I have trouble with this schematic and all like it - Page 1
Forum section: electronics / beginners

author: Hextejas
author_rank: Frequent Contributor
timestamp: May 03, 2018, 07:18:29 pm
content: when I breadboard it.In circuit A:The connector coming from the base of the transistor connects to:1) a capacitor2) a resistor3) a capacitor4) a 3-series resistor5) a terminatorWhereas in circuit B the connection to the resistor has been moved to be just before(after) the 3-resistor series.Will these 2 circuits perform the same ?Could the 2, 1u capacitors be connected anywhere along that connection between Q1 and P1My training was as a computer programmer and the processor would step sequentially through the code, one instruction after the other.I am afraid that that training colors my understanding of electronic schematics. So that the electrons will be stepping through this schematic and will hit each component at a different time.This is very confusing to me and a buddy of mine said to stop thinking like that. The electrons will hit every-place at the same time.Help !!

author: barry14
author_rank: Regular Contributor
timestamp: May 03, 2018, 08:52:55 pm
content: The two circuits are not identical. In the first circuit, the resistor is connected between the two capacitors. In the second circuit, the resistor has been moved to the left so that the two capacitors are now directly connected in parallel. Since I don't know what signals are involved, I can't tell you what difference that will make.

author: MarkF
author_rank: Super Contributor
timestamp: May 03, 2018, 09:08:36 pm
content: Quote from: Hextejas on May 03, 2018, 07:18:29 pmNo.Quote from: Hextejas on May 03, 2018, 07:18:29 pmNo. It would change the circuit response.Divide your two circuits into sections as shown below. In both circuits I show the internal source resistance of a battery or power source (not an actual resistor).Circuit 1: You have a parallel resistor capacitor followed by a series resistor capacitor.Circuit 2: You have a parallel resistor followed by a series resistor capacitor.Edit: Corrected part descriptors in circuit diagrams.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 03, 2018, 09:53:03 pm
content: Yes, circuits are not procedural. They are also reciprocal, for the most part, which may seem very strange coming from a procedural background.That is, suppose you had a function, which takes a parameter, which becomes modified by the function. Well, that'd be a function on a pass-by-reference parameter, which is pretty normal. But, consider if the function is executed not once, but continuously until its input parameter stops changing (reaches equilibrium). And the function's return value also is similarly affected by what it's connected to. That'd be a strange (and presumably, inefficient) language!But at least it's not inefficient, because it just happens naturally, it's physics. The challenge is to get your mind to see that simplicity.Perhaps going through the math connection will be more helpful? Study solutions of simultaneous equations, if you aren't up on it; see how they are evaluated computationally. Consider only simple linear functions: numbers come in, addition and multiplication is performed, then a number goes out; and how they evaluate when connected in a loop with other simple functions, and how you would optimize that process (hint: it comes down to matrix inversion).Then connect that system to physical circuits. Every element of the matrix represents an element of the circuit, a resistor or whatever; the matrix is always symmetrical, because of reciprocity (every element draws current from one node and delivers the exact same current to another node); Kirchoff's laws. This works identically at AC (with capacitors and inductors), but you use complex rather than real numbers in the matrix and solution.Or maybe you're not big on numerical analysis. Maybe you'd be more comfortable from a graph-theoretic angle? A circuit is a graph; mostly it's an undirected graph (current can flow either way between nodes; of course, once you've solved it, you know exactly which way it's going!). Though if you add nonlinear components (diodes, transistors..), you get clear directions to start with. The nodes are labeled with voltages, and the edges with currents. This again gives rise to a matrix form, which is easily solved with traditional linear algebra.If you've done much with graphs and matrices, the duality of matrices and graphs should be natural. Or with abstract data structures, where you might use an array, or array of arrays (a matrix), to store a graph. Or an array of allocated objects is used to store a multiply-linked list -- a sparse matrix, which is actually very practical in electronics simulators, because circuit matrices are usually on the sparse side.Graphs are nice to match patterns visually. If you wish to find the solution by hand, you start by picking out recognizable features, and working them separately -- anywhere you see a node with only two edges, those edges are in series, and the node can be optimized out. Or anywhere two nodes share multiple edges, those edges can be combined into a parallel equivalent. There are also series-parallel, wye-delta and other transformations, involving different characteristic arrangements of edges and nodes.So, regarding the circuits in question -- you are correct that they are not equivalent. In terminology that, uh, won't really help you much right now -- the first one has a two-pole filter (assuming the pin connects to a resistive signal source -- this is another theoretical staple, the Thevenin equivalent), while the second one simply has two capacitors in parallel, so their values add, and the equivalent circuit has only one pole. In both cases, the three resistors in series (R2-R4, R6-R8) can be reduced to a single 3k resistor. (See how MarkF has performed these simplifications!)Tim

author: rstofer
author_rank: Super Contributor
timestamp: May 05, 2018, 04:00:56 pm
content: What you could do is simulate the circuit in LTspice. Assume the driving impedance of the signal source is 1 KOhm and ignore the transistor. Just look at the output where it connects to the base.The 3 1K resistors serve as some kind of attenuator so I changed it to 100k to get it out of the way on the Bode' plot. All they do is make the magnitude start out a lot less than 0dB.Note that I buried the 1K driving impedance inside the voltage source. I didn't use an external resistor for this.LTspice is free

author: Hextejas
author_rank: Frequent Contributor
timestamp: May 06, 2018, 02:46:53 pm
content: T3s,,,,phew, just phew. and thanks

author: Hextejas
author_rank: Frequent Contributor
timestamp: May 06, 2018, 02:51:47 pm
content: Would one of you kind gentlemen or gentleladies do the breadboard for Circuit A and some how post a picture of it so I can see how it's supposed to be done. Truthfully, when I see a schematic like circuit A, my mind (what little of it there is left), fogs up and I struggle. It took me about an hour to get it right. It should not be that difficult but I have a mental block when I see some such.

author: MarkF
author_rank: Super Contributor
timestamp: May 06, 2018, 05:13:31 pm
content: Here are a couple videos you may find helpful. Additionally, google series and parallel RC circuits.

author: rstofer
author_rank: Super Contributor
timestamp: May 06, 2018, 05:54:51 pm
content: RC circuits are easier to think about in the time domain: what happens after I close a switch as a function of time. The capacitor charges with an exponential curve that is entirely predictable.Unfortunately, this is mostly meaningless in the frequency domain: what does the circuit do to my sinusoidal input as a function of frequency. Basically, it attenuates it more and more as the frequency gets higher. It is called a low pass filter, it passes low frequencies and attenuates high frequencies. The behavior is different between low and high frequencies where 'low' and 'high' are relative.I gave you the LTspice model above but I also said something like "It is meaningless if you don't have the driving impedance". Without the rest of the circuit, there is really no way of knowing what the low pass filter will do. But we do know it is a low pass filter.In a vague kind of way, we know that it attenuates the input but we don't know for certain where the critical points are located. We also know from the fact that the capacitor is large that the point at which it starts attenuating is fairly low. Somewhere down around 150 Hz. We get that from 1 / (2 * PI * R * C). But that's for the first circuit and only the second stage of the filter. The critical frequency of the first stage is totally dependent on the driving impedance.

author: technogeeky
author_rank: Frequent Contributor
timestamp: May 06, 2018, 07:07:51 pm
content: I have made a pair of small, simple, web-based simulations on the FalstadAnalog Filterapplet. In both cases, I have neglected the transistor both because I don't think it matters and because this version of the simulator can't have a transistor in it.In order to "start" the simulation, you need to select a frequency (a particular vertical line) on the bottom plot (which is frequency versus amplitude). You can then inspect (via mouse over) properties of the components, like their voltages and power dissipation. You can also get a sense by color of which components have high and low voltages at a given moment. You can also switch this (via options) to Power.This linkis a simulation of the resistor before the first capacitor.This linkis the simulation of the resistor after the first capacitor.I observe that the the so-called "3dB down" point of the resistor-before configuration is at about 80Hz. The "3dB down" point of the resistor-after configuration is at about 160Hz.

author: Hextejas
author_rank: Frequent Contributor
timestamp: May 07, 2018, 05:18:42 pm
content: Thank you all for these explanations and I will study them at length.BUT !You missed my question. I have trouble wiring the breadboard. Reading a schematic and wiring the breadboard when parallel circuits are involved cause me troubles.I wanted someone to do a breadboard of circuit A, then take a picture of it so I could see a good technique.Series connections are simple, it's the parallel ones that give me a headache. Maybe google can help me but so far no joy.Thanks again

author: BergRD
author_rank: Contributor
timestamp: May 07, 2018, 07:11:45 pm
content: Have you ever used the software Fritzing? If not it may be something you could experiment with and can be foundhttp://fritzing.orgIt's a virtual breadboard (among other things) that lets you lay components out on the breadboard and connect them with virtual wires.Just going on what you've stated about it might be that you're not sure how a breadboard works? To save time and worries of everyone I won't go there but I/we can if you want. It confuses many early on and myself included but once you know how the board layout corresponds to how you connect things; it becomes "easier" to follow diagrams to place and wire them.For example, in the first below image from Fritzing this is 2 resistors in series and in the second image they are considered parallel.

author: Hextejas
author_rank: Frequent Contributor
timestamp: May 07, 2018, 08:27:39 pm
content: thank you Berg, that looks useful and I will give it a try

author: MarkF
author_rank: Super Contributor
timestamp: May 08, 2018, 01:11:57 am
content: Just to expand on BergRD's breadboards -Each row of 5 holes are connected together as shown by the vertical green bars (I only did a few of them)The power bus holes are all connected together as shown by the horizontal green bars. In the 'new' breadboards, each row of buses are broken in the middle (i.e. 4 groups on the top and 4 groups on the bottom if the entire board was shown).