User talk:Circuit dreamer

Your request on my talkpage

Please check your e-mail, I have sent you a message. SpinningSpark 08:59, 1 December 2013 (UTC)

Gratitude for Circuit Idea

Hi, Circuit Dreamer. I'm glad to see that you're active again. I'm just writing to say that I've found the Circuit Idea book very enjoyable and enlightening. I've been teaching myself basic analog electronics lately (as a sort of, er, "carrot" from grad school). I never found the subject intuitive. Circuits of any complexity seemed to get very confusing very fast. Circuit Idea has given me some idea of how one thinks about circuit—seeing the simple ideas that get combined in many ways. I'm still a beginner, but now I see the main circuit ideas as having an elegance and fundamentalness somewhat like the basic theorems of Euclidean geometry. To take one illustration, your little section about "flow causes pressure" in Passive Current-to-Voltage Converter—that an impediment converts flow into pressure—enabled me to finally "get" what the collector resistor is doing in so many BJT circuits. I knew formulas and the like, but I hadn't gotten the really fundamental idea. The idea is so simple, just letting a current flow across a resistor to get a voltage by tapping before and after the resistor, it almost seems like nothing more than Ohm's law. But now I can perceive this little building block in many circuits, enabling me to see one part of a circuit as "input" and another as "output", whereas before, I just perceived components and connections. Resolving a CR differentiator into V-to-I differentiator and an I-to-V converter was wonderfully enlightening, even though it's just calling a capacitor and a resistor by different names! Having overcome these elementary hurdles with the aid of Circuit Idea, I got inspired to buy The Art of Electronics by Horowitz and Hill, and have been working steadily through it with surprising ease. Thanks for your fine work, and I hope to see more pages on Circuit Idea. —Ben Kovitz (talk) 19:10, 1 December 2013 (UTC)

Ben, what can I say after such nice words written brilliantly? For me this is the best reward for my efforts to reveal the truth about circuits, which is even more valuable in that it happens so rarely. Obviously, we have a similar heuristic way of thinking, which nowadays is not so popular as well as in the early 70's. Then I (during my 2-year military service in the army) first met with these powerful ideas about how to invent (lately - understand and present) from the famous books of Altshuller and especially Edward de Bono...

Have you visited my site of circuit-fantasia.com? In 2002, I created there a few animated stories about these circuits: V-to-I converter, I-to-V converter, Passive summer, Active summer, and of course, the most interactive Circuit builder. I suppose they will be interesting for you...

But I saw an interesting essay in your user page and will read it with interest. It is so wonderful that there are people like you in this world! Thanks again for the warm response!--Circuit dreamer (talk, contribs, email) 19:41, 1 December 2013 (UTC)

A discussion about NIC and TIA

Warning

I think you need reminding that your community ban on editing electronic articles extends to talk pages as well. I should really block you again for this, but I will extend good faith and assume you will stop now you have been warned. We have talked extensively offline about how you could go about getting the ban lifted, but you seem determined not to do that. Well so be it, but that then means you must not edit electronics - at all. SpinningSpark 17:07, 15 December 2013 (UTC)

Yes, I know that, Spinningspark; that is why I implicitly apologized for the breach in the beginning. I will not continue this discussion; I just shared some useful thoughts about NICs that can help improving the page. Circuit dreamer (talk, contribs, email) 17:50, 15 December 2013 (UTC)

copied from talk:Negative Impedance Converter

"Zen-in, you have really stressed me and I still can not realize that you have analyzed this simple circuit with a full screen of formulas while I have done it on one line! The difference between you and me is that you have "blindly" analyzed the circuit as a "black box" without seeing well-known sub-circuit building (functional) blocks. You have done it like a computer (a simulating program) while I have done it like a human being seeing familiar patterns inside the unknown circuit: voltage-to-current converter, current-to-voltage converter, voltage follower, non-inverting amplifier, non-inverting current source, current inverter... and even a balanced bridge. Yes, this circuit is exactly a bridge that is balanced by adjusting the supply voltage... and the result is VR1 = VS, VR3 = VR2. IMO the most useful way of presentation is as two cascaded devices - voltage-to-current converter and current inverter.

About your initial remarks, "there is an element that is both a current source and a voltage source but there is nothing that says what the relationship is between the two. There is an independent I and V so I suspect the analysis is incorrect"... The NIC circuit is a current source (a special kind of a voltage-to-current converter, voltage-controlled current source VCCS or a transconductance amplifier) that, in contrast to the ordinary current source, "pushes" its output current back to the input voltage source. This circuit does not use the op-amp output voltage VO; it is a 2-terminal (1-port) device. So, it is not represented by a transfer function; it is represented by an IV curve (with a negative slope). Circuit dreamer (talk, contribs, email) 15:17, 15 December 2013 (UTC)"

Starting the discussion

I think you will very soon be in deep doo-doo if you don't restrict your discussions about electronics to your own talk page. I'm sure you must realize by now that your method of analyzing electronic circuits, while it usually has the right answer and is intuitive, is not appreciated by people who are trained in the art of electronics. Horowitz and Hill's book "The Art of Electronics" presents circuits in a similar way, but without all the "over-helpers", stickmen, and colorful drawings. However H&H cover a lot of ground in their book and just present the results of the circuit analysis and an intuitive explanation. In your case you already know the answer. Quite often some of the statements you make to "support" your answer are incorrect. eg: "Your first statement I₂ = V_R1/R₁ = V_S/R₁ is incorrect. V_R1 = V_O - V_S." Your drawings, etc usually are correct but are not as useful as a few electronic formulae. When someone designs or improves a circuit they have to understand how it works. That is what incremental analysis is for. If someone wants to skip all that and look at your "circuit idea" explanations that is their choice. It may be more entertaining but they are deluding themselves if they think they are learning something they can use to design and build a circuit. I first came across this product of yours when I consulted the Transimpedance amplifier page. I needed to understand more about this for some work I was doing. Frankly I was very disappointed in Wikipedia after seeing what you had done to that page. There was nothing there I could use to solve the problem at hand. Now we have gone full circle and after researching tia circuits I have re-written the page so it is useful for someone like me. The casual reader can skim over the less technical sections and still learn something. I used graphs to explain the physics of transimpedance amplifiers and why they oscillate. That is something your version of the page never mentioned, despite all the colorful plots. Zen-in (talk) 16:52, 15 December 2013 (UTC)

Current-inversion NIC (INIC)

Well Zen-in, let's not argue about general things. But let's finish this issue by something concrete - to clarify your insertion, "Your first statement I₂ = V_R1/R₁ = V_S/R₁ is incorrect. V_R1 = V_O - V_S." I would like to know what is incorrect in I₂ = V_R1/R₁? This is just the Ohm's law (I = V/R). Then, what is incorrect in V_R1 = V_S? This is the basic property of the op-amp negative feedback configuration (the H&H "golden rule"). If this is true, then what is incorrect in V_R1/R₁ = V_S/R₁? Finally, what is correct in your statement "V_R1 = V_O - V_S"? IMO this is wrong (look at the picture on the right). Circuit dreamer (talk, contribs, email) 18:11, 15 December 2013 (UTC)

V_o is the output voltage of the op-amp and cannot be just ignored whatever the configuration of the circuit. What is wrong with V_R1/R₁ = V_S/R₁ is that V_R1 ≠ V_S because V_R1 ≈ V_o - V_S. Where does the current I₂ come from if V_O = 0 ? The output of an opamp is low impedance. It will sink as well as source current. Your idea implies that I_S just goes around the opamp, through R3, R2, and R1 to be sourced at the ground end of R1. That is not how opamps work. Your method, sadly, involves stating several "Ohms Law" truisms that may or may not be correct and then claiming they support an answer you already knew, because someone else did the analysis just as I did. Your intuitive explanations are sort of correct, but have limited usefulness. Electronics is very mathematical and there are rules for doing the math to produce a transfer function. You have stated this is just a blind and mechanical process but it does produce the correct transfer function you use. Some people prefer using Spice, where the computer program does the humdrum math for them. I think that anyone who wants to work in the electronics industry, especially in analog/analogue electronics, should know how to put pencil to paper and derive a formala that describes how a circuit operates, even if they rely on Spice most of the time. These two methods: electronic analysis on paper and/or spice (among many) are widely used in the electronics industry. Your circuit idea, graphi-analytical method doesn't teach anything about frequency response, why some circuits are unstable, the differences in large signal vs small signal response, etc, etc. It just restates simple circuit characteristics that have been derived elsewhere. I think you have a very good intuitive understanding of electronics. That is often not found in people who just rely on Spice. But you are selling your abilities short by not learning and using other methods Zen-in (talk) 22:32, 15 December 2013 (UTC)

I retract the first part of my statement above. I was reading R1 as R2. I'm sure you will agree V_R2 ≈ V_o - V_S. However when you derive transfer functions for opamp circuits V_o is always included. Zen-in (talk) 22:45, 15 December 2013 (UTC)

How NIC operates

How the INIC operates at positive input voltage (where currents flow)

Zen-in, it is obvious that both intuitive and mathematical approach are useful and necessary when analyzing circuits, and they should be applied in this sequence. I have a tendency to absolute the intuitive, and you - the formal approach, and so sometimes we clashed... but it was in the past... The fact that we discuss this circuit in favor of Wikipedia shows that we are benevolent.

Really, I have a very good intuitive understanding of these kinds of circuits and I want to benefit Wikipedia; that is why I joined this discussion temporarily violating the ban. I have considered step-by-step the INIC's operation in a Wikibooks story. I have implemented this story more like a "movie" with separate "frames" but the text explanations are few. I will do it here especially for you as a reward for your zeal.

To understand this circuit, it is crucial to see where and in what direction currents flow. Let's do it and you will see that I have not implied that "I_S just goes around the opamp, through R3, R2, and R1 to be sourced at the ground end of R1". Note that in this simple case (an ideal input voltage source with Ri = 0) the circuit is nothing else than a non-inverting amplifier. Of course, there is a resistor R connected between the op-amp output and its non-inverting input but it does not affect the operation of the non-inverting amplifier since there is no positive feedback and only the ideal voltage source sets the voltage of the non-inverting input. In practice, the input source has some internal resistance; then there is a positive feedback and the circuit becomes even more interesting.

In the beginning, the input voltage is zero... the op-amp output voltage is zero... all the circuit voltages are zero... so there are no currents flowing... Now we begin increasing the input voltage V_IN. The output voltage V_IN.(R₁ + R₂)/R₂ increases as well. A current begins flowing starting from the positive supply rail, then enters the positive supply pin of the op-amp and exits the op-amp output. Then this current splits into two currents flowing through the left and the right parts (legs) of this (bridge) circuit. The op-amp will stop changing its output voltage when reaching the equilibrium - V_(-) = V₍₊₎ (V_R1 = V_S in the article picture). In this simple case (ideal voltage source), the op-amp achieves its goal only changing the voltage at the inverting input; but when the source has some internal resistance, the op-amp changes both the voltages (of the non-inverting and non-inverting inputs).

Now I hope you are already convinced that the direction of I_S in the article picture is misleading - this current has to enter the input source when its voltage is positive... and exactly this is the INIC idea - to obtain negative resistance by inverting the current. Thus both the "half currents" exit the op-amp output and flow down through the two legs to the ground...

I hope I was helpful. Circuit dreamer (talk, contribs, email) 10:37, 16 December 2013 (UTC)

How to present circuits in Wikipedia

It might be entertaining for some people, but most electronics people want to know what the system function is. Just because you are looking at this circuit as a 1-port doesn't mean mean there is no system function. The DC characteristic is the starting point for most electronics people, just like in my re-write of tia. When you add poles and zeros to it the frequency response can be understood. So how would you demonstrate the frequency response of a simple opamp circuit with your circuit idea method? Isn't it easier to just use the GBW formula or to create a Bode plot? It's very likely this negative resistance circuit will self-oscillate like a lot of opamp circuits so understanding it's behavior in the frequency domain is important.Zen-in (talk) 04:52, 17 December 2013 (UTC)

I think we should present circuits in the encyclopedia in this order: first, intuitive explanations showing the basic circuit idea, structure and operation; then, a DC analysis and calculations based on an ideal op-amp and closely related to the previous intuitive explanations; finally, a detailed AC analysis (bandwith, frequency response) based on the real op-amp characteristic that you have considered above. In contrast to your opinion, I think the first and second stages are absolutely necessary for an encyclopedia while the third is not obligatory; it belongs to more specialized sources.

Zen-in, your problem is that you are more mathematician than an engineer. Engineers are not (only) mathematicians; they are something more mathematicians... they are technicians using math as a tool. Engineers do not blindly analyze circuits; instead they first grasp (invent) the basic circuit idea, then they assemble the circuit structure based on the idea and imagine how the circuit operate. Only then they analyze the circuit in detail... and they do not do it blindly as "pure mathematicians" do not understanding what they analyze... they do it based on the intuitive notion about the circuit... they first understand circuits and then analyze them...

So, my remark is that you have made your Transfer function derivation as a "pure mathematician" without seeing the basic circuit idea, its implementation and the circuit operation; that is why your derivation is so lengthy and clumsy... I have made it as an engineer; that is why I have settled my derivation on one line... Circuit dreamer (talk, contribs, email) 16:19, 17 December 2013 (UTC)

There is an old saying which I am sure you have heard "When in Rome do as the Romans". Your concept of what an Engineer does may be quite valid where you live and the same goes for me. I have worked as an Engineer in what is known as Silicon Valley for a long time. Most of the people I currently work with have PhDs in some field or another and are more mathematically inclined. Intuition and mathematical skills are not mutually exclusive. I have met and listened to many pioneers of this industry: Robert Noyce, Gordon Moore, Robert Pease, etc, etc. The discoveries they made in advancing the art of electronics came about because they understood the physics of semiconductors. If you were to read some of the papers from that era (50's - 70's) you would see a lot of math, data from IC characteristic measurements, and references to prior art. We would not have all the wonderful electronic appliances we have today otherwise. That is how engineering is done here in Silicon Valley.

Your circuit idea thing has its merits but most electronics people don't need that to understand how a circuit operates. It really can't be used in Wikipedia because there are no references. If you had a book published then a reference would exist. But I think your circuit idea thing always comes after you have seen the schematic and formula and if you were given a new circuit problem you would not be able to derive a solution. Maybe you can prove me wrong on this. Zen-in (talk) 05:06, 18 December 2013 (UTC)

An interesting discussion... I highly appreciate your experience as an Engineer in Silicon Valley and fully agree that people who create these technological wonders, in addition to their intuition, have very solid theoretical background in the physics of semiconductors... and they are very good mathematicians also... Once we started talking about great circuit designers I can say that I admire truly only one - Bob Widlar. With his unbridled imagination, originality, creativity and artistry, he rises above all others; in comparison with them, he was an artist in the art of electronics and they (were) are handicraftsmen... He was a circuit genius as unlikely to occur ever...

As for me, I have only intuition and imagination... but I have tremendously developed and use them skillfully for quality handling of circuits at the initial stage of understanding... and nothing more than that. And here I can not agree with you that "most electronics people don't need that to understand how a circuit operates"; this statement is simply absurd. Every human being (including Wikipedia visitors) want to first understand (qualitatively) what is the idea, how it is implemented, why it is implemented in this way and how the device operates... and only then to find out what its quantitative characteristics are...

The transimpedance amplifier - so simple but so unintelligible circuit solution...

What transimpedance amplifier is

For example, put yourself in the place of someone who does not know what "transimpedance amplifier" is and, to understand, visits your Wikipedia article. Seeing your picture he/she will ask him/herself: "What is this resistor? What is its function? Why is it needed? What is this operational amplifier? What does it do? Why the resistor is connected exactly in this way - between the op-amp output and the inverting input?" You have not answered these questions. You have said in the beginning "a transimpedance amplifier, (TIA) is a current to voltage converter". But it is quite possible that the visitor already knows (from the Ohm's law - V = I.R) that the simple resistor is such (but passive) current-to-voltage converter when passing the input current through it and taking the voltage drop across it as an output... and he/she discerns this resistor in your circuit diagram. But have you told him/her that this resistor is exactly such a passive current-to-voltage converter? And that the attached op-amp converts it into an active current-to-voltage converter? And that the op-amp does it in an extremely simple, clear and perspicuous way - by adding the same voltage V = I.R which is lost across the resistor? If you had done this, the reader would make the connection with the familiar simpler passive device and would grasp the meaning of this active circuit solution. You have not mention even a word about the virtual ground - the basic idea on which this circuit is based (even you have removed the link to this article). Instead, you have filled the whole introductory part with all kinds of photo applications and common phrases...

But do not be upset by this because there are no answers to these questions even in the introductory part of the famous Bob Pease's article. He has only said, "let's connect a feedback resistor across it, from the output to the -input". But why? What is the problem? How is it solved by adding the op-amp and connecting the resistor in this way? However, the article is great for readers who want to know in detail the problems with the bandwidth and stability... and this makes unnecessary much of this section... Who cares will go in the Pease's article... the encyclopedia is not the place for such details...

Finally, I would ask you a question (although I know that you do not like this subject). Do you see any connection between TIA and circuits with negative resistance, particularly the NIC? Circuit dreamer (talk, contribs, email) 18:32, 18 December 2013 (UTC)

You have 2 outstanding questions from me. Zen-in (talk) 06:13, 19 December 2013 (UTC)

If these were the questions, "So how would you demonstrate the frequency response of a simple opamp circuit with your circuit idea method? Isn't it easier to just use the GBW formula or to create a Bode plot?", here is my answer.

The "circuit idea method" (i.e., the intuitive, heurustic approach) is suitable for an initial quality analysis where we should reveal and show the basic idea, structure, causality and operation of an unfamiliar circuit. It is also a basic method for inventing new circuits (see de Bono's books). But it is not suitable at the next quantitative stages where we have to calculate, design, implement, produce, etc. the device based on this circuit solution. The GBW formula, Bode plot, locus analysis, etc. formalize and make easier these quantative procedures but they hide the basic circuit idea, structure, causality and operation. Circuit dreamer (talk, contribs, email) 08:37, 19 December 2013 (UTC)

The connection between the transimpedance amplifier and circuits with negative resistance

Now let'sanswer my question about the probable connection between the transimpedance amplifier and circuits with negative resistance (first see the picture of a TIA above). We can see at least two connections ..

VNIC is a transimpedance amplifier with an additional positive feedback

1. The op-amp of a TIA is a negative "resistor". The input current I_IN flows through the resistor R and creates a voltage drop V_R = I_IN.R. The op-amp produces the same but negative voltage V_OUT = - I_IN.R that compensates the voltage drop V_R and the result is the zero voltage (virtual ground) at the input. In other words, the circuit consists of two equal resistors connected in series: the first is positive with resistance V_R/I_IN; the second is negative with resistance -V_R/I_IN. They mutually neutralize and the result is zero resistance. Looking at the circuit from the side of the input source, it is just a point with zero voltage and resistance towards the ground; we can think of it as of a neutralized positive resistor.
So the first conclusion is:

The op-amp of a transimpedance amplifier acts as a (current-controlled) negative resistor that neutralizes the positive resistor R; thus the whole circuit acts as a zero resistor.

2. The VNIC is a TIA with additional positive feedback. The transimpedance amplifier is a circuit with (only) a negative feedback. If we introduce additional positive feedback by connecting a voltage divider R₁-R₂ between the op-amp output and its non-inverting input (now see the figure on the right), we actually increase the negative resistance of the op-amp; now it dominates over the positive resistance R one and the whole circuit acts as a (current-controlled) negative resistor (VNIC). So the second conclusion is:

The negative impedance converter (VNIC) is a transimpedance amplifier with an additional positive feedback.

If we compare the two circuits (TIA and VNIC), we can make a final conclusion: TIA is a "compensated resistor" while VNIC is an "overcompensated resistor"; so, as a whole, TIA is a "zero resistor" while VNIC is a "negative resistor".

Circuit dreamer (talk, contribs, email) 17:24, 21 December 2013 (UTC)

Fig. 4. Bode plot of uncompensated transimpedance amplifier

1. The op-amp of a TIA is a negative "resistor".Not really. Where is the negative resistance? The negative sign indicates an inversion or 180 degree phase shift in the signal with respect to the input. Does a common emitter amplifier have a negative resistor and a common collector not have one? Zen-in (talk) 00:23, 22 December 2013 (UTC)

The output of the TIA op-amp is exactly a "voltage inversion negative resistor" since it produces a voltage which is shifted by 180 degrees with respect to the input current. To see it, compare the signs of the two resistors with respect to the passing through them current. First, the entering input current creates a positive voltage (+ V_R -) across the resistor R, and then - a negative voltage (- V_OUT +) across the op-amp output.

The output (the collector-emitter junction) of the common emitter amplifier is not a negative resistor since the input current does not flow through it. Circuit dreamer (talk, contribs, email) 00:47, 22 December 2013 (UTC)

2. The VNIC is a TIA with additional positive feedback. You could say that about any opamp circuit except a comparator or any that just have positive feedback. However it is just a superficial association. The feedback resistor of a tia is very large in comparison to the - feedback resistor of other opamp circuits. And since it is the only feedback source it creates a dominant pole at the roll-off frequency where the capacitance is the input capacitance of the opamp and that of the sensor. So the behavior of a tia is very different. This is where the system function is important in analyzing a circuit. With opamp circuits that don't have a lot of gain (small feedback R), the poles do not result in oscillation (effectively positive feedback at f_i. The Bode plot shows this very well. Zen-in (talk) 00:23, 22 December 2013 (UTC)

The two resistors R1 and R2 (forming a voltage divider) introduce a DC positive feedback in this voltage-inversion NIC. The input capacitances of TIA introduce an AC positive feedback that occurs over a certain frequency. Circuit dreamer (talk, contribs, email) 00:59, 22 December 2013 (UTC)

A circuit puzzle

Design puzzle: Box A converts these pulses into a slowly changing signal with an amplitude proportional to the time between pulses. This signal is fed into box B. Box B converts the signal into a 0 - 240 mA current for the milliammeter on the right as follows: If the pulse repetition rate is .5 /Sec. the meter will indicate 30 mA. 1 /Sec. -> 60 mA, etc.

"It is also a basic method for inventing new circuits" I never thought of it that way but it does relate to a question I asked. Here is a design problem you can use to demonstrate this claim. There are 2 "black boxes" labeled A and B. Each is an electronic circuit that can be implemented with opamps, transistors and other components. You are to design both circuits. The input signal to the left consists of a series of pulses with a rate of .5 - 4 per Sec. They are just narrow pulses going 0 to 5 V. Box A converts these pulses into a slowly changing signal with an amplitude proportional to the time between pulses. This signal is fed into box B. Box B converts the signal into a 0 - 240 mA current for the milliammeter on the right (imagine it to be a 250 mA model) as follows: If the pulse repetition rate is .5 /Sec. the meter will indicate 30 mA., 1 /Sec. -> 60 mA, 4/Sec -> 240 mA etc. Zen-in (talk) 06:49, 20 December 2013 (UTC)

Well, Zen-in... As I can see, this device should continously convert (measure) the time interval (period) between the pulses; it is a "period-to-current converter". We can build it reasoning in the following order.

1. Designing the block A.

Time-to-voltage converter (integrator). First, we have to "feel" ("materialize") the elusive time by converting it into some real physical quantity. The voltage seems to be the most suitable electrical quantity; so we begin charging a capacitor by a constant current and take the voltage across it as a voltage output.

Sample and hold circuit. When an input pulse is detected, the present voltage should be sampled and hold until the next pulse; so we connect a S&H circuit. Immediately after the sample, the input capacitor should be discharged and then released to charge again; so, we connect a switch in parallel to the capacitor and a controlling (hmmm... а pretty crummy) circuit...

2. Designing the block B.

Voltage-to-current converter. I don't know why you have chosen an ammeter as an indicator; a voltmeter is more suitable. But well... we should convert the stored voltage into a current. So we need the voltage-to-current converter that was removed (redirected) by your adherent three years ago (now you followed his lead by removing the no less useful current-to-voltage converter. We can build a voltage-to-current converter simply by connecting a resistor in series to the ammeter (a passive voltmeter) or as a non-inverting amplifier with the ammeter connected in the place of R2 (an active voltmeter)...

This was only one possible but conventional design procedure... There is no invention here since it is a logical sequence of well-known ideas. To make an invention, we need some new, unexpected, illogical, original idea... an insight... I begin thinking about it but this needs a time and chance... Circuit dreamer (talk, contribs, email) 18:19, 20 December 2013 (UTC)

Your answer for device A is partly correct but also incomplete. The puzzle requires a schematic. It doesn't have to be a completely debugged design; just a schematic that would mostly work. It is also more than just an integrator and a sample and hold because these elements need logic controls; ie: when to integrate, when to sample, and when to hold.

Your answer to part B (just a resistor) is incorrect. Let's assume you designed part A so that a 1 pulse/Sec input generated a 1 Volt output. Your answer for part B would require a 16.7 Ohm resistor, to display 60 mA. Now if the input to part A was 4 pulses/Sec it's output would be .25 V and the meter would display 15 mA instead of 240 mA. Similarly if the input to part A was .5 pulses/Sec the meter would display 120. So the solution to part B is unexpected and not so simple. Your circuit idea thing did not detect this. Zen-in (talk) 22:47, 20 December 2013 (UTC)

OK, then connect the ammeter to the positive rail instead to ground... Circuit dreamer (talk, contribs, email) 17:24, 21 December 2013 (UTC)

"OK, then connect the ammeter to the positive rail instead to ground" Wrong again. You didn't state what the mA meter + voltage would be, but assuming the condition I stated earlier, for this new configuration to indicate 60 mA for 1 pulse/Sec = 1 V from part A, the meter + voltage would have to be +2V. With 4 pulses/Sec -> .25V from part A the voltage across the meter and 16.7 Ohm resistor would be 1.75 V, resulting in an indication of 107 mA instead of the required 240 mA. Also for .5 pulse/Sec -> 2V from part A the meter would indicate 0 mA instead of 30. I assume you do not want to continue with this puzzle and will therefore assign your final grade. For part A you got the basic idea but did not complete the design or produce a schematic; so you get 25 out of a possible 100 on part A. For part B you, unfortunately flubbed it entirely. However because I am generous today and because part B would no doubt include at least one resistor I will give you 10 out of a possible 100 for part B. So your total score on this puzzle is 17.5%. If you had completed it I would have given you part C for extra credit and you would then know what this electronic instrument is. However you should be able to figure out it's properties and functions once you derive a response function for part B. Maybe deriving a response function at the start would have been a better approach for you. Zen-in (talk) 18:43, 21 December 2013 (UTC)

Zen-in, I have no idea what the problem is. If this device should be a "period meter", the configuration consisting of a cascaded integrator, S&H circuit, V-to-I converter and ammeter should be perfect (linear) since its components are linear. Another way to reverse the transfer characteristic is if the integrator integrates in the opposite direction (a capacitor discharge)... Circuit dreamer (talk, contribs, email) 00:26, 22 December 2013 (UTC)

You should think about the system function of part B before the resistor. An input voltage of 1V results in an output of 1 V. An input voltage of 2 V. (.5 pulses/Sec) produces an output of .5 V. An input of .25 V (4 pulses/Sec.) results in an output of 4 V. So the system function is:

$V_{o}={\frac {1}{V_{i}}},$ where $V_{i}$ is the input to part B and $V_{o}$ is the output. This is the function part B has to perform.

Plug the values above into the system function and you will see the correct output. Add a 16.7 Ohm resistor and the meter will display the counts/Min. This is what a section of a heart tachometer does. Of course the EKG signal is not a clean 0-5 set of pulses. It has to be amplified and detected accurately. This is another problem. But now that you know what the system function is for part B you can try designing it. How would you approach the design of this if you were going to use opamps? It could also be done with transistors. Which is better? Zen-in (talk) 05:23, 22 December 2013 (UTC)

So we need a non-linear functional converter Y = 1/X. Maybe the Ohm's arrangement will help us if we keep up the voltage constant, change the resistance as an input and take the current as an output? To implement it, we need a voltage-controlled resistor (e.g., a FET operating in its linear part of the output IV curve). Circuit dreamer (talk, contribs, email) 12:00, 22 December 2013 (UTC)

I don't think that would work. I would have to see the schematic to know for sure. You need a non-linear circuit to make another kind of non-linear circuit. It also should be very accurate and with no dead zones, etc. It's use as the output stage of a heart tachometer requires this. When I designed this circuit I used a new approach for this circuit. It was done for a bioelectronics course. There are a few ways of creating this circuit but I think my method produced the most accurate results. Zen-in (talk) 21:34, 22 December 2013 (UTC)

Another suggestion: in an op-amp inverting amplifier with a transfer ratio K = -R2/R1, replace R2 with the ammeter; apply a a reference voltage, and change R1 as a function of V_IN (FET). Do you see any connection between this and the previous my suggestion? Circuit dreamer (talk, contribs, email) 11:53, 23 December 2013 (UTC)

The problems with that answer are: You want to use coefficients (R2, R1, etc) as variables in the system function, putting a meter in the feedback loop of an opamp is about the same as a short circuit, the mA meter is not included in part B. The connection between this and the previous suggestion is both are guesses. I'm not going to give you the answer to this problem. You should solve it yourself. Good luck. Zen-in (talk) 03:38, 27 December 2013 (UTC)

Don't remember that "it is also a basic method for inventing new circuits". So, by "putting a meter in the feedback loop of an opamp", I have combined two components (the part B + ammeter) into one... what is a kind of invention:)

The connection between this and the previous suggestion is... they are the same:)

In the first case, "the Ohm's arrangement helps us by keeping up the voltage constant, changing the resistance as an input and taking the current as an output. To implement it, we use a voltage-controlled resistor (e.g., a FET operating in its linear part of the output IV curve)." Here, we measure the output current by an ordinary (passive) ammeter.

In the second case, we do the same but we measure the output current by an active ammeter (an op-amp with an ammeter connected between the output and the inverting input); here is an animated story. Do you see any connection between this circuit and your favorite transimpedance amplifier? Circuit dreamer (talk, contribs, email) 19:56, 27 December 2013 (UTC)

Just saying that you would use the resistance properties of a FET to implement the function required for part B does not solve the problem. Also FETs are not linear voltage controlled resistors so you have the additional problem of making the FET variable controlled resistor linear. According to R.T Stone & Howard M. Berlin, Design of VMOS circuits with experiments, page 12, "A characteristic of "long" channel FETs is that the square root of the output current is proportional to the gate drive". So your R_FET = K(V_g)². There are methods of linearizing variable FET resistors but the solution to that is more complicated than the actual problem of implementing the function for part B. As far as putting a meter in the feedback path of an opamp being an invention you should try it and see what happens (ie: a latched opamp) before calling it an invention. You can't replace R_f of a transimpedance amplifier with a mA meter and get a useful circuit. What you are thinking about is a FET voltmeter. Your idea is missing a few resistors, etc. You are not very close to solving this puzzle. So much for the circuit idea thing. It is just an unnecessary elaboration on well-known circuits and not useful for solving real problems. Zen-in (talk) 02:54, 28 December 2013 (UTC)

"As far as putting a meter in the feedback path of an opamp being an invention you should try it and see what happens (ie: a latched opamp) before calling it an invention. You can't replace Rf of a transimpedance amplifier with a mA meter and get a useful circuit." A latched opamp? Can you even realize how ridiculous is that said with respect to this as simple as possible circuit with negative feedback? Zen-in, only this text is enough to show that you do not at all understand what а transimpedance amplifier is... but you are the guy who has rewritten entirely the Wikipedia article about it... a sad fact about Wikipedia and its administrators... Circuit dreamer (talk, contribs, email) 05:40, 28 December 2013 (UTC)

Incremental model of an op-amp circuit with a mA meter as a feedback element (??) R_M ≈ 50Ω L_M ≈ 10 mH

You forget that a mA meter has a significant inductance and small resistance. If you don't have any other resistors in the circuit the feedback current could be quite large and out of phase. This may result in an unstable oscillating circuit. The FET voltmeter (a simplified version) has to be over-damped to prevent this and it also has a series resistor. At any rate that is not a solution to the puzzle because if you were going to use a variable FET resistor you would want to put it in the feedback path of the op-amp instead of the mA meter. That would be a solution except for the R_FET = K(V_g)² problem mentioned earlier. Zen-in (talk) 04:11, 29 December 2013 (UTC)

Zen-in, first of all, this device is an ammeter, not voltmeter. So, the input source should be a current source, not a voltage source as shown in your figure on the right. This means we should somehow limit the input current (simply by connecting a resistor in series to the input voltage source).

I suggested to keep up the voltage across a resistor constant, to change the resistance as an input and to measure the current as an output; actually, this is a kind of an ohmmeter. Both the arrangements (inverting and non-inverting) can be used for this implementation, and in the both, the voltage-controlled resistor should replace the input (or grounded) resistor R1 while the ammeter is connected between the op-amp output and inverting input.

I agree that this circuit resembles an inductive differentiator and its bandwith should be limited to ensure the stability... maybe by connecting a resistor or capacitor in parallel to the ammeter...

Another suggestion to implement the non-linear function can be a diode linear approximation... Also, you can put a multiplier in the feedback loop to obtain a divider... Circuit dreamer (talk, contribs, email) 21:33, 29 December 2013 (UTC)

You have been doing some research. Next you can present a schematic using one of these methods. I used a simple multiplier configuration in the feedback loop of an op-amp to implement the 1/V_i function. By trimming offsets, etc I was able to cancel out the nonlinearity of the meter. The prof. showed how to derive an approximation of the transfer function with diodes. The more diodes, the more accurate the approximation. Today it would be trivial to implement most of a heart tachometer with an Arduino or DSP eveluation board. Implemented completely as an analog design it is more interesting. Zen-in (talk) 21:59, 29 December 2013 (UTC)

Remember that we have discussed once this unique property of the negative feedback.

Happy New Year! Cyril (Circuit dreamer (talk, contribs, email) 09:54, 31 December 2013 (UTC))

Topic ban

It appears that you are still under a topic ban for editing electronics-related articles (see WP:Editing restrictions, but you are now making substantial edits to Talk:Pearson–Anson effect and Load line (electronics). You must first get your topic ban lifted. Glrx (talk) 23:43, 27 April 2014 (UTC)

• The terms of your topic ban were set out here. If you make any further edits in violation of this restriction, anywhere on Wikipedia – be it on articles, on article talk pages or within your own userspace – I will block your account. At the end of 2011, you were warned that any further violations would result in longer blocks. SuperMarioMan ( talk ) 20:14, 30 April 2014 (UTC)

May 2014

You have been blocked from editing for a period of one week for continuing to test the limits of your topic ban – on top of previous, long-standing violations ([1], [2]), and despite warnings and an earlier block. Once the block has expired, you are welcome to make useful contributions. If you think there are good reasons why you should be unblocked, you may appeal this block by adding the following text below this notice: {{unblock|reason=Your reason here ~~~~}}. However, you should read the guide to appealing blocks first.  SuperMarioMan ( talk ) 15:06, 4 May 2014 (UTC)

A more conventional view of Diode Logic

In the hopes of enhancing your article on Diode logic I plan to add the following introduction and then at the end a more conventional description. I hope you approve!

The original portion of this article is a philosophical description of how one might think about diode AND and OR logic circuits and their relationship to each other. The section to be added has been added at the end that provides a more conventional description of these fundamental computer logic functions. Thingmaker (talk) 13:24, 7 August 2014 (UTC)

Hi Thingmake, I would be glad to see your creation since you are a great professional! Where can I see it? BTW I am going for a week to the country and maybe I will be not able to answer you during this time... Regards, Cyril (Circuit dreamer (talk, contribs, email) 08:21, 8 August 2014 (UTC))

It will probably take a few days till I can complete it. Have a nice vacation! Thingmaker (talk) 12:50, 8 August 2014 (UTC)

Thanks! Cyril (Circuit dreamer (talk, contribs, email) 17:56, 8 August 2014 (UTC))

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