Art of Electronics - Study support - Page 1
Forum section: electronics / beginners

author: alex_palvai
author_rank: Contributor
timestamp: February 26, 2020, 02:34:27 am
content: Hi Everyone,Glad I found a place where i can find some like minded people . Happy to see all the forums and discussions here.I am a hobbyist in electronics who want to improve my knowledge and expertise in subject. I started following Art of Electronics and LAB and managed to finish few pages in 1st Chapter. I started studying since a week and progress has been relatively slow as i want to ensure i understand the concepts properly before turning pages.I am stuck in sinusodal signals in theories of physics , differential, formula's and trigonometry and would like to start my first question from this . How necessary is this knowledge to proceed ? Do i need to go back to math & physics , rehearse those and come back ?

author: rstofer
author_rank: Super Contributor
timestamp: February 26, 2020, 04:04:06 am
content: AoE tends to minimize the math. If you have specific questions ask here. Somebody will know the answer.

author: munkeyman1985
author_rank: Contributor
timestamp: February 26, 2020, 04:34:47 am
content: Quote from: alex_palvai on February 26, 2020, 02:34:27 amWe can't read your mind to see if the needed information in there. If you took engineering in college, you might be stating this book at the same time calculus 1 and calculus based physics. Maybe that helps.

author: GerryR
author_rank: Frequent Contributor
timestamp: February 26, 2020, 12:09:01 pm
content: There is no silver bullet, magic pill or the like. You need the basics to advance. If you just want to touch the surface as a hobbyist, there are many cookbook style texts available. Math is essential, and there are a multitude of texts available with applications of calculus and trigonometry to electronics. Get the specialized books geared to electronics. My feeling is that if you master the math, the rest is a knock. JMO.

author: rstofer
author_rank: Super Contributor
timestamp: February 26, 2020, 04:11:52 pm
content: I am of two minds on the math thing. First of all, electronics is applied math; simple as that! It's all math and, yes, the math is important. However...Do I need to be able to derive the differential equation for charging a capacitor (Question 13 in the attachment) or just accept that it is correct and use the exponential equation just above and including Figure 7? Let's face it, the derivation is pretty if you have the math but it's the resultant exponential equation that gets used every day.http://web.mit.edu/molly/Public/circuits-b.pdfIntegral calculus is usually taught in the second semester of EE school assuming, of course, that the student had competence with Pre-Calculus before starting college and this isn't always true. If they don't have competence, Pre-Calc is usually a two semester course so add a year to your graduation date. Not paying attention in high school comes at a cost!For the hobbyist, accept Figure 7, know the exponential result and be absolutely certain you know how to plug the numbers into a calculator.Take 10k Ohms and 0.1 ufd. Tau (the time constant) = R * C (R in Ohms, C in Farads) and this is a number you need to calculate and you better get 1 ms (0.001 on the calculator). Then when somebody says, what percent charge is on the capacitor at t = 1 Tau (or 1 ms) the result should be 63% - this is just an exercise in calculator operation but it is an important equation. Assume Qf = 1As to sines and cosines, you usually run across cos(wt) where the lower case w is supposed to be the Greek symbol for Omega. It is pronounced 'cos omega t'. You need to remember that omega = 2 * PI * frequency (in Hz). It's just shorthand, EEs are a lazy bunch.https://en.wikipedia.org/wiki/Radian_per_secondWhen you get to Kirchhoff's Laws with AC circuits, the arithmetic gets a little hairy. I CAN solve the equations by hand but I don't. MATLAB is my tool of choice but the free program Octave is a work-alike. Maybe this will help, maybe it will just add to the overload:https://www.amazon.com/gp/product/0982497008What you need to decide is at what level do you want to know electronics? The hobbyist is usually doing a cut and paste operation with minimal actual design. Serious math may not be required.A couple of my favorite resources:For solving stuff:www.symbolab.comFor graphing stuff:www.desmos.com

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 26, 2020, 05:07:59 pm
content: Quote from: rstofer on February 26, 2020, 04:11:52 pmI have a similar mindset.Being able to analyze a circuit in the S domain is important, but rarely do I need to do the same analysis in the time domain (where the work load goes up). Even so, if I need to analyze something that has more than a couple of poles/zeroes, I will just shove the netlist into a symbolic solver to get the transfer function.

author: rstofer
author_rank: Super Contributor
timestamp: February 26, 2020, 05:34:25 pm
content: Quote from: Wimberleytech on February 26, 2020, 05:07:59 pmDoes the average hobbyist even need to know there is an S domain? Unless they get to filter design or control systems, I'm not sure how important that concept actually is. Laplace Transforms and Fourier Series, good luck with that! I don't even want to think about Field Theory and Maxwell's Equations.But, yes, by all means, stuff the equations into MATLAB. I got through college with a slide rule and I'm pretty sure I don't remember how I made it past the math. Maybe that is why I have never played in the analog circuits corner of the electronics sandbox. I have stayed in my little corner over in the digital domain.As Bob Widlar (very famous analog circuits designer for National Semiconductor) said:

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 26, 2020, 05:45:02 pm
content: Quote from: rstofer on February 26, 2020, 05:34:25 pmI am an analog guy, so, yes I think in terms of poles and zeroes. I dont know what an average hobbyist is but you may be correct. The easiest way to know is to survey all of the questions on the Beginner board and see how many reference bandwidth, frequency response, poles, zeroes, etc. That would be an interesting exercise.

author: GerryR
author_rank: Frequent Contributor
timestamp: February 26, 2020, 07:49:55 pm
content: I don't think you have to be an expert at everything, but you should at least know what has to be used where, and then know where to look it up. Will Rogers put it best, "Everybody is ignorant, just on different subjects." So, in the areas you are lacking, know where to look it up. I'm stuck in the middle between analog and digital and have done a fair amount in each area, but mostly related to the automation world involving sensors, signal conditioning, etc. I have an extensive library, so when I get in trouble, I go to my references. But, if I didn't have the initial exposure and understanding, how would I know what needed to be researched? Like I said before, there's no magic pill, just good old fashioned hard work. JMO.

author: ebastler
author_rank: Super Contributor
timestamp: February 26, 2020, 08:05:09 pm
content: Quote from: Wimberleytech on February 26, 2020, 05:45:02 pmI have not done that exercise (and don't intend to do so), but can venture a guess:You will find a fair share of questions about bandwidth, and some about frequency response. But hardly any on Laplace transforms, poles, and zeroes -- and the few you find won't be from hobbyists looking for help with a project, but from students looking for help with their homework.

author: rstofer
author_rank: Super Contributor
timestamp: February 26, 2020, 08:31:51 pm
content: Quote from: GerryR on February 26, 2020, 07:49:55 pmBut to what end? My point is that, as I see it, most electronics practiced by hobbyists seems to be related to using an Arduino and then realizing that everything that can be done with an Arduino has been done and it's on the Internet. There are clearly exceptions...A step up would be to understand Ohm's Law, Kirchhoff's Laws, Thevenin's Theorem and Norton's Theorem. That is probably enough for most hobbyists at even a fairly advanced level. They are not likely to be designing 3 pole filters. At a minimum, Ohm's Law, in all its manifestations, needs to be understood from day 1. We go nowhere with op amps without Kirchhoff's Current Law. But these are just ordinary arithmetic, nothing esoteric.You're absolutely right about resources. I have a huge library but my real resource is Google. All of human knowledge is on Google, all I need to do is come up with the right search string.There's a lot of space in the electronics sandbox and people can move to a spot that is comfortable. I like digital, the only reason I went to college was to learn digital and that's the portion of the sandbox where I play. Can I go back and recapture what I learned about Laplace Transforms? Sure! But I would need an application that generated enough interest to bother re-reading my textbooks. Of course, having MATLAB will make the actual work a lot easier. I really don't like partial fraction expansion of polynomials if I have to do them by hand.I should put in a plug for Khan Academy for math and the Electrical Engineering program at Khan Academy for the more focused stuff. Khan Academy is free! Or you can contribute - I do... I highly recommend these programs!For a fee, CalcWorkshop.com is excellent. There is a free series on Limits just to show how the lecture series work. It's a GREAT program!Here is the Limits program:https://calcworkshop.com/online-limits-course/I highly recommend taking the entire program starting from wherever seems appropriate. Jenn does a great job as an instructor. She starts with Algebra I and works up through Differential Equations in 8 major blocks and about 450 videos! If a person were to spend $150 for a year and they got through the entire series, it would work out to about $0.33/video.All of the education required is right there in one very long program. It isn't free but it is well worth the money. Subscribe for a year and go through the program as fast or slow as you wish. Education has never really been free.BTW: You may wonder why you even care about Limits when they are just a prelude to the derivative. It turns out the concepts are actually useful in certain conditions when you have a discontinuity in a function and it can't be differentiated at that point.

author: rstofer
author_rank: Super Contributor
timestamp: February 26, 2020, 08:37:01 pm
content: Quote from: ebastler on February 26, 2020, 08:05:09 pmHow Not To Blow Up Your Scope is another popular topic. I actually like working on some of the homework problems.I would think the poles and zeros would be over in the Projects forum, if they were to show up at all. People who understand what these are probably don't need to ask.

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 26, 2020, 08:45:39 pm
content: OK. I admit defeat.

author: GerryR
author_rank: Frequent Contributor
timestamp: February 27, 2020, 12:55:35 pm
content: Quote from: rstofer on February 26, 2020, 08:31:51 pmWhat you say about Arduino is "cookbook engineering." Having a more in-depth knowledge allows you to change the recipe and adds a little creativity to a design. I have electronic cookbooks, as I expect most engineers have; they save a lot of time in a design process, but it would be a rare occasion for me to find a design that didn't need a "recipe change" to suit my application. Without a more in-depth (not necessarily extensive) knowledge, I couldn't make these changes. Of course, there are the unique project requirements where you have to start from scratch for the entire design, and cookbooks are of very little help.I don't know how anyone can even be considered an electronics hobbyist without knowing the basics you describe in your second paragraph. I would merely describe them as "kit builders." That is evidenced by the types of questions I see asked when something they have built doesn't work. But hey, if they are having fun, each to his or her own.I have had to make a living at what I do.

author: rstofer
author_rank: Super Contributor
timestamp: February 27, 2020, 03:41:28 pm
content: Quote from: GerryR on February 27, 2020, 12:55:35 pmRegardless of labeling, if the hobbyist is having fun, that is sufficient. If they need to know more, they can take a side trip and study up on a subject. But it is clear to me that a lot of hobby projects can be built without differential equations or Maxwell. There is no need for a 5 year degree just to play in the sandbox and they probably aren't making a living out of electronics. For me, electronics has always been a hobby. I spent a lot of time in school just to support a hobby...

author: GerryR
author_rank: Frequent Contributor
timestamp: February 27, 2020, 05:23:15 pm
content: I think we are on the same page here!

author: alex_palvai
author_rank: Contributor
timestamp: February 28, 2020, 12:50:07 pm
content: Thank you for all the responses. Appreciate suggestion from rstofer about khan academy. Few guesses are right, i didn't pay attention to calculus , derivatives , integrals in my college (finished 10 years ago) as i couldn't relate to them in real life then, they did not interest me or i didn't know the importance of it (guess many have similar stories ) , just studied to pass my exams that's it. I don't rely on electronics for living but lately in last few years felt passionate to learn. I enjoyed working on Arduino, its probably best thing for prototyping something quick in days regardless of what other thoughts are it gives a practical education, not in theories only . I am in no rush to finish things fast , if math & physics are must to move forward my journey will be a bit slow as i expected anyways .My basis will be Art of Electronics & I will use references to progress and ask questions on this forum if i am stuck ( Really appreciate your responses ) . My next questions will be more specific. I am happy1. Because its interactive ( Am not just interfacing with books and artificial intelligence " google" ) - What i wanted toand2. There is no risk of me assuming something right ( I can clarify if there are any questions ) .I don't want to ask questions on Math as i'll research and figure them out myself , but i want to ask questions on learning tricks in practical ways.I want to start off with my first question here,While working on voltage divider circuits , i was able to use bread board and experience Thevenin's theorem using resistance and load using few resistors and DMM, I was able to relate those results to the math i did and relate to .I am studying signals now . Theory exercises in this phase are relating to comparing signals , amplitudes , measuring them and converting to volts and power.The only analog experiments i experimented so far are using Arduino (PWM simulator ) . I understand Analog is more harder to experience than basics hoping not impossible. Wondering if there any way i can start doing small lab to practically do things and verify theories.I want to be able to measure signals (with something ) , build a basic analog circuit with (input and output ) and test linearity , scalier properties , make changes and perceive same same in output (say led or sound similar to what i did for voltage divider & Thevenin's equivalent learning .Thank You in advance.

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 28, 2020, 01:54:30 pm
content: Quote from: alex_palvai on February 28, 2020, 12:50:07 pmDo you have an oscilloscope? If not, I suggest that you learn to use a simulator (LTSpice). You will be able to measure voltages and currents very easily (easier than on a breadboard).

author: alex_palvai
author_rank: Contributor
timestamp: February 28, 2020, 05:50:23 pm
content: Wimberleytech: Thanks for the quick response and reference.

author: rstofer
author_rank: Super Contributor
timestamp: February 28, 2020, 05:52:01 pm
content: If you are trying to learn electronics, there is no more suitable measuring system than the Digilent Analog Discovery 2. It has a 2 channel scope, a 2 channel arbitrary waveform generator (advanced signal generator) plus a lot of other gadgets.https://www.eevblog.com/forum/beginners/learning-the-art-of-electronics-signal-attenuation-(figure-in-34)/Read through the link, I make the case for the AD2 and attach some plots.

author: alex_palvai
author_rank: Contributor
timestamp: February 28, 2020, 06:11:16 pm
content: Thank You rstofer.

author: alex_palvai
author_rank: Contributor
timestamp: February 29, 2020, 06:32:31 pm
content: Just finished understanding topic 1.3 (Signals ) - Sinusoidal Sinusoidal Signals , signal amplitude and decibels topics in AoE.I would like to contribute references i used to make my understanding clear, perhaps it will help other folks who are studying AoE .I wanted to get a better understanding of Waves ( From physics PoV ) and i studied Chapter 8 ( Alternating currents ) from "Physics Demystified - Stab Gibilisco" book which i am using as reference . This helped me to get a clear idea of types of waves are , Why do we need RMS , what does peak to peak mean and answered questions like why we have degrees and rad's representing same thing , similarly cycles per second and Hertz. After getting my mind cleared on these topics i went back to AoE reread things which then became more clear ;1.3.2 I also finished the next part "Signals and amplitude and decibels":it took me 1 video to understand the whole topic explained on Decibels , dbm, dbu, dbV (what they all are and how they are originated. ) . The way this guy explained the topic is an Epic of explanations.Hope it helps.

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 29, 2020, 07:01:31 pm
content: DecibelsThis is from my files. It is from an article retyped by my secretary (back when you called them that) back in 1985. I seem to recall some typos somewhere but am not checking it here before posting.

author: rstofer
author_rank: Super Contributor
timestamp: February 29, 2020, 09:32:47 pm
content: Quote from: alex_palvai on February 29, 2020, 06:32:31 pmAlan Wolke, w2aew, has a bunch of very good videos. It is worth the time to peruse his site. Dave also has some excellent videos in his Fundamentals Fridays series. Dave's Op Amp video remains a personal favorite. EEV Blog #600 - What Is An Op Amp?FWIW, there is a very long Oscilloscope tutorial linked at the top of this forum. It, too, is Alan's work. He is an application engineer at Tektronix so that's why he has all the cool test equipment. It's interesting to see him use old Simpson 260 V-O-Ms instead of DMMs for some of his videos.

author: rstofer
author_rank: Super Contributor
timestamp: February 29, 2020, 09:55:50 pm
content: As explained in the video above, when comparing voltage (say the input and output voltages of a low pass filter), the form isdBV = 20 log (Vout/Vin)That -3dBV breakpoint in the Bode' Plot has an Vout/Vin ratio calculated as-3dBV = 20 log (Vout/Vin). Divide through by 20 and you get -3/20 = log (Vout/Vin). Then take the antilog of both sides10(-3/20)= Vout/Vin which results in 0.707 = Vout/Vin If Vin is 1V then Vout at the -3dBV point is 0.707V, less than the input voltage, as expected.

author: alex_palvai
author_rank: Contributor
timestamp: March 07, 2020, 08:17:18 pm
content: Friends,I realized AoE has everything in shortcut and i am referring other references for how those equations are derived and those conclusions in AoE made. Currently stuck on how time (t) equation(1.22) in RC Circuits is derived from Vout Equation (1.21). I understand how Vout was derived but am trying to connect the gap from Vout to t. Hoping someone can help me filling this gap step wise. Thanks in advance.

author: rstofer
author_rank: Super Contributor
timestamp: March 08, 2020, 12:35:41 am
content: I'm not sure my AoE is the same version as yours. However...Start with Vout = Vapplied(1-e-(t/RC))Divide out VappliedVout / Vapplied = 1-e-(t/RC)Swap things arounde-(t/RC)=1-(Vout/Vapplied)Take the ln() of both sides-t/RC = ln(1-(Vout/Vapplied)Change signst/RC = -ln(1-(Vout/Vapplied))Multiply both sides by RCt=-RC*ln(1-(Vout/Vapplied))Not quite the same form as AoE, let's see if it works. We know that the output will be 95% of the input in 3 time comstants. Let's just make Tau = 1 (the time constant)t = -1*ln(1-0.95) = -1 * ln(0.05) = -1 * (-2.99) = 3 time constants to reach 95%We also know the output reaches 63% in 1 Taut = -1*ln(1-0.63) = -1*ln(0.37) = -1*(-0.994) which is just about 1 time constant.I'm not entirely sure why my form is different but I'm pretty sure it is right. If it isn't, everybody will jump in and I'll know about the problem real soon.I know I am missing a step to clean up the output because the AoE equation is also correct. At the moment, my algebra isn't strong. Maybe it will come to me later.

author: alex_palvai
author_rank: Contributor
timestamp: March 08, 2020, 12:19:51 pm
content: A big very thanks rstofer,. Your derivation sat is my brain easily .I am using 3rd Edition of AoE . Trying to verify the equation given in book .I tried substituting in the formula derived in AoEt=RC Log ( Vapplied / Vapplied - V)Say , Let's Say Resistor is 1 k and Capacitor is 1 micro Farads , Vapplied = 1 and Vout 0.63 ( from what we know ) . Putting them in Equation1 Log ( 1 /1-0.63) => 1 Log ( 1 / 0.37) => 1 Log ( 2.7028) = 0.4318 Seconds( Incorrect )Is it possible that the equation given in book is incorrect or Am i doing something wrong .

author: ebastler
author_rank: Super Contributor
timestamp: March 08, 2020, 12:38:42 pm
content: You need to use the natural logarithm (log base e, or ln), not the decadic logarithm.

author: Siwastaja
author_rank: Super Contributor
timestamp: March 08, 2020, 01:42:08 pm
content: Art of Electronics is both a great reference and a good study book, but it isn't sufficient as-is learning material, IMHO - no book is. Its approach is very pragmatic; so you will greatly benefit from adding practical prototyping and testing. When learning a new thing, if you just read 50 pages of the heavy, concentrated stuff like AoE, your head starts to spin. If you iteratively build, test, measure, read, understand, rinse and repeat, you'll do much better.Understanding the phenomenon should come first, then refine it with math. For example, understand an RC circuit at this level first:All the voltage is over the resistor: larger current flows (I = U/R), capacitor charges fasterNow capacitor has some voltage; voltage over resistor is smaller: smaller current flows. Capacitor charging slows down.Increase C: it charges slower from the same current, obviously.Increase R: capacitor charges slower, because charging current is smaller.Understand that the time to charge to some level is relative to R times CThen add math: refine what the time constant actually means...Refine the curve shape: you already intuitively understand it slows down, but find the exact logarithmic shape.Finally, understand the 1-e^-(t/RC) formula.The last bit is, actually, non-critical. You can go far by understandingconcepts, not exact math, then fill in gaps when you actually need it the first time. IMHO, combining different ways to learn is the key.

author: rstofer
author_rank: Super Contributor
timestamp: March 08, 2020, 03:44:57 pm
content: Quote from: Siwastaja on March 08, 2020, 01:42:08 pmI sometimes wonder why we seem to gush over the AoE and LtAoE as much as we do. I still have my 1970 textbook and I'm pretty convinced it is a better source of information. If has some diagrams, nothing in color, and is written as old textbooks are - dry. But the information seems to be in more of a natural progression. Just an opinion...I have played with this problem in MATLAB as a demonstration I added a couple of lines of output just to verify both my solution and the book solution. Both forms give the correct result. This code runs in Octave as well.I want to lodge a formal complaint, somewhere... In the world of math, the natural logarithm function is ln(x) and the logarithm base 10 is log(x). But not in the programming world and certainly not in MATLAB. Here the natural logarithm function is log(x) and the logarithm base 10 is log10(x). Somewhere I have seen the natural logarithm as alog(x). Come on people, let's get it together! Log(x) is base 10, ln(x) is the natural logarithm - let's standardize!In any event, in the code below, log(...) means natural logarithm despite what is on the buttons of my calculator. The code plots the charge and discharge curves for a 10k resistor and a 0.1 ufd capacitor (or any other combination that has Tau = 0.001 seconds). It also prints the percent charge and a back calculation of 't' using the two forms of the 't' equation. I copied the output and pasted it into the source.Note that when I print the results, I start the line with a comment character '%' so I can copy and paste the results into the source as a series of comments.Make friends with Octave, you'll need it.in Octave: File -> New Script, Copy and Paste code in Editor window, click "Gear" icon to run (you will have to name and save the file) and you should get a plot. In the Command Window (tabs at bottom of pane), you will find the printed results. This is a perfect example for starting with Octave. It's real, it demonstrates how to code for Octave (and MATLAB) and it makes a nice plot. The printed results are just there to verify the calculations.Code:V0 = 1; % assume a supply voltageR = 10000; %10k OhmsC = 0.1*10^-6; % 0.1 ufdTau = R*C;t = linspace(0,8*Tau); % set up the x axis to run from 0 to 8*Tau 100 intervalsVchg = V0 * (1 - exp((-t/Tau))); % charge equation 100 valuesVdis = V0 * (exp((-t/Tau))); % discharge equation 100 valuesplot(t, Vchg)grid onhold onplot(t,Vdis)ylabel("Voltage")xlabel("Time - seconds")%there is a known bug in 'legend', the [h,~] is a kludge to work around it[h,~]=legend("Charge Voltage","Discharge Voltage");figure(gcf) % or shg command - pull plot to topfor Tau = 0:6pct_charge = V0 * (1 - exp((-Tau))); %percent charget1 = -log((1-pct_charge)); % my version of t equationt2 = log(1/(1-pct_charge));% book version of t equationfprintf("%% Tau = %.0f Percent Charge = %5.2f,\t t1 = %7.3f, t2 = %7.3f\n", ...Tau, 100*pct_charge,t1,t2)end% output% Tau = 0 Percent Charge = 0.00,t1 = -0.000, t2 = 0.000% Tau = 1 Percent Charge = 63.21,t1 = 1.000, t2 = 1.000% Tau = 2 Percent Charge = 86.47,t1 = 2.000, t2 = 2.000% Tau = 3 Percent Charge = 95.02,t1 = 3.000, t2 = 3.000% Tau = 4 Percent Charge = 98.17,t1 = 4.000, t2 = 4.000% Tau = 5 Percent Charge = 99.33,t1 = 5.000, t2 = 5.000% Tau = 6 Percent Charge = 99.75,t1 = 6.000, t2 = 6.000

author: tkamiya
author_rank: Super Contributor
timestamp: March 08, 2020, 05:36:00 pm
content: My advice is, if you are just getting started, don't worry so much about theory. I started when I was 7 years old. I did not even know what sine waves were and my math was pretty much non-existent. I went quite far this way. Only in last 10 years or so, I got serious and re-learned and extended use of mathematics. Took about 10 math courses at college. Also, for every complex math modeling, there are simpler version that's sufficiently approximate the first.That said, understanding electronics mathematically will take you much further. Eventually, your curiosity and needs will take you there.

author: alex_palvai
author_rank: Contributor
timestamp: March 08, 2020, 07:20:18 pm
content: rstofer: Thank you for taking time to do the exercise in MATLAB . Your detailed instructions helped me get started on the tool . Language used by the tool seems very close to Python which i am good at , this tool is perfect for me , will definitely help and make my learning days more interesting and Enjoyable . Your the best! .

author: alex_palvai
author_rank: Contributor
timestamp: March 08, 2020, 07:47:29 pm
content: tkamiya, Siwastaja : Thanks for sharing your insights and experiences . I definitely understand how capacitors work but my goal is to learn subject to a level i can tune circuit performance and interpret same in language like a pro ( want to develop in multi dimension ) . For this reason, when i don't understand how equations are derived it gets me disconnected and unsatisfied. Even if its hard and slow process , when i learn it completely this way i feel satisfied and confident :-) .eblaster: Thanks for pin pointing where the problem was in equation .

author: alex_palvai
author_rank: Contributor
timestamp: April 19, 2020, 04:55:19 pm
content: Hello All, Hope you all are doing good.I am stuck trying something on LTspice , hoping I can find some help. Given the Circuit with 2 resistors , capacitor and voltage source. The as was to find V(t) and graph it.In my mind first 5 ms will charge the capacitor to 10 V ( as its connected through voltage divider circuit) . Then my pulse kicks in creating a step voltage "0" for another 5 ms . then it spikes up again to 20 V till 50 ms. I consider the Voltage input (green) from 0 ms to 10 ms as 1 cycle . What should i do to have this cycle repeated ?

author: rstofer
author_rank: Super Contributor
timestamp: April 19, 2020, 05:39:07 pm
content: Change Ncycles from 1 to something else?You may need to use the period parameter as well if there is time left over in each cycle.Lower left on Pulse dialog.

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 19, 2020, 05:52:21 pm
content: Quote from: rstofer on March 08, 2020, 12:35:41 amln(x) = -ln(1/x)

author: iMo
author_rank: Super Contributor
timestamp: April 19, 2020, 06:12:02 pm
content: Quote from: alex_palvai on April 19, 2020, 04:55:19 pm

author: alex_palvai
author_rank: Contributor
timestamp: April 19, 2020, 07:08:24 pm
content: Thanks rstofer.Your suggestion brought me quiet close. Vout was for some reason 10% curvy instead of flat(rstofer's suggestion.PNG)Further below reference helped understand issue.https://electronics.stackexchange.com/questions/237108/ltspice-default-rising-falling-times-of-pulse-edgesI had set trise and tfall to non-default (0.001ms) and started seeing flat consecutive graphs.Thank You once again.

author: alex_palvai
author_rank: Contributor
timestamp: April 20, 2020, 11:22:13 pm
content: Hello all,I am working on Integrator circuit on Ltspice. I am satisfied with the outcome of graph on LtSpice (its matching with the reference I have on AoE) .Now i am trying to calculate V using formula "V(t) = 1/RC* (Integral of t) [Vin(t)dt] + C "( attached picture )at 0.4ms i am expecting to get an answer of 12 Volts ( as per graph) but i am getting 1.6 V ( incorrect answer ) . Can someone help pointing where i am doing wrong.V(.4ms) = 1/RC* (Integral of t) [Vin(t)dt]+CTaking Vin (20v) to left , and integrating for t.dt i get t^2/2 :=> 1/1* 20 *( t^2/2)substituting 0.4 ms for 't':=> 1*20( 0.16/2) => 10*.16 = 1.6 volts/ms(attaching screenshot incase the math part not clear ) .

author: alex_palvai
author_rank: Contributor
timestamp: April 21, 2020, 12:39:48 am
content: I tried slightly different approach by substituting Vin(t) = 20VV(t) = 1/RC* (Integral of t) [Vin(t)dt]+C=> 1/1 * (Integral) [ (20) dt +C]taking 20 to left:=> 20 (Integral) 1dt +C=> 20 (t)Substituting time 't' = 0.4ms=> 20 ( 0.4 ms ) = 8 Volts / ms (still wrong . expecting 12V )(See attachment on Math on whiteboard. )

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 21, 2020, 01:30:49 am
content: Quote from: alex_palvai on April 21, 2020, 12:39:48 amYou are not taking into account initial conditions...you are ignoring CUse this:

author: alex_palvai
author_rank: Contributor
timestamp: April 22, 2020, 10:10:42 am
content: Wimberleytech,Thanks for the response.I might be doing something wrong in application but i am still getting wrong answer while applying that formula:vFinal = 20 VVInit = 0RC = 1 msApplying in formula , i get 6.96 while it should be 12 .------------------------------------------------------------------------Vc(t) = Vfinal + ( Vinit - Vfinal ) . e^ (-t/rc)Vc(0.4) = 20 + ( 0 - 20 ) . e^ ( -0.4 / 1 )=> 20 - ( 20 ) . e^(-0.4)=> 20- 13.04 => 6.96

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 22, 2020, 02:32:18 pm
content: Quote from: alex_palvai on April 22, 2020, 10:10:42 amOnce more or less steady state, Vinit is not zero. It is about six volts

author: alex_palvai
author_rank: Contributor
timestamp: April 23, 2020, 12:37:55 am
content: I guess i figured out the issue. Formulae is only working when voltage across capacitor is very low than input voltage . Attaching the references of perfected circuit and Math . The mistake i was doing was following the guidelines of RC differentiation circuit for integrator circuit . In differentiatior circuit the ask was to keep RC value low so that Vout is always low than Vin. in Integrator circuit its opposite as capacitor is connected to ground RC value must be large enough so that Vout is negligible .

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 23, 2020, 02:06:32 am
content: Quote from: alex_palvai on April 23, 2020, 12:37:55 amNo, you have not figured it out.

author: alex_palvai
author_rank: Contributor
timestamp: April 23, 2020, 11:06:26 pm
content: Hello All,In my mind battery is a voltage source and current is bi product of voltage source caused by potential difference between +ve & -ve terminals . My idea of capacitors are kind of same that they are voltage source and when discharged produces current (similar to battery) . I am reading about inductors in 1st chapter of AoE which refers inductors store current (not voltage ) in the form of magnetic flux and my understanding that inductors release current as they discharge and voltage is bi-product of current in this case(opposite of capacitor) .Confusion starts at a point where there was a mention "the defining equation V= LdI/dt requires average voltage must be zero with explanation "otherwise magnitude of its average current is rising without limit" .Does this sentence mean the input to inductor must always be a pulse ? ( figured its incorrect )I ran a test in LTspice to input constant voltage of 1v to circuit with inductor. Observed two things 1. voltage across inductor is zero ( I don't yet understand what's happening to voltage after it crosses Resistor ). And my second observation was that current measured on inductor in this configuration is stable at 1.0 mA. ( So my assumption of what i read about infinite current is incorrect) .I guess these two are interlinked and if i know the reason or logic for "why voltage across inductor is zero" i would probably understand why magnitude of current will be infinite if voltage across inductor is zero . Appreciate if someone can shed some light from their experience or direct me to some reference where i can find the best explanation .Attaching the tests i ran on LTspice .

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 24, 2020, 01:17:54 am
content: Quote from: alex_palvai on April 23, 2020, 11:06:26 pmAn ideal inductor has a resistance of zero (0). So, if di/dt =0 (current is constant) then there can be no voltage drop across an inductor. Capacitors are similar in terms of voltage. If dv/dt = 0, no current flows in a capacitor and thus the resistance is infinite. For capacitors and inductors, they only become interesting when terminal conditions are dynamic.

author: alex_palvai
author_rank: Contributor
timestamp: April 25, 2020, 06:13:42 pm
content: Hi Wimberleytech,I am not sure if i understand your response correctly. What i am trying to understand is why the voltage on the wire between inductor and resistor is zero.

author: alex_palvai
author_rank: Contributor
timestamp: April 25, 2020, 07:06:07 pm
content: Wimberleytech,Please disregard my last response. I understand what you mean now.if di/dt = 0 then voltage across inductor is becoming zero. Thank You.

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 25, 2020, 07:06:18 pm
content: Quote from: alex_palvai on April 25, 2020, 06:13:42 pmIt is because the resistance of the inductor is zero ohms. Thus even with 1mA flowing through it (dc) there will be no voltage drop.An ideal inductor only opposes current flow if the current is changing.Does that help?

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 25, 2020, 07:07:17 pm
content: Quote from: alex_palvai on April 25, 2020, 07:06:07 pmoops I already answered...looks like you got it now!! Great!

author: iMo
author_rank: Super Contributor
timestamp: April 25, 2020, 08:36:03 pm
content: FYI.

author: alex_palvai
author_rank: Contributor
timestamp: April 25, 2020, 09:48:07 pm
content: Thanks Imo,. Your circuit is depicting clearly initial resistance and change in voltage for few nano sec's (innitial resistance of inductor to change in current ) .Thinking loud:And i guess what else is interesting is voltages comes back to zero in a nano second and current stays there at 10 mA ( I am guessing that's a charge stored in magnetic field of L (incorrect I realized its a current passing through inductor) ) .. And as input voltage drops from 10 to 0v , Inductor is starting to sense the change the current and starts resisting the change which creates voltage spike . ( Is it appropriate to assuming inductor is current sensing device as a general idea ? - I guess so that's appropriate , i answered my own question )Actually One more thing . I see the current across resistor is going negative -10 mA as source voltage is 10 V (I am still trying to find why this is happening ) .Attaching my LT spice file.

author: alex_palvai
author_rank: Contributor
timestamp: April 26, 2020, 01:43:37 am
content: I am trying to build circuit 1.52 A in LTSpice. Having trouble with switch. Is there a way i can configure this switch to switch between two circuits with 50% duty cycle?

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 26, 2020, 02:27:00 pm
content: Quote from: alex_palvai on April 26, 2020, 01:43:37 amYes, use two switches driven 180° out of phase of each other. What out for a glitch at each transition. I hacked one together and it seems to work.

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 28, 2020, 03:11:13 am
content: For some reason "quote" is not working for me..."When current is reading 0 mA , why is voltage on node l1 reading (1.9950) volts instead of zero."Because there is an "off" resistance of your switch. So, the initial operating point is calculated with a resistance connected to the 2V source.So then, why isn't the initial voltage 2V? It is simply due to the simulator convergence accuracy. When the iteration got to 1.9950 volts, the tolerances in the simulator were met so it stopped.Change your off resistance to something lower, like 10Meg and see how the results change.

author: alex_palvai
author_rank: Contributor
timestamp: May 01, 2020, 10:18:09 am
content: Quote from: Wimberleytech on April 26, 2020, 02:27:00 pmI successfully created two switches which switch between voltage source and ground at 50% duty cycle.I am stuck on this circuit for a week clueless of whats going wrong . As per the attached instructions in page "Inductor Voltage Diverer.PNG" Circuit Fig (1.52 A) must Vout have 1/2 input voltage and (Fig. 1.52B )must have Vout equal to 2 times Vin.I am experimenting on first circuit now "1.52A" and expecting to get Vout be 1/2 Vin without a success.Loosing the sense of volt-second balance rule, I started to go trial and error method for different combinations . Hope someone helps me with a fix and help me understand how to bring the voltage and current into steady state .Details of trials :Ind :.1 u Cap: .1u -- Vout is square wave with voltage spikes at switch open and close peaks at 4V and -2V reduces back to 2 and 0.Ind : 1 Cap: .1u -- Vout rises endlessly 0 to 20V ++Switched source voltage from 2v DC to pulse - Ended up with very wierd result Vout (unexplainable shapes) in nano volts.Disconnected second switch to ground with pulse voltage input ---- Vout is sign wave between 2 and -2 volts .Switch back from pulse input to DC input of 2 V - Vout is abnormal signwave peak to peak from ( 2.00006 v to 1.999993 v )

author: Wimberleytech
author_rank: Super Contributor
timestamp: May 01, 2020, 01:01:31 pm
content: Quote from: alex_palvai on May 01, 2020, 10:18:09 amYou are clocking at a very low frequency compared to the filtering from the LC. L and C need to increase in value. To avoid ringing, you need to add some resistance to the inductor model as that would be real-world anyway (I put 10 ohms, but a more realistic value is closer to 1 ohm). You will need to run more cycles of your pulse, and run the simulation out longer. See attachments

author: alex_palvai
author_rank: Contributor
timestamp: May 01, 2020, 03:52:45 pm
content: Quote from: Wimberleytech on May 01, 2020, 01:01:31 pmGood Morning Wim,Hope you are doing well.Thanks a lot for showing me that this actually works. I am sure your years of expertise played nailing the issue very quickly. For someone who just started learning like me , what would be the key elements to consider and indicators to realize the right frequency and time for a specific configuration of capacitor and inductor combination .

author: Wimberleytech
author_rank: Super Contributor
timestamp: May 01, 2020, 06:08:10 pm
content: QuoteGlad to help. Think of the L and C as an LC filter that filters a square wave input.

author: alex_palvai
author_rank: Contributor
timestamp: May 06, 2020, 12:31:15 pm
content: I am reading about Center-tapped full-wave rectifier explanation in AoE section 1.6.4 (B) - Attached references .I find it really hard to construct/visualize what he is trying to explain in this section .Here's what i struggle with ; typing this paragraph and the questions in parenthesis . kindly please someone guide me through this .Center-tapped full-wave rectifier :The circuit in Fig 1.63 ( attached ) is called center-tapped full-wave rectifier . The output voltage is half what you get if you use bridge rectifier . It is not most efficient circuit in terms of transformer design, because each half of the secondary is used only half the time ( ME:clear to point ) .To develop some intuition on this subtle point , consider two different configurations that produce same rectified DC output voltage ( ME:clear to point , he'z going to compare two circuits )cont..(a) circuit in Fig 1.63( attached) and(b) the same transformer (( ME: which same transformer) , this time secondary cut at the center tap and rewired with two halves in parallel ( ME: i am thrown off already at this point , where did you cut and how did you join) cont.. the resultant combined secondary winding connected to full wave bridge ( ME: What is that resultant circuit , how does that look which is connected to full wave bridge) .

author: Wimberleytech
author_rank: Super Contributor
timestamp: May 06, 2020, 01:14:43 pm
content: Quote from: alex_palvai on May 06, 2020, 12:31:15 pmYou have reason to be confused. I think this is a poorly-written section. They should have drawn both configurations and given a little better explanation. I will spend a little time and try to draw something up that makes a little more sense. Stay tuned.

author: alex_palvai
author_rank: Contributor
timestamp: May 10, 2020, 09:46:59 pm
content: For those like me "Starters struggling with shortcut's in Art of electronics" , What's helping me other two references I haveI also have Grob's Basic electronics by Mitchel E Shultz - 10th edition --- Provides a great break down explanation progressively.My other reference is "Practical electronics for Inventors - —Paul Scherz and Simon Monk - 4th Edition .Some topics are well written by one writer than others .Although Expensive , Grob's Basic electronics by Mitchel E Shultz is probably great reference (well written and easier to understand )

author: alex_palvai
author_rank: Contributor
timestamp: May 12, 2020, 08:54:34 am
content: Quote from: NeverDie on May 11, 2020, 07:14:07 pmThanks for feedback, it motivates me more towards analog whenever someone says that, reasons below:I played with(arduino , rasberry pi) for years before i jumped into Analog :-) (they are only good for starter's , i cannot imagine to create a product of my own and do a mass production with Adriuno / rasberry pi . ) , they are simply not ideal. My perspective, plug and plays are not challenging & lack in power and limited- not customization(don't make me contended or give me any sense of "Yes i know electronics" ) .

author: alex_palvai
author_rank: Contributor
timestamp: May 12, 2020, 09:27:20 am
content: Wanted to share with you all , i ordered my first oscilloscope(Rigol DS1054Z 4-channel 50 Mhz) and function generator(Koolertron Upgraded DDS Signal Generator Counter) and excited to explore those two .back to question:I tried quiet sometime trying to decode the values "3/8 and 8/3" in Irms solving section of attached jpg. Can someone help me understand how he got those two values based on figure 3N.30( attached ) .I just learnt few concepts yesterday about phase angle, phasor drawing and tried to apply what i learnt to decode but still couldn't .On a high level i understand what instructor is trying to do ( calculate rms current ) based on heating in order to pick the right capacity of transformer for unregulated power supply.

author: Wimberleytech
author_rank: Super Contributor
timestamp: May 12, 2020, 03:50:40 pm
content: Quote from: alex_palvai on May 12, 2020, 09:27:20 amI believe the 3/8 is derived empirically from the timing diagram. Nothing magic about that number. The 8/3 number comes from 3/8 * 8/3 = 1.The duty cycle for the current delivered during recharge is factored by 3/8 so whatever current from the transformer is multiplied by 3/8. In this case a full 1A must be delivered to the load, so more current must be supplied...thus 8/3 * 3/8.Does that help?

author: alex_palvai
author_rank: Contributor
timestamp: May 12, 2020, 06:38:57 pm
content: Quote from: Wimberleytech on May 12, 2020, 03:50:40 pmThank's a ton for giving me the logic Wim. Your the best teacher. I read through that paragraph at least 20 times and it didn't strike me . It makes sense, and i could relate that explanation to other worked example(3W) in AoE Lab course. That fraction is actually a percentage of current against total cycle . I think where he is getting at with all this is estimate your best and take a transformer with a current rating 2 or 3 Amp's bigger than your recorded estimate.I wonder if this is AoE specific method for calculating transformer current rating or industry standard practice . I am asking this because, i couldn't find this formula or type of explanation in any other books i was referring or anywhere on internet.

author: Wimberleytech
author_rank: Super Contributor
timestamp: May 12, 2020, 09:29:59 pm
content: You might get some additional insight by downloading the National Semiconductor Power Supply Handbookhttps://ia800703.us.archive.org/0/items/NationalSemiconductorVoltageRegulatorHandbook1980/NationalSemiconductorVoltageRegulatorHandbook1980.pdf

author: Simon_RL
author_rank: Regular Contributor
timestamp: May 14, 2020, 01:44:57 am
content: Quote from: GerryR on February 26, 2020, 12:09:01 pmTotally agree @GerryR, as a newbie I had all these great vision of learning electronics and making students, man was I delusional. I very quickly learned that I needed the maths skills as well. At 53 I soon realised that I don’t remember much math from uni and have spent more time brushing up on algebra, trig and calculus over the last 4 months than actual electronics.

author: alex_palvai
author_rank: Contributor
timestamp: May 14, 2020, 11:00:33 am
content: Quote from: Simon_RL on May 14, 2020, 01:44:57 amThanks for sharing that story Simon_RL, Part of your story matches mine, I've taken GerryR's comments seriously and learnt math .In order to remove fear of Math , i did a course of "Algebra", "PreCalc", "CalC1" and "Calc2" in thattutorguy.com. It took me a month+ to recap , but i thankfully i am much comfortable now seeing differential equations , integration and other things than I was before .

author: alex_palvai
author_rank: Contributor
timestamp: May 14, 2020, 12:09:32 pm
content: I am testing signal rectification circuit Fig 1.70 Chap 1 , attached.Per what i learnt so far , Diode must block a reverse voltage & current . This is conflicting with what i am seeing in Vout ( has negative spikes) unfiltered by diode. I am wondering if this issue is specific to LtSpice.

author: rstofer
author_rank: Super Contributor
timestamp: May 14, 2020, 05:28:42 pm
content: I have no idea what is causing the negative spike but I will say that there is a reverse voltage characteristic to the diode.You have chosen very fast edges - dV/dt approaches infinity. Therefore, it seems reasonable to assume that you can create a reverse voltage far in excess of the diode's rating.See what happens if you slow down the edges. Or add some series resistance to the voltage source. Try a 1us edge.Expand the scale so you can see what is happening at both spikes.A lot of people hammer away on LTspice and there doesn't seem to be a crowd yelling about the diode models. There is also an LTspice grouphttps://groups.io/g/LTspice/topicsThere used to be very active on Groups.Yahoo.Com but that went away. Bummer...

author: rstofer
author_rank: Super Contributor
timestamp: May 14, 2020, 05:45:00 pm
content: See about 1/2 way down here:https://www.electronics-tutorials.ws/diode/diode_3.htmlThe idea that a diode blocks reverse current flow is wrong for low levels of current.

author: alex_palvai
author_rank: Contributor
timestamp: May 14, 2020, 08:48:57 pm
content: Quote from: rstofer on May 14, 2020, 05:28:42 pmHi Rstofer,Good to see your response again. I tried your suggestions 1. added source resistance 2. slowed down edges (set trise and tfall to 1n) ... And tried changing diodes to different model , also tried increasing capacitor size to increase currents . It remained same .I received my new oscilloscope but not my function generator yet, assuming that's required to generate signals and test this practically . I strongly feel this particular module in LTspice is buggy.I read the suggestions , Its mentioned a small leakage in (uA) is possible through the junction , In my case the leakage is few 12 mA (not small )From "https://www.electronics-tutorials.ws/diode/diode_3.html" (thanks for sharing) :"This condition represents a high resistance value to the PN junction and practically zero current flows through the junction diode with an increase in bias voltage. However, a very small leakage current does flow through the junction which can be measured in micro-amperes, ( μA ).

author: alex_palvai
author_rank: Contributor
timestamp: May 14, 2020, 10:58:44 pm
content: I've done the same simulation in Falstad and Vout is accurate there( attached screenshot ) . Confirms LTspice is culprit.

author: vk6zgo
author_rank: Super Contributor
timestamp: May 14, 2020, 11:54:21 pm
content: OP:-Rectifier circuits are normally explained using sine waves.The moment you start hitting them with square waves with a fairly fast risetime, the facts of life intervene.Diodes are not perfect, and a reverse biased diode looks like a capacitor .You can approximate this, in an (unachievable) circuit using perfect components, by imagining a small capacitor connected in parallel with the diode.The result will be that the diode does not "clip" the high frequency component of the fast rising square wave.In this respect, it looks as if LTspice is actually doing its job.That said, there are different types of diode optimised for different applications---with some, the capacitance in the reverse biased direction is minimal, & others it is quite high, because other parameters are regarded as more important in tne suggested applications.You say you have a DSO &.function generator----- look at real world circuits rather than simulations, think about how circuits should work, why they are a bit different from what you expected, & so on.Another thing which would be nice would be if you could find some old transformer type ac "wall warts" & using their low voltage secondaries, build a few rectifier circuirs using real Mains frequenciesThe explanations in the text you have shown are horribly "clunky".I would suggest reading the "ARRL Handbook" (especially the older editions) to get a better idea of what your existing text is trying to communicate.

author: alex_palvai
author_rank: Contributor
timestamp: May 15, 2020, 02:07:30 am
content: Quote from: vk6zgo on May 14, 2020, 11:54:21 pmvk6zgo ,Nice meeting you. Thanks for good explanation with lot of tips for learning , its all very motivating . I ordered but haven't received my function generator yet :-) (hopefully will get it next week ) . I worked a simulation of power supply circuit before trying this one and it didn't give me much trouble or WimberleyTech saved me a bit of trouble I had understanding some theory. Yes , I realized very soon that simulations are not giving me real world practical sense therefore going to rely more on hands on going forward.And thank you for feedback on explanation in text and suggestions on "ARRL Handbook" . Will add it into my library.

author: Wimberleytech
author_rank: Super Contributor
timestamp: May 15, 2020, 04:20:52 pm
content: Quote from: alex_palvai on May 14, 2020, 10:58:44 pmI would not trust Falstad for anything but a simplistic analysis.Consider the following.When a diode is forward biased, the depletion region shrinks to a very small (relative value). When you instantaneously reverse bias it, those charges take time to move away from the pn junction. In the mean time, the capacitance across the diode is much higher than when steady state reverse bias. I think that the reverse recovery time specification of a diode captures this feature.So, the glitch has nothing to do with reverse current and all to do with a very non-linear capacitance.

author: Wimberleytech
author_rank: Super Contributor
timestamp: May 15, 2020, 09:16:32 pm
content: Quote from: Wimberleytech on May 15, 2020, 04:20:52 pmThere you go! 1n4148 in this test.

author: alex_palvai
author_rank: Contributor
timestamp: May 16, 2020, 11:09:00 pm
content: Quote from: Wimberleytech on May 15, 2020, 09:16:32 pmThanks to all of you for correcting me. I almost went into wrong assumptions and thought those were right.Wim. Thanks for good explanation of what's happening and taking time to run trace for me. Really greatful.With all your guidance i moved forward and perfected my circuit on LtSpice by correcting ( Trise & Tfall) - Rstofer pointed it right in his first response but it didn't strike me that moment why its behaving that way.I have a better understanding of diodes and signal rectifier circuits now than yesterday. I documented all my test cases and notes . One thing didn't strike me still and am trying to figure out since an hour why my circuit is not displaying forward voltage drop of .7V . Appreciate if someone can throw me light . I am trying to test the trick he gave wherein diode D1 compensates D2's forward voltage drop in this circuit . But without seeing the forward drop its a dead end.Thanks in Advance for help.

author: alex_palvai
author_rank: Contributor
timestamp: May 30, 2020, 03:03:44 am
content: Hello All,Hope you all are doing good. I moved forward from the last question (the answer was when applied DC voltage you see clearly .6v drop and when AC voltage is applied its hard to see as but there is .545 volt drop ) . Was busy learning controls on my newly purchased oscilloscope and function generator past week.I landed on a new topic now " Frequency analysis of reactive circuits " and RC filters. From what i read , low pass filters allow low frequencies and block high frequencies , opposite for high pass filters. My understanding is (0.001 - 50hz ) are low frequencies and high frequencies are (50 - above ) . And i tested same , i constructed a low pass filter with 100 ohm resistor and 100uF capacitor and noticed it passed frequencies from 1 - 50hz and blocked anything above 60hz.now mathematically i want to prove that point :Reactance (Xc = 1/ ωC)ω= 2πfand my resistance R=100 ohmat what point reactance Xc will be equal to equal to value of resistor ?I know from my lab at f=60hz approx. is when i almost see 99% signal is blocked .Substituting same ,=> 1/(2x(3.14)x60x100x10^-6) => 26.53 ( I was expecting reactance this would show a value close to 100 )Is my assumption correct that i should that value of reactance (Xc ) should be 100(=resistance) when capacitor blocks ?

author: gcewing
author_rank: Regular Contributor
timestamp: May 30, 2020, 11:50:22 am
content: No, the frequency at which 99% of the input is blocked is considerably higher than the one where R = Xc.Think of it as a voltage divider made from two impedances Z1 and Z2.Then:Vout/Vin = Z2 / ( Z1 + Z2)Here Z1 = 1/jwC and Z2 = R.So if 1/wC = R, we haveVout/Vin = R / (R/j + r) = 1/(1 - j)Draw a diagram of 1-j and you will see that it is a triangle with a hypotenuse of length sqrt(2), soVout / Vin = 1/sqrt(2)which is approximately 0.7. So the filter only blocks about 30% of the signal at this frequency.What frequency is it? For the component values you gave,1/(2 pi f * 100uF) = 100 ohmswhich givesf = 100 / 2 piwhich is about 16Hz.Incidentally, the frequency at which Xc = R is known as the "-3dB point", because 20 * log10(1/sqrt(2)) is fairly close to 3.It's also sometimes called the "cutoff frequency", but keep in mind that an RC filter doesn't suddenly cut off a this frequency, there is a fairly broad curve. It's just a mathematically convenient point to choose.

author: rstofer
author_rank: Super Contributor
timestamp: May 30, 2020, 02:28:46 pm
content: See Reply 52 in this thread for an example of a low pass filter and using an Analog Discovery 2 to draw the Bode' response. There are 4 examples in that thread (starting with reply 52) that use the various features of the AD2 and just a simple resistor and capacitor.The link takes you to Reply 50, scroll down a couple...https://www.eevblog.com/forum/testgear/starter-scope/50/

author: alex_palvai
author_rank: Contributor
timestamp: June 06, 2020, 07:08:23 pm
content: having hard time understanding dB plotting .the left side scale shows 0.001 to 1 on y-axis . Wondering how 6dB / octave or 20dB / decade were converted from dB to range on y-axis from (0.001 to 1 ) .

author: Wimberleytech
author_rank: Super Contributor
timestamp: June 06, 2020, 08:29:38 pm
content: Quote from: alex_palvai on June 06, 2020, 07:08:23 pmDoes this help?What I am really showing is dBV which is dB relative to 1 volt.

author: alex_palvai
author_rank: Contributor
timestamp: June 06, 2020, 09:09:47 pm
content: Quote from: Wimberleytech on June 06, 2020, 08:29:38 pmHello Wim. Thank You . It helped 50% in understanding 20 dB per decade . Can you help me understanding 6dB / Octave on this same plot please.

author: Wimberleytech
author_rank: Super Contributor
timestamp: June 06, 2020, 09:21:07 pm
content: Quote from: alex_palvai on June 06, 2020, 09:09:47 pmAn equation may make more sense.

author: alex_palvai
author_rank: Contributor
timestamp: June 06, 2020, 10:20:18 pm
content: Quote from: Wimberleytech on June 06, 2020, 09:21:07 pmUnderstood now. If what i am thinking is right , you nailed my real question without directly answering(Don't know how to thank you for that) . I was under impression decade to octave will change the x-axis (from currently its a factor of 10 to factor of 2) . So 6dB/ Octave and 20dB/ Decade is same reference point . We actually make changes to y-Axis not x-axis is what i understand .I am typing this just for my future reference , so what i get is , irrespective of whether its octave or decade scales , slope will not change. If I want to read it in octave scale , i put 6dB per vertical block instead of 20dB ( attached what's in my mind ) .

author: robsims
author_rank: Frequent Contributor
timestamp: June 07, 2020, 12:36:53 am
content: If you want to study electronics, you need at least high school level math. You don't need to derive formulas with calculus etc., but you have to know how they work of course. I am a mechanical engineer and i am self studying electronics. A book that is not heavy on math and i think is best for starters is Electronic Principals from Malvino and Bates. I am currently at the chapter where they explain JFET's

author: Wimberleytech
author_rank: Super Contributor
timestamp: June 07, 2020, 01:34:54 am
content: Quote from: alex_palvai on June 06, 2020, 10:20:18 pmBINGO!! You got it!

author: Wimberleytech
author_rank: Super Contributor
timestamp: June 07, 2020, 01:37:30 am
content: Quote from: robsims on June 07, 2020, 12:36:53 amIt is useful to know how a JFET works, but I would hazard to guess that JFETs appear in 1% or less of modern circuits. Understand MOSFETs and BJTs. Of the three/four terminal devices, these will suit you well.

author: alex_palvai
author_rank: Contributor
timestamp: June 14, 2020, 05:21:23 pm
content: Driving issues : its given that when choosing R and C values for blocking capacitor application I need to pick R value to be reasonable load i.e. not so small that its hard to drive .Circuit loading:My understanding of circuit loading is that the circuit being driven will add more load on driving circuit. So what will happen to circuits if more load is added , assuming the straight forward answer is smaller components with least resistance will blow up correct ?question:Why does picking small R value makes it hard to drive ? These statements align with what I read about source resistance and circuit loading that I need to pick larger load resistance (10x times) the source resistance to avoid circuit loading . But its not explained even there why would it cause circuit to load and how ?In water analogy if you allow more water on top (less resistance)and restrain at bottom(more loax resistance)there is problem , pipes may burst due to pressure.Thinking loudSmaller resistance passes more current and larger passes less current . Suggestion is small resistance on top (allow more current to pass) then 10x bigger resistance at bottom ( limit current ) through heat .What happens if I don't?More current is passed on the top and even more at bottom due to even smaller resistance. So what? How does that load the circuit ?

author: gfmucci
author_rank: Regular Contributor
timestamp: June 14, 2020, 08:52:18 pm
content: Interesting question about "how much math, calculus, differentials, etc." is essential to advance in the hobby or profession of electronics.It reminds me of the question I had in seminary: Should we be required to learn to read and speak Greek and Hebrew, or is it ok to just learn how to use the tools that express the meanings? I much preferred the second path because I suck at languages, and two semesters of that torture was more than I could stomach. I went back to my day job. The tools can be challenging enough, never mind trying to learn how the tools were made.If you want to teach it, that may be another matter.

author: Wimberleytech
author_rank: Super Contributor
timestamp: June 14, 2020, 11:12:09 pm
content: Quote from: gfmucci on June 14, 2020, 08:52:18 pm...for it is God at work in you, both to will and to work for His good pleasure...

author: alex_palvai
author_rank: Contributor
timestamp: June 16, 2020, 10:41:20 pm
content: I am assuming something , but want to make sure that's right.For the problem in attached reference " Imaging putting 1uF Cap across 115 Volt (rms) 60 Hz powerline ? what's current flow was the question.I understand how the current was derived . But i don't understand how based on the result one can say which one is leading what "Current leading voltage or voltage leading current and by what angle "My assumption is Given "Voltage" was taken as cosine i.e. 90 degrees before time t=0 begins ( please correct me if i am wrong ) , and the resultant current I came out to be sine (wt) i.e. crosses point t=0 . Therefore voltage is leading current . Is that correct ? (Given in text is opposite i.e. " Current leading voltage". Or is my assumption that cost will always start at 90 degrees before t=0 is incorrect ?I tried to draw phasor diagram of this current and voltage, I was a bit skeptical to about the current on positive x-axis as the result i received was negative -0.6sine(wt). I made two representations , Appreciate if some one can guide to say which one is right one..C=1Fv(t) = Acos(wt)A= 115/root(2) rms voltsv(t) = 163cos(wt)Ic= C*dv/dt=> 10^-6 * d ( 163cos(wt) )=> 10^-6 *w*A*-sin(wt)=> 10^-6 * 6.2*60*163 ==>-0.06sin(wt)

author: vk6zgo
author_rank: Super Contributor
timestamp: June 29, 2020, 10:18:39 am
content: Quote from: alex_palvai on June 16, 2020, 10:41:20 pmNo, you can't go before t=0.In the Universe where this question lives, there is no time before t=0.

author: MuseChaser
author_rank: Contributor
timestamp: February 22, 2022, 08:46:17 pm
content: Greetings, all. I just registered because...well... I haven't felt this stupid in a long time, maybe ever, and need some help I guess. I've been building kits, doing some limited speaker and crossover design and building, board repopulations, etc., for decades, and can kind of read a schematic to the point where I know the components I'm looking at and how they're connected, but I know precious little, if anything, about actual electronic engineering or design theory. Sooo.. after getting stuck on trying to solve an issue on a Dynaco SCA80Q on my bench right now, I figured it's never too late to learn. Bought a copy of AoE 2nd ed from Goodwill in great shape, and downloaded the accompanying student handbook and started the journey. I have a headache. At one time, I got as far as Calc BC, but that was back in the early 80s and I haven't done anything since other than routine "How many sheets of plywood do I need to build a 4x16 table?" .. Ok, maybe a bit more than that. I'm embarrassed by how much I've forgotten.Anyway... Here's today's conundrums, still in the very basic parts of the books, dealing with Ohm's law (that, at least, I'm comfortable with), Kirchhoff's voltage and current laws (pretty OK there), Thevenin equivalences (so far, so good)... but I'm hitting a wall with some examples in the student handbook.For instance...I thought I had this.. the original diagram on the left becomes a Thevenin reduction to a 10v source into a 50K resistor. The meter's 10,000ohm/1V spec returns a current of 0.1mV. Using R=V/I to solve for internal resistance when reading 10v, we get an R value of 100K. Since that is 2/3 of the now total resistance (150K ...50k resulting from the Thevenin depiction of the divider circuit plus the 100K internal resistance), then the voltage at X when measured with the VOM labeled 10,000 ohms/V would be 6.6666 or, 6.7 as depicted. Check. I get it.HOWEVER.. when connected to the scope w/ 1 MOhm input resistance, I a can't figure how they get the 9.5V result. Since the input resistance is given w/out respect to voltage, is it constant on a 'scope, or do you still use I=V/R to calculate I=1V/1Mohm, resulting in 0.001mA? Then... as before... R=V/I = 10/0.000001 = 10M resistance, which is 0.995 of the total resistance, 10,050,000. Wouldn't that yield a voltage reading of 9.95, and not 9.5 as suggested in the answer below?Another question that stumped me -Reducing the figure on the left to a Therevin equivalent easily yields the 11K result for the series resistors on top and .91 for the parallel on the bottom... all good there. However, isn't the resulting voltage at the dividing point equivalent to the results obtained from R1*R2/(R1+R2) for the resulting resistances? 11k*0.91k/11.91k yields 0.84V, not 0.74 as indicated. If I use the ratio method, lower resistance against total resistnace, or 0.91/11.91, I DO get .076, which then multiplied by the 10V input DOES yield the expected 0.76, but I don't understand why the other formula doesn't work in this case.I know.. for all of YOU folks, this is like trying to explain to a two-year-old why 2+2 does not equal five, but it's been a VERY long time. Hopefully, the rust will dissipate quickly. Thanks in advance!

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 22, 2022, 09:35:20 pm
content: In the case of the scope, it is just a fixed 1Mohm load. So your voltage divider is 1M/(50k+1M) = 0.9524For your other problem, do not rely on a formula. Go back to first principles of ohms law. Add the total resistance in series. Divide that into the 10 volts. That gives you current. Then multiply that current by the lower resistance to get the voltage across it. If you do this exercise symbolically, you will arrive at the correct symbolic formula.

author: rstofer
author_rank: Super Contributor
timestamp: February 22, 2022, 10:13:19 pm
content: When a V-O-M says 10k Ohms/Volt, it means 'volts full scale'. If you are on a 1V scale, the input resistance is 10k * 1v or 10k Ohms. If you are on a 20V scale, the input resistance is 10k * 20 or 200K Ohms. So, without being told what scale the meter is set to, it isn't really possible to get the right answer.A better explanation is here:https://www.allaboutcircuits.com/textbook/direct-current/chpt-8/voltmeter-impact-measured-circuit/

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 23, 2022, 12:41:39 am
content: Quote from: rstofer on February 22, 2022, 10:13:19 pmBut they screwed up R4 in the schematic.