Suppose I run the following configuration for one effect device:
Regulator in its own box > cable > RC filter (10R-220u) inside the effect box > effect circuit
Using the RC filter as additional HF filtering that the cable may have picked up. Does this filter have any effect on the regulator's performance? Or does it have no effect other than reducing any external HF noise?
Does the fact that the capacitor of the RC filter will be used as energy storage for the circuit have any effect on the use of the regulator? I mean, is performance of the circuit totally characterised by that cap, or by the regulator?
Regulator in its own box > cable > RC filter (10R-220u) inside the effect box > effect circuit
Using the RC filter as additional HF filtering that the cable may have picked up. Does this filter have any effect on the regulator's performance? Or does it have no effect other than reducing any external HF noise?
Does the fact that the capacitor of the RC filter will be used as energy storage for the circuit have any effect on the use of the regulator? I mean, is performance of the circuit totally characterised by that cap, or by the regulator?
From a voltage stand point It's better to place the regulator on the load end of the cable.. Certainly it must come after any series resisters on in an RC filter. THe 10R resister will drop the voltage depending on the load.
Also from a noise standpoint it is better to place the regulator on the load end of the cable. Regulators are good, far better then RC filters are attenuating noise.
Finally theat last filter cap that follows the regulator is best if it is physically close to the load. If you are designing a PCB, place it close to whatever uses the most current and bypass it with a small film cap. For purposes of analysis you can assume that your device is powered by the last filter cap and if you have done as suffext above the power supply will appear to have a very low output impedance.
If you must do it your way use a shielded twisted pair for the power cable. Put the supply and return on the twisted pair and ground ONE end of the shield. (An XLR mic. cable could work.)
Thanks for your answer ChrisA!
There is no exact way that I "must" do something, I am just thinking about what would be the proper way.
Facts:
(1) I want to avoid the run af AC currents close to the pedals, in order to avoid interference to my pickups. I play close to them.
(2) I am willing to build good regulators, but if the scenario I placed above will actually make them useless, I think it is not that clever to regulate before reaching the loads (pedals).
Saying useless, of course, I mean that I cannot exploit their full potential. If I am to degrade their performace, I'd better not build them.
Another option I am thinking about is this (following your comment): rectifying and filtering the AC voltages from the transformer in an aluminum box, and then running twisted cables into the pedals. Inside the pedals then, a regulator will be placed to regulate the voltage, filter noise etc.
How about that solution?
EDIT: The meaning of rectification and smoothing will result in low ripple current being present in the cables, which means less trouble for AC pickup interference.
There is no exact way that I "must" do something, I am just thinking about what would be the proper way.
Facts:
(1) I want to avoid the run af AC currents close to the pedals, in order to avoid interference to my pickups. I play close to them.
(2) I am willing to build good regulators, but if the scenario I placed above will actually make them useless, I think it is not that clever to regulate before reaching the loads (pedals).
Saying useless, of course, I mean that I cannot exploit their full potential. If I am to degrade their performace, I'd better not build them.
Another option I am thinking about is this (following your comment): rectifying and filtering the AC voltages from the transformer in an aluminum box, and then running twisted cables into the pedals. Inside the pedals then, a regulator will be placed to regulate the voltage, filter noise etc.
How about that solution?
EDIT: The meaning of rectification and smoothing will result in low ripple current being present in the cables, which means less trouble for AC pickup interference.
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Yes, a DC umbilical with a screen connected to chassis at both ends.
Do you need:
+ & -
or
+, 0 & -
Do you need:
+ & -
or
+, 0 & -
10 of the power supplies will be positive (0 to +) and one of them will be symmetric (-,0,+).
audiostrat,
Regarding your "EDIT" comment (and PSUs in general):
(ChrisA and AndrewT and others have already stated this, here and elsewhere. I just want to try to amplify and maybe clarify a few of the ideas.)
It's not mostly the rectification and smoothing that will result in low ripple current in the umbilical. The main ripple is caused by the load, not the AC mains.
Power supplies are there to provide current, including (especially) dynamically-changing current, while keeping the voltage within a reasonable range as required. The accurately-provided dynamically-changing current is what we want and need. It is the signal. (That "should" be obvious, since we are always trying to keep the voltage constant.)
For example, the PSU reservoir capacitor "ripple current" is almost exactly the signal that a power amplifier is sending to the speakers (and IS exactly the signal, between charging pulses, if we disregard the tiny transistor base currents). [The ripple VOLTAGE is just due to the load (signal) current from the caps causing their voltage to sag between charging pulses, and can be made as small as we want by increasing the reservoir and/or decoupling capacitance.]
So if you want low ripple current in your umbilical cabling, which is a very good idea, then what will probably matter most is the reservoir (actually "decoupling") capacitance at the LOAD end of the cable. (I don't call them "filter caps" or "smoothing caps" because that belongs more to the voltage-centric view of power supplies, which often promotes misunderstanding, since the current is the signal and is usually, or should be, the quantity of main interest.)
Having a good decoupling cap configuration at the load end of the cable (right at each point of load) is very important and will provide at least two kinds of "baked-in goodness": The fast-transient currents that each load needs will be mostly confined to small local loops, which will give the lowest radiated and conducted interference, because the radiated magnetic field strength is directly proportional to the enclosed geometric loop area (see Faraday's Law), and because trying to pull the fast time-varying transient currents all the way through the umbilical would cause corresponding voltage spikes across the inductance of the cabling, with spike amplitudes proportional to the time-rate-of-change of the current (since V = L di/dt), so that even small but fast-changing currents could cause large voltage spikes. And those voltage spikes could then pervade everything else that is using the same power supply (part of the reason for the term "decoupling"). Additionally, due to the relatively-long umbilical's inductance, without the local decoupling caps the accuracy of the fastest components of the transient response of the load could be degraded, because the current simply could not arrive in time, at the correct amplitude.
Basically, you don't need to filter out noise, you need to not create it in the first place, by using proper decoupling capacitor configurations, very close to each load.
Twisting and shielding and minimizing the areas enclosed by all loops will help to prevent both external and internally-generated fields from inducing unwanted currents in your circuits. But don't forget to also implement RF filtering. "Everything is an input, for RF."
Cheers,
Tom
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Many thanks for your detailed response Tom!
I realise eveything you say. I was thinking of adding a regulator to the load, and close to it. So, if I get you correctly, should I preceed the regulator with a good decoupling capacitor, of say 220uF? I will need 20mA maximum of "audio current" in class A pedals.
I think this solution will keep everyone happy:
Secondaries > Bridge + Reservoir > Filter > Cable > 220uF decoupling > Regulator > Circuit's final power supply
Am I right?
I realise eveything you say. I was thinking of adding a regulator to the load, and close to it. So, if I get you correctly, should I preceed the regulator with a good decoupling capacitor, of say 220uF? I will need 20mA maximum of "audio current" in class A pedals.
I think this solution will keep everyone happy:
Secondaries > Bridge + Reservoir > Filter > Cable > 220uF decoupling > Regulator > Circuit's final power supply
Am I right?
You're welcome!
Looks pretty good. But what is "circuit's final power supply"?
Not sure about the 220 uF. To calculate the minimum C required there, or worst-case ripple amplitude versus C value, we would need to know more, such as nominal cap voltage and max average current draw, and regulator output voltage. Since it's at a regulator input, we would need to ensure that the bottom of the ripple voltage waveform could never (ever) cause the regulator input voltage minus the regulator output voltage to dip below (or too near) the regulator's dropout voltage spec for the given current.
I also think that the active load devices, after the regulator, would also need decoupling caps. They would also each need one very-small-sized cap for both bypassing and decoupling for the very high frequencies, maybe mainly bypassing, for stability, i.e. to prevent ringing or oscillation due to the hidden positive feedback loop through the power rail that is inherent in most transistor amplifier-type circuits. Physical size (low self-inductance) and mounting configuration (short paths for low inductance) are important, for that one, since oscillation frequencies could be in the tens of MHz.
20 mA. Roger that. But at what max and min slew rates over what voltage range and into what load resistance?
Looks pretty good. But what is "circuit's final power supply"?
Not sure about the 220 uF. To calculate the minimum C required there, or worst-case ripple amplitude versus C value, we would need to know more, such as nominal cap voltage and max average current draw, and regulator output voltage. Since it's at a regulator input, we would need to ensure that the bottom of the ripple voltage waveform could never (ever) cause the regulator input voltage minus the regulator output voltage to dip below (or too near) the regulator's dropout voltage spec for the given current.
I also think that the active load devices, after the regulator, would also need decoupling caps. They would also each need one very-small-sized cap for both bypassing and decoupling for the very high frequencies, maybe mainly bypassing, for stability, i.e. to prevent ringing or oscillation due to the hidden positive feedback loop through the power rail that is inherent in most transistor amplifier-type circuits. Physical size (low self-inductance) and mounting configuration (short paths for low inductance) are important, for that one, since oscillation frequencies could be in the tens of MHz.
20 mA. Roger that. But at what max and min slew rates over what voltage range and into what load resistance?
First of all, "final supply" isn't anything - I just ended the series like this! I mean that power will be drawn from the regulator's output, sorry for the confusion. 🙂 Which will indeed carry decoupling caps.
Final voltage should be 9V. I was thinking about implementing a Sulzer regulator beacuse I need low currents, and I also think that it needs quite low dropout voltage.
Slew rate should not be of high concern since guitar signals don't even reach the whole audio spectrum. I assume that something like 10V/us should be fine.
The application is not demanding (the load, to be precise). In other words, maybe just one decoupling cap could do the trick, should I feed the pedal with 9V regulated. But I don't think this has any benefits other than not needing to fit the regulator inside the pedal.
EDIT: I forgot to mention load resistance. Typical effect circuits include "volume" pots as final stage of the design, that could be as high as 500k. In any case, I don't think this could drop below 1k. So say that load is in the kilo range.
Final voltage should be 9V. I was thinking about implementing a Sulzer regulator beacuse I need low currents, and I also think that it needs quite low dropout voltage.
Slew rate should not be of high concern since guitar signals don't even reach the whole audio spectrum. I assume that something like 10V/us should be fine.
The application is not demanding (the load, to be precise). In other words, maybe just one decoupling cap could do the trick, should I feed the pedal with 9V regulated. But I don't think this has any benefits other than not needing to fit the regulator inside the pedal.
EDIT: I forgot to mention load resistance. Typical effect circuits include "volume" pots as final stage of the design, that could be as high as 500k. In any case, I don't think this could drop below 1k. So say that load is in the kilo range.
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For low-power, low-frequency ICs and transistor circuits, such as audio circuits usually use, you're probably safe with blind guessing for the decoupling caps, which means that each active device should have a 10 uF or higher electrolytic and a 0.1 uF X7R ceramic in parallel, from each power pin to ground. The 0.1 uF needs to be connected as closely as possible to the body of the device; not more than about 2 mm away. Each board would usually have a larger electrolytic where each power rail and ground enter the board, unless there is only about one active device on the board.
Note that slew rate is not the slew rate of the audio signals. It should be the maximum slew rate that an active device is capable of, based on the datasheet data. For example, even if your audio signals will only slew at up to 2 V/us, any internal (or external) feedback loops might need to employ slew rates that are at least 10X faster, as they try to keep up with high-frequency harmonics, or whatnot. Most kinds of audio processing circuits might employ slew rates of up to 20 V/us, and Bob Cordell recommended that a power amplifier should be able to slew its output at least 50 V/us.
Note, too, that the maximum slew rate for a particular frequency also changes with signal amplitude. Taking a single sine wave as an example, the maximum slew rate (steepest part) is where it crosses through zero. The slew rate, there, is
slew rate in V/us = 2 π f A / 1000000
where f is frequency in Hz and A is amplitude in Volts.
So with 8-Volt signal peaks, a 22 kHz sine would have a maximum slew rate of about 1.1 V/us. But 40-Volts peaks would have to be able to slew 5X faster.
Anyway, if it ever becomes an issue, you will want to make sure that the decoupling networks have low-enough inductances to be able to supply the current accurately at the required rate. That will usually equate to the connection lengths, and might require using more smaller caps in parallel, to try to divide the inductance by the number of paralleled caps.
Note that slew rate is not the slew rate of the audio signals. It should be the maximum slew rate that an active device is capable of, based on the datasheet data. For example, even if your audio signals will only slew at up to 2 V/us, any internal (or external) feedback loops might need to employ slew rates that are at least 10X faster, as they try to keep up with high-frequency harmonics, or whatnot. Most kinds of audio processing circuits might employ slew rates of up to 20 V/us, and Bob Cordell recommended that a power amplifier should be able to slew its output at least 50 V/us.
Note, too, that the maximum slew rate for a particular frequency also changes with signal amplitude. Taking a single sine wave as an example, the maximum slew rate (steepest part) is where it crosses through zero. The slew rate, there, is
slew rate in V/us = 2 π f A / 1000000
where f is frequency in Hz and A is amplitude in Volts.
So with 8-Volt signal peaks, a 22 kHz sine would have a maximum slew rate of about 1.1 V/us. But 40-Volts peaks would have to be able to slew 5X faster.
Anyway, if it ever becomes an issue, you will want to make sure that the decoupling networks have low-enough inductances to be able to supply the current accurately at the required rate. That will usually equate to the connection lengths, and might require using more smaller caps in parallel, to try to divide the inductance by the number of paralleled caps.
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Many thanks for your detailed responses!
Do you think that the following solution could be also applied?
Regulators in aluminum box > Cables > Good decoupling cap in effects box
This is the easiest thing to do for me. Bearing in mind that the application is not very demanding, of course.
This does not allow good filtering from anything cables pick up I suppose, but in conjuction with a film cap where the cable enters I think I can ameliorate the situation.
Pedal effects usually run on 9V batteries, and such batteries are regarded as one of the best solutions available. But cost. I think these have an output resistance of 100-300 milliohms if I am not mistaken. In that manner, could a fully stabilised 9V power supply fed through cable to the pedal that uses a good decoupling cap probably have better performance than a battery, or not?
Do you think that the following solution could be also applied?
Regulators in aluminum box > Cables > Good decoupling cap in effects box
This is the easiest thing to do for me. Bearing in mind that the application is not very demanding, of course.
This does not allow good filtering from anything cables pick up I suppose, but in conjuction with a film cap where the cable enters I think I can ameliorate the situation.
Pedal effects usually run on 9V batteries, and such batteries are regarded as one of the best solutions available. But cost. I think these have an output resistance of 100-300 milliohms if I am not mistaken. In that manner, could a fully stabilised 9V power supply fed through cable to the pedal that uses a good decoupling cap probably have better performance than a battery, or not?
Something I forgot to ask: so if I understand well, the capacitor after the regulator is there to provide the circuit with the current signals it needs. And the regulator ensures that whenever the voltage of the cap tries to fall because charge has been provided to the load, the regulator ensures an accurate charge-up to the specified voltage. Like a bridge feeds the reservoirs, but in a better way.
So, given that a good decoupling cap is used, can we say that this cap can be responsible for the perceivable performance? To put it another way, what if no uF's are used after the regulator? This is also what ChrisA has actually proposed.
Now I think the following: suppose that between the regulator and load is a feet of cable. Take a Belden type 1883A, it shall have approximately 25 milliohms of resistance, 180nH of inductance and 40pF of capacitance. And then comes the load, using a 220uF good electrolytic bypassed with a 100nF film cap. Assume the load is Class A operated, 20mA DC max.
Doesn't all this sound adequate? Even at 200mA the voltage regulation until we reach the decoupling cap will be 0.01 V at worse. I have not calculated whether this LC circuit shall be critically dumped though.
If the above is not ideal (it isn't) how much can it degrade the regulator's performance? And in what way other than voltage sag?
So, given that a good decoupling cap is used, can we say that this cap can be responsible for the perceivable performance? To put it another way, what if no uF's are used after the regulator? This is also what ChrisA has actually proposed.
Now I think the following: suppose that between the regulator and load is a feet of cable. Take a Belden type 1883A, it shall have approximately 25 milliohms of resistance, 180nH of inductance and 40pF of capacitance. And then comes the load, using a 220uF good electrolytic bypassed with a 100nF film cap. Assume the load is Class A operated, 20mA DC max.
Doesn't all this sound adequate? Even at 200mA the voltage regulation until we reach the decoupling cap will be 0.01 V at worse. I have not calculated whether this LC circuit shall be critically dumped though.
If the above is not ideal (it isn't) how much can it degrade the regulator's performance? And in what way other than voltage sag?
I have seen people literally solder the decoupling cap to the power lead of the IC. This is where surface mount components have an advantage, no lead length.
You think you are dealing only with audio frequencies but radio from lights, computers and what not finds it's way into everything.
Actually in real life, it is not so hard, but getting that last 0.1% of performance is hard. Most people just live with being 99.9% perfect.
You are thinking only about average DC current and voltage. Audio is not DC. Think of the signal in your power cable.
Think also about the 0.01v drop in the return or "ground" conductor. Worst case is if the ground move 0.01v away from ground, that is a 10millivolt signal imposed over your music.
This is why ground loops are bad. ANY current at all in a grounded conductor means there is a voltage difference between the ends of the ground wire.
Actually I don't think just about DC. Read again what I say, I will explain my thoughts again. 🙂
According to the view that gootee proposed (and I think it is correct), a decoupling cap supplies the "audio" current to the load. So it is just fed by the power supply to replenish its charge. This is what gootee said in the first place, and I think it is correct.
Mind the graphs he kindly uploaded - have a look at diode currents and reservoir currents. The audio currents are supplied by the reservoirs to the load, and the bridge supplies the lost charge. The changing audio current does not appear on the diode current.
So my view is the following: if we say that the decoupling cap acts like a the reservoir for my load (the effects pedals) and the regulator + cable as the power supply that feeds that cap with charge, like the way the transformer + bridge did in the graphs, then I assume that we should have the same behaviour. Of course, on condition that the decoupling configuration is able to handle the load much as the reservoir configuration does.
At least this is what I have understood. If there is something wrong in my words, please help me realise it! 🙂
According to the view that gootee proposed (and I think it is correct), a decoupling cap supplies the "audio" current to the load. So it is just fed by the power supply to replenish its charge. This is what gootee said in the first place, and I think it is correct.
Mind the graphs he kindly uploaded - have a look at diode currents and reservoir currents. The audio currents are supplied by the reservoirs to the load, and the bridge supplies the lost charge. The changing audio current does not appear on the diode current.
So my view is the following: if we say that the decoupling cap acts like a the reservoir for my load (the effects pedals) and the regulator + cable as the power supply that feeds that cap with charge, like the way the transformer + bridge did in the graphs, then I assume that we should have the same behaviour. Of course, on condition that the decoupling configuration is able to handle the load much as the reservoir configuration does.
At least this is what I have understood. If there is something wrong in my words, please help me realise it! 🙂
Yeah, it should be perfectly adequate. And the decoupling caps will dominate the performance, at least down to a frequency determined by their impedance (size, and distance from load, relative to load current demands). I would add as much as I could fit. (Note too that using more smaller electrolytics in parallel would give lower impedance than one larger one.)
All that matters is the impedance that the power and ground pins of the load see. Nothing before that point matters, except as it affects that. That's why you want both larger and smaller paralleled caps, as close to those pins as you can get them. They will better-cover the frequencies of interest.
It's really just Ohm's Law. The impedance seen by the device is literally the amount of voltage sag that will occur per unit of current change that the device needs to draw.
The impedance will vary with frequency. But at any not very low frequencies, where it might matter, the current will come from the decoupling caps, if they are sufficiently configured. So the inductance and resistance will be very small (in the low tens of nanoHenries and a milliohm or two) because the caps will be very close to the point of load.
You can make the impedance as low as you want, almost, by using more capacitance, with the lower limit probably limited mostly by the ESRs of the caps and the lengths of the connections and the lead-spacing.
I would not usually use zero uF after the regulator, unless the regulator output was less than 5 mm from the point of load. Even then, I think that many regulators require a small cap from their output to their sense or ground pin. Check the datasheet. By the way, don't leave 10 mm of conductor in that path. Mount the cap(s) right at both pins, if possible. Usually, three-terminal regulators like at least a small cap, e.g. 0.1 uF, from both their input pin and their output pin to their sense or ground pin. Leave no excess conductor length on either side of those caps. With most three-terminal regulators I would still also want an electrolytic after the output. Note that many of them also require greater than some value of ESR, for damping, to keep them stable.
You can keep most of the external nasty stuff out of the cables by using twisted pair, or shielded twisted pair. If there is more than one subsystem or circuit, I would use multiple pairs, because not sharing power or ground conductors between circuits that don't have to would be a good idea. And multi-pair shielded networking cable is relatively cheap. Don't use the shield's drain wire for your circuit ground. It should only connect to the chassis, and probably at one end only.
All that matters is the impedance that the power and ground pins of the load see. Nothing before that point matters, except as it affects that. That's why you want both larger and smaller paralleled caps, as close to those pins as you can get them. They will better-cover the frequencies of interest.
It's really just Ohm's Law. The impedance seen by the device is literally the amount of voltage sag that will occur per unit of current change that the device needs to draw.
The impedance will vary with frequency. But at any not very low frequencies, where it might matter, the current will come from the decoupling caps, if they are sufficiently configured. So the inductance and resistance will be very small (in the low tens of nanoHenries and a milliohm or two) because the caps will be very close to the point of load.
You can make the impedance as low as you want, almost, by using more capacitance, with the lower limit probably limited mostly by the ESRs of the caps and the lengths of the connections and the lead-spacing.
I would not usually use zero uF after the regulator, unless the regulator output was less than 5 mm from the point of load. Even then, I think that many regulators require a small cap from their output to their sense or ground pin. Check the datasheet. By the way, don't leave 10 mm of conductor in that path. Mount the cap(s) right at both pins, if possible. Usually, three-terminal regulators like at least a small cap, e.g. 0.1 uF, from both their input pin and their output pin to their sense or ground pin. Leave no excess conductor length on either side of those caps. With most three-terminal regulators I would still also want an electrolytic after the output. Note that many of them also require greater than some value of ESR, for damping, to keep them stable.
You can keep most of the external nasty stuff out of the cables by using twisted pair, or shielded twisted pair. If there is more than one subsystem or circuit, I would use multiple pairs, because not sharing power or ground conductors between circuits that don't have to would be a good idea. And multi-pair shielded networking cable is relatively cheap. Don't use the shield's drain wire for your circuit ground. It should only connect to the chassis, and probably at one end only.
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Concur! It's always nice to see someone else harping on the same things I like to harp on. <grin>
I believe that RF incursion is a bigger problem than many DIYers think, if they think about it at all.
Another thing that strikes me: in the case of bridge/reservoir, the current pulses through the diodes that charge the reservoirs cannot mimic the signal, since they have to pass an equal (roughly) amount of energy into the cap during the time the diodes are on.
But if a regulator feeds the decoupling cap, assuming it has the capability of great DC stability, it can always feed the decoupling cap with energy. Not like before, that we had the conduction time for that purpose. So, is there a possibility that the charging currents coming from the regulator can mimic the audio currents drawn by the load? Or is this just the case for high frequencies, where the inductance will smooth any current into the decoupling cap?
Just not sure about that.
But if a regulator feeds the decoupling cap, assuming it has the capability of great DC stability, it can always feed the decoupling cap with energy. Not like before, that we had the conduction time for that purpose. So, is there a possibility that the charging currents coming from the regulator can mimic the audio currents drawn by the load? Or is this just the case for high frequencies, where the inductance will smooth any current into the decoupling cap?
Just not sure about that.
You have to model the caps as filters. If you do it is easy to understand.
The "signal" comes from the final audio power transistor. OK in a guitar pedal if might only be 100mW of power. But still that is the signal and it flows to the regulator. But between the two there is a low pass filter. What the regulator "sees" depends on the filter.
Yes, the regulator can do that, and might have to. Some are better than others. If you used only a tiny high-frequency decoupling cap, then the regulator would have to supply almost all of the music current. The current would have to precisely mimic the music waveform (since it IS the music waveform!), as the output transistor modulated its channel resistance according to the music waveform from the input, letting the current flow down from rail to ground (also through the load, on its way, of course). If there's no cap there, to supply the current, then the regulator would have to try to do it. I've never tried it that way but I'm guessing that most of them could follow an audio-frequency-range signal quite well. I have seen people use regulators instead of chip amps, to make small power amplifiers.
Now that I think about it, a regulator that was very close to a larger decoupling cap would probably provide the current whenever it was capable of it, instead of the cap, because it's doing just what a cap does, i.e. if the voltage falls even a tiny amount, it provides more current (as it tries to raise the voltage back up). A cap also can't release more current until the voltage falls by a tiny amount, at least (but it doesn't raise the voltage back up, like a regulator). But in both cases, the slight drop in voltage that happens when the output transistor "opens its valve a little wider" would trigger the release of more current. So whichever one was the lower impedance would "win", and supply more of the current.
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