Support in learning the art of electronics

[stoica adrian]

Hello all,
My name is Adrian and im 28 years old . I just received the AoE + Learning AoE and i want to study these books.
The plan for learning  is to use the schedule used by the author Paul Horowitz and Winfield Hill.
There are 26 chapters  ( 26 days of class in Learning the art of electronics, one chapter has Notes, Lab, Problems, and Suppliments)  in the book. I will make one  chapter per week ( 2 when its hard the chapter) , starting with one chapter per weekend using  ,, Learning the art of electronics,,  followed by five days focusing on the AoE. I am planning to respect autor schedule.
I want to do this because, the  idea it is to start designing my own stuff.  Design the schematic, create the layout pcb, test, debug, programming, as Dave says.


I created this topic for Q&A session coming in my help. If some question seemed stupid i am sorry
i am sorry fore my english.

[rstofer]

That's an ambitious schedule but you clearly have a plan so go for it!  You have come to the right place for help.  There are some folks here who really understand the topic.  I'm sure they will be glad to help.

[Blaffetuur]

1. i don't understand how to calculate the power dissipation of  the resistance value  shown. Or i don't understand the question.

They ask you to calculate the resistance you can put across a 5V supply so that the maximum powerdissipation in that resistor doesn't exceed .25W.
P=V*I and I=V/R => P=V*(V/R) = V²/R

plug in the numbers et voilà

[orolo]

In your drawing you will measure essentially the 10 volts of the battery, and your 2 Volts measurement range will overflow. Instead of using the multimeter in parallel with the big resistor, try using it in series. What will happen?

[MattSR]

The clue here is that the DMM has an internal resistance of 10Mohm, so by putting it in series with the "load" you are forming a voltage divider.

[kt11]

I was also planning on doing the same thing, but with the previous version of the books which I had to purchase while in school. However, I just learned about this new version today. For those that have experienced both the second and third edition, what are your opinions on whether or not it's worth it to pick up the most recent version?

[MrSlack]

The new one is bad enough to find parts for as well...

[Gripnook]

Parts lists are available on their website with distributors.
http://learningtheartofelectronics.com/parts-lists/

[MrSlack]

Glad they uploaded that. I emailed them about it as I am supporting a friend going through the book.

The kit wouldn't be cheap. Really expensive actually.

I already provided some cheaper substitutions/sources here if anyone is interested: https://www.eevblog.com/forum/beginners/learning-the-art-of-electronics-parts-bom/

[BobsURuncle]

The spreadsheet while more convenient than the listing in the book is still a bit of a mess. They did add quantities needed - they don't say whether these are suggested Q or the actual number you would need for a project/s. I suspect  a mix of both.   

They have not dealt with many of the availability or min order quantity issues.  For example at Rochester you have to order at least $100 of each part and a minimum order of $250. This makes any part only offered by Rochester a nogo for individual buyers.    The 1N5294J constant current diode,  min. order 25 at $4.43 each at Newark, totaling over $110. The extended cost put down in their spreadsheet for that part was $0.00,  they put zero in for a number of parts so the total extended cost they show in the spreadsheet is well under the actual cost - even if you didn't have to buy 25 pieces to get a part.

They should also indicate where this spreadsheet differs from the published list in the book. Are there errors in the book?  Is the text still valid for their substitute parts?  Trying to figure out what they have changed is a nightmare.

They should buy these parts that require expensive min orders and sell them as a kit.  Drop the unavailable parts and publish mods to the text using substitutes parts. And they need to log the changes.

[MrSlack]

1N5294J is obsolete and commands a high price. It's just a JFET and a resistor in a diode package. You can read AoE and make your own in about 10 mins flat with a resistor and a stock JFET. The book also describes how to make a current mirror instead of a limiting diode.

YMMV but I'd ignore the shopping list and buy a chapter at a time with suitable substitutes.

[BobsURuncle]

YMMV but I'd ignore the shopping list and buy a chapter at a time with suitable substitutes.

To each his own, but personally I'd risk buying a few wrong or less than cost efficient parts for which I may or may not find some other use rather than pay huge  shipping and handling charges.  I don't know what you mean by a chapter at a time in this context as the book is divided into Labs. There are 25 labs and a half dozen or so suppliers.  Assuming an average of 2.5 suppliers per lab and $6.00 shipping charge each that adds up to $375.  Even if you buy supplies for one of the 6 "Parts" of the book at a time with an average of 3 suppliers each,  that is  over $100.  And you are going to be waiting months for some of the parts which makes it impossible to go through the book in a reasonably orderly and timely fashion.

Actually this discussion ought to be in the "Learning The Art of Electronics - Parts BOM" post since this thread was supposed to be about helping people who get stuck on certain technical topics in LTAOE. So I desist.






 

[MrSlack]

You missed the $1000+ of test gear as well...

This is still pretty cheap for an education. University here in the UK costs ~$12,745 a year.

[rstofer]

In your drawing you will measure essentially the 10 volts of the battery, and your 2 Volts measurement range will overflow. Instead of using the multimeter in parallel with the big resistor, try using it in series. What will happen?

At their heart, voltmeters are measuring current through their internal resistance (plus scaling resistance).  In the case of an analog meter, the current in the meter winding deflects the needle.

So, putting the meter in series with the 10V battery and the 1GOhm resistor will cause 9.9 nA to flow (10V / (1GOhm + 10MOhm) ).  The meter itself has a full scale of 200 nA (2V / 10MOhm) so the meter should read 9.9/200 of full scale or about 0.05V.

Does that seem about right?

[orolo]

So, putting the meter in series with the 10V battery and the 1GOhm resistor will cause 9.9 nA to flow (10V / (1GOhm + 10MOhm) ).  The meter itself has a full scale of 200 nA (2V / 10MOhm) so the meter should read 9.9/200 of full scale or about 0.05V.

Does that seem about right?

Very good reasoning, save for a small error: 9.9/200 of a scale of 2 volts, is about 2*0.05 = 0.1V.

[rstofer]

So, putting the meter in series with the 10V battery and the 1GOhm resistor will cause 9.9 nA to flow (10V / (1GOhm + 10MOhm) ).  The meter itself has a full scale of 200 nA (2V / 10MOhm) so the meter should read 9.9/200 of full scale or about 0.05V.

Does that seem about right?

Very good reasoning, save for a small error: 9.9/200 of a scale of 2 volts, is about 2*0.05 = 0.1V.

Yup!  Times 2V...

A little early in the morning, I guess!
Measuring in the region of 100 mV seems practical with many digital meters.  When I first saw the problem, the GOhm thing kind of put me off but the problem had to have a solution, it was just a matter of finding it.

[Chai]

It would be nice to have an ongoing thread! I just started the lab + aoe text combo and more than a few times you come across lab book sections that say something along the lines of, "you probably already have guessed the answer", without any further explanation! I was pretty bummed about this until I made it to the first Worked Examples section.

And the text box has example problems with no answers! 

[Aeternam]

It would be nice to have an ongoing thread!

I'd be happy to participate.

Although you'll see more questions from my end than answers, that's for sure

[rstofer]

So, putting the meter in series with the 10V battery and the 1GOhm resistor will cause 9.9 nA to flow (10V / (1GOhm + 10MOhm) ).  The meter itself has a full scale of 200 nA (2V / 10MOhm) so the meter should read 9.9/200 of full scale or about 0.05V.

Does that seem about right?

Very good reasoning, save for a small error: 9.9/200 of a scale of 2 volts, is about 2*0.05 = 0.1V.

Of course, you get the same answer if you work out a series voltage divider.  The 1 gigaohm resistor connects to the + battery and the top of the 10 megohm resistor (the meter).  The bottom of the 10 megohm resistor connects to the - battery.  We are interested in the voltage across the 10 megohm resistor.  So:  10V / (1 gigaohm + 10 megohm) * 10 megohm = 0.0909V or just about 0.1V

In broad numbers, we would expect the 1 gigaohm resistor to drop 100 times as much as the 10 megohm resistor which would see approximately 0.01V (1/100) if we started with 1V but we start with 10V so 10 * (1/100) = 10 / 100 = 0.1V.  Nearly...

[Chai]

Was anyone able to find out more information about the mystery boxes used in 1L.5 page 30? The footnote says to see their website for details but I couldn't find something. I guess one would just be  a resistor? and the other might contain a resistor in series with diode? 

[MrSlack]

They are an RC combination I understand, usually provided by the lab staff. Actual values and configurations are not discussed.

[Chai]

They are an RC combination I understand, usually provided by the lab staff. Actual values and configurations are not discussed.

Oh, OK. Maybe I'll pick out a few RC combos found in next section and test those. 

Also, what to do about the need for a 6.3 VAC transformer in 2L and beyond? What's the safest way for a beginner hobbyist to hook this guy up? I guess I could just use the 2ch function generator I ordered to emulate this instead of dealing with mains (?).

[rstofer]

They are an RC combination I understand, usually provided by the lab staff. Actual values and configurations are not discussed.

Oh, OK. Maybe I'll pick out a few RC combos found in next section and test those. 

Also, what to do about the need for a 6.3 VAC transformer in 2L and beyond? What's the safest way for a beginner hobbyist to hook this guy up? I guess I could just use the 2ch function generator I ordered to emulate this instead of dealing with mains (?).

I wouldn't do that to my function generator.  Transformers usually drive rectifiers that drive large capacitors and such.  The lowly transformer can take a lot of abuse.

Just buy a transformer, connect a power cord and be done.  There should be a switch.  You can buy any kind of switch and put it in a box.  Maybe put the transformer in the same box and just run the power cord out the back and have 2 or 3 binding posts in the front.  I assume you will be using this for a long time.

[Chai]

They are an RC combination I understand, usually provided by the lab staff. Actual values and configurations are not discussed.

Oh, OK. Maybe I'll pick out a few RC combos found in next section and test those. 

Also, what to do about the need for a 6.3 VAC transformer in 2L and beyond? What's the safest way for a beginner hobbyist to hook this guy up? I guess I could just use the 2ch function generator I ordered to emulate this instead of dealing with mains (?).

I wouldn't do that to my function generator.  Transformers usually drive rectifiers that drive large capacitors and such.  The lowly transformer can take a lot of abuse.

Just buy a transformer, connect a power cord and be done.  There should be a switch.  You can buy any kind of switch and put it in a box.  Maybe put the transformer in the same box and just run the power cord out the back and have 2 or 3 binding posts in the front.  I assume you will be using this for a long time.

OK. That sounds like a good idea! 

[MrSlack]

You can buy AC wall warts as well. Cheaper, ready packaged and does the job. They're useful for doing curve tracing as well. 9v one should be fine if you can't find a 6.3v one.

You will drive the transformer via your function generator later - this is completely fine to do.