7.1VGS = VDD = 1.8 VVDS > VGS − VTH (in order for M1 to operate in saturation)VDS = VDD − ID(1 k)= VDD −12μnCoxWL(VGS − VTH)2 (1 k)> VGS − VTHWL< 2.04

7.3VGS = VDD − ID(100 )VDS = VDD − ID(1 k + 100 )> VGS − VTH (in order for M1 to operate in saturation)VDD − ID(1 k + 100 ) > VDD − ID(100 ) − VTHID(1 k + 100 ) < ID(100 ) + VTHID(1 k) < VTHID < 400 μASince gm increases with ID, we should pick the maximum ID to determine the maximum transconduc�tance that M1 can provide.ID,max = 400 μAgm,max =2ID,maxVGS − VTH=2ID,maxVDD − ID,max(100 ) − VTH= 0.588 mS

7.5ID1 = 0.5 mAVGS = VTH +s2ID1μnCoxWL= 0.612 VVGS =

110ID1R2R2 = 12.243 k

VGS = VDD −110ID1R1 −1110ID1RSR1 = 21.557 k

7.6ID = 1 mAgm =2IDVGS − VTH=1100VGS = 0.6 VVGS = VDD − IDRDRD = 1.2 k

7.8 First, let’s analyze the circuit excluding RP .VG =20 k

10 k + 20 k

VDD = 1.2 VVGS = VG − IDRS = VDS = VDD − ID(1 k + 200 )ID = 600 μAVGS = 1.08 VWL=2IDμnCox (VGS − VTH)2 = 12.9758 ≈ 13Now, let’s analyze the circuit with RP .M110 k

VDD20 k

ID + IRP 1 k

IRP RPRS 200

VG = 1.2 VID + IRP =VDD − VDS

1 k + 200

VGS = VG − (ID + IRP )RS = VDS + VTHVG −VDD − VDS1 k + 200

RS = VDS + VTHVDS = 0.6 VVGS = 1 VID =12μnCoxWL(VGS − VTH)2= 467 μAID + IRP = ID +VDSRP=VDD − VDS1 k + 200

RP = 1.126 k

7.9 First, let’s analyze the circuit excluding RP .VGS = VDD = 1.8 VVDS = VDD − ID(2 k) = VGS − 100 mVVDD −12μnCoxWL(VGS − VTH)2 (2 k) = VGS − 100 mVWL= 0.255Now, let’s analyze the circuit with RP .M1RPIRP30 k 2 k

VDDVGS = VDD − IRP (30 k)IRP =VGS − VDSRP=50 mV

RPVGS = VDD − (ID − IRP ) (2 k) + 50 mVVDD − IRP (30 k) = VDD −�12μnCoxWL(VGS − VTH)2 − IRP�(2 k) + 50 mVVDD − IRP (30 k) = VDD −�12μnCoxWL(VDD − IRP (30 k) − VTH)2 − IRP�(2 k) + 50 mVIRP = 1.380 μARP =50 mVIRP= 36.222 k

7.12 Since we’re not given VDS for the transistors, let’s assume � = 0 for large�signal calculations. Let’salso assume the transistors operate in saturation, since they’re being used as current sources.IX =12μnCoxW1L1(VB1 − VTH)2 = 0.5 mAW1 = 3.47 μmIY =12μnCoxW2L2(VB2 − VTH)2 = 0.5 mAW2 = 1.95 μmRout1 = ro1 =1�IX= 20 k

Rout2 = ro2 =1�IY= 20 k

Since IX = IY and � is the same for each current source, the output resistances of the current sourcesare the same.7.13 Looking into the source of M1 we see a resistance of 1gm. Including � in our analysis, we have1gm=1μpCoxWL (VX − VB1 − |VTH|) (1 + �VX)= 372

7.17 (a) Assume M1 is operating in saturation.ID = 0.5 mAVGS = VTH +s2IDμnCoxWL= 0.573 VVDS = VDD − IDRD = 0.8 volt > VGS − VTH, verifying that M1 is in saturation(b)Av = −gmRD= −2IDVGS − VTHRD= −11.557.18 (a) Assume M1 is operating in saturation.ID = 0.25 mAVGS = VTH +s2IDμnCoxWL= 0.55 VVDS = VDD − IDRD = 1.3 V > VGS − VTH, verifying that M1 is in saturation(b)VGS = 0.55 VVDS > VGS − VTH (to ensure M1 remains in saturation)VDD − IDRD > VGS − VTHVDD −12μnCoxW

L(VGS − VTH)2 RD > VGS − VTHWL<2 (VDD − VGS + VTH)μnCox (VGS − VTH)2 RD= 366.67= 3.3200.18Thus, W/L can increase by a factor of 3.3 while M1 remains in saturation.Av = −gmRD= −μnCoxWL(VGS − VTH)RDAv,max = −μnCox�WL�max(VGS − VTH)RD= −227.19P = VDDID < 1 mWID < 556 μAAv = −gmRD= −r2μnCoxWLIDRD= −5WL<200.18RD > 1.006 k

7.20 (a)ID1 = ID2 = 0.5 mAAv = −gm1 (ro1 k ro2)= −s2μnCox�WL�1ID1�1�1ID1 k1�2ID2

�= −10�WL�1= 7.8125(b)VDD − VB = VTH +s2 |ID2|μpCox�WL�2VB = 1.1 V

7.22 (a) If ID1 and ID2 remain constant while W and L double, then gm1 ∝p(W/L)1 ID1 will not change(since it depends only on the ratio W/L), ro1 ∝ 1ID1will not change, and ro2 ∝ 1ID2will notchange. Thus, Av = −gm1 (ro1 k ro2) will not change .(b) If ID1, ID2, W, and L double, then gm1 ∝p(W/L)1 ID1 will increase by a factor of √2, ro1 ∝ 1ID1will halve, and ro2 ∝ 1ID2will halve. This means that ro1 k ro2 will halve as well, meaningAv = −gm1 (ro1 k ro2) will decrease by a factor of √2 .

7.26 (a)ID1 = ID2 = 0.5 mAVGS1 = VTH +s2ID1μnCox�WL�1= 0.7 VVDS1 = VGS1 − VTH (in order of M1 to operate at the edge of saturation)= VDD − VGS2VGS2 = VDD − VGS1 + VTH = VTH +s2ID2μnCox�WL

�2 �WL�2= 4.13(b)Av = −gm1gm2= −q2μnCox�WL�1 ID1q2μnCox�WL�2 ID2= −vuut�WL�� 1WL�2= −3.667(c) Since (W/L)1 is fixed, we must minimize (W/L)2 in order to maximize the magnitude of the gain(based on the expression derived in part (b)). If we pick the size of M2 so that M1 operates at theedge of saturation, then if M2 were to be any smaller, VGS2 would have to be larger (given thesame ID2), driving M1 into triode. Thus, (W/L)2 is its smallest possible value (without drivingM1 into saturation) when M1 is at the edge of saturation, meaning the gain is largest in magnitudewith this choice of (W/L)2.7.27 (a)Av = −gm1gm2= −q2μnCox�WL�1 ID1q

2μnCox�WL�2 ID2= −vuut�WL�� 1 WL�2= −5�WL�1= 277.78(b)VDS1 > VGS1 − VTH (to ensure M1 is in saturation)VDD − VGS2 > VGS1 − VTHVDD − VTH −s2ID2μnCox�WL�2>s2ID1μnCox�WL�1ID1 = ID2 < 1.512 mA7.28 For this problem, recall that looking into the drain of a transistor with a grounded gate and sourcewe see a resistance of ro, and looking into either terminal of a diode�connected transistor we see aresistance of 1gm k ro.(a)Av = −gm1�ro1 k1gm2 k ro2�(b)Av = −gm1�

ro1 k ro2 k1gm3 k ro3�(c)Av = −gm1�ro1 k ro2 k1gm3 k ro3�(d)Av = −gm2�ro2 k ro1 k1gm3 k ro3�(e)Av = −gm2�ro2 k ro1 k1gm3 k ro3�(f) Let’s draw a small�signal model to find the equivalent resistance seen looking up from the output.−vt+it+vgs2−RDgm2vgs2 ro2it = gm2vgs2 +vt − itRDro2vgs2 = vtit = gm2vt +vt − itRDro2it�1 +RDro2�= vt�gm2 +1ro2�vtit=1 + RDro2

gm2 + 1ro2=ro2 + RD1 + gm2ro2Av = −gm1�ro1 kro2 + RD1 + gm2ro2�7.30 (a) Assume M1 is operating in saturation.ID = 1 mAIDRS = 200 mVRS = 200

Av = −RD1gm+ RS= −RD1 √2μnCoxWL ID+ RS= −4WL= 1000VGS = VTH +s2IDμnCoxWL= 0.5 VVDS = VDD − IDRD − IDRS= 0.6 V > VGS − VTH, verifying that M1 is in saturationYes , the transistor operates in saturation.(b) Assume M1 is operating in saturation.WL=500.18RS = 200

Av = −RD1 √2μnCoxWL ID+ RS= −4RD = 1.179 k

VGS = VTH +s

2IDμnCoxWL= 0.590 VVDS = VDD − IDRD − IDRS= 0.421 V > VGS − VTH, verifying that M1 is in saturationYes , the transistor operates in saturation.

7.42 (a)Rout = RD = 500

VG = VDDVD > VG − VTH (in order for M1 to operate in saturation)VDD − IDRD > VDD − VTHID < 0.8 mA(b)ID = 0.8 mARin =1gm=1 q2μnCoxWL ID= 50

WL= 1250(c)Av = gmRDgm =150SRD = 500

Av = 107.43 (a)ID = I1 = 1 mAVG = VDDVD = VG − VTH + 100 mVVDD − IDRD = VG − VTH + 100 mVRD = 300

(b)RD = 300

Av = gmRD=r2μnCoxWLIDRD= 5WL= 694.47.44 For this problem, recall that looking into the drain of a transistor with a grounded gate and sourcewe see a resistance of ro, and looking into either terminal of a diode�connected transistor we see aresistance of 1gm k ro.(a) Referring to Eq. (7.109) with RD = 1gm2and gm = gm1, we haveAv =1gm21gm1+ RS(b) Let’s draw a small�signal model to find the equivalent resistance seen looking up from the output.−vt+it+vgs2−RDgm2vgs2it = gm2vgs2vgs2 = vtit = gm2vtvtit=1gm2Av =gm1gm2(c) Referring to Eq. (7.119) with RD = 1gm2, R3 = R1, and gm = gm1, we haveAv =R1 k 1gm1RS + R1 k 1gm1gm1gm2(d)Av = gm1

�RD +1gm2 k ro3�(e)Av = gm1�RD +1gm2�7.45 (a)vXvin= −gm1�RD1 k1gm2�voutvX= gm2RD2voutvin=vXvinvoutvX= −gm1gm2RD2�RD1 k1gm2�(b)limRD1!1−gm1gm2RD2�RD1 k1gm2�= −gm1RD2This makes sense because the common�source stage acts as a transconductance amplifier witha transconductance of gm1. The common�gate stage acts as a current buffer with a currentgain of 1. Thus, the current gm1vin flows through RD2, meaning vout = −gm1vinRD2, so thatvoutvin= −gm1RD2.This type of amplifier (with RD1 = ∞) is known as a cascode and will be studied in detail inChapter 9.7.40ID = 0.5 mA

Rin =1gm=1 q2μnCoxWL ID= 50

WL= 2000VD > VG − VTH (in order for M1 to operate in saturation)VDD − IDRD > Vb − VTHRD < 2.4 k

Since |Av| ∝ RD, we need to maximize RD in order to maximize the gain. Thus, we should pickRD = 2.4 k . This corresponds to a voltage gain of Av = −gmRD = −48.

7.42 (a)Rout = RD = 500

VG = VDDVD > VG − VTH (in order for M1 to operate in saturation)VDD − IDRD > VDD − VTHID < 0.8 mA(b)ID = 0.8 mARin =1gm=1 q2μnCoxWL ID= 50

WL= 1250(c)Av = gmRDgm =150SRD = 500

Av = 107.43 (a)ID = I1 = 1 mAVG = VDDVD = VG − VTH + 100 mVVDD − IDRD = VG − VTH + 100 mVRD = 300

(b)RD = 300

Av = gmRD=r2μnCoxWLIDRD= 5WL= 694.47.44 For this problem, recall that looking into the drain of a transistor with a grounded gate and sourcewe see a resistance of ro, and looking into either terminal of a diode�connected transistor we see aresistance of 1gm k ro.(a) Referring to Eq. (7.109) with RD = 1gm2and gm = gm1, we haveAv =1gm21gm1+ RS(b) Let’s draw a small�signal model to find the equivalent resistance seen looking up from the output.−vt+it+vgs2−RDgm2vgs2it = gm2vgs2vgs2 = vtit = gm2vtvtit=1gm2Av =gm1gm2(c) Referring to Eq. (7.119) with RD = 1gm2, R3 = R1, and gm = gm1, we haveAv =R1 k 1gm1RS + R1 k 1gm1

gm1gm2(d)Av = gm1�RD +1gm2 k ro3�(e)Av = gm1�RD +1gm2�7.45 (a)vXvin= −gm1�RD1 k1gm2�voutvX= gm2RD2voutvin=vXvinvoutvX= −gm1gm2RD2�RD1 k1gm2�(b)limRD1!1−gm1gm2RD2�RD1 k1gm2�= −gm1RD2This makes sense because the common�source stage acts as a transconductance amplifier witha transconductance of gm1. The common�gate stage acts as a current buffer with a currentgain of 1. Thus, the current gm1vin flows through RD2, meaning vout = −gm1vinRD2, so thatvoutvin= −gm1RD2.This type of amplifier (with RD1 = ∞) is known as a cascode and will be studied in

detail inChapter 9.

7.48 For small�signal analysis, we can short the capacitors, producing the following equivalent circuit.M1R4vinR2 k R3 k RDvoutAv = gm (R2 k R3 k RD)7.49VGS = VDSVGS = VDD − IDRS = VDD −12μnCoxWL(VGS − VTH)2 (1 + �VGS)RSVGS = VDS = 0.7036 VID = 1.096 mAAv =ro k RS1gm+ ro k RSgm =r2μnCoxWLID = 6.981 mSro =1�ID= 9.121 k

Av = 0.86287.50Av =RS1gm+ RS=RS1μnCoxWL (VGS−VTH) + RS= 0.8VGS = 0.64 VID =12μnCoxWL

(VGS − VTH)2= 960 μAVG = VGS + VS = VGS + IDRS= 1.12 V

7.55 For this problem, recall that looking into the drain of a transistor with a grounded gate and sourcewe see a resistance of ro, and looking into either terminal of a diode�connected transistor we see aresistance of 1gm k ro.(a)Av =ro1 k (RS + ro2)1gm1+ ro1 k (RS + ro2)(b) Looking down from the output we see an equivalent resistance of ro2 + (1 + gm2ro2)RS by Eq.(7.110).Av =ro1 k [ro2 + (1 + gm2ro2)RS]1gm1+ ro1 k [ro2 + (1 + gm2ro2)RS](c)Av =ro1 k 1gm21gm1+ ro1 k 1gm2(d) Let’s draw a small�signal model to find the equivalent resistance seen looking down from theoutput.R2+vgs2−R1gm2vgs2 ro2−vt+itit =vtR1 + R2+ gm2vgs2 +vtro2vgs2 =R2R1 + R2vtit =

vtR1 + R2+ gm2R2R1 + R2vt +vtro2it = vt�1R1 + R2+gm2R2R1 + R2+1ro2�vtit= (R1 + R2) k�R1 + R2gm2R2�k ro2Av =ro1 k (R1 + R2) k�R1+R2gm2R2�k ro21gm1+ ro1 k (R1 + R2) k�R1+R2gm2R2�k ro2(e)Av =ro2 k ro3 k 1gm11gm2+ ro2 k ro3 k 1gm1(f) Looking up from the output we see an equivalent resistance of ro2 + (1 + gm2ro2) ro3 by Eq.(7.110).Av =ro1 k [ro2 + (1 + gm2ro2) ro3]1gm1+ ro1 k [ro2 + (1 + gm2ro2) ro3]

7.58P = VDDID = 2 mWID = 1.11 mARDID = 1 VRD = 900

Av = −gmRD= −r2μnCoxWLIDRD= −5WL= 69.44

7.60 Let’s let R1 and R2 consume exactly 5 % of the power budget (which means the branch containing RD,M1, and RS will consume 95 % of the power budget). Let’s also assume Vov = VGS − VTH = 300 mVexactly.IDVDD = 0.95(2 mW)ID = 1.056 mAIDRS = 200 mVRS = 189.5

Vov = VGS − VTH = 300 mVID =12μnCoxWLV 2ovWL= 117.3Av = −RD1gm+ RS= −RD1 √2μnCoxWL ID+ RS= −4RD = 1.326 k

V 2D DR1 + R2= 0.05(2 mW)R1 + R2 =V 2

D D0.1 mWVG = VGS + IDRS = Vov + VTH + IDRS = 0.9 VVG =R2R1 + R2VDD=R2V 2DD0.1 mW= 0.9 VR2 = 29.16 k

R1 = 3.24 k

7.61 Let’s let R1 and R2 consume exactly 5 % of the power budget (which means the branch containingRD, M1, and RS will consume 95 % of the power budget).RD = 200

IDVDD = 0.95(6 mW)ID = 3.167 mAIDRS = Vov = VGS − VTHRS =VovIDgm =2IDVovAv = −RD1gm+ RS= −RDVov2ID+ VovID= −5Vov = 84.44 mVRS = 26.67

WL=2IDμnCoxV 2ov= 4441V 2DDR1 + R2= 0.05(6 mW)R1 + R2 =V 2DD

0.3 mWVG = VGS + IDRS = Vov + VTH + IDRS = 0.5689 VVG =R2R1 + R2VDD=R2V 2DD0.3 mW= 0.5689 VR2 = 6.144 k

R1 = 4.656 k

7.62Rin = R1 = 20 k

P = VDDID = 2 mWID = 1.11 mAVDS = VGS − VTH + 200 mVVDD − IDRD = VDD − VTH + 200 mVRD = 180

Av = −gmRD= −r2μnCoxWLIDRD= −6WL= 2500VGS = VTH +s2IDμnCoxWL= 0.467 VVGS = VDD − IDRSRS = 1.2 k

12�fC1 ≪ R112�fC1=110R1f = 1 MHzC1 = 79.6 pF12�fCS k RS ≪1gm

12�fCS=1101gmgm =r2μnCoxWLID = 33.33 mSCS = 52.9 nF

7.64 (a)Av = −gm1 (ro1 k RG k ro2)(b)P = VDDID1 = 3 mWID1 = |ID2| = 1.67 mA|VGS2| = |VDS2| = VDS =VDD2|ID2| =12μpCox�WL�2(|VGS2| − |VTH|)2 (1 + �p|VDS2|)�WL�2= 113Av = −gm1 (ro1 k RG k ro2)RG = 10 (ro1 k ro2)ro1 =1�nID1= 6 k

ro2 =1�p|ID2|= 3 k

RG = 10 (ro1 k ro2) = 20 k

Av = −s2μnCox�WL

�1ID1 (ro1 k RG k ro2)= −15�WL�1= 102.1VIN = VGS1 = VTH +s2IDμnCox�WL�1 (1 + �nVDS1)= 0.787 V

7.66P = VDDID1 = 1 mWID1 = |ID2| = 556 μAVov1 = VGS1 − VTH =p2IDμnCox�WL�1= 200 mV�WL�1= 138.9Av = −gm1gm2= −q2μnCox�WL�1 ID1q2μnCox�WL�2 |ID2|= −vuut�W

L�� 1WL�2= −4�WL�2= 8.68VIN = VGS1 = Vov1 + VTH = 0.6 V7.67P = VDDID = 3 mWID = I1 = 1.67 mARin =1gm=1 q2μnCoxWL ID= 50

WL= 600Av = gmRD =150

RD = 5RD = 250

7.68P = VDDID = 2 mWID = 1.11 mAVD = VG − VTH + 100 mVVDD − IDRD = VG − VTH + 100 mVVG = VDDAv = gmRD =2IDVGS − VTHRD = 4RD = AvVGS − VTH2IDVDD − IDAvVGS − VTH2ID= VDD − VTH + 100 mVVGS = 0.55 VRD = 270

VS = VDD − VGS = IDRSRS = 1.125 k

WL=2IDμnCox (VGS − VTH)2 = 493.8

7.73P = VDDID1 = 3 mWID1 = ID2 = 1.67 mAAv =ro1 k ro21gm1+ ro1 k ro2=ro1 k ro21q2μnCox(WL )1ID1+ ro1 k ro2= 0.9ro1 = ro2 =1�ID1= 6 k

�WL�1= 13.5Let Vov2 = VGS2 − VTH = 0.3 V. Let’s assume that VOUT = VDS2 = Vov2.VGS2 = Vb = Vov2 + VTH = 0.7 V�WL�2=2ID2μnCox (VGS2 − VTH)2 (1 + �VDS2)= 161VGS1 = VTH +s2ID1μnCox�WL�1 (1 + �VDS1)VDS1 = VDD − VDS2 = 1.5 VVGS1 = 1.44 V

VIN = VGS1 + VDS2 = 1.74 V