Errata
Below are a listing of known errors in the book’s first printing.
| Page No. | Section | Example/ Problem | Figure | Equation reference | Original | Correction |
| 23 | 1.2.3 | 1.8 | none | none | Using (1.67), we find for Vds = Veff = 0.4V, |
Using (1.67), we find for Vds = Veff = 0.35V, |
| 23 | 1.2.3 | 1.8 | none | none | In the case where VDS = Veff + 0.5V = 0.9V, we have | In the case where VDS = Veff + 0.5V = 0.85V, we have |
| 33 | 1.2.7 | 1.13 | none | none | Cov = 2.0 x 10-4 pF/μm | Cov‘ = 2.0 x 10-4 pF/μm |
| 47 | 1.4.2 | none | 1.31 | none | In the plot of Id vs. Vgs, the slope of Id vs. Vgs in mobility degradation looks same as or larger than that of square law, which contradicts the plot of gm vs. Vgs |
Eliminate curvature on Id vs Vgs plot in mobility degradation. |
| 53 | 1.5.3 | none | 1.33 | none | Two text labels include the “section end” character |
Replace with “/” |
| 62 | 1.7.1 | none | none | 1.146 | Dn | Dp |
| 70 | 1.10.2 | P1.12 | none | none | The technology node and the values of Vdb, Vsb, and Cj-sw are not given |
Assume the 0.35μm CMOS technology, Vdb = Vsb = 0, and ignore sidewall capacitances. |
| 101 | 2.3.3 | none | none | none | gm/ID = 1/(2Veff) | gm/ID = 2/Veff |
| 102 | 2.3.3 | none | none | 2.19 | σ2(ΔVGS) = 1/WL[AVt02(gm/ID)2 + AK’2] | σ2(ΔVGS) = 1/WL[AVt02 + (ID/gm)2AK’2] |
| 106 | 2.4.2 | 2.6 | 2.31 | none | left and right sides of the figure have been cropped |
left and right layout labels should indicate 10μm and 19.6μm dimensions respectively |
| 124 | 3.4 | none | none | none | …, the input impedance, rout, is found to be 1/gm… | …, the input impedance, rin, is found to be 1/gm… |
| 140 | 3.11.2 | P3.12 | none | none | … shown in Fig. P3.11. |
… shown in Fig. P3.12. |
| 145 | 4.1 | none | none | none | Further, the real parts of all the poles will be positive… | Further, the real parts of all the poles will be negative… |
| 148 | 4.1.2 | 4.2 | none | none | … for the case f = 10 Hz, where ω = 2πf = 628.3 rad, … | … for the case f = 10 Hz, where ω = 2πf = 62.83 rad, … |
| 154 | 4.1.2 | 4.5 | none | 4.60 | ∠A(ω) = -tan(ω/ω-3dB)-1 | ∠A(ω) = -tan-1(ω/ω-3dB) |
| 155 | 4.1.3 | none | none | 4.70 | Φ = -tan-1(ω/ωp1) – tan(ω/ωp2) | Φ = -tan-1(ω/ωp1) – tan-1 (ω/ωp2) |
| 161 | 4.1.4 | 4.7 | none | 4.93 | 1.05 x 108 rad/sec | 1.05 x 1010 rad/sec |
| 161 | 4.1.4 | 4.7 | none | 4.94 | 5 x 106 rad/sec | 5 x 108 rad/sec |
| 170 | 4.2.3 | 4.8 | none | none | … to the case where Y(s) = 1/sC. | … to the case where Y(s) = sC. |
| 175 | 4.2.4 | 4.9 | 4.19 | none | gm1 labels should be beside the dependent current sources |
|
| 176 | 4.2.4 | 4.10 | none | none | use the transistor parameters listed in Table 1.5 for the 0.8-μm CMOS technology in Table 1.5. |
use the transistor parameters listed in Table 1.5 for the 0.8-μm CMOS technology. |
| 176 | 4.2.4 | 4.10 | none | 4.158 | the denominator of equation (4.152):Rin[Cgs1 + Cgs1(1 + gm1R2)] + R2(Cgd1 + C2) |
the denominator of equation (4.152):Rs[Cgs1 + Cgs1(1 + gm1R2)] + R2(Cgd1 + C2) |
| 177 | 4.2.5 | 4.11 | 4.13 | none | L1=|Ao| l L1 Veff,1 | L1=(|Ao| l L1 Veff,1) / 2 |
| 179 | 4.2.5 | 4.12 | none | none | 2 x 355mA x 0.18μm / (270mA/V2 x (0.1V)2) |
2 x 350mA x 0.18μm / (270mA/V2 x (0.1V)2) |
| 191 | 4.4 | 4.16 | none | 4.212 | = (4.27 x 109 rad/s = 2π x 680 MHz) |
= 4.27 x 109 rad/s = 2π x 680 MHz |
| 194 | 4.5.1 | none | 4.31 | none | capacitor symbol overlaps the label, Cdb |
move capacitor symbol to the left |
| 195 | 4.5.3 | none | none | 4.226 | negative sign in the numerator | no negative sign in the numerator |
| 196 | 4.5.3 | none | 4.34(a) | none | label “Vout” is mislocated | move “Vout” label to the output below “RD“ |
| 196 | 5.4.3 | none | 4.35 | none | dependent current source on the right points down | dependent current source on the right points up |
| 197 | 5.4.3 | none | 4.36 | none | dependent current source on the right points down | dependent current source on the right points up |
| 197 | 5.4.3 | none | 4.37 | none | dependent current source on the right points down | dependent current source on the right points up |
| 205 | 5.1.2 | none | none | 5.7 | ΔA / [(1+L)2+ ΔAβ] | ΔA / [(1+L)2 + ΔAβ(1+L)] |
| 209 | 5.2 | 5.3 | none | none | In practice, the voltage at v– will increase above vin | In practice, the voltage at v+ will increase above vin |
| 210 | 5.2.1 | none | none | none | Key Point: “… -180 degres …” |
Key Point: “… -180 degrees …” |
| 215 | 5.3.1 | none | none | 5.29 | Second equals sign: “=” |
Second equals sign should be approximately equals |
| 216 | 5.3.1 | 5.6 | 5.9(b) | none | expression in the figure: “s/ωta“ |
expression in the figure: ” ωta/s” |
| 217 | 5.3.1 | 5.6 | none | none | … is well approximated by s/ωta. | … is well approximated by ωta/s. |
| 217 | 5.3.1 | 5.6 | none | 5.32 | first term on right side: “(s/ ωta)” |
first term on right side: “(ωta /s)” |
| 221 | 5.4 | none | none | none | Key Point: “… desired gain, β …” |
Key Point: “… desired gain, 1/ β …” |
| 225 | 5.4.1 | 5.9 | 5.13(b) | none | arrow points down on the dependent source gm,1 | arrow points up on the dependent source gm,1 |
| 225 | 5.4.1 | 5.9 | 5.13(c) | none | line connecting the tops of Zx and Cgs,5 is broken | line connecting the tops of Zx and Cgs,5 should be connected |
| 225 | 5.4.1 | 5.9 | 5.13(c) | none | Cgs,1/2 | Cgs,1 /2 |
| 229 | 5.4.2 | 5.11 | none | none | 2π 2.86MHz | 2π 28.6MHz |
| 235 | 5.5 | none | none | none | Key Point: “… -180 degres …” |
Key Point: “… -180 degrees …” |
| 235 | 5.5 | none | none | none | Key Point: “… desired gain, β…” |
Key Point: “… desired gain, 1/β …” |
| 244 | 6.1.1 | 6.1 | none | none | We can calculate the transconductances of Q1, Q2, Q7 and Q8. |
We can calculate the transconductances of Q1, Q2, and Q7. |
| 245 | 6.1.1 | 6.1 | 6.3 | none | Av1 Av2 = -1043 V/V or 60.4dB |
Av1 Av2 = 475 V/V or 53.5dB |
| 248 | 6.1.2 | 6.3 | none | unnumbered | ωp2 = 3.12mA/V / (5pF + 1pF) = 2π 82.8 MHz | ωp2 = 3.12mA/V / 5pF = 2π 99.3 MHz |
| 249 | 6.1.2 | 6.3 | none | unnumbered | |ωp2| = 3.12mA/V / 5pF = 2π 99.3 MHz | ωp2 = 3.12mA/V / 1pF = 2π 497 MHz |
| 254 | 6.2.1 | none | none | none | …introduce an additional approximately +20° phase shift… | …introduce an additional approximately +20° to 30° phase shift… |
| 258 | 6.2.2 | 6.7 | none | none | the opamp and ωz, ωp, and ω2 are the frequencies of a zero, the dominant pole, and the equivalent second pole, respectively. |
the opamp and ωz, ωp1, and ω2 are the frequencies of a zero, the dominant pole, and the equivalent second pole, respectively. |
| 259 | 6.2.3 | none | none | none | Repeating equations (6.9) and (6.20) here, we have | Repeating equations (6.42) with ß=1 and(6.20), we have |
| 261 | 6.2.3 | none | none | 6.71 | lefthand side, denominator: Veff16 | lefthand side, denominator: Veff9 |
| 268 | 6.4 | none | 6.19 | none | Dependent current source, Gmavi points down | Dependent current source, Gmavi points up |
| 279 | 6.5 | 6.11 | none | none | the unity-gain frequency is increased by 150% |
the unity-gain frequency is increased by 100% |
| 292 | 6.8 | none | 6.38 | none | Grounded nodes in the schematic | These are AC grounds, and should be connected to a common-mode reference voltage |
| 304 | 7.1.1 | 7.1(b) | none | none | Vb = 1.1(450mV)+(200mV)/(0.9 √0.9) = 729mV | Vb = 1.1(450mV)+(200mV)/(√0.9 √0.9) = 717mV |
| 307 | 7.2.1 | none | 7.6 | none | Q13 size is 100 |
Q13 size is 25 |
| 309 | 7.2.1 | none | none | none | effective gate-source voltages increase by 27 percent | effective gate-source voltages increase by 39 percent |
| 309 | 7.2.2 | none | none | none | Q13 is five times greater … gate voltage of Q13 … | Q14 is five times greater … gate voltage of Q14 … |
| 310 | 7.2.2 | none | 7.8 | none | Resistor label “RB” |
Resistor label should use subscript: “RB“ |
| 311 | 7.3.1 | none | none | none | Vbe will have approximately a -2mV/°K | Vbe will have approximately a -2mV/K |
| 313 | 7.3.1 | 7.4 | none | none | For the special case of T0 = 300°K |
For the special case of T0 = 300K |
| 313 | 7.3.1 | 7.4 | none | none | at 300°K | at 300K |
| 313 | 7.3.1 | 7.4 | none | none | Recalling that 0°K | Recalling that 0K |
| 314 | 7.3.1 | 7.4 | none | 7.29 | 8 mV/°K / 1.24V = 6.5×10-6 parts/°K = 6.5ppm/K | 8 mV/K / 1.24V = 6.5×10-6 parts/K = 6.5ppm/K |
| 314 | 7.3.2 | none | none | none | the same collector currents and collector-emitter voltages | the same collector currents and collector-base voltages |
| 317 | 7.3.2 | none | 7.13(b) | none | opamp positive terminal on top | opamp positive terminal on bottom |
| 318 | 7.3.2 | 7.5 | none | 7.51 | Now, recalling from (7.26) that |
Now, recalling from (7.25) that |
| 320 | 7.3.3 | none | 7.15 | none | Leftmost resistor is Ra | Leftmost resistor is Ra/M |
| 320 | 7.3.3 | none | 7.15 | none | Polarity of the opamp is incorrect with the positive terminal on the left | Switch the polarity of the opamp so that positive terminal is on the right |
| 325 | 7.4.3 | 7.6 | none | 7.74 | RL/(rds1 + RL) | -gm1R’L |
| 326 | 7.4.3 | 7.6 | 7.21 | none | vgs,1 labeled incorrectly | Should be labeled so that vgs,1 = v1 – vDD |
| 326 | 7.4.3 | 7.6 | 7.21 | none | current source gm1 has wrong polarity | reverse polarity of current source gm1 |
| 326 | 7.4.3 | 7.6 | none | 7.76 | RL/(rds1 + RL) | -gm1R’L |
| 329 | 7.7.2 | P7.7 | none | none | Assume (W/L)12,13 = (2μm/0.2μm) and (W/L)12,13 = (8μm/0.2μm). | Assume (W/L)14 = (2μm/0.2μm) and (W/L)15 = (8μm/0.2μm). |
| 329 | 7.7.2 | P7.10 | none | none | an dc input | a dc input |
| 329 | 7.7.3 | P7.12 and P7.13 | none | none | 320°K | 320K |
| 330 | 7.7.4 | P7.20(d) | none | none | … so that ωpL ≈ ωpa/100? |
… so that ωpL ≈ ωpa/10? |
| 361 | 8.9.1 | 8.2 | P8.2 | none | Vdd=5V | Vcc=5V |
| 361 | 8.9.1 | 8.9 | P8.9 | none | Vdd=5V | Vcc=5V |
| 371 | 9.2.3 | 9.4 | none | none | Kv2=(50nV2) | Kv2=(50nV)2 |
| 378 | 9.3.1 | none | none | none | temperature (300°K) | temperature (300K) |
| 383 | 9.3.6 | 9.8 | none | none | temperature of 300°K | temperature of 300K |
| 392 | 9.4.2 | 9.11 | none | none | i.e. 300°K | i.e. 300K |
| 400 | 9.5.2 | none | none | none | Δω = ω2 – ω | Δω = ω2 – ω1 |
| 403 | 9.5.3 | 9.14 | none | none | ID3 = 45 dB | ID3 = -45 dB |
| 408 | 9.8.2 | P9.11 and P9.12 | P9.11 and P9.12 | none | y-axis title: Vrms | y-axis title: Vrms2 |
| 420 | 10.3 | 10.3 | none | none | VDD = 1V | VDD = 2V |
| 420 | 10.3 | 10.3 | none | none | -2.8mV charge injection; total of -56mV | -28mV charge injection; total of 80mV |
| 467 | 11.9.3 | P11.11 | none | none | Equation denominator: “Chld + Cbe6“ |
Equation denominator: “Chld + Cbe6 + CF“ |
| 471 | 12.1.2 | none | none | none | independent of the feed-in gain, K1,2,3. | independent of the feed-in gain, K0,1,2. |
| 471 | 12.1.2 | none | none | none | determined by the feed-in gain, K1,2,3. | determined by the feed-in gain, K0,1,2. |
| 472 | 12.1.2 | 12.1 | 12.3 | none | 1/Q = √2 | 1/Q = -√2 |
| 477 | 12.2.4 | none | none | 12.26 | s(Gm3 / Cx+ Cb) | s(Gm3 / Cx +CB) |
| 478 | 12.2.4 | none | none | 12.31 | Gm3 / Cx+ Cb | Gm3 / Cx +CB |
| 482 | 12.3 | 12.5 | none | none | For the case re,max < 0.01Re | For the case re,max < 0.01Rs |
| 482 | 12.3 | none | 12.13 | none | Gm = 1/Re | Gm= 1/Rs |
| 482 | 12.3 | none | 12.14 | none | Gm = 1/Re | Gm = 1/Rs |
| 510 | 12.8.1 | 12.10 | 12.41(b) | none | Vno | Vpo |
| 510 | 12.8.1 | 12.10 | 12.41(b) | none | Vpo | Vno |
| 536 | 12.13.4 | P12.20 | none | none | Sketch the resulting magnitude response |Vin(ω)/Vout(ω)| … |
Sketch the resulting magnitude response |Vout(ω)/Vin(ω)| … |
| 554 | 13.6 | 13.8 | none | none | at 40kHz and 60kHz is 0.2399 and 0.1559, |
at 40kHz and 60kHz is 0.2339T and 0.1559T, |
| 568 | 14.2.3 | none | 14.13(a) | none | vco(nT-T) | vco(nT-T/2) |
| 568 | 14.2.3 | none | 14.13(b) | none | vci(nT-T/2) and vco(nT-T/2) | vci(nT) and vco(nT) |
| 571 | 14.3 | none | none | none | … a rms voltage Vc12(rms)= ÖkT/C … | … a rms voltage Vc1 (rms)= ÖkT/C … |
| 571 | 14.3 | none | none | none | … capacitor noise voltage Vc22(rms)= ÖkT/C. |
… capacitor noise voltage Vc2(rms)= ÖkT/C. |
| 573 | 14.4 | 14.3 | none | none | the bilinear transform p = (z-1) / (z+1) |
the bilinear transform s = (z-1) / (z+1) |
| 574 | 14.4 | none | none | none | As in Example 14.2, | As in Example 14.3, |
| 577 | 14.4.2 | none | 14.21 | none | Φ1, Φ2 | swap clock phases near C3: Φ2, Φ1 |
| 605 | 14.12.6 | P14.20 | none | none | ensure a dc offset of less than 5mV |
ensure a output dc offset of less than 5mV |
| 605 | 14.12.6 | P14.21 | none | none | Find the resulting dc offset |
Find the resulting output dc offset |
| 619 | 15.5.3 | 15.4 | none | none | For the gain error, from (15.25) we have |
For the gain error, from (15.24) we have |
| 627 | 16.1.2 | none | 16.4 | none | The LSB decoder at the bottom of the figure must also accept as input the digital signal b2 since the folded resistor-string changes direction. | |
| 642 | 16.5 | none | none | none | … employ an array of 2N-1 unitary … |
… employ an array of 2N-1 unitary … |
| 669 | 17.4.1 | 17.7 | none | none | V3 = 2(V2-Vref/4) = -540V |
V3 = 2(V2-Vref/4) = -540mV |
| 669 | 17.4.1 | 17.7 | none | none | Vin = 375mV |
Vin = 312.5mV |
| 669 | 17.4.2 | none | none | 17.42 | Vin = Vref/2 (S(bi,0 + bi,1 – 1)2-i + bn-1,12N-2 +bN-1,02N-1) |
Vin = Vref/2 (S(bi,0 +bi,1 – 1)2-i + bn-1,12-N+2 +bN-1,02-N+1) |
| 671 | 17.4.2 | 17.8 | none | none | V2 = 2(V1 – V1x) = -200mV |
V2 = 2(V1 – V1,x) = -200mV |
| 671 | 17.4.2 | 17.8 | none | none | left-align the variables “b1,0“, “b1,1“, and “b2,1 b2,0” in the table |
|
| 681 | 17.7 | none | 17.31 | none | The bottom of the resistor string is grounded. | The bottom of the resistor string should be connected to an additional amplifier output that compares the input, Vin, to ground. |
| 693 | 17.12.5 | P17.22 | none | none | the reference voltage supply Vref. | the reference voltage supply Vref = 0.5V. |
| 711 | 18.3.1 | none | none | none | f0 = 44.1 kHz | 2f0 = 44.1 kHz |
| 720 | 18.5.2 | none | none | 18.44 | RHS of 1st line: ŜTF2(z)X1(z) + N̂TF1(z)X2(z) | ŜTF2(z)X1(z) – N̂TF1(z)X2(z) |
| 720 | 18.5.2 | none | none | 18.44 | RHS of 2nd line: all terms are positive | RHS of 2nd line: 3rd and 4th terms (those including N̂TF1) should be negative |
| 720 | 18.5.2 | none | none | 18.45 | RHS: ŜTF2(z)STF1(z)E1(z) + N̂TF1(z)NTF2(z)E2(z) | ŜTF2(z)STF1(z)E1(z) – N̂TF1(z)NTF2(z)E2(z) |
| 722 | 18.6 | none | 18.23 | none | Modulator as drawn is unstable |
Modify H(z) so that z-2 appears in the feedback loop and there is no delay in the feedforward path. |
| 729 | 18.8.4 | none | none | 18.55 | first term on the right-hand-side of the equals sign: U(z)z-1 | U(z)z-1/(2-z-1) |
| 729 | 18.8.4 | none | none | 18.55 | exponent in the 3rd and 4th terms on the right-hand side: (1-z-1)-2 |
(1-z-1)2 |
| 730 | 18.8.4 | none | 18.29 | none | Gain “1/2” appears between 2nd and 3rd summers at the top left of the figure |
To make it consistent with the reference [Hairapetian, 1994], the gain “1/2” should be moved to between 1st and 2nd summers at the top left of the figure |
| 730 | 18.8.4 | none | 18.29 | none | box at lower-right hand corner, with Us(z) as its input, contradicts the reference [Hairapetian, 1994]: 2(1-z-1)2 |
2-z-1 |
| 742 | 19.1.3 | 19.1 | 19.3 and 19.4 | none | waveform labels: Vosc | waveform labels: Vdiv |
| 750 | 19.2.1 | none | 19.9 | none | caption: “(a) time-domain step response; (b) frequency domain jitter transfer function magnitude response” | caption: “(a) frequency domain jitter transfer function magnitude response; (b) time-domain step response” |
| 760 | 19.3 | 19.11 | none | none | 1st and 3rd equations, second bracketed term includes “gm“ |
1st and 3rd equations, second bracketed term, “gm1“ |
| 760 | 19.3 | 19.11 | none | none | 2nd equation, right-hand side: “nk / |Vs|” |
2nd equation, RHS: “nk / |Vs| gm (Rd||ro1)” |
| 760 | 19.3 | 19.11 | none | none | 3rd equation, term preceding the brackets: “(Rd||ro1)2 / |Vs|2“ |
3rd equation, term preceding the brackets: “1 / gm2 |Vs|2“ |
| 778 | 19.5.2 | 19.19 | none | 19.95 | denominator is incorrect |
 |
| 778 | 19.5.2 | 19.19 | none | 19.96 | denominator: “2ωpll“ | denominator: “4ωpll “ |
| 784 | 19.8.3 | P19.14 | none | none | zero crossing times: “… 40.2ns, 41.0ns, 49.0 ns, …” |
zero crossing times: “… 40.2ns, 49.0 ns, …” |
| End papers | none | none | none | none | g = 2ID / Veff | gm = 2ID / Veff |
| End papers | none | none | none | none | Endpaper CMOS technology parameter values differ slightly from those in Table 1.5. |
Posted problem solutions use the values in Table 1.5. |
Section 18.3.1 (p.710) specifies fs=5.6448 MHz and f0=44.1 kHz to give an OSR=128. But per the definition of OSR used in the book (eq. 18.8), this should give an OSR=fs/(2*f0)=64. I assume the intention here might be to say 2*f0=44.1 kHz ?
Yes you are correct – thanks for the comment. I will add it to the errata.
Section 7.3.3, Fig. 7.15, and Eq. (7.57):
The parameter “N” is not defined in the text.
Yes, this parameter is not explicitly defined. It is hopefully obvious to the reader that “N” refers to the number of diodes in parallel.
There is an error in the errata. For the correction of the mistake in page 320 regarding opamp polarity, it is mentioned that the ref figure is “none”. The ref figure is Fig. 7.15. Please correct.
Fixed – thanks!
Page 314, Section 7.3.2:
“both transistors have the same collector currents and collector-emitter voltages”
Should be “and collector-base voltages”.
Agreed – I have added this to the errata.
Section 7.2.1, page 309, 2nd paragraph:
1) The reduction in mobility is 28%, not 27%.
2) If the mobility decreases by 28%, then Veff should be: Veff/0.72 = 1.39Veff –> Veff should increase by 39%, not 27%.
1) My calculation gave 0.2786 – a minor point!
2) Agreed – Veff should increase by 39%, not 27%. I have added this to the errata – thanks!
Example 7.1-b)
ID is in inverse proportion to R.
BUT Veff is proportional to sqrt of ID rather than ID.
Thus, “0.9” should be replaced by “sqrt(0.9)”.
The correct answer should be 717mV
Yes – thank you! I’ve put this on the errata.
In Section 7.2.2, second paragraph, third line, TWO occurrences:
Q13 should be replaced by Q14.
Right. The current density of Q14 is five times greater than that of Q4. Hence, the gate voltage of Q14 is suitable for use as the PMOS cascode bias Vcasc-p. Added to the errata – thank you!
page 420,Example 10.3, solution part:second line,”using (10.8),this …=-2.8mV”,i think it is “-28mV”,dose it mean: 2Vx1.4fF/101.4fF=2.8mV ?
Yes – you are correct. The charge injection due to overlap capacitance should be 27.6mV and so the total charge is found by adding 27.6mV and 52.7mV for a total of 80mV. I have added this to the errata.
On page 24(second ed.), you state that for a p-channel MOS to conduct, it must have VSG>Vtp, where Vtp is a negative quantity. So as a consequence, VSG=0 gives a conducting transistor since 0 is greater than any negative number, but this cannot be true ??
It is easy to get the signs confused in this context. Without having to get into the weeds of it myself, it is clear that your understanding of the matter is correct!
Dear Professors,
The equations 5.2 and 5.3 in Chapter 5 should have a negative sign for the loop-gain term. If we set input to zero for loop-gain calculations, and follow the loop in Fig.5.1, v=-ABx and L=-AB, due to terminal v multiplying with -1. This also implies that there is a negative feedback. These equations should be included in the list of errata.
This book is one of my favorites and I liked this time in 2nd edition, topic of feedback was given more attention.
Thanks
The book adopts the more common convention of defining the loop gain of a negative feedback amplifier as positive (at low frequency). The signal x experiences gains of A and then B to the point V so the gain from X to V is AB not -AB. If one defines L=-AB, then all the equations of 5.4 and onward must also be modified.
V=ABX=AB(U-V) so V=U.AB/(1+AB) & Y=A(U-V) so Y=U.A/(1+AB)
In the solution of Example 5.1 on page 206 the change in the closed-loop gain should be less than 5%, because |4.9751-4.9975|=0.0224, which is 2.24% change.
The change in gain is expressed as a percentage of the originally computed gain. Hence,
|4.9751-4.9975|/4.9975 * 100 = 0.00448*100 = 0.45%
I believe 4.133 is incorrect. It should be
wp2 = 1/(RS*(Cgs1+Cgd1))
4.133 is derived assuming C2 is large directly on 4.127,
However we cannot not use 4.127 directly (4.127 is based on the assumption that Cdg1*gm1*R2 >> R2*C2 which is in contradiction with the case we are considering in 4.132.
There is no contradiction in the most common cases.
The assumption of widely-spaced real poles which leads to (4.126) & (4.127) is tantamount to assuming the Miller capacitance, Cgd1*gm1*R2 >> C2. This is reasonable in the common case where Cgd1 includes a compensation capacitor Cc so that the Miller effect makes it much larger than C2.
Then, in order to arrive at (4.133) from (4.127), we assume C2 >> Cgs1, also reasonable assuming the opamp is loaded by a capacitance much bigger than Cgs1. This then allows us to neglect the term Cgs1*Cgd1 in the denominator of (4.128) as being much smaller than C2*Cgd1.
Looking at equation 1.90 and example 1.13，I find Cov having different definitions.
Yes – you are correct. Example 1.13 should more properly refer to Cov‘, the overlap capacitance normalized per unit length of gate width. Hence, Cov = Cov‘ W. Thanks for catching this!
In EXAMPLE 1.16,I have a question about the definition of Ioff. Only when VGS=0,we have Ioff. Then Ioff has nothing to do with the changes of VGS,and it is a function of vth. Is that Right,Dr Carusone?
Yes – your interpretation is consistent with equation (1.121). This definition is particularly suitable for CMOS digital logic circuits where the MOSFETs act as switches that are OFF when VGS = 0. In analog circuits, sometimes the source (and even the body terminal) may not be at ground. It may be possible to reduce subthreshold leakage in these cases by decreasing VGS even further, below zero. Thanks for raising this confusing point!
The second pole and zero frequency is incorrect in example 4.7 on page 159-163. The circuit shows 1pF, but the calculations must have used 100pF, because the frequency is off by 100.
Thanks very much for catching this. It’s now reflected in the errata.
On page 182,Equation4.168 Rout is rds2//RL.I think Rout is gm2rds2rds1//RL;
Equation4.170,gs2=1/rs2=gin2+gds1,gin2 I think is gm2;
Equation4.171 I think gin2 is gm2 too.
Equation (4.168) refers to rd2 which is the small-signal resistance seen looking into the drain of Q2, not simply rds2 as you suggest. The substitution for rd2 is then made in (4.174). In equations (4.170) and (4.171), note that gin2 is approximately gm2, but is more accurately given by the expression in (4.160) which takes into account the circuit’s finite load impedance.
Page 124, last paragraph, first sentence:
“…, the input impedance, rout, is found to be 1/gm…” -> rout should be rin.
Equation 6.71, page 261. It says Veff7 / Veff16 = …
I assume it should be Veff7 / Veff6? (since there’s not even a Q16).
No, not quite, although of course there is a typo there. In fact, in equation (6.71) Veff16 should be replaced by Veff9, the triode device realizing Rc. Thanks.
On page 420, example 10.3.
There’s a print error on Vdd=1V which should be 2V according to the solution below.
Equation 1.146: I think there should be Dp instead of Dn
Page 197 Figure 4.37: Dependent current source arrow direction should be up in order to model + gain of the active load diff. pair.
Page 196 Fig. 4.35 right-hand side dependent current source arrow should be upwards since vgs2=-vg1/2. Then page 195 Eqn. 4.226 becomes positive.
Figure 7.13 – Based on Ye’s Paper, the terminals of OP-AMP are incorrectly connected. Should they not be reversed.
Yes – for Fig. 7.13(b) only they should be reversed in order to provide negative feedback, and in accordance with Ye’s paper. Fig. 7.13(a) is correct as is. Thanks!
Page 510, Ex. 12.10, Fig. 12.41(b): Output differential voltage polarization should be reversed in order to have a negative gain of (-1/(sRC)) as stated in the solution.
Page 225 example 4.9 solution:
Fig. 5.13b: simplified small-signal model of the diff pair’s dependent current source direction should be reversed (Av=+gm*Rout).
Then Fig. 5.13c becomes correct for dependent current source:
gm1(Vin-Vout)”up arrow”
= gm1(0-Vt) “up arrow”
=-gm1*Vt “up arrow”
= gm1*Vt “down arrow”
Page 170 example 4.8 solution:
In the first sentence, Miller feedback admittance is stated as “Y(s)=1/sC”,which should be Y(s)=sC.
polarity of opamp on fig 7.15 is wrong!!!
on page 446 says Since this charge is negative, it will cause the junction voltages to have negative glitches.
i think positive charge is right since switch of Fig.11.3 is PMOS transistor.
I think you are confused by the symbol in Fig. 11.3. In that figure, Q1 is in fact an NMOS transistor, as indicated by the arrowhead.
In p.145 sec. 4.1 last sentence in last paragraph.
“Further, the real parts of all the poles will be positive for stable transfer functions.”
The term “positive” should be modified as “negative”!
I noticed that on page 148 that with f=10Hz, ω = 2πf = 62.83 rad/s and not 628.3 rad/s
The caption on Fig 4.1 on page 151 says lowpass circuit but the circuit is a highpass.
In fact, the transfer function for Vin/Vout of that circuit is, indeed, lowpass. You may find the analysis in Example 4.3.
The H(s) in the Ex. 4.3 is a high-pass filter, sRC/(1+sRC). Which part of the example shows a low-pass transfer function? I got a little confused by this. The basic formula looks like low-pass, but as the circuit transfer function derives, it is clear a high-pass filter, doesn’t it? Thank you.
The capacitor acts like open circuit at DC and short circuit to ground at very high frequencies. So the circuit will pass the DC content and the capacitor will attenuate the signal at very high frequencies. This result can also be confirmed with mathematical transfer function of the circuit:
The transfer function is just a voltage divider between the capacitor and the resistor impedances:
H(s) = (1/Cs) / (R+1/Cs) = 1/(1+RCs)
which is a lowpass transfer function. (If you replace s=0 for DC: H=1; and if you replace s=infinity: H=0)
Equation (2.19)
It should be identical with Eq. (2.18)
σ^2(∆VGS) = … = 1/WL[Avt0^2+(ID/gm)^2*(Ak’^2)]