Support in learning the art of electronics - Page 1
Forum section: electronics / beginners

author: stoica adrian
author_rank: Contributor
timestamp: April 09, 2016, 01:57:36 pm
content: 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 sorryi am sorry fore my english.

author: rstofer
author_rank: Super Contributor
timestamp: April 09, 2016, 02:12:36 pm
content: 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.

author: Blaffetuur
author_rank: Regular Contributor
timestamp: April 09, 2016, 02:35:59 pm
content: Quote from: stoica adrian on April 09, 2016, 02:14:54 pmThey 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/Rplug in the numbers et voilà

author: orolo
author_rank: Frequent Contributor
timestamp: April 14, 2016, 09:14:12 pm
content: 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?

author: MattSR
author_rank: Regular Contributor
timestamp: April 15, 2016, 07:26:30 am
content: 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.

author: kt11
author_rank: Newbie
timestamp: April 17, 2016, 02:45:24 am
content: 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?

author: MrSlack
author_rank: Frequent Contributor
timestamp: April 17, 2016, 03:41:33 pm
content: The new one is bad enough to find parts for as well...

author: Gripnook
author_rank: Contributor
timestamp: April 17, 2016, 04:07:46 pm
content: Parts lists are available on their website with distributors.http://learningtheartofelectronics.com/parts-lists/

author: MrSlack
author_rank: Frequent Contributor
timestamp: April 17, 2016, 04:52:56 pm
content: 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/

author: BobsURuncle
author_rank: Regular Contributor
timestamp: April 17, 2016, 06:57:14 pm
content: 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.

author: MrSlack
author_rank: Frequent Contributor
timestamp: April 17, 2016, 08:18:01 pm
content: 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.

author: BobsURuncle
author_rank: Regular Contributor
timestamp: April 18, 2016, 12:46:10 am
content: Quote from: MrSlack on April 17, 2016, 08:18:01 pmTo 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.

author: MrSlack
author_rank: Frequent Contributor
timestamp: April 18, 2016, 05:40:08 am
content: 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.

author: rstofer
author_rank: Super Contributor
timestamp: April 18, 2016, 02:33:20 pm
content: Quote from: orolo on April 14, 2016, 09:14:12 pmAt 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?

author: orolo
author_rank: Frequent Contributor
timestamp: April 18, 2016, 02:48:07 pm
content: Quote from: rstofer on April 18, 2016, 02:33:20 pmVery good reasoning, save for a small error: 9.9/200 of a scale of 2 volts, is about 2*0.05 = 0.1V.

author: rstofer
author_rank: Super Contributor
timestamp: April 18, 2016, 05:28:34 pm
content: Quote from: orolo on April 18, 2016, 02:48:07 pmYup! 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.

author: Chai
author_rank: Regular Contributor
timestamp: April 18, 2016, 06:35:24 pm
content: 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!

author: Aeternam
author_rank: Unknown
timestamp: April 18, 2016, 07:19:58 pm
content: Quote from: Chai on April 18, 2016, 06:35:24 pmI'd be happy to participate.Although you'll see more questions from my end than answers, that's for sure

author: rstofer
author_rank: Super Contributor
timestamp: April 18, 2016, 08:52:38 pm
content: Quote from: orolo on April 18, 2016, 02:48:07 pmOf 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 0.1VIn 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...

author: Chai
author_rank: Regular Contributor
timestamp: April 21, 2016, 05:49:12 pm
content: 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?

author: MrSlack
author_rank: Frequent Contributor
timestamp: April 21, 2016, 05:54:06 pm
content: They are an RC combination I understand, usually provided by the lab staff. Actual values and configurations are not discussed.

author: Chai
author_rank: Regular Contributor
timestamp: April 21, 2016, 07:05:24 pm
content: Quote from: MrSlack on April 21, 2016, 05:54:06 pmOh, 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 (?).

author: rstofer
author_rank: Super Contributor
timestamp: April 21, 2016, 07:12:28 pm
content: Quote from: Chai on April 21, 2016, 07:05:24 pmI 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.

author: Chai
author_rank: Regular Contributor
timestamp: April 21, 2016, 07:19:46 pm
content: Quote from: rstofer on April 21, 2016, 07:12:28 pmOK. That sounds like a good idea!

author: MrSlack
author_rank: Frequent Contributor
timestamp: April 21, 2016, 08:51:17 pm
content: 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.

author: rstofer
author_rank: Super Contributor
timestamp: April 21, 2016, 09:34:12 pm
content: Driving the transformer from a function generator isn't a problem, true enough! As long as things are linear, no big deal. But hanging a bridge rectifier, filter capacitor and load on the other end of the transformer might cause the function generator a little stress.A PC through-hole mount 120/240 : 6.3/12.6 transformer is about $5 at Digikey.A similar chassis mounted transformer (my preference) is about $12.Here's a chassis mount for $11http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_102163_-1

author: rstofer
author_rank: Super Contributor
timestamp: June 02, 2016, 03:53:39 pm
content: The important things are the value and ratings (voltage, current, etc) and don't worry if all you can find is a higher voltage (current) model. I would substitute as much as possible.Here is one version of the shopping lists:http://learningtheartofelectronics.com/parts-lists/Almost everything can be sourced a lot closer than the US. One thing you can do to help with shipping is to order as much as possible in one shipment.Pay attention to that 555 timer - you must make certain that you get a CMOS version, not one of the TTL versions. Check the datasheet for the particular device.You might want to do as the list author suggests and buy a resistor assortment and maybe even a capacitor assortment. You will need the values eventually.http://www.ebay.com/sch/i.html?_from=R40&_trksid=p2050601.m570.l1311.R7.TR10.TRC2.A0.H1.Xresistor+assortment.TRS0&_nkw=resistor+assortment+1%2F4w&_sacat=0http://www.ebay.com/sch/i.html?_odkw=resistor+assortment+1%2F4w&_osacat=0&_from=R40&_trksid=p2045573.m570.l1311.R4.TR9.TRC2.A0.H0.Xcapacitor+assor.TRS0&_nkw=capacitor+assortment+kit&_sacat=0http://www.ebay.com/sch/i.html?_odkw=capacitor+assortment+kit&_osacat=0&_from=R40&_trksid=p2045573.m570.l1313.TR3.TRC2.A0.H0.Xtransistor+assortment.TRS0&_nkw=transistor+assortment&_sacat=0You might check eBay for your location and see if these are available.

author: orolo
author_rank: Frequent Contributor
timestamp: January 02, 2017, 02:32:01 pm
content: Quote from: stoica adrian on January 01, 2017, 08:19:35 pmThink about your circuit and try to form a criterion about the basic characteristic you need in your amplifier, starting with must haves (see below) without forgetting compromises (ease of construction, cost, availability, and your own strong and weak points). Most of the different circuits you saw could work for you.At the input of the amplifier you should have a selective resonator, probably an inductor in parallel with a variable capacitor. You don't want to load the resonator, for it will lose selectivity: the amplifier must have a high input impedance. Since AM broadcast has a carrier frequency from 700kHz to 1.8MHZ (with country variations), your amplifier must have a bandwidth from 500kHz to 2MHz or so. You should also consider what voltage and power gain you are looking for: if an AM station is close enough, you might need none. At this stage, gain is not so important, just what your demodulator needs. Once you got the audio, you can easily amplify it.I'd aim for an amplifier with high input impedance, 2MHz bandwidth, and at least unity gain. So review the circuits you got based on that. For example,- Using an op-amp amplifier is ok if the IC has got the bandwidth (no 358 or 741, but TL072 does fine) and if you don't abuse the gain. Very high input impedance, fair bandwidth, fair power gain, low voltage gain.- Using an emitter follower, if you bias it right, gives you high input impedance, lots of bandwidth, good power gain but no voltage gain. Coupling the input via a small capacitor will reject low frequency interference. This is a very good simple option for close stations. A jfetcommonsourcefolloweramplifier is very easy to bias and has lots of input impedance, but a bit less power gain.- A common emitter amplifier is a compromise: medium input impedance, ok bandwidth, but you get voltage and power gain. You can't abuse the voltage gain, watch for the input capacitance. Probably best is an emitter follower and then a common emitter.It is a good idea to know what kind of signals you will be amplifying. Probe the resonator with the oscilloscope, at the lowest voltage level; you should easily see the AM signal bursting there, in the tens of millivolts range.

author: akos_nemeth
author_rank: Contributor
timestamp: January 23, 2017, 10:10:28 pm
content: Quote from: stoica adrian on January 01, 2017, 08:19:35 pmHi Adrian,I found this schematic for the short antenna problem:http://learn.mikroe.com/ebooks/radioreceivers/chapter/simple-radio-receiver-with-lm386-ic/It uses the LM386 jellybean audio amplifier IC.Do you have a tuning variable capacitor? They are not sold by Farnell, TME, Mouser, but I have seen in amazon.co.uk:https://www.amazon.co.uk/Spiratronics-CR3-016-Miniature-Tuning-Capacitor/dp/B0093Z0VP2/, also ebay has many tuning capacitors, but the real oldschool ones are not cheap...Regards,Ákos

author: Mule
author_rank: Contributor
timestamp: July 21, 2017, 12:15:16 pm
content: I need help in setting up the very first circuit in 2L. I have 2 BNC leads, a function generator and a scope.I know this seems very basic but how can I drive my RC circuit with the function generator?What do I need to be careful of? I don't want to go breaking anything.Cheers,

author: rstofer
author_rank: Super Contributor
timestamp: July 21, 2017, 02:21:40 pm
content: Quote from: Mule on July 21, 2017, 12:15:16 pmHook it up like they show it...On the left, connect the signal generatorinputoutput leads - hot (usually red) to the resistor, ground to the common ground (3 connections should be shown tied together at this point in the book).On the right, connect the scope probe to the junction of the resistor and capacitor, scope ground to the common ground.At this stage in the learning process, I would be happier if they showed the 3 ground connections tied together.

author: Mule
author_rank: Contributor
timestamp: July 27, 2017, 11:10:15 am
content: Ahh yes all so obvious now. Thank you very much.

author: coolyota
author_rank: Contributor
timestamp: January 09, 2019, 01:35:42 am
content: Hello, I am attempting to do the lab 1L.3 on page 27 but very confused of what to do. It says we are making a voltmeter but doesn't have instructions or parts I need. What am I supposed to do for this section of LAoE?

author: rstofer
author_rank: Super Contributor
timestamp: January 09, 2019, 03:30:16 am
content: 1) you need a meter - say 0-1 mA full scale2) you need to calculate the internal resistance by building a voltage divider with, say, a 1k divider resistor and a variable voltage source 0-2V (or 10k 0-12V). Anything that is adjustable and can provide 1 mA while including the unknown internal resistor3) set your variable power source to just move the needle to full scale and measure the output of the voltage source4) calculate the internal resistance from the voltage divider equations by first calculating the total resistance from your measured voltage divided by 1 mA. Then subtract off the divider resistor and what you have left is the internal resistance5) now that you know the internal resistance, you can compute voltage divider resistors for various ranges like 0-1V, 0-10V, 0-100V and so on.

author: rstofer
author_rank: Super Contributor
timestamp: January 09, 2019, 03:43:22 am
content: If you don't have a meter, you can use a DMM on some low current measurement range. I can set my Aneng 8008 (cheap meter) for a range of 0-9.999 mA so I can know exactly where 1.000 mA comes up.Personally, I would use the DMM even though the instructions recommend a bare meter. That's fine, if you happen to have one. Not so fine if you have to go buy one.If you buy a quality meter, like a Simpson, you can go to their catalog and find out what the internal resistance is supposed to be. Kind of a cross-check on the calculation. If you buy a meter from Yuan-Hung-Low, you're on your own.Page 5 shows a 0-1 mA meter as having 43 Ohms of internal resistance. If I want to measure 0-100V then I need to figure a total resistance of 100 V / 1 mA = 100,000 Ohms. My meter accounts for 43 Ohms so I need a 99.957K resistor. 99.95 seems to be available at DigiKey Part number Y145399K9500A9L-ND 0.05%Available but not in stock and it looks like you have to buy 2500 at a time. I would use 100k and call it good enough.https://www.simpsonelectric.com/images/File/datasheets/widevue_datasheet.pdf

author: coolyota
author_rank: Contributor
timestamp: January 09, 2019, 08:06:13 am
content: Quote from: rstofer on January 09, 2019, 03:30:16 amCorrect me if I understood it.Adjust power source until the DVM reads 1mA on the display. Then determine the RTby Vin/ I.Internal resistance = RT- R (in this case I picked a 10k Ohms Resistor).

author: rstofer
author_rank: Super Contributor
timestamp: January 09, 2019, 03:34:40 pm
content: Quote from: coolyota on January 09, 2019, 08:06:13 amYup! That concept is all the experiment is supposed to show. As additional information, you can review the schematics of the popular Simpson 260 VOM starting with the original version manufactured in 1940. This is some serious history! This schematic doesn't show the switch but it is pretty easy to step through the voltage divider chain for the various voltage ranges.http://www.simpson260.com/downloads/downloads.htmLook specifically at this version and you can see the protection diodes that are discussed in a note to the instructor in the lab manual.http://www.simpson260.com/downloads/simpson_260-7_and_7m_schematic.pdfThere are applications where VOMs are still the preferred tool. Think about the starter in a car and the internal condition of the battery. Suppose the battery has developed a relatively high internal resistance. When the starter cranks the voltage drops. This is hard to see on a DMM but it's very easy on a VOM. The meter will read 12V, like normal, and you could swear the battery is just fine. Hit the starter while watching the needle and everything will be explained.

author: rstofer
author_rank: Super Contributor
timestamp: January 09, 2019, 06:47:33 pm
content: There is one other consideration: The voltage source has an internal resistance and we didn't account for it. Usually, this will be a small number of Ohms but it is not zero. It might be essentially zero when compared to a 10k divider.The higher the applied voltage and the higher the dropping resistor, the less important this internal resistance will be. You can calculate the voltage source internal resistance by first measuring the open circuit voltage and then measuring the voltage with a known (and low valued) resistor. This is an application of Thevenin's Theorem, also a topic for this experiment.Maximum power transfer (P = I * E) will occur when the load resistor is exactly equal to the internal resistance of the source. Also a result of Thevenin's Theorem. And it might smoke the battery, do be careful!

author: khatus
author_rank: Regular Contributor
timestamp: January 10, 2019, 11:40:35 pm
content: I have the book The art of electronics (3rd edition) . But I don't think it's a good book for theoretical analysis.it's more related to practical

author: rstofer
author_rank: Super Contributor
timestamp: January 11, 2019, 04:41:16 am
content: I think 'practical' was the authors' intent. They tend to not use deeper math and spend more time talking about actual circuits. I have the book but I haven't done anything with it.

author: rhb
author_rank: Super Contributor
timestamp: January 12, 2019, 11:23:57 pm
content: Quote from: kt11 on April 17, 2016, 02:45:24 amI bought and read the 2nd ed years ago. I was on the verge of writing a letter to the authors begging them to update the book when the 3rd came out. It's well worth the money. But I'd keep the 2nd as you might well find that some useful information is mentioned there but not in the 3rd.A lot of the parts mentioned in AoE 2 are *very* hard to find. And there are much better options for a lot of things. I'd like to suggest "Electronic Principles" by Malvino and Bates. Don't get the current edition as it is insanely expensive. AoE is pretty advanced for a novice. The original target audience was physics students embarking on a PhD in experimental physics. So H&H wanted the book to focus on what the mathematics *didn't* tell you. But they really do presume you know all the math already.I worked through the 2nd ed of "Electronic Principles". Unlike many authors, Malvino went to the trouble to actually construct *every* circuit published to make sure there were no typos.FWIW It is my understanding that the the AoE lab manuals are riddled with typos.

author: rhb
author_rank: Super Contributor
timestamp: January 13, 2019, 12:29:40 am
content: Quote from: akos_nemeth on January 23, 2017, 10:10:28 pmThat circuit provides gain at audio. A common way to get gain at RF in a simple radio is to attach the gate of a JFET to the antenna.http://www.techlib.com/electronics/antennas.htmlThis is what you want:[You can reduce it to a single transistor by breaking the circuit at the base divider of the 3904. But the circuit shown is worth building in a small box for future use as a bench tool. It's *very* basic, but a lot of times that is all you need. Use a pair of female BNCs and a barrel connector and add an LED to indicate that it has power. Leave off the switch unless you choose to use a battery. But do yourself a favor. Put labels on the box. I've found a bunch of such things I built 30+ years ago and which I now have no idea what they are for.You should analyze the circuit to determine bias points, currents, etc.FWIW I'm 65 and recently allowed myself the indulgence of buying over $10K of T&M gear. So what have I built lately?A continuity tester powered by a AAA cell in a small box with a 1.5-3 V buzzer from allelectronics.com and two banana jacks so I can safely trace 1.8 V logic level circuits. My HP 34401As supply ~7 V for the continuity test. That's a recipe for causing serious damage trying to trace modern logic.The next project was a level shifter in a box with 8 banana jacks using this board:https://www.allelectronics.com/item/llc-46/logic-level-converter-bi-directional/1.htmlA pair for each logic level and then the 4 logic signals. $3 of electronics in $10 of box.Let's suppose you have a piece of test gear with a CPU and ROM and it's not working because one of the ROM address lines is a bit flakey and ROM images are not available. So you *really* need to read the flakey ROM. By connecting the problem address line Vcc to a bench supply you can bump up the voltage of that one address line a few percent above the normal logic level and see if you can dump the ROM with a logic analyzer.I'm doing another one with DB-9 connectors for RS-232 conversions. That will require special cables, so it will all eventually go into a small storage case with instructions on using it. I was rather dismayed to go though a box of stuff I'd built 30 years ago and realize I had no clue what they did.I've designed more complex stuff, but I needed these gadgets. The 1.5 V buzzer won't trace into a BGA, so I'm contemplating a pulsed LF signal and a detector version to handle that. A test bench version of the signal tracers they sell for sorting out telephone wiring.Ask good questions and you will get good answers. Ask poorly thought out questions and you will get answers of equal quality. There are some *really* smart people who hang out here because they enjoy talking about this stuff.

author: rstofer
author_rank: Super Contributor
timestamp: January 13, 2019, 01:17:51 am
content: There are some much simpler radios, some as simple as a single diode and a 30' antenna.https://electronics.howstuffworks.com/radio9.htmI couldn't find the lab in my 3rd edition - I must have just been looking in the wrong chapter.

author: rhb
author_rank: Super Contributor
timestamp: January 13, 2019, 02:37:32 am
content: The lab manual is a separate book.The OP's question was about adding an RF (high frequency) amplifier, not an AF amplifier. so the prior answer was going in the wrong direction.As I have suitable parts on hand, I think I'll build the circuit I posted for my own use. It's *very* basic, but for a general purpose RF amplifier for LF to VHF it's certainly satisfactory if carefully built Manhattan style.

author: rstofer
author_rank: Super Contributor
timestamp: January 25, 2019, 08:30:11 pm
content: Well, I don't think so. We have some voltage drop across the ballast resistor based on collector current, we have Vfof the LED and that's going to be a couple of volts (usually), we have VceSatfor the transistor (on the order of 0.2V) and, once this is all worked out, we can figure IcLet's take some 'pretend' numbers like Vf = 2.2V. Then 3.3V - 2.2V - 0.2V - (330 * I) = 0330 * I = 0.9V so I = 0.9 / 330 = 2.7 mA - probably not going to be enough to light up the LED - check the datasheet.Looking at the base,we have 0.7VbeSatso we're thinking about 2.6V drop across 10k or 0.26 mA so we need a minimum Beta of 10.4. Ic/ IbNow, we need to know the real numbers: What is Vfof the LED? What is VbeSatof the transistor? What is VceSatof the transistor?Anyway, that's the way I would approach it. I have no idea what the books expects.