Newbie and really dumb question! - Page 1
Forum section: electronics / beginners

author: deano
author_rank: Newbie
timestamp: August 08, 2014, 03:22:22 pm
content: Hello, I am new to this site but hope to spend some time here! I am just getting into electronics having built a guitar effects pedal from a kit and getting a bit hooked. I simply followed the instructions and it seems to be working but why it works is beyond me, but I want to learn, so I've been digging around the net a bit.The stumbling block for me at the moment is pretty basic. If you look at the schematic of the next circuit I want to build (below), then I have no problems with any of it except at the junctions. So when the signal flows through R3 I'm struggling to figure out where it goes next. I assume it is some combination of R4 and C3 but I don't know why! How does the signal "know" to turn left or go straight on through R4 and then to ground, or even whether or not to turn right and go into R5?Again, once the signal has gone through C3 where does it go next and why? Why would it go left into C2 instead of going on to R6 and C4?I appreciate there is actually no decision made by the current and it is following some laws of electronics but what are they? What are the laws that state that a current will flow this way at a junction instead of that way?I told you it was a dumb question but I a have read a few books and even followed some schematics with breakdowns but none seems to actually state that "at this junction the current will take this path because...." Or if they do then I'm probably a bit to thick to get it.I don't mind following maths as that is what my degree is in but the physics side of it escapes me.

author: Creep
author_rank: Regular Contributor
timestamp: August 08, 2014, 06:37:05 pm
content: I suggest you have a read on three things: (Voltage and Current) divders, Kirchoff's laws and Ohms law. The current doesn't decide where to go: it goes in every direction it can. How it splits (how much current goes in one way and how much in the other) depends on the resistance of the actual path to ground. The more resistance, the less current going down that path.I guess you could imagine it this way: imagine a swarm of people walking in a tunnel when suddenly the path is blocked by a wall. The wall has two oppenings: a large one and one not so large. In order to get to the other side, the swarm will split up with more people deciding to stick with the large opening. You could say that the size of the opening coresponds to electrical resistance: the larger the opening (smaller the resistance), the more people (current) can go through during a time interval.Now i do realise that this analogy is pretty far from what actaully happens and I might get a lot of flack for this, but it should give the OP a general idea as to how the current splits when faced with multiple pathways.

author: jeremy
author_rank: Super Contributor
timestamp: August 08, 2014, 11:38:26 pm
content: Welcome!On top of Creep's reply, I think it would be useful to learn about "impedance". It is like resistance, but it also takes into account energy storage in some components.The interesting thing about audio circuits from a learning perspective is that you usually don't want DC signals (defined as 0Hz), only sine waves somewhere between 20Hz and 20kHz because that's what we can hear. This is quite different from your typical LED-computer-case-mod kind of electronics.The truth is that for a single fixed frequency, a capacitor can be modelled as a resistor. So that means that a capacitor is kind of like a frequency dependent resistor! Generally speaking, the rule of thumb is that "low frequencies" are blocked by capacitors (open circuit) and "high frequencies" are passed (closed circuit). So for example, C1 and C4 are DC block capacitors.Because of this frequency dependent complexity, I would probably try to do something called "phasor analysis". If you already have a maths degree, it should be pretty straightforward for you to do this; it's just ohms law with basic complex numbers, where the complex component of the impedance represents the frequency dependent component of the resistance, and the real part is the true ohmic resistance. There is lots of information on google about this. You can also try building the circuit in LTSpice and probing the various bits to see what is happening. Fortunately you can't blow up a transistor in software, so try whatever you like.Hope this helps.

author: IanB
author_rank: Super Contributor
timestamp: August 09, 2014, 12:55:21 am
content: Quote from: deano on August 08, 2014, 03:22:22 pmEvery circuit element has a mathematical model that describes its physical behaviour.Here is a resistor:Rv1(t) ----/\/\/\/----- v2(t)i(t)The model variables are R (the resistance), v1and v2(the voltages at each terminal of the resistor), and i (the current through the resistor).The resistor model says:v1(t) - v2(t) = i(t) RThis (ideal) model is true for every (assumed ideal) resistor in the circuit.A capacitor:Cv1(t) -----| |------ v2(t)i(t)The (ideal) capacitor model says:i(t) = C d/dt ( v1(t) - v2(t) )This is a linear first order differential equation.Any circuit consisting of resistors and capacitors can be modelled using these equations.To complete the model two other equations are needed.At any junction, the voltages on each branch are equal:v1(t) = v2(t) = v3(t) = ...At any junction, the sum of all currents entering and leaving is zero (with appropriate sign convention):sum( i1(t), i2(t), i3(t), ... ) = 0With these equations, any circuit consisting of resistors and capacitors can be represented as a system of differential and algebraic equations that can be solved simultaneously with the addition of suitable boundary conditions. One voltage must be assigned a value of zero as a reference voltage (often the negative supply rail in the circuit, but it doesn't have to be). Any input voltages must be provided with a forcing function (the input signal the circuit will respond to).Transistors can also be given a model. It will be slightly more complicated, but a similar principle applies.The answer then to the question, "how does the signal know where to go?" is answered by the solution to this system of equations.In reality, circuit designers have come up with various simpler ways to estimate what circuits will do when given AC signals, but ultimately it all comes down to a system of equations involving voltage, current and time.

author: Kremmen
author_rank: Super Contributor
timestamp: August 09, 2014, 05:45:23 am
content: Quote from: deano on August 08, 2014, 03:22:22 pmThen you already speak the language - good. Now to get you on the map. Luckily, Coursera to the rescue: there is an intro course in linear circuits beginning in September, just 3 weeks away. By all means do attend as it is free and it _will_ create understanding.https://www.coursera.org/course/circuits. Your background in math should make that part of the course a breeze.

author: deano
author_rank: Newbie
timestamp: August 09, 2014, 10:45:53 am
content: Thanks for the replies everyone. I was hoping to avoid the maths to be honest as I did it alltwentythirty -- years ago! Oh well, if it comes down to it I'll trawl through it.I just thought there would be some simple rules of thumb that would enable me to figure it out at a high level rather than getting down and dirty into differential equations. I am only building guitar effects for me and my son after all.Cheers anyway.

author: David Hess
author_rank: Super Contributor
timestamp: August 09, 2014, 11:46:09 am
content: Quote from: deano on August 08, 2014, 03:22:22 pmC2 and C3 bootstrap the base of the transistor from the emitter raising the input impedance. Google for "bootstrap emitter follower".The gain of the emitter follower is very close to 1 so the emitter side of those capacitors closely follow the base side. R5 and R4 set the DC level at the base but C3 pulls this voltage up and down to follow the input so the voltage across R3 is always very low so the current through R3 is low. To the input, this makes R3 look like a much higher value. C2 does the same thing at higher frequencies and for the base of the transistor.

author: IanB
author_rank: Super Contributor
timestamp: August 09, 2014, 03:30:07 pm
content: Quote from: deano on August 09, 2014, 10:45:53 amThere are some simple rules of thumb that work at a high level, but these rules of thumb are derived from how those differential equations behave.The main rule of thumb is based on impedance and frequency response. If you provide a circuit with a forcing function of a sine wave at various frequencies you find that each component has an impedance to current flow based on the frequency of the signal. For a resistor the impedance is the same at all frequencies, while for a capacitor the impedance is infinite at DC and tends towards zero as the frequency increases.This means that a network of resistors and capacitors can be used to divide up and separate a signal, sending high frequencies one way and low frequencies another.

author: Seekonk
author_rank: Super Contributor
timestamp: August 09, 2014, 04:06:33 pm
content: I'm going to go for the really simple explanation. You have not chosen a very good schematic to diagnose. The signal level that comes in one side is pretty much the same as goes out. The output does have higher current capacity. If you were to drive a long cable that would make sense. As said before, the voltage on each side of R3 will be about the same so the input is a much higher impedance than the resistor represents. A lot of components and it does about nuttin.

author: deano
author_rank: Newbie
timestamp: August 09, 2014, 05:09:03 pm
content: Quote from: Seekonk on August 09, 2014, 04:06:33 pmYou got it about right really with the cable comment. It is actually the main schematic of the Pete Cornish LD-1 buffer, which is supposed to be about the best buffer there is for guitars. Of course there are lots that claim that.So it goes before a long cable or a series of true bypass effects, and therefore, yes, it wont do "much" but the bit it does do is supposed to be very good by all accounts.I'm getting there with some of the comments, so thank you very much for the help. I'm on holiday in Spain at the moment so can't do much except read up, but when I get back I'll breadboard it and stick the multimeter in various points to see what is happening.

author: Creep
author_rank: Regular Contributor
timestamp: August 09, 2014, 05:43:04 pm
content: Maybe a bit off topic, but i wouldn't suggest spending your time digging and researching during your travels in holidays. You can always rush into in when you return home. You won't however be able to enjoy your visit at Spain (you know, sightseeing and stuff) when you return home.