Talk:Circuit Idea/Simple Op-amp Summer Design

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Contents 1 The challenge 2 Revealing the idea 3 What does "the ground resistor RG is not necessary" mean? 4 fully differential op amp

[edit] The challenge

A month ago, DavidCary sent to me a link to the EDN's article Single-formula technique keeps it simple by Dieter Knollman. Initially, I had a look with reluctance at the html version of the article (I don't like calculations and formal presentations). First, I was totally irritated by the typical for EDN awkward article arrangement (a rough text without figures and links pointing to the same pdf file); then, I calmed down when I found out the pdf version. The material drew my attention because my pursuit is to reveal the secrets behind passive and basic op-amp amplifying and summing circuits.

In the beginning, I was not understanding what the author wanted to say. The material seemed to be a scientific falsification. I was not understanding what a ground resistor is and why there was a need of it, who were these Plato and Daisy, etc. Thus, I took the material as a challenge and began trying to reveal the idea behind the procedure. Doing that, I have spent an almost a month!

[Djhk] I propose that we change this book from Circuit Fantastic to Circuit Fantasies. New ideas, even bad ones, can be good. Before moving any idea to an unproven level, I propose that the idea be tested via simulation or a breadboard. The next level would need a formal proof or multiple tests and agreement from all. For example, I consider Plato's gain formula proven. Daisy's theorem as stated unproven, but proven under special conditions. Much of the discussion here should fall into the Fantasy classification.

Djhk 18:16, 24 November 2007 (UTC)

[edit] Revealing the idea

Gradually, I began figuring out the connection between the novel procedure and the "legacy" one; I began seeing the brilliantness of the solution. I even began creating this new circuit story where to show the reason of the procedure.

...Daisy's theorem. In the beginning, it seemed quite strange to sum gains; I thought that voltage summers sum voltages. Then, I quessed to assume that the input voltages are equal to 1V. As a result, the summer began summing gains! So, in the common case, this kind of summer (a parallel one) sums actually products of gains and voltages. In case of Gi = 1, it sums voltages; in case of Vi = 1V, it sums gains. By the way, I use similar explanations when my students and I build a DAC. Really, a DAC is just a summer with digital-controlled binary-weighted input voltages (at constant input resistances) or gains (at constant input voltages equal to the reference voltage). Thus I began gradually figuring out the role of the ground resistor - to ensure the gains needed in the case when their sum does not constitute 1. It serves actually as a complementing, supplemental resistance that adjusts the overall gain to 1 (in the beginning, I thought that the author has added it to equalize the equivalent input resistances).

[Djhk] Daisy's theorem is very different from any electronics concept. It applies to linear circuits and voltages. In a linear circuit the voltage at any node is simply the sum of gains times voltage inputs. If ALL gains are included, then the sum is equal to one. This applies to any circuit node. The key is ALL. Electronics rarely considers the gain from ground or power supplies. If you leave these out, the gain sum may not be equal to one.

Djhk 18:35, 24 November 2007 (UTC)

...Plato's formula. Then, I guessed to impose the requirement for equal input resistances (minimal bias-current error) and replaced the term 1 + RF/RI with K; this simplified the formulas for all the circuits. Interesting... imposing a restriction simplifies the formula. I suppose that the author has first managed to see that the particular gains are proportional to the ratio Rf/Ri; then, he has had the good fortune to establish that the coefficient of proportionality is just equal to 1, if we impose the requirement for equal input resistances (or, maybe v.v.?)

I think that supplementing the gains and imposing the requirement for equal equivalent input resistances lead to the magic expression Ri = Rf/|Gi|. Circuit-fantasist 18:26, 1 September 2007 (UTC)

{Djhk] Plato's formula was derived via circuit analysis. It only applies the the [General Summing amplifier circuit]:[1]. Many circuits, but not all, can be analyzed with Plato's gain formula. The equal impedance constraint comes from the analysis technique used. Note that the 1 + RF/RI formula only applies to a non-inverting amplifier. Plato's formula applies to more circuits, but not all.

Djhk 18:35, 24 November 2007 (UTC)

[edit] What does "the ground resistor R_G is not necessary" mean?

Djhk, I have noted your comment about the resistor Rg: "Not true. The circuit will not function without Rg. It may be a zero value resistor."

Sorry, saying "the ground resistor R_G is not necessary" I mean that it may have zero resistance.

Well, let's discuss what the phrase "there is no resistor" means: "the resistance is zero (shorted circuit)" or "the resistance is infinite (open circuit)." What do you think about this? Circuit-fantasist 15:32, 30 September 2007 (UTC)

Yes, a reader might misinterpret "not necessary" to mean "I don't need to connect anything to the "+" op amp input, I can leave it floating". Doing that causes a real op-amp to not merely show "degraded performance", but to oscillate or show other undesired behavior. So I changed the text to explicitly state: 0 resistance wire. --DavidCary 17:55, 5 November 2007 (UTC)

[Djhk] We need to show complete circuits. This includes inputs. Circuits respond differently if the input is an ideal versus a non-ideal source. To move an idea to not proven, I propose that you test the idea. This will require a complete circuit. Without a complete circuit, any idea remains a Fantasy.

Djhk 18:42, 24 November 2007 (UTC)

[edit] fully differential op amp

TI manufactures several 24-bit ADCs. All of their data sheets recommend using the OPA1632 "fully differential amplifier". (No surprise that it is also manufactured by TI).

A " fully differential amplifier" is very different from a " difference amplifier" or a "w:instrumentation amplifier".

Can Dieter's procedure be applied to a fully differential amplifier? At first glance, I'm guessing that there's one minor and one major change:

• Use equal feedback resistors Rf, one from the + output to the - input, and one from the - output to the + input. Connect the "common" input to the appropriate voltage source. (minor change to the " difference amplifier" )
• The sum of the gains = zero in a properly-designed fully differential amplifier circuit. (Right? major change)

[Djhk] Wrong! Daisy says that the gain sum is equal to one at the +out node, the -out node, or any other node.

How can you make the sum of two sets of gain equal to zero, if each set has a sum of one?

Is 1 + 1 = 0 ?

Djhk 18:46, 24 November 2007 (UTC)

• Calculate resistor values for each input, using Ri = Rf / |desired gain|.
• For inputs with positive gain, connect the calculated resistor value between that input and the + input on the amplifier.
• For inputs with negative gain, connect the calculated resistor value between that input and the - input on the amplifier.
• If all your inputs are differential pairs, then the sum of all the gains is now zero. Done. Otherwise add a ground resistor to bring the total gain to zero.

If that process looks good, please move it to Electronics/Op-Amps#fully_differential_amplifier. (Should we mention fully differential amplifiers on this Circuit Idea/Simple Op-amp Summer Design page? ) --DavidCary 23:15, 5 November 2007 (UTC)

Don't insult Daisy! For any linear circuit the sum of ALL gains is equal to one.

You need to analyze the circuit and derive the gain equation.

Speculation should remain on the talk page. User:Djhk