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A Common uses section would be useful. I'm trying to figure out if a voltage divider is a good way to step down a 12V power source to 5V, or if there's a better way... ~MDD469603:17, 7 July 2006 (UTC)
Maybe an advantages/disadvantages section? Like power issues (current drain).[reply]
This would be a really bad way to do it, as the voltage is dependent on load resistance. You could do it with a voltage divider and a transistor, though.Flying fish23:52, 24 March 2007 (UTC)[reply]
I've moved "Voltage divider rule" back to "Voltage Divider", with the blessing of the original mover Karol Langner. The article is more about the device than the equation, and there are more pages that link to voltage divider anyway. Flying fish15:09, 8 April 2007 (UTC)[reply]
It might be useful for this page to include an example where more than 2 resistors are used in the voltage divider to show how the voltage at various nodes is calculated. e.g. {(R1+R2)/(R1+R2+R3)}* Vin to provide the voltage at the first node after the first resistor in a circuit with 3 resistors and a device across the 2nd resistor (it would be the voltage at one of the device inputs). I realize it would be easier if I could provide a graphic of what I mean but I'm not able to do that right now so I hope my description is clear enough to explain what I mean. Thanks. Radiantbutterfly20:12, 2 August 2007 (UTC)[reply]
Do you mean this section? It was put there to demonstrate a practical application of voltage dividers. There's also a similar section in the current divider article. Don't delete them, but you could combine them both into a new article (probably the best thing to do). Otherwise just add a cleanup tag or something. -Roger (talk) 22:03, 12 April 2008 (UTC)[reply]
OK, I have added some explanatory text. I agree that the section should be moved to a separate article but I don't know enough about the subject to write it. Biscuittin (talk) 18:57, 16 April 2008 (UTC)[reply]
Hello. Should not we replace the impedance "Z" by ? Newbies might get confused by the the symbol Z that is used for both for the complex value and the modulus. --Youssef (talk) 12:19, 11 May 2010 (UTC)[reply]
Hello. See for example the first paragraph of Electrical impedance. The impedance is normally a complex number. And complex number can be expressed by imaginary and real part or by modulus and phase. When we write "Z" without anything above the letter, conventially it means the modulus. However, some engineers do not find it easy to add that symbol so may write "Z" also for complex impedance. My point is that voltage divider rule, generalized Ohm rule, etc works only with complex impedance. Regards-- Youssef (talk) 12:03, 6 June 2010 (UTC)[reply]
I've reverted these changes several times because the actual resistive divider still obeys the usual rule. The argument that the combination of the op-amp and the voltage divider is now a "modified voltage divider" is one point of view, but if so, it's no longer a "resistive voltage divider", and so no longer belongs in this section. Oli Filth(talk|contribs)11:10, 5 June 2010 (UTC)[reply]
If this page was dedicated to the pure voltage divider from Ohm's times, it would consist only of a few sentences. But, as all we can see, it is enlarged with many other aspects concerning various applications in circuitry. Some of them are not so appropriate and they should be removed or moved to other pages.
For example, have you noticed this strange MOS subsection? What is this? Maybe, the author wants to say something about non-linear "voltage dividers"? But then we have to include many other non-linear circuits as well... because almost all the analog circuitry is based on this circuit of two series connected elements.
Then, do we need two subsections about resistive divider saying almost the same and repeating this R2/(R1 + R2) expression?
Now see the whole loading effect section. Although it is valuable and the picture is very expressive, I would like to ask you, the authors and the other wikipedians: Is this article the most suitable place for considering in details (in separate subsections) all the amplifier's interfacing problems (input loading, output loading and the overall gain? See the talk above (Roger, please, join the discussion).
You have said that the combination "voltage divider and op-amp" is no longer a "resistive voltage divider". But then the combination "voltage divider and op-amp buffer" (buffered resistive divider) or "voltage divider and INIC" (helped resistive divider) will be no longer a voltage divider... BTW, in the inverting and the non-inverting op-amp circuits, the op-amp serves as a buffer as well (besides the other functions). All these op-amp circuits show typical useful voltage divider applications where the imperfect passive circuit is incorporated in a perfect electronic circuit.
Let's finally say the simple truth: it is important to show the bare voltage divider but it is more important to show its useful applications in electronic circuits, to show the connection between passive and active versions, to follow the evolution, the metamorphose of this humble passive circuit into various perfect op-amp circuits. Circuit dreamer (talk, contribs, email) 14:00, 5 June 2010 (UTC)[reply]
In fact, on reflection, the combination of the op-amp and voltage divider does not form another voltage divider. The topoology is wrong; where is the intermediate point that forms the voltage output? Furthermore, the terms "inverted voltage divider" and "reversed voltage divider" are your invention. I would remove this again, but that would take me past 3RR, so it'll have to wait until tomorrow. Oli Filth(talk|contribs)15:05, 5 June 2010 (UTC)[reply]
An arrangement representing Miller theorem
I have noticed from your edits (e.g., about op-amp differential input resistance and various Miller theorem manifestations) that you do not accept this manner of expressing when some impedance is artificially modified. In all these cases, when an impedance Z is connected between the amplifier's output and input we say that the impedance is increased, decreased, etc. Indeed, the very impedance Z is the same; actually, the combination "impedance Z and amplifier" (the new virtual impedance) is increased, decreased, etc. (or, as you will probably say, the amp input impedance is increased, decreased, etc.)
I will say it in another way. Imagine the right end of the impedance in the picture is initially grounded (there is no amplifier connected); so, the input source "sees" an impedance Z. Now let's "insert" the amplifier in this arrangement (to "shift" the right end of the impedance by a proportional voltage Vo). As a result, the input source will "think" that the impedance Z is increased, decreased, etc.; it does not suppose the presence of the additional voltage source Vo. So, in the both situations (with and without amplifier) we see only the impedance Z; we do not notice the additional amplifier in the second case.
My advice is to be not so strict, precise and punctilious since this prevents from understanding circuit concepts and seeing the connection between apparently different circuit phenomena. Think of all these arrangements as of clever circuit tricks, circuit illusions. Circuit dreamer (talk, contribs, email) 16:51, 5 June 2010 (UTC)[reply]
Now let's see where, in the "inverting voltage divider" (i.e., an inverting amplifier with "gain" of R2/R1 < 1), the intermediate point that forms the voltage output is. For this purpose, let's build the circuit step-by-step by converting the humble resistive divider into the perfect op-amp inverting version.
The resistive voltage divider may be considered as two consecutively connected voltage-to-current and current-to-voltage converters. The problem is that they are not separated - the current-to-voltage converter (R2) is a part of the voltage-to-current converter (R1 + R2) and the transfer function R2/(R1 + R2) is not "tidy":) So, we want to separate them: only R1 to act as a pure voltage-to-current converter and R2 to continue acting as a pure current-to-voltage converter.
Miller's idea helps us to find a remedy. We add an additional voltage Vo = -VR2 in series with R2 that compensates the voltage drop across it. As a result, a virtual ground appears in the intermediate point. The actual output voltage VR2 and the resistance R2 are removed; only the resistance VR1 continues existing in the circuit and only it determines the current I = Vi/R1 (R1 acts as a pure voltage-to-current converter as we wanted).
The output voltage (in the intermediate point) is destroyed but we guess to take the compensating voltage Vo (the mirror "copy") as an inverse output. So, the lower (right) end of R2 (the former ground) begins acting as an "intermediate" point and the transfer function -R2/R1 becomes "tidy":) The use of this pretty solution is obvious - the output is buffered (the load consumes current from Vo instead from Vi). Circuit dreamer (talk, contribs, email) 17:56, 5 June 2010 (UTC)[reply]
That was a rather long-winded way of getting to the obvious conclusion: the intermediate point is, of course, the virtual ground. It's always at zero voltage, and is not dictated by R1 or R2. Therefore, the overall circuit is not a "voltage divider" in any meaningful sense. Unless you can find a source that describes it as such, I will remove this material again tomorrow. Oli Filth(talk|contribs)18:04, 5 June 2010 (UTC)[reply]
Maybe, we have first to define what the name "voltage divider" means. If it originates from the transfer ratio expression, both the R2/(R1 + R2) and -R2/R1 versions may be considered as "dividers". So, the overall op-amp inverting circuit is a "voltage divider" (inverting, buffered and having a "pure" transfer ratio) in whatever meaningful sense. Circuit dreamer (talk, contribs, email) 18:24, 5 June 2010 (UTC)[reply]
Well, that definition would cover any circuit that produces a voltage output based on a voltage input. I would stick with the definition given in the article: "A voltage divider ... is created by connecting two electrical impedances in series...". I'm comfortable with the idea that one of those impedances might be a complicated linear network, or a negative-impedance simulator, for instance. However, the op-amp circuit cannot be decomposed into two impedances. So if you'd like to extend that definition, then you'll need to find a source for it! Oli Filth(talk|contribs)18:30, 5 June 2010 (UTC)[reply]
I just have decided to develop further the sentence "...using resistors alone it is not possible to either reverse the voltage or increase Vout above Vin..." and to show that it is possible to do the impossible.
It's clear from your latest edit that you've accepted that this isn't a voltage divider, and that you've just invented the terminology. Therefore, I really don't see why you continue to add this material! An amplifier obviously allows voltage gain; it has nothing to do with a resistive voltage divider. Oli Filth(talk|contribs)12:04, 6 June 2010 (UTC)[reply]
Well, there is no problem if the names were the problem; I have removed them although they were useful for understanding. Note that I put such synthetic names in quotes to show that they are descriptive, not real. The names are not so important; they are only a form, the envelope for the content. But I insist on keeping the text since it makes the connection between the passive resistive divider and the basic op-amp amplifying circuits; it shows how this imperfect passive circuit may be modified by applying negative feedback and thus converted into these perfect op-amp circuits. To understand circuits, it is extremely important to show their evolution from the imperfect passive versions to the perfect active versions, to build and reinvent circuits step-by-step. In this connection, it would be useful even to build the humble resistive divider from this page presenting it as two consecutively connected voltage-to-current and current-to-voltage converters or by reinventing it from the Ohm's experiment.
Let's finally say some words about the definition given in the article - "A voltage divider ... is created by connecting two electrical impedances in series...", and your assertion - "...the op-amp circuit cannot be decomposed into two impedances..." In the inverting configuration, we have the exactly same two elements as in the passive one. Only, the second element (R2) is not grounded; it is connected to an additional voltage source and the circuit of two elements connects as a "bridge" the two voltage sources. So, this configuration is a "shifted voltage divider" whose ground is "moved" by voltage VR2 to an opposite direction in comparison with the input voltage direction. Actually, this is no longer a resistive voltage divider; it is already a resistive summer (the virtual ground in the intermediate point is the result of summing). From another viewpoint, this is a resistive divider with "annihilated" second element (R2). Only the first element (R1) exists in this configuration; only it determines the current and the input impedance. The load does not affect the current since it consumes current from the additional voltage source (op-amp output).
As a conclusion, I would like to say that we have exchanged interesting thoughts about voltage dividers and basic op-amp amplifying stages that may be used in the corresponding pages. So far, this was a constructive dialog but as I can see, you continue removing my edits regardless of my compromissary edits. At the same time, you do not notice the obvious faults in the article. So, I have the feeling that you fool around with me. As you can see, I have nothing against since you are clever opponent stimulating me to expose gradually my philosophy on the talk pages and to write my insights into history... but remember one must draw the line somewhere...
For example, I have not understood what you think about the non-linear MOS "divider". Is it a divider? If so, what is its transfer ratio? Is it constant? Will we develop this extremely interesting topic here? If so, we will reach dynamic resistors, parametric voltage stabilizers, dynamic loads, cascode circuits and even differential negative resistors and the discussion will become very intriguing... Are you ready for such a heroism:)? Circuit dreamer (talk, contribs, email) 14:08, 6 June 2010 (UTC)[reply]
Because of the unique properties of the negative feedback, these op-amp circuits may be thought as voltage dividers that are modified by the attached amplifier. Indeed, there is something paradoxical in this viewpoint since, in these arrangements, the ubiquitious op-amp serves the humble resistive divider; but the negative feedback makes the op-amp be in service of the divider. This is the great idea behind all the negative feedback circuits - an op-amp with an enormous amplification accommodates to some passive circuit so that the overall gain depends only on the circuit's attributes and does not depend on the op-amp's gain. Circuit dreamer (talk, contribs, email) 14:15, 6 June 2010 (UTC)[reply]
I'm afraid there is no need to compromise at the moment. You don't have a source for your viewpoint that the inverting amplifier acts as a voltage divider (and you won't find one, because it isn't!), so it cannot go into the article.
But you have not clarified what "voltage divider" means. I need this definition to substantiate my assertion. For now, I consider that a circuit with a transfer function R2/R1 is (more than obviously) a divider. Circuit dreamer (talk, contribs, email) 14:35, 6 June 2010 (UTC)[reply]
I've already stated that I consider the definition in the article to be correct, i.e. a network of two or more impedances in series across the input, with the output taken from the intermediate point. As I've already said, to extend the definition as you are proposing is ludicrous; it would mean that all amplifiers could be considered voltage dividers. Oli Filth(talk|contribs)14:42, 6 June 2010 (UTC)[reply]
No, I have not said this; please, read all my thoughts about negative feedback circuits above. These two circuits (inverting and non-inverting) are not only amplifiers; they are amplifiers with negative feedback in a form of a voltage divider. I have said that the combination voltage divider + inverting amplifier is again a voltage but inverting divider and the combination voltage divider + non-inverting amplifier is again a voltage but reversed divider. The whole op-amp circuits (voltage divider + op-amp) are again voltage but modified dividers! Negative feedback has the unique property to modify circuit attributes: in the first case, it zeros the impedance (Miller effect); in the second case, it swaps the circuit input and output (see the Deboo's book). Circuit dreamer (talk, contribs, email) 14:53, 6 June 2010 (UTC)[reply]
You continue to flip between two contradictory viewpoints:
You state that the circuit is a divider simply because its input-output relationship can be expressed as a ratio of impedances. This is clearly a ludicrous definition.
You continue to hold the view that the overall circuit is somehow still in a standard divider topology. It's not, for reasons that I've already covered. The only way to view them in the same way as the "standard" divider (Z1 and Z2) topology is to consider Z1 to consist of the input resistor, and Z2 to consist of the parallel combination of the amplifier and the feedback resistor. But again, the "output" of this network is the virtual ground, which is not the output of the amplifier.
I do not think of this discussion as of wasting of time since, in such disputes, valuable insights conceive in human minds. Honestly, it is not interesting for me if these circuits are voltage dividers or not. I am interested in circuit phenomena, concepts, fundamental ideas...
As I can see, you decide if a circuit is a divider depending on its topology. But I decide if it is a divider depending on the fact that the output voltage is a part of the input one. The lede says the same: "...a voltage divider is a simple linear circuit that produces an output voltage that is a fraction of its input voltage..." Then, "...a simple example of a voltage divider consists of two resistors in series..."; so, this topology is only an example. The conclusion is that there is a need of convention what a divider is. IMO is too primitive to use only the topology as a criterion and to fix the output in the intermediate point. Who has said that the output has to be only there? And, if you want to use this point as an output, please, use it - connect the (floating) load in parallel to Z2. I.e., the "standard" voltage divider exists and its genuine output voltage exists as well and it has the same polarity as the input voltage and you can use it... but it is a bad idea since this "standard" output is not buffered. This is the great idea behind negative feedback inverting circuits - the amp output voltage destroys and replaces the genuine output voltage drop across Z2.
My conclusion is what I have said is true and useful (no matter if these circuits are dividers) since it establishes a connection between these old-fashioned passive circuits and up-to-date op-amp circuits. It shows a useful application of Miller effect; it is another viewpoint at basic op-amp amplifying circuits and it reveals the main negative feedback properties.
BTW, Z2 and the amplifier (output) are connected in series, not in parallel combination (think of the op-amp output as of a voltage source) so that the two voltages are in series; "...the parallel combination of the amplifier and the feedback resistor..." is a ludicrous sentence here. Actually, in this (Miller) configuration (see the figure above) all the three components (the input voltage source, the element Z2 and the output voltage source) are connected in series... but this is a circuit with parallel negative feedback:) Circuit dreamer (talk, contribs, email) 16:12, 6 June 2010 (UTC)[reply]
(outdent) As you are well aware, the purpose of talk pages is to discuss changes to articles, not to hold meandering discussions about circuit theory! Nor is it a forum for you to lecture others on your views (you don't need to keep repeating the "great idea" mantra).
From your latest reply, it is clear that you are back to Argument #1. There is indeed a convention; it's the one that everyone (apart from you, apparently) uses, i.e. the one given in this article. If you can find a source that agrees with you, then we can discuss. Otherwise, it's time to leave it alone, because the article isn't going to be changed without a source.
And no, from the point of view of a standard divider topology, Z2 and the amplifier are in parallel; the voltage across Z2 is the voltage between the amplifier's terminals. Oli Filth(talk|contribs)16:21, 6 June 2010 (UTC)[reply]
Well, it's time to have a rest to us and to wikipedians keeping an eye on our conversation... But now I am trying to imagine how a 4-terminal (2-port) device may be connected in parallel to a 2-terminal element... and what would be the use of this presentation if it would possible... Because, if you say that the amp's output (2-terminal element, a voltage source) is connected in series to Z2 (2-terminal element, impedance), the use is obvious: the output voltage compensates the Z2 voltage drop... and the basic idea behind all op-amp inverting circuits with negative feedback becomes obvious as well. Maybe, the 4-terminal op-amp is connected in parallel (output to input and input to output) to the 4-terminal voltage divider? But is this really a parallel connection? I think they are connected in a ring forming the negative feedback loop... Thank you for the discussion. Regards, Circuit dreamer (talk, contribs, email) 16:48, 6 June 2010 (UTC)[reply]
-these will depend on your application-- how much resolution do you need for your code or circuit to work? how much current waste is ok? 0.5mA is too much, going to drain battery? then use 100k. >100k not usually used for a divider, usually like 10-50k I want to say...24.98.133.72 (talk) 22:59, 18 August 2014 (UTC)[reply]
I dont think that an RC circuit can be properly called a voltage divider as its opartion depends on ac of a certain frequency. I intend to remove the RC network unless there are convincing arguments to keep it.86.187.170.222 (talk) 22:33, 15 December 2016 (UTC) — Preceding unsigned comment added by 86.187.170.222 (talk)
I think the formulas laid out in the current version should take into account whether the source is current (DC) or alternating (AC). AXONOV(talk)⚑08:57, 30 May 2023 (UTC)[reply]
I arrived at this page as a user. At the time I was using Ubuntu 22.04 and Chrome Version 122.0.6261.57 (Official Build) (64-bit). The LaTex math formulae did not render. I see this: "$ V_{\mathrm {out} }={\frac {Z_{2}}{Z_{1}+Z_{2}}}\cdot V_{\mathrm {in} } $". I mistakenly thought it was a formatting issue on this page. I tagged it, which tag was gently reverted.
I have since looked at other pages with the same OS/browser combo, which render in similar fashion. So the problem is not unique to this page. I used the same machine with Firefox browser and the LaTex renders perfectly. At this point, I don't know whether the problem is in Chrome or in Wikipedia's delivery of LaTex to Chrome. Where IS the proper place to report this issue? Rhadow (talk) 15:00, 30 March 2024 (UTC)[reply]
1) On my computer, it has 64bit Chrome 123.0.6312.86 and this article seems to display fine. Before reporting, you should try some of the following: A) update to most recent browsers, B) clear your chrome browser cache at chrome://settings/clearBrowserData , C) temporarily disable all browser extensions and/or only for Wikipedia domain. 2) Afterward, please look at the following too. https://en.wikipedia.org/wiki/Wikipedia:FAQ/Problems - • Sbmeirow • Talk • 15:42, 30 March 2024 (UTC)[reply]
