Fundamentals Of Microelectronics Bahzad Razavi Chapter 10 Solution Manual

10.3 (a) Looking into the collector of Q1 , we see an infinite impedance (assuming IEE is an ideal source). Thus, the gain from VCC to Vout is 1 . (b) Looking into the drain of M1 , we see an impedance of ro1 + (1 + gm1 ro1 ) RS . Thus, the gain from VCC to Vout is ro1 + (1 + gm1 ro1 ) RS RD + ro1 + (1 + gm1 ro1 ) RS (c) Let’s draw the small-signal model. + rπ1

gm1 vπ1

vπ1

ro1

vcc

−

vout

vout = −vπ1   vcc − vout rπ1 vout = gm1 vπ1 + ro1   vcc − vout = −gm1 vout + rπ1 ro1   rπ1 rπ1 = vcc vout 1 + gm1 rπ1 + ro1 ro1 vout rπ1   = vcc 1 + β + rπ1 r o1

=

ro1

rπ1 ro1 (1 + β) + rπ1

(d) Let’s draw the small-signal model. + vgs1

gm1 vgs1

−

vout RS

ro1

vcc

vout = −vgs1   vcc − vout RS vout = gm1 vgs1 + ro1   vcc − vout RS = −gm1 vout + ro1   RS RS vout 1 + gm1 RS + = vcc ro1 ro1 vout RS   = vcc ro1 1 + gm1 RS + RS ro1

=

RS ro1 (1 + gm1 RS ) + RS

10.8

VX (t) VY (t) 2I0 RC 1.8I0 RC

I0 RC 0.8I0 RC

−π/ω

π/ω −0.2I0 RC

X and Y are not true differential signals, since their common-mode values differ.

t

10.9 (a) VX = VCC − I1 RC VY = VCC − (I2 + IT ) RC

VX (t) VY (t) VCC

VCC − IT RC VCC − I0 RC

VCC − (I0 + IT ) RC VCC − 2I0 RC

VCC − (2I0 + IT ) RC −π/ω

π/ω t

(b) VX = VCC − (I1 − IT ) RC VY = VCC − (I2 + IT ) RC

VX (t) VY (t) VCC + IT RC

VCC − (I0 − IT ) RC VCC − IT RC VCC − (2I0 − IT ) RC VCC − (I0 + IT ) RC

VCC − (2I0 + IT ) RC −π/ω

π/ω t

(c)   VX − VY VX = VCC − I1 + RC RP     VY RC = VCC − I1 − RC VX 1 + RP RP   VY RC VCC − I1 − R P VX = RC 1+ R P VCC RP − (I1 RP − VY ) RC R + RC   P VY − VX RC VY = VCC − I2 + RP     VX RC = VCC − I2 − RC VY 1 + RP RP   VX VCC − I2 − R RC P VY = RC 1+ R P =

= VX = = VX

1−

2 RC

(RP + RC )2 2

VX

2 (RP + RC ) − RC RP + RC

VCC RP − (I2 RP − VX ) RC RP + RC   2 RP −VX )RC RC VCC RP − I1 RP − VCC RP −(I RP +RC RP + RC

VCC RP − I1 RP RC +

VCC RP RC −I2 RP R2C +VX R2C RP +RC

RP + RC

!

=

!

= VCC RP − I1 RP RC +

VCC RP − I1 RP RC +

VCC RP RC −I2 RP R2C RP +RC

RP + RC 2 VCC RP RC − I2 RP RC RP + RC


2 VX RP2 + 2RP RC = VCC RP (RP + RC ) − I1 RP RC (RP + RC ) + VCC RP RC − I2 RP RC

2 VCC RP (RP + RC ) − I1 RP RC (RP + RC ) + VCC RP RC − I2 RP RC 2 RP + 2RP RC VCC RP (2RC + RP ) − RP RC [I1 (RP + RC ) + I2 RC ] = RP (2RC + RP )

VX =

Substituting I1 and I2 , we have: VCC RP (2RC + RP ) − RP RC [(I0 + I0 cos (ωt)) (RP + RC ) + (I0 − I0 cos (ωt)) RC ] RP (2RC + RP ) VCC RP (2RC + RP ) − RP RC [I0 (2RC + RP ) + I0 cos (ωt) RP ] = RP (2RC + RP ) RC RP = VCC − I0 RC + I0 cos (ωt) 2RC + RP

VX =

By symmetry, we can write: VY = VCC − I0 RC − I0 cos (ωt)

RC RP 2RC + RP

VX (t) VY (t) RP VCC − I0 RC + I0 2RRCC+R P

VCC − I0 RC

RP VCC − I0 RC − I0 2RRCC+R P

−π/ω

π/ω t

(d) VX = VCC − I1 RC   VY RC VY = VCC − I2 + RP   RC VY 1 + = VCC − I2 RC RP VCC − I2 RC VY = C 1+ R RP =

VCC RP − I2 RC RP RP + RC

VX (t) VY (t) VCC

VCC − I0 RC P VCC RPR+R C

VCC RP −I0 RC RP RP +RC

VCC − 2I0 RC VCC RP −2I0 RC RP RP +RC

−π/ω

π/ω t

10.11 Note that since the circuit is symmetric and IEE is an ideal source, no matter what value of VCC we have, the current through Q1 and Q2 must be IEE /2. That means if the supply voltage increases by some amount ∆V , VX and VY must also increase by the same amount to ensure the current remains the same. ∆VX = ∆V ∆VY = ∆V ∆ (VX − VY ) = 0 We can say that this circuit rejects supply noise because changes in the supply voltage (i.e., supply noise) do not show up as changes in the differential output voltage VX − VY .

10.23 If the temperature increases from 27 ◦ C to 100 ◦ C, then VT will increase from 25.87 mV to 32.16 mV. Will will cause the curves to stretch horizontally, since the differential input will have to be larger in magnitude in order to drive the current to one side of the differential pair. This stretching is shown in the following plots.

IC1 , T IC1 , T IC2 , T IC2 , T

= 27 ◦ C = 100 ◦ C = 27 ◦ C = 100 ◦ C

IEE

IEE 2

Vin1 − Vin2

Vout1 , T Vout1 , T Vout2 , T Vout2 , T

= 27 ◦ C = 100 ◦ C = 27 ◦ C = 100 ◦ C

VCC

VCC − IEE RC /2

VCC − IEE RC

Vin1 − Vin2

Vout1 − Vout2 , T = 27 ◦ C Vout1 − Vout2 , T = 100 ◦ C IEE RC

Vin1 − Vin2

−IEE RC

10.33 (a) Treating node P as a virtual ground, we can draw the small-signal model to find Gm . iout + vin

rπ

vπ

gm vπ

ro

−

RE

vin − vπ vπ + rπ RE vπ = vin − (−iout + gm vπ ) ro

iout = −

vπ (1 + gm ro ) = vin + iout ro vin + iout ro vπ = 1 + gm ro vin + iout ro vin + iout ro vin iout = − − + rπ (1 + gm ro ) RE RE (1 + gm ro )     ro 1 1 ro 1 = vin iout 1 + − + − rπ (1 + gm ro ) RE (1 + gm ro ) RE rπ (1 + gm ro ) RE (1 + gm ro )     rπ (1 + gm ro ) − RE − rπ rπ RE (1 + gm ro ) + ro (rπ + RE ) = vin iout rπ RE (1 + gm ro ) rπ RE (1 + gm ro ) iout rπ (1 + gm ro ) − RE − rπ Gm = = vin rπ RE (1 + gm ro ) + ro (rπ + RE ) Rout = RC k [ro + (1 + gm ro ) (rπ k RE )]

Av = −

rπ (1 + gm ro ) − RE − rπ {RC k [ro + (1 + gm ro ) (rπ k RE )]} rπ RE (1 + gm ro ) + ro (rπ + RE )

(b) The result is identical to the result from part (a), except R1 appears in parallel with ro . Av = −

rπ (1 + gm (ro k R1 )) − RE − rπ {RC k [(ro k R1 ) + (1 + gm (ro k R1 )) (rπ k RE )]} rπ RE (1 + gm (ro k R1 )) + (ro k R1 ) (rπ + RE )

10.36 VDD −

ISS RD > VCM − VT H,n 2 VDD > VCM − VT H,n + VDD > 1 V

ISS RD 2

10.38 Let JD be the current density of a MOSFET, as defined in the problem statement. ID 11 = µn Cox (VGS − VT H )2 W 2 L s 2ID = W L µn Cox s 2JD = 1 µ L n Cox

JD = (VGS − VT H )equil

The equilibrium overdrive voltage increases as the square root of the current density.

10.39 Let id1 , id2 , and vP denote the changes in their respective values given a small differential input of vin (+vin to Vin1 and −vin to Vin2 ). id1 = gm (vin − vP ) id2 = gm (−vin − vP ) vP = (id1 + id2 ) RSS = −2gm vP RSS ⇒ vP = 0 Note that we can justify the last step by noting that if vP 6= 0, then we’d have 2gm RSS = −1, which makes no sense, since all the values on the left side must be positive. Thus, since the voltage at P does not change with a small differential input, node P acts as a virtual ground.

10.41 P = ISS VDD = 2 mW ISS = 1 mA VCM,out = VDD −

ISS RD = 1.6 V 2

RD = 800 Ω |Av | = gm RD s   W = 2 µn Cox ID RD L 1 

W L



1

=5   W = = 390.625 L 2

Let’s formulate the trade-off between VDD and W/L, let’s assume we’re trying to meet an output common-mode level of VCM,out . Then we have: ISS =

P VDD

ISS RD 2 P RD = VDD − 2V   DD VDD − VCM,out RD = 2VDD P

VCM,out = VDD −

|Av | = gm RD r W µn Cox ISS RD = L r    W VDD − VCM,out P = 2VDD µn Cox L VDD P To meet a certain gain, W/L and VDD must be adjusted according to the above equation. We can see that if we decrease VDD , we’d have to increase W/L in order to meet the same gain.

10.55 Let’s draw the half circuit. vout Q3

vin

Gm = gm1 RP 2

= gm1

RP 2

Q1

k ro1 k rπ3

k ro1 k rπ3 + RP 2 k gm3 R2P

gm3

RP /2

1 gm3

ro1 k rπ3





k ro1 k rπ3   RP Rout = ro3 + (1 + gm3 ro3 ) rπ3 k k ro1 2
   RP gm3 2 k ro1 k rπ3 RP
Av = −gm1 k r r + (1 + g r ) r k o1 o3 m3 o3 π3 2 1 + gm3 R2P k ro1 k rπ3 1+

10.60 Assume IC =

IEE 2

for all of the transistors (since β ≫ 1).

Av = −gm1 {[ro3 + (1 + gm3 ro3 ) (rπ3 k ro1 )] k [ro5 + (1 + gm5 ro5 ) (rπ5 k ro7 )]} h   ih   i VA,n VA,p βn VT VA,n βp VT VA,p V + 1 + V + 1 + A,n A,p VT βn VT +VA,n VT βp VT +VA,p 1 h   i h   i =− V V β V V βp VT VA,p A,n A,p n T A,n VT VA,n + 1 + + VA,p + 1 + VT

= −800

VA,n = 2.16 V VA,p = 1.08 V

βn VT +VA,n

VT

βp VT +VA,p

10.61     1 Av = −gm1 [ro3 + (1 + gm3 ro3 ) (rπ3 k ro1 )] k ro5 + (1 + gm5 ro5 ) rπ5 k k rπ7 k ro7 gm7 This topology is not a telescopic cascode. The use of NPN transistors for Q7 and Q8 drops the output resistance of the structure from that of the typical telescopic cascode.

10.73 (a) VN = VDD − VSG3 s = VDD −

I
SS W L 3 µp Cox

− |VT Hp |

(b) By symmetry, we know that ID for M3 and M4 is the same, and we also know that their VSG values are the same. Thus, their VSD values must also be equal, meaning VY = VN . (c) If VDD changes by ∆V , then both VY and VN will change by ∆V .

10.83 P = VCC IEE = 1 mW IEE = 0.4 mA Av = −gm1 (ro1 k ro3 k R1 ) = −100 R1 = R2 = 59.1 kΩ

10.92 P = VCC IEE = 3 mW IEE = 1.2 mA Av = gm,n (ro,n k ro,p ) = 200 VA,n = 15.6 V VA,p = 7.8 V