Practical Electronics for Inventors, 3rd

[JoeO]

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

Exactly.
The dichotomy is that the technical editor of a book should know MORE about the topic and be MORE exacting than the author.

 

[robrenz]

For the average nitwit like me, Practical Electronics for Inventors has so much broader a range of information than AoE that it is worth checking the errata when necessary.  The way things are presented fits my learning style.

[Rigby]

Same.  Great book.

[timb]

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

You've never opened a college or high school textbook I see.


Sent from my Smartphone

[Bored@Work]

Now that I realize what a gem this book is, I think I will be getting much more familiar with it.

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.

Some people take glass beads, thinking they got real gems.

[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?

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.

Why is it 75v and not 100v?

I don't understand

[TomC]

Why is it 75v and not 100v?

I don't understand

Hi the_phet,

The answer has to do with the location of the reference point or ground. Try to visualize using a voltmeter to read the voltages with the common of the voltmeter connected to the junction of R3 and R4. If you read the voltage across R4 you get a negative voltage because its bottom end is connected to the negative side of the 100V V_S. The voltage in this case is -25V. On the other hand, if you read the voltage at the top of R1 you get a positive voltage. However, since 25V were dropped across R4, you'll only see 100V -25V = 75V. To see the full 100V the voltmeter's common would have to be connected to the bottom of R4, but that's not the reference point.

Although the example deals with calculating the voltage divider resistors required to supply specific voltages and currents to three loads, it is also possible to just use ohms law and the parallel resistors formula to simplify the circuit using the given values. That's what I did to come up with the attached Simplified Figure 2.50 illustration in an attempt to help visualize the above. For example, Load 3 is equivalent to 25V/0.100A = 250 ohms, and R4 parallel to Load 3 is equivalent to (1562 x 250)/(1562+250) = 215.5 ohms.

Let me know if this helps with your question!

[the_phet]

Why is it 75v and not 100v?

I don't understand

Hi the_phet,

The answer has to do with the location of the reference point or ground. Try to visualize using a voltmeter to read the voltages with the common of the voltmeter connected to the junction of R3 and R4. If you read the voltage across R4 you get a negative voltage because its bottom end is connected to the negative side of the 100V V_S. The voltage in this case is -25V. On the other hand, if you read the voltage at the top of R1 you get a positive voltage. However, since 25V were dropped across R4, you'll only see 100V -25V = 75V. To see the full 100V the voltmeter's common would have to be connected to the bottom of R4, but that's not the reference point.

Although the example deals with calculating the voltage divider resistors required to supply specific voltages and currents to three loads, it is also possible to just use ohms law and the parallel resistors formula to simplify the circuit using the given values. That's what I did to come up with the attached Simplified Figure 2.50 illustration in an attempt to help visualize the above. For example, Load 3 is equivalent to 25V/0.100A = 250 ohms, and R4 parallel to Load 3 is equivalent to (1562 x 250)/(1562+250) = 215.5 ohms.

Let me know if this helps with your question!

yes it does, thank you very much

[the_phet]

Hi, I have another question about the book and the unofficial errata.

page 75 in the book, I1 and I2. I see why I2 is Va+IbR1 but why I1 is negative (Va-IbR2)

is it not calculated the same why? Why I1 is I11 - I12?

[TomC]

Hi, I have another question about the book and the unofficial errata.

page 75 in the book, I1 and I2. I see why I2 is Va+IbR1 but why I1 is negative (Va-IbR2)

is it not calculated the same why? Why I1 is I11 - I12?

Hi,

Oversimplifying a bit, I think we can look at it this way: V_A and I_B are both positive at their top end, so they are generating opposite currents. As a result, both V_A and I_B contribute to the current flowing through R2 (I2). So for I2 we add the currents of both sources. On the other hand, the current through R1 depends on the difference of potential between V_A and I_B, for example, if both were generating the same voltage no current would flow through R1. So for the current through R1 (I1) we subtract I22 from I21 to find the difference between the two sources.

However, in reality, the current through R1 could flow either from left to right or right to left depending on whether the voltage generated by the I_B source is lower or greater than V_A. But to try and help visualize the situation, let's assume that these two voltages are equal. In this case, without any math, we can see that I1 = 0 and I2 = I_B. To test the equations with simple numbers let's assume that R1 = 0.5 ohms, R2 = 0.5 ohms, V_A = 5V, and I_B = 10A. Note that by choosing 10A for I_B the current source will need to output 5V for 10A to flow through 0.5 ohms, so since V_A is also 5V this should make both voltages equal. With these values we get:

I21 = V_A = 5A

I22 = I_B x R1 = 10 x 0.5 = 5A

I2 = I21 + I22 = 5 + 5 = 10A

I1 = I21 - I22 = 5 - 5 = 0A



Let me know if this helps with your question!

[the_phet]

yes, thank you!

I think they need this explanations in the book

[the_phet]

I am having problems with the exercise that spreads between page 78 and 79 with 4 circuits

in circuit B, my Inorton is 9V / 400A so 0.0225, while they give 0.02A (I assume nothing goes through the resistance with 600 ohms)

in circuit D,  my I norton is 0.011A (3/266) while they give 0.007A

Can anyone help me solve this example? thanks!

[TomC]

I am having problems with the exercise that spreads between page 78 and 79 with 4 circuits

in circuit B, my Inorton is 9V / 400A so 0.0225, while they give 0.02A (I assume nothing goes through the resistance with 600 ohms)

in circuit D,  my I norton is 0.011A (3/266) while they give 0.007A

Can anyone help me solve this example? thanks!

Hi the_phet,

It seems that you are calculating the total current instead of the Inorton. For the Inorton you can imagine that you have an ammeter connected between points A and B. Keep in mind that the ammeter itself looks like a short between points A and B. Alternately, if you already know Vthev and Rthev, you can simply get Inorton by dividing Vthev by Rthev. The attached image shows one way to solve part d of the example, part b can be solved in a similar way.

Let me know if this helps with your question!

[the_phet]

When you calculate Inorton I don't see where the 0.01/3*2 comes from, especially the part "/3*2"

[TomC]

When you calculate Inorton I don't see where the 0.01/3*2 comes from, especially the part "/3*2"

It's just a shortcut for calculating the current through parallel resistors. For example, if you have two parallel resistors of equal value you get half of the current through each one. You can easily use the same principle if they are exact multiples, like in this case 200 ohms and 100 ohms. The 100 ohms carries 2/3 of the current and the 200 ohms 1/3 of the current.

[the_phet]

When you calculate Inorton I don't see where the 0.01/3*2 comes from, especially the part "/3*2"

It's just a shortcut for calculating the current through parallel resistors. For example, if you have two parallel resistors of equal value you get half of the current through each one. You can easily use the same principle if they are exact multiples, like in this case 200 ohms and 100 ohms. The 100 ohms carries 2/3 of the current and the 200 ohms 1/3 of the current.

I understand that R4 is ignored because it is shorted, but then I don't understand where you get the 200ohms, because there is one with 100 ohms and two with 200 ohms

[TomC]

I understand that R4 is ignored because it is shorted, but then I don't understand where you get the 200ohms, because there is one with 100 ohms and two with 200 ohms

The total current is 0.01A, this flows from the positive side of the battery through the first 200 ohm (R1), then it flows through the second 200 ohm (R2) in parallel with the 100 ohm (R3). So 0.01A is flowing through the parallel combination of R2 and R3. We need to find out what part of this 0.01A flows through R3 to get the Inorton. The classical way of doing this is to first find the voltage drop across one of the voltage divider resistors and then use that to find the current. In this case, the voltage divider consists of R1 = 200 ohms, and R2 || R3 = 67 ohms. The voltage drop across 67 ohms is 0.01A x 67 ohms = 0.67V. Now the current through R3 can be calculated with ohms law: I_R3 = 0.67V / 100 ohms = 0.007A, which is also Inorton. However, instead of calculating it this way, I used a shortcut that is easier to use in some cases when trying to determine the portion of the current flowing through individual parallel resistors when you know the total current flowing through them in combination. Because the resistors in parallel are R2 (200 ohms) and R3 (100 ohms), it was easier to mentally figure out that R3 would carry 2/3 of the current. So I simplified the calculation to 2/3 of 0.01A = 0.007A.

[the_phet]

I understand that R4 is ignored because it is shorted, but then I don't understand where you get the 200ohms, because there is one with 100 ohms and two with 200 ohms

The total current is 0.01A, this flows from the positive side of the battery through the first 200 ohm (R1), then it flows through the second 200 ohm (R2) in parallel with the 100 ohm (R3). So 0.01A is flowing through the parallel combination of R2 and R3. We need to find out what part of this 0.01A flows through R3 to get the Inorton. The classical way of doing this is to first find the voltage drop across one of the voltage divider resistors and then use that to find the current. In this case, the voltage divider consists of R1 = 200 ohms, and R2 || R3 = 67 ohms. The voltage drop across 67 ohms is 0.01A x 67 ohms = 0.67V. Now the current through R3 can be calculated with ohms law: I_R3 = 0.67V / 100 ohms = 0.007A, which is also Inorton. However, instead of calculating it this way, I used a shortcut that is easier to use in some cases when trying to determine the portion of the current flowing through individual parallel resistors when you know the total current flowing through them in combination. Because the resistors in parallel are R2 (200 ohms) and R3 (100 ohms), it was easier to mentally figure out that R3 would carry 2/3 of the current. So I simplified the calculation to 2/3 of 0.01A = 0.007A.

I understand now sorry for all these questions... but I have another one.

Same example, part d, the way you calculated Vthev I don't understand. To solve this I read some tutorials elsewhere and used the transformation between Norton and Thevenin equivalent circuits, but it seems you are doing something else.

I understand how the I_tot = 0.01 is, but then, why divided by 1 multiplied by 100 ohms.

THe 0.01 amps go through R1 and then arrive to the junction between R2, R3 and R4. I don't know how you followed your calculation

[TomC]

THe 0.01 amps go through R1 and then arrive to the junction between R2, R3 and R4. I don't know how you followed your calculation

It's the same shortcut for calculating the current. Since R2 is 200 ohms and is in parallel to R3 & R4 (100 ohms + 100 ohms) which are in series, we have the equivalent of two 200 ohms resistors in parallel. So the current through each branch is half the total current or 0.01A/2 = 0.005A. Since R4 is on one of these branches, you also have 0.005A going through it. Now to find Vthev you find the voltage across it using Ohms law: Vthev = 0.005A x 100 ohms = 0.5V.

[socratidion]

New member here, stumbled upon this post when googling for errata for Monk/Scherz 3rd edition.
I read 'Practical Electronics' over the summer/autumn of 2013: I was a complete beginner, so I read very carefully. I collected errata as I went, which (since there seemed to be a lot of them) made me read more carefully. At some point in chapter 2, I contacted the publisher, who put me in touch with the editor, Roger Stewart; and I began sending packets of errata at intervals as I read. I am told that Paul Scherz has submitted his corrections for a fourth edition (which would include some of my suggestions).

Being a beginner, there were obviously a large number of things I didn't know anything about: what I could spot was errors in logic, inconsistencies, poor labelling, and maths; and as time went on, some little knowledge I had got from practical experience.

I haven't had time to look carefully at the errata list from TomC, nor to compare it with mine, but a brief scan suggests he's done a comprehensive, admirable job, which I will study with great interest. Many things there that I didn't spot! I only waste everyone's time with this post because there was at least one error I happened to remember from my list which I didn't see on his:

p. 689, the diagram showing the internal layout of the 555 timer, there is a connection between pin 6 and the junction of the internal voltage divider. This diagram is a kind of repeat of the one on p. 687, where they are not connected. I know enough about the workings of this chip, at least, to know that the p. 689 version is wrong.

While I'm on it, I did offer a corrected version of the paragraph at the bottom of p.687, and see that TomC has done so too. However, I can add that at the top of p. 687, in the small text accompanying the diagram, line 7, the words '- input to comparator 2' should be changed to '- input to comparator 1'.

I'm so fearfully pressed for time I can't write any more. I would post my complete collection of errata but for two reasons: 1) No point in duplicating what you've already go, and 2) there are some of my suggestions that I now realise are wrong. But maybe in due course I'll be able to weed out the bad stuff and find a few more things to add to TomC's rather wonderful assembly.

[TomC]

Hi socratidion,

Welcome to the EEVblog!

I agree with you about the error on page 689 figure 10.9 and the error on page 687 on the caption for figure 10.6. I will add both errors to the next version of the unofficial errata.

I'm very interested on any additional errors that you have uncovered! I haven't really read the entire book yet, just the areas I needed to refresh while working on other projects, and the areas where errors have been reported by others, specially, the Bucknell errata. So I really think the errata could benefit from someone that has read the entire book while collecting errata!

[TomC]

The latest revision of the Unofficial Errata is now available:

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:

687
689

[RogerL]

I have just joined this community having found this post while searching for the official errata.  I am an electronics newbie and have been dabbling with Arduinos for a few months.  I soon came to the conclusion that I needed a decent grounding in electronics to achieve what I wanted out of my projects, so bought the book last week from Amazon UK and have started flogging through it.

The reason for my post is that although my copy of the book is still the third edition, it seems to have had at least some of the corrections mentioned in TomC's errata as being in the official errata (I haven't checked them all yet).  I mention this in case the errata issues have put anyone off buying the book ----- the publishers do appear to be making an effort.

[TomC]

Hi Rogert,

Welcome to the EEVblog!

That's interesting news!

Can you post more details regarding the errors you can confirm have been corrected?

[JoeO]

I have just joined this community having found this post while searching for the official errata.  I am an electronics newbie and have been dabbling with Arduinos for a few months.  I soon came to the conclusion that I needed a decent grounding in electronics to achieve what I wanted out of my projects, so bought the book last week from Amazon UK and have started flogging through it.

The reason for my post is that although my copy of the book is still the third edition, it seems to have had at least some of the corrections mentioned in TomC's errata as being in the official errata (I haven't checked them all yet).  I mention this in case the errata issues have put anyone off buying the book ----- the publishers do appear to be making an effort.

Roger:  Thanks for letting us know.  How about giving us some info about which printing it is.  Could you please document here the numbers that are on the page before the first Content page.  It would be page viii although mine does not have this Roman Numeral on it.

If you could scan and paste a copy of this page it would be better and very helpful.
Thanks