# Analog IC Design Errata

Below is a listing of known errors in the first printing of Analog Integrated Circuit Design, 2nd edition.

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Amr Tawfik
9 months ago

HI, in ANALOG INTEGRATED CIRCUIT DESIGN ,2nd edition ,in ch 10 example 10.1 , page 414
you said that ” The 90% confidence interval for a normal distribution is at 1.2816 times its standard deviation ”
from erfc sheet it is around 1.6

10 months ago

Hi, I think on page 549, Fig 13.12 is wrong in the international student edition 2nd edition. The text does not match the figure. I did not see it in the errata, so I thought I’d better let you know.

Shuang Liu
1 year ago

Hi Professor,

The text of last paragraph on page 567 does not correspond to Fig. 14.13. (“With this arrangement . . . when $\phi_2$ turns on . . . as illustrated in Fig. 14.13(b).” “Now when $\phi_1$ is turned on in Fig. 14.13(a) . . . .”) Either the reference to Figs. 14.13(a) and (b) in the text should be changed, or the indices of the figures should be changed. Thanks!

Last edited 1 year ago by Shuang Liu
Shuang Liu
1 year ago

Hi Professor,

On page 231, the three replacement in the middle, I believe it should be “replace Z1 with Zs,” and “replace Z2 with Zf,” which is also consistent with equation (5.66) on page 232. Thank you!

Shuang Liu
1 year ago

Hi Professor,

On page 223, in Example 5.8, the line starting with “The desired gain of this circuit…” is incorrect. Although the desired gain is -Z2/Z1, this gain does not equal to 1/beta. Beta here should be Z1/(Z1+Z2). The inconsistency between 1/beta and the desired gain is due to the fact that Vin here is not u in Fig. 5.1, and u here corresponds to -Vin * Z2/(Z1+Z2). Therefore, equation (5.6) cannot be applied directly to Example 5.8.

I take it back. But maybe it’s better not to define beta as a feedback factor and 1/beta as a desired gain? Since these two definitions are not consistent in some cases.

Thanks!

Last edited 1 year ago by Shuang Liu
Shuang Liu
1 year ago

Hi Professor, on page 24, the fourth line from the bottom, should the condition for turning on a PMOS be “Vsg > – Vtp”? Thank you!

Odin Østvedt
1 year ago

In example 1.1 on page 2 (2. edition) the solution uses ni = 1.1×10^16 /cm^3. This is exactly the intrinsic value for /m^3. Is the value of ni just supposed to be high, or should it be /1.1×10^10 /cm^3 so that it fits with the intrinsic values on page 1? The anwer given in example 1.1 uses 1.1×10^16 /cm^3 as the value, giving a really low value of n_0.

Z
1 year ago

Hello Professor,

On page 492, eq. 12.90, a factor of 2 is missing? Currently, the equation is not consistent with eq. 12.89.

Thanks!

John
1 year ago

Sect. 9.4.4, page 397, equation 9.122:
I believe the algebra is such that the second term should be 16/9, not 4/9, after setting Cgs = Cin from equ. 9.121.

Thanks, great book.

2 years ago

Pg. 190, Eq. 4.203, Sec. 4.4

Shouldn’t we have Gin(Cgs1 + Cs) instead of Gin(Cs) in denominator for expression for Q?

Jacob
2 years ago

on page 244, equation 6.1 says:

Av1 = -gm1(rds2||d*rds4)

where it should just say:

Av1 = -gm1(rds2||rds4)

In other words, there’s a minor typo which adds in a ‘d’ next to rds4 that should not be there.

Trisha
2 years ago

There is one sentence in page 130 which does not make sense. Whoever was the author of this sentence is grossly incompetent in English. There are many many other mistakes but this mistake is very severe since this disrupts understanding of a very important part

“Since the smallest output voltage,VD4
, can be without Q4 entering the triode region is given by Vds2+Veff , the minimum allowed voltage for Vout is give by”

Chris
2 years ago

figure 1.30(a) on page 46 (second edition): the current Id seems unusually large versus Vgs… should the unit for Id be mA rather than A? If using this plot, by plugging in theta = 1.7V^-1, Vgs = 1V, Id = 0.3A, Vth = 0.45V (gives muCox(W/L)=2.72mA/V^2), one will get m = -0.106 by using equation 1.126. But if changing Id = 0.3A->0.3mA, one will arrive m = 2 that is more reasonable.

Phae
2 years ago

P.764, example 19.15
equation of fj(t):

Should the denominator of the power of e be 5*10^(-23) instead of 5*10^(-25)?

Thanks

Anonymous
2 years ago

P.764, the equation fj(t) in Example 19.15:

Should the power of e be 5*10^(-23) instead of 5*10^(-25)?

Mario Iannazzo
2 years ago

hi there,

(4.175) is Cout=Cgd2+Cdb2+CL+Cbias=20pF+25fF+7pF+25pF=52.025pF

(4.175) should be Cout=Cgd2+Cdb2+CL+Cbias=20fF+25fF+7pF+25pF=32.045pF

br,

mario

RyanQ
3 years ago

Page 281, third line:
The only difference [between] them is…

Jan Nissinen
4 years ago

Dear professor,

About the problem 10.4 on page 442. The problem can be solved and that is clear for me. However, if the input signal Vin can be only positive value because supply voltages are gnd and 5V so can the comparator output go to high? So do we have the think that the input Vin is just relative signal like positive or negative?

Best regards,

Jan

allen
4 years ago

Page670,the comparator offset should less than Vref/8, but the book in first paragraph is less than Vref/4

Hesham
4 years ago

Page 111 and Fig. 2.36
Why not connecting the metal shield to VDD?
Last three lines in 2nd paragraph mention that the nwell acts as a bypass cap. This is true if the shield is VDD rather than GND.
Also in this sentence VDD should probably be replaced by GND: “this helps minimize noise in the substrate, which is connected to VDD”

viga
4 years ago

In p.250 exp6.4, For a 10-mv step, the slope by (6.23) is 0.05 V/us, not 0.05 V/s.

Hesham
4 years ago

I have one question about a circuit proposed several times in the book in Fig. 14.34, Fig. 14.35, Fig. 14.36, and Fig. 16.12. In this circuit there is no DC path for the inverting terminal of the op-amp. This node is always floating, and it is not set by any of the switches. I would appreciate if you can let me know how you set the DC bias at this point.

jsalinas
4 years ago

chapter 7, fig15. (Banba Bandgap reference)

to make valid eq(7.57)
resistor Ra in the right side should be Ra/M
and I2a_right = I2a_left/M.

Hesham
4 years ago

Page 732: second key point: “LSB2” => “2” should be superscript.

Hesham
4 years ago

Page 711: typo in “often an desirable trade-off in integrated circuits”. “an” should be “a”.

Hesham
4 years ago

Page 708, there is a typo in the key point: 1.5 bit/octave should be 2.5 bit/octave

Hesham
5 years ago

There is an error in the errata for the correction of page 249. This correction is for wz not wp2.

Øyvind
6 years ago

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 ?

Yayla
6 years ago

Hi, in eqn 6.43, zero is on the left-half plane if Rc=0. It has to be on the RHP. Actually, if you look at the sentence below also, it mentions that if Rc is too large, zero will move from RHP to LHP. It think there is an extra (-) sign in that eqn.

Hesham
6 years ago

Section 7.3.3, Fig. 7.15, and Eq. (7.57):
The parameter “N” is not defined in the text.

Hesham
6 years ago

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.

Hesham
6 years ago

Page 314, Section 7.3.2:
“both transistors have the same collector currents and collector-emitter voltages”
Should be “and collector-base voltages”.

Hesham
6 years ago

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%.

Hesham
6 years ago

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

Hesham
6 years ago

In Section 7.2.2, second paragraph, third line, TWO occurrences:
Q13 should be replaced by Q14.

charlie Li
7 years ago

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 ?

Anders
7 years ago

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 ??

Leevi Karjalainen
3 years ago

I don’t see why this hasn’t been added to the errata. It is an obvious mistake and it seems silly to have to find it in the comments.

Rahul
7 years ago

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

Atanas Ganchev
7 years ago

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.

Claudio Talarico
8 years ago

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.

PHAT
8 years ago

Looking at equation 1.90 and example 1.13，I find Cov having different definitions.

phatonic
8 years ago

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?

Travis
9 years ago

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.

Michael
9 years ago

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.

zeynep
9 years ago

Page 124, last paragraph, first sentence:
“…, the input impedance, rout, is found to be 1/gm…” -> rout should be rin.

Simon
9 years ago

Equation 6.71, page 261. It says Veff7 / Veff16 = …
I assume it should be Veff7 / Veff6? (since there’s not even a Q16).

Chi
9 years ago

On page 420, example 10.3.
There’s a print error on Vdd=1V which should be 2V according to the solution below.

awuesiak
9 years ago

Equation 1.146: I think there should be Dp instead of Dn

sevil
10 years ago

Page 197 Figure 4.37: Dependent current source arrow direction should be up in order to model + gain of the active load diff. pair.

sevil
10 years ago

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.

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