Practical Electronics for Inventors, 3rd

[TomC]

Page 52, Figure 2.40, the middle plot shows a non-linear curve for Current vs. Resistance (I = V/R).  Shouldn't this be linear for resistors?

Hi cdwilson,

The x axis appears to be from 100 ohms to 1000 ohms. If you apply the formula using those values the graph appears to be pretty close to me.

[cdwilson]

   doh! you are right, when I saw it I was thinking about V=IR being linear where R is fixed, but this plot is I=V/R where V is fixed (i.e. plot y=a/x).  Thanks for putting me right!

[cdwilson]

On page 53, the second paragraph states:

Factoring out V_total, we get: I_total = V_total(1/R1 + 1/R2 + 1/R3 ... 1/R4).  We call the term in brackets R_total.

Should it be changed to the following?

Factoring out V_total, we get: I_total = V_total(1/R1 + 1/R2 + 1/R3 + ... + 1/RN).  We call the reciprocal of the term in brackets R_total.)

[TomC]

Hi cdwilson,

Thanks for the input!

I see a problem with the formula on that sentence but a little different than what you mentioned. It seems that the author missed a division sign between V_total and the term in brackets. So I think the formula should be:

I_total = V_total / (1/R1 + 1/R2 + 1/R3 + ... 1/Rn)

I think the rest of the sentence would be correct with this change.

[cdwilson]

Whoops, yeah, you're right, thanks for catching that.

[cdwilson]

Wait, I don't think that's right.  R_total = 1/(1/R1 + 1/R2 ...), but if you just add a "/" sign, then (1/R1 + 1/R2 ...) is 1/R_total right?

[TomC]

Wait, I don't think that's right.  R_total = 1/(1/R1 + 1/R2 ...), but if you just add a "/" sign, then (1/R1 + 1/R2 ...) is 1/R_total right?

Right, but I think that's what you need to get the correct answer. On the next page he shows the formula in a simpler form:

I_in = V_in / R_total

let me know what you think!

[cdwilson]

I think the equation is correct as is, but the "We call the term in brackets R_total." is wrong.

I_in = V_in / R_total = V_in / 1/(1/R1 + 1/R2 + ...) = V_in * (1/R1 +1/R2 + ...)

This equation is the same as what's printed in the book.  However, saying that (1/R1 +1/R2 + ...) = R_total is wrong.

If you wan to put the "/" sign into the equation, you need to change (1/R1 +1/R2 + ...) to 1/(1/R1 +1/R2 + ...).

[TomC]

I think the equation is correct as is, but the "We call the term in brackets R_total." is wrong.

I_in = V_in / R_total = V_in / 1/(1/R1 + 1/R2 + ...) = V_in * (1/R1 +1/R2 + ...)

This equation is the same as what's printed in the book.  However, saying that (1/R1 +1/R2 + ...) = R_total is wrong.

If you wan to put the "/" sign into the equation, you need to change (1/R1 +1/R2 + ...) to 1/(1/R1 +1/R2 + ...).

I see what you are saying now! and I agree, the last sentence should say something like: The term in brackets is the reciprocal of R_total.

[cdwilson]

Yeah, he basically skips the step factoring out the V_total variable from "I_total = V_total/R1 + V_total/R2 + ... + V_total/RN" and skips directly to I_total = V_total * (1/R1 +1/R2 + ...).  It's confusing because most people expect to see something that looks like I_total = V_total / R_total.

[TomC]

Yeah, he basically skips the step factoring out the V_total variable from "I_total = V_total/R1 + V_total/R2 + ... + V_total/RN" and skips directly to I_total = V_total * (1/R1 +1/R2 + ...).  It's confusing because most people expect to see something that looks like I_total = V_total / R_total.

Yeah, it's pretty simple fractions math though, invert the divisor and multiply, for some reason I didn't see that initially. It's a good thing you were alert and caught it!

So I think the last two sentences should then look like:

Factoring out V_total, we get: I_total = V_total (1/R1 + 1/R2 + 1/R3 + ... 1/Rn). The term in brackets is the reciprocal of R_total.

Thanks a lot for the input! I'll add this to the next revision of the errata.

[Bored@Work]

The term in brackets is the reciprocal of R_total.

Why beating around the bush? It is the conductance.

[TomC]

Hi Bored@Work,

You are right, it's the conductance!

However, that concept should probably be introduced in more detail before using it to describe a term in an equation. Although the conductivity of materials was introduced earlier, I don't think that conductance in the way that it could be portrayed here is introduced until page 247.

[TomC]

The latest revision of the Unofficial Errata is uploading and should be available in about 30 minutes:

https://onedrive.live.com/redir?resid=967A90CA47FD025B!172&authkey=!ACEbpvA4f9gUlxc&ithint=file%2c.pdf

Thanks to the EEVblog members that reported errors or problems with the text the following pages have been added or modified:

53

[khendar]

I've just downloaded a preview of the Kobo version of this book, and it appears as if the illustrations are of reasonable quality in this version. Those in the first two chapters at least are easily readable on my 7" tablet (some of the smaller labels on the schematics are a little hard to make out but their meaning can be easily determined). Does anybody else have the Kobo version of this and can verify  that the rest of the book is okay ?

[TomC]

Hi khendar,

It doesn't look like any other members have the Kobo version!

Did you buy it anyway?

I'm also curious as to the quality of the illustrations, so if you have it, would you mind sharing your impression?

[cdwilson]

Quick question about Figure 2.54, c. on page 65 ("Practical Current Source").

Figure C labels the the 2nd transistor base "B" and the emitter "E".  When I first read the equation "I_0 = V_BE/R2", I thought that V_BE was referring to the base-emitter voltage on the 2nd transistor (because that was the one that was labeled with the "B" and "E" nodes).  However, I think the V_BE in the equation is actually referring to the base-emitter voltage on the 1st transitor (the left one).  Is that right?

[Fantome]

Hello, I am new to the EEVblog forums and I have to say this is the most interesting tech thread I have ever read. I already had the 3rd edition of the book, but I hadn't spent much time with it yet. Now that I realize what a gem this book is, I think I will be getting much more familiar with it.

@TomC...you are an inspiration man! I thank you profusely for the work you have done and your willingness to share it. 

I think I am really going to enjoy EEVblog forums!

P.S.

For those of you who would like to make your books lie flat, here is a solution (though not cheap):
Cutter - shears the binding right off the book: http://www.amazon.com/QCM-8200M-Heavy-Desktop-Cutter/dp/B0010KTIL6
Comb punch & binder (up to 2" books): http://www.amazon.com/Akiles-MegaBind-2-Plastic-Binding-Spiral-O/dp/B002RX1DFY
Example Combs (there are many sizes): http://www.amazon.com/Fellowes-Plastic-Binding-Capacity-52501/dp/B000YD17U0

The blade in that cutter can actually shear the binding off of a book as thick as Practical Electronics For Inventors 3rd edition in one pass, but unfortunately the book is about 50 pages too thick for the clamp that holds it in place. So it would require manually slicing the binding between two pages (say in the middle of the book) with a razor blade and then shearing the binding off for each half. The book is about 1.5 inches thick, so a 1.5 or 1.75 inch comb would fit. Or since it is somewhat unweildy, you bind it in two volumes (or more!). I'll probably do mine soon, perhaps I will take some pictures.

I've never tried it, but Wire binding is another option that may allow the book to lie flat even better than combs, but may not work for thicker books.
Example Wire punch & binder: http://www.amazon.com/gp/product/B004KU0V6Y

[TomC]

Quick question about Figure 2.54, c. on page 65 ("Practical Current Source").

Figure C labels the the 2nd transistor base "B" and the emitter "E".  When I first read the equation "I_0 = V_BE/R2", I thought that V_BE was referring to the base-emitter voltage on the 2nd transistor (because that was the one that was labeled with the "B" and "E" nodes).  However, I think the V_BE in the equation is actually referring to the base-emitter voltage on the 1st transitor (the left one).  Is that right?

Hi cdwilson,

I agree with you that the formula refers to the first transistor's V_BE. This can be significant depending on the type of transistors used since the V_BE voltage drops may be different. However, in most cases, the V_BE is basically a diode's voltage drop. So if both transistors have a similar base emitter drop, say 0.6V, the answer would be the same regardless of which V_BE you use.

While looking at the circuit I remembered seeing something similar in "The Art of Electronics". So I checked my copy and there is an almost identical circuit on page 76, except that it uses PNP's. The formula is the same, and the text clearly indicates that the V_BE in the formula refers to Q1, which is the equivalent to the first transistor on Fig. 2.54. The example in "The Art of Electronics" is a 10mA current source where R2 is 62 ohms. They don't state the V_BE value in the text, but 62 ohms x .010A = 0.62V.

[TomC]

Hello, I am new to the EEVblog forums and I have to say this is the most interesting tech thread I have ever read. I already had the 3rd edition of the book, but I hadn't spent much time with it yet. Now that I realize what a gem this book is, I think I will be getting much more familiar with it.

Hi Fantome,

Welcome to the EEVblog!

I'd like to see how your book looks like once you rebind it with a comb! I did mine, very slowly, with a box cutter, then just three hole punched it. It came out pretty good, but required a lot of patience!

Again, welcome, and hope to see you in this thread often!

[the_phet]

Hi, first comment here, great community.

I have a question, third edition, page 62, at the very end of the example, when it calculates R1 = 25V/0.116A = 216ohms

Why the voltage through R1 is 25V and not 50?

the source is 100V, and the first junction, when it goes to load1, it is 50V, shouldn't it be 50V?

[TomC]

Hi, first comment here, great community.

I have a question, third edition, page 62, at the very end of the example, when it calculates R1 = 25V/0.116A = 216ohms

Why the voltage through R1 is 25V and not 50?

the source is 100V, and the first junction, when it goes to load1, it is 50V, shouldn't it be 50V?

Hi the_phet,

You are right the source is 100V total, however, in respect to ground, the voltage at the top  of R1 is +75V, although it's not labeled. If you look just below R4 it's labeled -25V, so that's a clue that at the top of R1 we have +75V. Following your reasoning, with 75V at the top of R1 and 50V at the first junction the voltage across R1 is 75 - 50 = 25V.

[GK]

Why do so many keep singing praises for a book that need a companion volume of errata? Just put the thing into the toilet and do a big sloppy #2 on it.

[timb]

Find me any electronics book that doesn't have egregious errors. Go on. I'll wait. At least with this it's fairly minor stuff for the most part. Even the Holy Sacred Art of Electronics had some pretty big errors in places.


Sent from my Smartphone

[GK]

The number of errors for what is an elementary text are a bit excessive, don't you think?