What is the function of these capacitor? - Page 1
Forum section: electronics / beginners

author: huu_tri0101
author_rank: Contributor
timestamp: November 19, 2013, 07:46:21 am
content: I am about to replace all the capacitors in my amplifier with new better caps (for better sound). So I need to know the functions of these caps in this picture.When replace them with new capacitor, what parameters in new capacitor I should care for better sound ( more/less uF , ESR...)? Can I replace them with non-polar cap for better sound?The function of 2 resistors ?Thank you guys.

author: andtfoot
author_rank: Unknown
timestamp: November 19, 2013, 08:01:06 am
content: The capacitors there function as high-pass filters which block the DC part of the signal (only letting the AC part through).Increasing the value (more uF) would lower the frequency that the signal would be cut off, but whether this is an issue in the first place is determined by the rest of the circuit and what the amp is connected to.I'm not sure what would make better sound though sorry.I did a quick Google for "audio input capacitors" and found this that may be useful:http://www.wolfsonmicro.com/documents/uploads/misc/en/WAN0176.pdf

author: garak
author_rank: Contributor
timestamp: November 19, 2013, 10:51:35 am
content: What andtfoot said.These capacitors are just DC blocking capacitors. They make sure that no DC reaches the amplifier so that you just get the AC signal (i.e. the audio).He's also right in saying that the larger the value of the cap, the more weighted towards the low end it'll be. Ideally your ESR will be as low as physically possible to avoid attenuating your signal too much.Though, if I may say, replacing caps in an amplifier for the hunt for better sound seems like Audiophoolery to me. I'd say you'd be far better off looking at things like the PSU, shielding, OP-Amps etc.Just my $0.02

author: John Coloccia
author_rank: Super Contributor
timestamp: November 19, 2013, 11:37:01 am
content: Once input and output caps are "big enough", there's little to be gained by changing them...for generic, full range audio curcuits, anyway. That JRC4558, on the other hand, is ancient technology. You might try an LM833. A TL072 would probably brighten things up a bit, due to it's very high input impedance.You do need to be careful with caps in audio circuits. Sone have lousy responsev and have to be larger than you might expect. Many are microphonic.

author: Monkeh
author_rank: Super Contributor
timestamp: November 19, 2013, 11:42:37 am
content: *facepalm*There is no simple component swapping you can do which will improve the sound of your amplifier.

author: madires
author_rank: Super Contributor
timestamp: November 19, 2013, 11:57:04 am
content: The resistors are there to set the input impedance and please use film capacitors, not electrolytics! For audio frequency ranges 2.2µF are more than sufficient. BTW, I agree with Monkeh, unless unsuitable components are used in the original design.

author: Monkeh
author_rank: Super Contributor
timestamp: November 19, 2013, 12:03:12 pm
content: Electrolytics can work perfectly well if properly applied. They're actually very, very common..

author: John Coloccia
author_rank: Super Contributor
timestamp: November 19, 2013, 12:34:41 pm
content: Quote from: Monkeh on November 19, 2013, 11:42:37 amOf course there is, just like there is simple component swapping that can make the amp sound worse. The general assumption that everything is optimized for everyone and ever situation is a bit unrealistic, and I doubt much thought goes into the vast majority of consumer audio designs beyond making things big enough so it doesn't sound BAD, and making things beefy enough so they don't blow up. Ship it!I also wonder if that schematic is drawn correctly. As drawn, the input cap will be reverse biased half the time. It would make more sense if the input and opamp were ground referenced, and the second 100k resistor biased the amp mid voltage. As it is now, the amp will have to have positive and negative rails, with the input ground referenced and that will reverse bias the cap. Ditto for the output cap. Maybe I'm having a brain fart?

author: madires
author_rank: Super Contributor
timestamp: November 19, 2013, 01:07:00 pm
content: Quote from: Monkeh on November 19, 2013, 12:03:12 pmSure, a 10µF electrolytic is less expensive than a 2.2µF MKS. Old measurements of different cap types for usability in audio applications showed following rank list:1. KP/MKP/Styroflex/KS/MKS2. MKT/KC/MC (about 3 times worse than 1.)3. electrolytics (about 100 times worse than 1.)

author: Zero999
author_rank: Super Contributor
timestamp: November 19, 2013, 01:29:22 pm
content: John is right, the non-inverting input should be biased at half the supply and you should look into upgrading the op-amp before worrying about the capacitors.

author: John Coloccia
author_rank: Super Contributor
timestamp: November 19, 2013, 02:48:32 pm
content: What is this amp for, anyway? Right now, it's really just acting like some sort of isolation amp...maybe taking an input from a high impedance source and just turning it into something that can drive the real filtering and amplifier stages further down the line.

author: huu_tri0101
author_rank: Contributor
timestamp: November 19, 2013, 04:08:13 pm
content: Quote from: John Coloccia on November 19, 2013, 02:48:32 pmthis amp is used for audio mussic.

author: Monkeh
author_rank: Super Contributor
timestamp: November 19, 2013, 04:28:01 pm
content: Quote from: madires on November 19, 2013, 01:07:00 pmAnd still massively swamped by speaker and room characteristics. Also, 2u2 might be okay with a 10k load, but not much lower. Use of electrolytics is not always avoidable, and in many systems it's not worth the space and expense to.

author: huu_tri0101
author_rank: Contributor
timestamp: November 19, 2013, 06:13:34 pm
content: Quote from: madires on November 19, 2013, 01:07:00 pmso what you mean that I should use the bipolar-cap ? and what is KP/MKP/....? the name of some seri cap form some manufacturer like Visay, epCos? Can you explain more.Thank you.

author: madires
author_rank: Super Contributor
timestamp: November 19, 2013, 08:50:00 pm
content: Quote from: huu_tri0101 on November 19, 2013, 06:13:34 pmFilm capacitors are bipolar anyway ;-)The types I mentioned describe the dielectric material used:KP/MKP = polypropylenStyroflex/KS/MKS = polystyreneKC/MKC = polycarbonateMKT = polyethylene

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: November 19, 2013, 09:42:44 pm
content: The input capacitor could be replaced with 100n film, and be 3dB down at 16Hz. 220n gives 7Hz. But replacing it for a "posh" low-ESR or film type will make no obvious difference - suppose the original has an ESR of 10 ohms; as it works into 100k, the loss will be 0.00087dB. Doubt you'd measure that, let alone hear it.We can't give any advice about the output capacitor without knowing the impedance this circuit has to drive.If you bias the op-amp away from ground, you will reduce the maximum available voltage swing - never a good plan. Nothing is reverse-biased to any extent*, as the capacitors will always have 0V across them, whatever the input audio signal is doing.* The input bias currents will multiply by 100k, so there will be a non-zero potential at the inputs (and hence output), but it'll be millivolts. The polarity depends on whether the input stage is NPN or PNP. For example, an NE5532 in that position would have something in the -20mV region at the inputs and output. That won't trouble the caps...Replacing the opamp with an NE5532 would lower the distortion, but it takes about 2mA more than the 4558 (probably not an issue). But while the change might be measurable with the right gear, you'd never be able to identify it in a controlled ABX test scenario with any statistical validity. The audibility of (correctly implemented) op-amps is massively over-stated, to say the least.Be aware that some op-amps will need more attention to power supply decoupling. And not all are unity-gain stable.In short, leave it alone. If you want to learn about audio electronics, design something yourself; it's easy, and there is plenty of help out there. Learn to filter out the nonsense - of which there is plenty. Rod Elliot's site is an excellent place to start. Then progress to Douglas Self's writings.

author: John Coloccia
author_rank: Super Contributor
timestamp: November 19, 2013, 10:51:53 pm
content: Quote from: Mark Hennessy on November 19, 2013, 09:42:44 pmAudio is an AC signal. Once you send the signal through the first coupling cap, you end up with a signal that is +/- referenced to ground. If you leave it like that, and you don't have a positive and negative rail op-amp, you've clipped the negative part of the signal. Biasing the op-amp gives you MORE headroom in that case, and that's the general case for audio anytime you send it through a coupling cap into an opamp without a negative rail.On the other hand, if it's an opamp with negative and positive rails, then the output capacitor will be reverse biased half the time...and not necessarily by a little bit. It will be reverse biased by whatever the negative voltage swing happens to be for the input signal, up to and including however close the opamp can get to it's negative rail.The only way that schematic makes sense is if the opamp input is biased to something other than ground, through the second 100K, and the opamp's negative rail is ground. Then the negative half of the input AC audio signal will not be clipped and the capacitors are always biased correctly.

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: November 19, 2013, 11:29:10 pm
content: Quote from: John Coloccia on November 19, 2013, 10:51:53 pmThe op-amp will have dual power supplies - operation from a single supply is not feasible with the input returned to earth as shown. Single-supply operation is relatively rare in audio gear, portable stuff being an obvious exception (but the pro gear I deal with often does implement dual-supplies, perhaps with DC-DC converters, or just lots of cells).QuoteNo, a capacitor passing an audio signal will always have the same DC potential at either side, whether it's passing an audio signal or not. That's the definition of a coupling capacitor; the capacitor does not change its charge during the period of the lowest frequencies encountered in the audio signal - if it does, then it's acting as a high-pass filter, and that's why the turn-over frequency is always seriously sub-sonic when electrolytics are used.Also, it's well-known that using electrolytics in low-pass filters is a bad idea - quite apart from their poor tolerance, it will result in a rise in LF THD. So, in an intelligently designed audio system, thesystemLF point is set by a non-electrolytic capacitor that sees a constant impedance (hence constant F3) irrespective of external connections and/or user control settings; once done, any other coupling capacitors are set to turn over at least an octave below that point under worst-case conditions, so that they play no part in the system LF point.

author: John Coloccia
author_rank: Super Contributor
timestamp: November 20, 2013, 01:20:24 am
content: If the opamp has positive and negative rails, and the input is biased at ground potential, the DC output of the opamp is 0V. Presumably, it's looking into something that also has a 0V potential. When the output swings negative, the + terminal is negative with respect to the - terminal. Normally, you'd use a non-polarized EL in that case, or you design the circuit so the EL has a DC bias on it, as would be the case with a mid voltage biased opamp.Again, I might be having a brain fart, but I'm having trouble following the logic of your post.

author: bgsteiner
author_rank: Contributor
timestamp: November 20, 2013, 05:55:15 am
content: If you want to improve the quality of the amplified sound you will want to change out your op-amp as different types of op-amps have different sound charecteristics

author: Monkeh
author_rank: Super Contributor
timestamp: November 20, 2013, 07:03:12 am
content: Quote from: bgsteiner on November 20, 2013, 05:55:15 am

author: M0BSW
author_rank: Frequent Contributor
timestamp: November 20, 2013, 07:53:30 am
content: Audio foolery time, you could try some of those wun hung lo pure gold cables on e-bay pair for $10000000000000000000000000.00 free postage guaranteed to improve quality use less power and tell you , your beautiful every time you switch it on, serious note I used to work in a audio hifi shop when I first left school, when tubes were the order of the day, all these fancy ideas of putting better parts, would make it sound better went on, the engineers used to have belly laughing sessions, the boss was happy selling expensive tubes , that made no difference, apart from the hole in your pocket which got bigger every visit.

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: November 20, 2013, 09:19:24 am
content: Hi John,Quote from: John Coloccia on November 20, 2013, 01:20:24 amI'm in a rush today, but two things that might help:1. For all practical purposes, the op-amp inputs are always at the same potential because of negative feedback.2. The decoupling capacitors will charge up when the circuit is first powered up: for a single-supply circuit, they might charge up to around half the supply voltage - for a dual supply, there might be very little potential - perhaps only millivolts. However, and this is the important point, once this has happened, the charge on them will stay the same while passing audio frequencies. You can think of the capacitors as being like DC voltage sources.If the charge varied during the cycles of the audio signal - and hence the voltage across them - then the audio signal is not being passed/coupled correctly. In that scenario, you have a high-pass filter, not a coupling cap. Which, as mentioned before, you'd want to implement with non-electrolytic capacitors.All the best,Mark

author: BravoV
author_rank: Super Contributor
timestamp: November 20, 2013, 10:47:48 am
content: You guys are scaring the OP off with those advance discussions, looking at the original question and also OP's previous posts , I suspect he is just a novice in electronics.

author: John Coloccia
author_rank: Super Contributor
timestamp: November 20, 2013, 12:58:30 pm
content: Quote from: Mark Hennessy on November 20, 2013, 09:19:24 amThe op amp inputs have nothing to do with it. If you "charge up" the capacitor to a couple of millivolts, and you slam it with -5V, you're going to damage the capacitor.

author: madires
author_rank: Super Contributor
timestamp: November 20, 2013, 01:14:35 pm
content: Quote from: Mark Hennessy on November 19, 2013, 09:42:44 pmJust a brief supplement on how to calculate the bottom frequency at -3dB:f = 1 / (2 pi R C)

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: November 20, 2013, 01:46:06 pm
content: Quote from: John Coloccia on November 20, 2013, 12:58:30 pmBut how, under normal circumstances, are you going to get -5Vacrossthe capacitor?And if you did put -5V DC at the input, would that input capacitor actually be damaged?(for the purposes of understanding, let's restrict ourselves to the input capacitor only - that is totally deterministic because we have the whole circuit in front of us. We can think about the output later if you like)(I am trying to help. And I know from my day job that people can struggle with this topic)All the best,Mark

author: John Coloccia
author_rank: Super Contributor
timestamp: November 20, 2013, 02:01:24 pm
content: The input cap has the - terminal looking at the input, so if you put +5 on it....or +2, or whatever, you reverse bias the cap. At a minimum, it will behave terribly as a cap when the voltage swings positive. The + terminal is sitting, more or less, at ground potential. I don't know what you're asking how I get the voltage "across" the cap. One end is looking at ground, and the other end is looking at something other than ground.

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: November 20, 2013, 02:29:18 pm
content: Quote from: John Coloccia on November 20, 2013, 02:01:24 pmNo.Whatever voltage you put on the input, the other end of the cap will have thesame. Hence, the voltageacrossthe cap will be 0V.OK, after a while, the charge on the capacitor will change. But how long is a while? Well, to get an idea, let's calculate the time constant: 10uF times 100k is1 second.What is the period of the lowest audio frequency? Let's say 50ms (20Hz). So, with a healthy 20Hz sine wave playing into this, the capacitor doesn't havetimeto significantly alter its charge during the positive-going part of the waveform.To prove this, get a dual input scope and connect probes either side of the cap. You will see exactly the same waveform on each side - perhaps with a small DC offset. Case closedAbove, when working out the time constant, I said 100k. Was that fair? Yes, it is. The positive end of the capacitor sees 100k to ground, plus whatever the op-amp presents.Earlier I quoted the first "golden rule" of op-amps: the inputs are always at the same potential. The second "golden rule" is that the inputs draw no current. In other words, the input impedance is extremely high (and is further raised by bootstrapping because of NFB).To answer the second question I posed: if you place 5V DC at the input such that the capacitor "saw" the wrong polarity (which - having just re-checked the schematic - would be +5V, not -5V - sorry!), then that would be in series with 100k. The worst-case current would be tiny: just 50uA. Would that damage the cap? Based on my experience, I very much doubt it.I've seen dozens of 40 year old capacitors with 3 or 4 volts of incorrect polarity (caused by a silkscreen error), and they've always checked out fine - I don't even bother to change or reverse them... (I can provide more details if you like)

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: November 20, 2013, 02:37:49 pm
content: Quote from: BravoV on November 20, 2013, 10:47:48 amYes, I agree that the OP is probably new to this, but speaking as someone who teaches electronics, I do think that this is something that is pretty basic; the idea of separating out the DC conditions and ACsignalis fundamental to understand any circuit that deals with signals. And, as we're seeing, it is a source of confusion, even among experienced people. In my experience, the earlier you deal with this topic, the easier everything else becomes.Perhaps the mods should move these posts to a new thread? I'm new to this forum, so don't know the "form"Mark

author: John Coloccia
author_rank: Super Contributor
timestamp: November 20, 2013, 03:11:13 pm
content: If you had a pure audio signal, such as when coupling one stage into another, you wouldn't need the input cap.

author: sync
author_rank: Frequent Contributor
timestamp: November 20, 2013, 03:29:54 pm
content: The world is not perfect. There are small DC offsets. But 1V reverse bias doesn't damage an electrolytic cap. This circuit is used billion times in audio gear and the caps survive.

author: ejeffrey
author_rank: Super Contributor
timestamp: November 21, 2013, 05:56:26 am
content: Quote from: madires on November 19, 2013, 01:07:00 pmThat depends strongly on what you are using it for. You don't want electrolytic capacitors in feedback networks or filters because their value can change dramatically with time and temperature, changing the frequency response. For input coupling caps the limitations of electrolytic caps are not important and they will work fine.

author: madires
author_rank: Super Contributor
timestamp: November 21, 2013, 02:20:48 pm
content: Quote from: ejeffrey on November 21, 2013, 05:56:26 amI found an old Elektor article about that topic (http://www.die-wuestens.de/iz/audiokondensatoren.pdf). Unfortunately it's in German but you'll see the measurement setup they have used. I'm not an audio expert but I've done a internship in a high-end audio EE company (no audiophoolery!) while studying and learned a lot. They had two amazing EEs, one is an absolute expert in analog audio stuff and electrolytic caps for input coupling was a no-no. If you took some audio circuit diagram from an electronics magazine to him, he made some changes whithin 5 minutes and told you to take measurements of the original design and then of his modified one. Guess which one always won!

author: sync
author_rank: Frequent Contributor
timestamp: November 21, 2013, 02:50:20 pm
content: Thanks for the PDF. The distortion of electrolytics was too low to measure (<0.005%) . But there was a difference using the square wave. Unfortunately it lacks information. Rise and fall times. What does 1% difference mean. What are the frequencies of the difference. And I'm missing a inductance compensation.

author: huu_tri0101
author_rank: Contributor
timestamp: November 22, 2013, 03:59:38 am
content: Thank all of you for helping me out.First, this is really a working circuit, I draw from a Pioneer amplifier A-N702.It's use +/-12V for opamp 4558s. This circuit is just apart of big circuit. the output of this 4558 will be the input of another opamp (JRC2068 for volume adjust) then another 2 JRC4558 (for bass and treble adjust) then final opam JRC2068 to drive some small BJTs, and BJTs drive 4 mosfets.After reading all your post. I learn a lot of things. Thank again guys.So, I should change the electrolytic capacitor to film capacitor for lower the THD, reliability and better high pass filter ?Can I change the value of capacitor from 10uf to 2.2uf ? what will change when I do that(cut off frecquency, input current,...) ?And why when we calculate the cut off frequency, we calculate at 3 dB?how can we calculate/measure the impedence of input and output ?Sure I will change the opamp, any suggestion ?Thank you

author: amyk
author_rank: Super Contributor
timestamp: November 22, 2013, 04:53:25 am
content: Unless the electrolytics have dried out significantly and lost most or all of their capacitance you will not hear a difference. Otherwise the bass will be gone.

author: Monkeh
author_rank: Super Contributor
timestamp: November 22, 2013, 07:25:12 am
content: Quote from: huu_tri0101 on November 22, 2013, 03:59:38 amYes, leave them alone. There's nothing wrong with the opamps you have.

author: madires
author_rank: Super Contributor
timestamp: November 22, 2013, 11:17:12 pm
content: Quote from: huu_tri0101 on November 22, 2013, 03:59:38 amActually you could use any value down to 220nF for that circuit.QuoteFor a lot of stuff the frequency limits are specified for -3dB.QuoteIt's based on the in/output circuitry and the OPamp. The input impedance of your buffer amp is about 50k (set by the input resistors).

author: Orpheus
author_rank: Regular Contributor
timestamp: November 23, 2013, 04:15:48 am
content: +/- 3dB is an arbitrary cutoff but there is *some* rationale behind it. +3dB voltage or current means 2x (-3dB means a factor of 1/2). To be more precise: +/- 3.0103 dB = (10 log 2) dB = 2x the voltage.In general situations, engineers may not want to worry much about factors of one-point-something, so they often set the limit at "this difference is not just 1.x". Hence 3dB = factor of 2 = the cutoff

author: Hideki
author_rank: Frequent Contributor
timestamp: November 23, 2013, 02:04:32 pm
content: There is more rationale for it. Here's an image of a (first order) lowpass filter response. Highpass looks the same, except you have to flip the curve left to right.If you look at the blue curve, the response of the first order filter falls off at 20 dB per decade (or ~6 dB per octave) once you get past the cutoff frequency. Following the slope of the curve up from the bottom right corner, the "ideal" curve would hit the 0 dB point at exactly the given cutoff frequency.However, a real filter doesn't behave like two straight line segments. There is a smooth curved transition between no attenuation and attenuating 20 dB per decade. At the cutoff frequency the response is at the half-power pont, ~3 dB down.

author: Mark Hennessy
author_rank: Frequent Contributor
timestamp: November 23, 2013, 03:00:27 pm
content: Quote from: Orpheus on November 23, 2013, 04:15:48 amNot quite. -3dB is half thepower, not the voltage.A halving of the voltage would be -6dB. You'd get that with two equal value resistors.The -3dB point occurs when the capacitor has the sameimpedanceas the resistor in the filter. To calculate the impedance, use Zc = 1 / (2pi times FC)So on the face of it, it might seem like the voltage would half when the capacitor has the same impedance as the resistor. But, when you factor in thephase anglecaused by thereactanceof the capacitor, the change is actually 0.707 (1/sqr-root(2)), which is -3dB.Oh, and the input impedance of this circuit is 50k, as there are two 100k resistors in parallel.Take it from me - I've been doing this stuff for all my life, and professionally all my working life: there is no point whatsoever changing the op-amp in this circuit. It won't sound any different (at best), and you won't learn anything from doing it. Build your own gear - and then debug it. Correctly selected and correctly implemented op-amps do not have a "sonic signature".