors.)

vi

VG2

VG3

VG4

R1

R2

Ri2

2V−

vo

Q1

Q2

Q4

Q3

2V

100k

Bias:

ID1 = ID2 = ID3 = ID4 =P

2− (−2)= 1 mA

ID1 = 0.5µnCox(W/L)V 2

OV 1

10−3 = 0.5× 200× 10−6(20/0.18)V 2

OV 1

VOV 1 = 0.30 V

ID2 = ID1 → VOV 2 = VOV 1 = 0.30 V

ID3 = 0.5µpCox(W/L)V 2

OV 3

10−3 = 0.5× 50× 10−6(20/0.18)V 2

OV 3→ VOV 3 = 0.60 V

ID4 = ID3 → VOV 4 = VOV 3 = 0.60 V

gm =2IDVOV

& ro =1

λID

gm1 = gm2 =2× 10−3

0.3= 6.67 mA/V & gm3 = gm4 =

2× 10−3

0.6= 3.33 mA/V

ro1 = ro2 =1

0.1× 10−3= 10 k & ro3 = ro4 =

1

0.2× 10−3= 5 k

From elementary R forms:

R1 = ro3(1 + gm3ro4) + ro4 = 5× 103(1 + 3.33× 10−3 × 5× 103) + 5× 103 = 93.3 k

R2 = ro2(1 + gm2ro1) + ro1 = 10× 103(1 + 6.67× 10−3 × 10× 103) + 10× 103 = 687 k

Q2 (CG Amp): R′

L = RL ‖ R1 = 100 k ‖ 93.3 k = 48.3 k

Av2 ≈ gm2(ro2 ‖ R′

L) = 6.67× 10−3(10 k ‖ 48.3 k) = 55.2

R′

L1 = Ri2 =ro2 + (RL ‖ R1)

1 + gm2ro2=

10 k + 48.3 k

1 + 6.67× 10−3 × 10× 103= 862 Ω

Q1 (CS Amp): Av1 = −gm1(ro1 ‖ R′

L1) = 6.67× 10−3(10 k ‖ 862) = −5.29

AV = vo/vi = Av1 × Av2 = −292

If we had used gmro ≫ 1 in calculations, we would have got R1 = 83.3 k, R2 = 667 k,

R′

L = 45.4 k, Av2 = 54.6, R′

L1 = 831 Ω, Av1 = −5.54, and Av = −302.

Problem 5. Find vo/vi in the circuit below if ID3 = 1 mA and the DC bias at the

gate of Q1 is zero. (Transistors are identical with µnCox = 200 µA/V2, λ = 0.1 /V, and

W/L = 10/0.18.)

−VSS

VG5

vi

vo

VDD

vx

Q2Q1

Q5

1k

vo

VDD

vx

R i2

Q2

1k

vi

VDD

vx

r o3 i2R

Q1

Note that circuit is NOT symmetric, This is a two-stage

amplifier, Q1 is a source follower and Q2 is a common-

gate amplifier.

Since VG1 = VG2 = 0 and VS1 = VS2, we have VGS1 =

VGS2 and VOV 1 = VOV 2 leading to:

ID1 = ID2 = 0.5ID3 = 0.5 mA

ID1 = 0.5µnCox(W/L)1V2

OV 1

0.5× 10−3 = 0.5× 200× 10−6 × (10/0.18)V 2

OV 1

VOV 1 = VOV 2 = 0.3 V

gm1 = gm2 =2ID1

VOV 1

=2× 0.5× 10−3

0.3= 3.33 mA/V

ro1 = ro2 =1

λID1

=1

0.1× 0.5× 10−3= 20 k

ro3 =1

λID3

=1

0.1× 10−3= 10 k

Q2 (CG Amp): vo/vx ≈ gm2(ro2 ‖ 1 k)

vo/vx = 3.33× 10−3(20 k ‖ 1 k) = 3.17

Ri2 =ro2 + 1 k

1 + gm2ro2

Ri2 =20 k + 1 k

1 + 3.33× 10−3 × 20× 103= 311 Ω

Q1 (CD Amp): R′

L1 = ro3 ‖ Ri2 = 10 k ‖ 311 = 301 Ω

vx/vi =gm1R

′

L1

1 + gm1R′

L1

= 0.5

vovi

=vovx

×vxvi

= 3.17× 0.5 = 1.59

Problem 6. Transistors in the circuit below all have gm = 1 mA/V2, ro = 10 k,

Csb = Cdb = Cgs = Cgd = 10 fF. Find the upper cut-off frequency of this amplifier.

vo

VDD

vsig

1k

Q1

Q2

100Cgs1

Cgd1

Cgd2

Cgs2

Cdb2

vo

vsig Csb1

Cdb1Q1

Q2

1001k Cin

vo

CLR1

R2

Rivsig

1k

Q1

Q2

100

=⇒ =⇒

In the above circuit (from left to right), we first include all internal capacitors of MOS.

Source of Q2 is grounded, thus Csb2 and Cgs2 are shorted out. Looking at the above

middle circuit, the following capacitors are attached between output and ground: Cdb1,

Cgd1, Cdb2, and Cgd2. Cgs1 and Csb1 are connected between the input and the ground, and

no capacitors appear between input and output. Thus, the circuit reduces to the right

circuit above with:

CL = Cdb1 + Cgd1 + Cdb2 + Cgd2 = 40 fF & Cin = Cgs1 + Csb1 = 20 fF

Cin

vsig

Ri1k

100

vo

CLR2 R1

Using the time-constant method (note that with vsig = 0, a resistance of 100 ‖ 1 k =

90.9 Ω appears in the source of Q1):

Ri =ro1 + ro21 + gm1ro1

= 1.82 k

Cin RCin = 100 ‖ 1. k ‖ 1.82 k = 87 Ω

τ1 = RCinCin = 87× 20× 10−15 = 1.74× 10−12

R1 = ro1(1 + gm1 × 90.9) + 90.9 = 11.0 k

R2 = ro2 = 10 k

CL RCL = R1 ‖ R1 = 10 k ‖ 11 k = 5.24 k

τ2 = RCLCL = 5.24× 103 × 40× 10−15 = 2.10× 10−10

b1 = τ1 + τ2 = 2.11× 10−10

fH =1

2πb1= 753 MHz