Talk:RC circuit

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[edit] English, Layman Version

This is great for eggheads who already know the subject up and down, but how about something for the layman, who, say, wants to build an RC circuit as 1/10th of a second a timer? More practical and less theoretical would be helpful. --68.97.208.232 14:13, 8 May 2006 (UTC)

Seconded. I agree that the physics and math involved are important, but I would also like to see a more practical approach to this article. As it is now, it seems straight from a physics textbook, and the raw information isn't really all that useful without a great deal of educational context. At the very least, I think this article could use an explanation of why such a circuit is useful, and why it is useful for those applications. --Ktims 07:52, 9 May 2006 (UTC)

Another request for a section which describes RC for dabblers. My question is: when a make an RC circuit and attach to my microcontroller, how exactly are the electrons flowing? What makes the electrons start and stop so that a series of peaks are created and can be timed? Thanks for what is here so far. 59.183.11.174 14:02, 29 November 2006 (UTC)John

[edit] Parallel RC Discussion

How about the same discussion of the parallel RC circuit?

Actually, the parallel RC circuit as shown in the article is incorrect. You would never place a voltage source across a parallel combination of a resistor and a capacitor. The input source should be a current source, in which case the circuit would function as the exact dual of the series circuit. -- Rdrosson 03:27, 6 November 2005 (UTC)
If you think about it, the series circuit is really just a Thevenin equivalent power source driving a capacitor. So the dual circuit would have to be a Norton equivalent power source driving an inductor. But that, of course, would be an RL circuit. So the parallel RC circuit, with a current source, is actually the dual of the series RL circuit. Likewise, the series RC circuit is the dual of the parallel RL circuit. -- Rdrosson 03:39, 6 November 2005 (UTC)

[edit] Complex Impedance Methods

[edit] if input signal not pure sinusoid

What if the input signal is not a pure sinusoid? Why is the article restricting the frequency domain analysis only to pure imagnary frequencies? There is a much more general form involving Laplace transforms where, instead of using

Z_C = { 1 \over j \omega C }

the complex impedance is

Z_C = { 1 \over Cs }

where s is a complex number

s \ = \ \sigma + j \omega


Sinusoidal steady state is then a special case where

\sigma \ = \ 0

and

s \ = \ j \omega


This approach then enables you to use some interesting and powerful techniques:

  • Solution of the differential equations using polynomial functions of s
  • Laplace transfomations of inputs and outputs to derive complex valued functions in terms of s
  • Analysis using complex valued transfer functions, also in terms of s, that are simple ratios of the complex valued input and output functions
  • Identification of poles and zeros of the transfer functions, and plotting the poles and zeros in the complex s-plane
  • Calculation of gain as the magnitude of the transfer function and phase angle as the argument of the transfer function.
  • Frequency domain analysis involving not only pure sinusoids but also damped sinusoids
  • Fourier decomposition and analysis of arbitrary (non-sinusoidal) signal inputs and outputs.
-- Rdrosson 12:45, 4 November 2005 (UTC)

[edit] Laplace domain stuff

If you look slightly further down the page, you'll see that Laplace domain stuff is included there. -Splashtalk 12:51, 4 November 2005 (UTC)
Yes, I see what you are saying. But actually, the article barely scratches the surface of these ideas, and everything prior to the mention of Laplace Transforms can be vastly simplified by a much more general and elegant set of techniques. If you read the the list of bullet points I created (above), I don't see any of these concepts other than one brief mention of Laplace Transforms in the article. Furthermore, even the discussion of Laplace misses the key point: you don't need to restrict the input signals to sinusoids -- you can represent and analyze the behavior for virtually any input signal . -- Rdrosson 20:09, 4 November 2005 (UTC)
I don't honestly see the utility of the section of the complex impedance of a capacitor, since it just duplicates that in capacitor. We don't need to repeat basic information. I've removed it. The bit about the pole/zero is also probably covered better elsewhere and doesn't actually have any context here at all. It would be better discussed elsewhere is context, with just a reference from here. This article, is after all, about RC circuits, not poles and zeros of transfer functions. The generalisation of the Laplace stuff is good, though, except that there are some undefined symobls that need fixing. I have limited internet seconds now, so can't doit myself.-Splashtalk 19:04, 7 November 2005 (UTC)
I also dislike the fact that jω stuff is not in the analysis at all anymore. It's the perspective from which everybody studies it first, and it should be the way we present it first. Laplace may be prettier, but it's less instructive from a fundamentals of circuitiry perspective. -Splashtalk 19:07, 7 November 2005 (UTC)

[edit] j ω

Are you looking at the same article that I am? I see j ω all over the place. Of course, now that you unilaterally removed the section on Complex Impedance, the fact that s = j ω is no longer in the article, so it's pretty difficult for people to make the connection. Why don't you simply revert the article back to the way it was before I started making any changes at all?

If you notice, I didn't really remove any information from the article. All I did was to add new information that was not already there.

Why would you want to have Pole-Zero diagrams in an article about RC circuits? Do you mean, besides the fact that they provide a phenomenally easy way to understand what's going on? Of course it was out of context -- Wikipedia is a work in progress -- you cannot expect someone to do it all in one sitting. So I guess it is better to do nothing than to at least get started moving down a path, even if it is not completed on Day 1. Well, good for you, thanks for undoing all of the hard work that I have done over the last few days. Oh, and congratulations on your open mind and willingness to consider someone else's point of view before trashing their ideas. -- Rdrosson 21:24, 7 November 2005 (UTC)

If all your good work had been in those few lines of mathematics, the article would be much the poorer. On the other hand, I left almost all of your edits in. There's really no need to get all angry about it — after all, you can revert me quite easily. The mathematics is not introduced from the j omega perspective, and my personal feeling is that it should be. My personal feeling is nothing to get annoyed with, surely? I didn't revert all your changes, because nearly all of them were good. Take a few deep breaths, and re-read my messages, and you'll find them much less offensive than you currently think they are. And as for unilateralism, just about every Wikipedia edit is unilateral: yours were too. -Splashtalk 16:33, 8 November 2005 (UTC)

[edit] Step response

Is the step response the same thing as the impulse response? Because if it is it should says so, and if its not - the step response should be added. Fresheneesz 23:29, 9 December 2005 (UTC)

No, it's not. The step response is in the Time domain considerations section. Unfortunately, the recent rearrangement of the page was a little haphazard and needs fixing. Jump in... -Splashtalk 23:31, 9 December 2005 (UTC)

[edit] Time Domain Plots

I've increased the line weight on my plots by 600%, it should make them much easier to see (you're right! the small thumbs were impossible to see [I set a large thumb size]). Rather than get in a revert war over it, if you're okay with them, give me permission to replace them or do it yourself. Here's how they look now.--Ktims 00:36, 1 April 2006 (UTC)

This formula \,\!V_C(t) = V\left(1 - e^{-t/RC}\right) \,\!V_R(t) = Ve^{-t/RC} give the same result as 0.999^1000 = 0.367695425 \tau \ = \ RC , The discharge time is 744,760591 \tau \ , example 0.999^744 760.590 = 4.94065646 × 10-324 , 0.999^744 760.591 = 0 on google calc. 1-result tau = charge tau. Pawem1 --213.199.225.33 12:15, 17 September 2007 (UTC)

[edit] Voltage source?

I don't think it has to be driven by a voltage source... why would it? Fresheneesz 06:05, 14 April 2006 (UTC)

[edit] Integrator Circuit

Should the statement: "...Consider the output across the capacitor at high frequency..." instead read "...at low frequency..."? I'm not certain about this, but doesn't the voltage across the capacitor go to 0 at high frequencies, so the integrating circuit, the one whose output voltage waveform closely matches the area of that of the input voltage, should only be observed at low frequencies? 134.226.1.229 20:19, 9 January 2007 (UTC)

[edit] Definition seems wrong, what about multiple caps and resistors?

The definition says that a RC circuit has only one cap and one resistor. An RC circuit should be any circuit comprised of only resistors and caps, of any complexity. In that case I'd say the article was too focused on the simple canonical case. Also I don't think its necessary to give the transfer function for the case where the output is across the resistor and when its across the cap. I'm gonna try making the series example a bit more concise if no one objects. Roger 01:12, 12 May 2007 (UTC)

Any circuit with multiple resistors, multiple sources and one capacitor can be put into the form of a circuit consisting of one source, one resistor, and one capacitor. While I agree with you that multiple capacitors would still make an 'RC' circuit in the broadest sense (plural Rs and Cs); in the stricter sense that is most common, RC refers to a resistor and a capacitor (the singular sense) such that we have a circuit with a single time constant.
I certainly don't object to making this article more concise. In fact, there is much repetition of this material in RL circuit so have a go at that one too. The fact is, I do believe that the majority of the 'filter' theory here is out of place. Go for it. Alfred Centauri 04:23, 12 May 2007 (UTC)
Well the definition says "It consists of a resistor and a capacitor, either in series or in parallel, driven by a voltage or current source", that excludes multiple resistors as well. I don't think the convention of calling a circuit a "RC" (Resistor/Capacitor) circuit is meant to literally mean one resistor and one capacitor since there's alot of network theory that applies to circuits having only resistors and capacitors (including multiples). I'll work on shortening the rest of the article, but I think the definition needs changing. Roger 04:38, 12 May 2007 (UTC)
Done. Alfred Centauri 13:43, 12 May 2007 (UTC)

[edit] Linear RC circuit with voltage gain

This is more of a curiosity than a comment on the article. I recently came across an RC circuit with voltage gain. The author of the book [1] says that the circuit has a maximum voltage gain of 1.15. I found this difficult to believe, so I simulated it in PSpice and analysed the circuit algebraically. It does indeed behave as advertised, with a Vo/Vs of about 1.15 V/V at about 1 kHz. I'm not trying to plug the book, and to prove it I have redrawn the circuit here so that you don't have to follow the link above.

I'm not claiming anything supernatural, since the circuit obviously doesn't have power gain, but I just can't understand intuitively how it works. I understand how LC circuits can have voltage gain, but I've never seen it happen in a linear RC circuit before. Is there any way to explain it, other than by saying "do the sums"? --Heron 10:15, 30 May 2007 (UTC)

Unless I've made an error, the transfer function is:
T(s) = \frac{1 + s[R_2 C_2 + (R_1 + R_2) C_1]}{s^2 R_1 R_2 C_1 C_2 + s[R_2 C_2 + (R_1 + R_2) C_1] + 1}
Comparing this to the standard form:
T(s) = \frac{1 + \frac{1}{Q} (\frac{s}{s_0})}{(\frac{s}{s_0})^2 + \frac{1}{Q}(\frac{s}{s_0}) + 1}
yields:
s_0 = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}}
Q = \frac{1}{\sqrt{\frac{R_2 C_2}{R_1 C_1}} + \sqrt{\frac{R_1 C_1}{R_2 C_2}} + \sqrt{\frac{R_2 C_1}{R_1 C_2}}}
Using the values specified yields:
f_0 = 1592 Hz \,
Q = \frac{1}{2.0001}
So, we have a sum of a 2nd order LPF and a 1st order BPF where both filters have the same Q (about 1/2) and the same cut-off / center frequency. The sum of these two low Q filters gives a magnitude greater than unity somewhat below the cut-off frequency.
Looking at this circuit in the time domain with a 'hand waving' argument, observe that C1 'sees' R1 in series with and impedance while C2 'sees' R2 in parallel with some impedance. Thus, the phase of the current through C2 lags the phase of the current through C1. This implies that, at points in time, there is a current 'down' through C2 and a current 'up' through C1. But, this is what we need for Vo to exceed Vs. The current 'up' through C1 causes a voltage across R1 that adds to Vs. Alfred Centauri 15:40, 30 May 2007 (UTC)
Thanks for that insight. I agree with your transfer function, but I couldn't have turned that into a filter type without your help. The hand-waving explanation is also ingenious, but one thing troubles me. You say that the current up C1 causes a voltage across R1 that adds to Vs, which I understand, but isn't that an instantaneous current? It might only be true at some points in the cycle, as it would be true of any two out-of-phase sinusoids. How do you generalise from that to saying that Vo(RMS) > Vs(RMS)? --Heron 21:10, 30 May 2007 (UTC)
Yes, it's instantaneous current. Take a look at the transient analysis screenshot. See that with the appropriate phase relationship, the two sinusoids add constructively. To answer your question about generalizing to the the AC case, as long as there is a current 'up' through C1 when Vs is at max positive peak, we are assured that Vo(RMS) > Vs(RMS). This seemed like the case to me as long as we are below the cutoff frequency. Alfred Centauri 22:29, 30 May 2007 (UTC)
Thanks. I've just played around with my version of the transient simulation, so I can see now how V(R1) adds to Vs below Fcutoff (right down to DC). Last time I only looked at the AC simulation, which didn't help me much. --Heron 12:59, 31 May 2007 (UTC)
Thanks for bringing this circuit to our attention. I didn't suspect that a passive RC filter could give a voltage gain. BTW, I just thought of a better 'hand waving' argument. First, see that the branch containing R1 and C1 has a much larger impedance than the branch with R2 and C2 so the R2 C2 branch behaves more or less like a standard 1st order LPF. At low frequencies, the voltage across R2 leads Vs by close to 90 deg. But, the voltage across R2 'drives' the R1 C1 branch. Thus, at low frequencies, the voltage across R1 leads the voltage across R2 by close to 90 deg. But this means that the voltage across R1 leads Vs by close to 180 deg which is what we need to add constructively. Alfred Centauri 13:52, 31 May 2007 (UTC)
That explanation is easier to grasp. I consider the mystery solved. --Heron 19:25, 31 May 2007 (UTC)
The book "Fast Analytical Techniques for Electrical and Electronic Circuits" by Vatche Vorperian discusses this circuit, and more complicated versions, quite extensively (making use of the extra element theorem). The reason why you get a >1 voltage gain is because you can design the zero to occur before the two poles. Roger 20:08, 31 May 2007 (UTC)
Roger, that is a good observation but I have to point out that having the zero come before the poles doesn't necessarily give insight as to why this passive RC filter has a gain exceeding unity. Up until Heron brought this circuit to my attention, I assumed that any passive RC filter with a zero before any poles would necessarily have a DC gain less than unity. Thanks for the reference, I'm going to take a look at it. Alfred Centauri 23:17, 31 May 2007 (UTC)

[edit] Make this page useful to someone who isn't a PhD or EE

Good teaching and education starts with tailoring information to audience. The disclaimer at the top of this article - "This article relies on knowledge of the complex impedance representation of capacitors and on knowledge of the frequency domain representation of signals " is comical. If one knew all those things, well, one probably wouldn't be looking for this page.

RC filters are one of the first things that someone exploring electronic theory runs into. They can be conceptually explained in a few short paragraphs before falling into all the obtuse and complex math, which perhaps some might come to a Wiki page for, but more likely would be the interest of the minority. —Preceding unsigned comment added by 38.119.114.42 (talk) 05:54, 17 March 2008 (UTC)

Since you were able to type this comment, I assume your fingers aren't broken and thus I wonder why you haven't made the changes yourself? The editors here are unpaid volunteers and don't really give a rat's ass for armchair editors. If you have the chops, fix it. If not, then I suggest you come down down off of your high horse and show some respect to those that have actually taken the time to edit Wikipedia. Alfred Centauri (talk) 02:25, 19 March 2008 (UTC)

You both have good points; by converting all caps heading to normal, I hope to have put OP's comment into a form less offensive, and nicely invite him to follow up on your suggestions. Dicklyon (talk) 04:49, 19 March 2008 (UTC)