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Old 2nd January 2013, 01:02 AM   #21
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Quote:
Originally Posted by sreten View Post
Your last point applies to nearly everything. There is nothing in the
standards to prevent silly levels of very low frequencies down to DC.
Got it, thanks.
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Old 2nd January 2013, 01:07 AM   #22
sreten is offline sreten  United Kingdom
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Quote:
Originally Posted by sreten View Post
Hi,

Your last point applies to nearly everything. There is nothing in the
standards to prevent silly levels of very low frequencies down to DC.

rgds, sreten.
Hi,

Just to add, AV systems rely on effective subsonic filtering of the sub
channel to its frequency limits, allowing encoding of very low frequencies
in the 0.1 channel, to be then properly managed by the AV system.

Music CD's are not AV, don't have that luxury, and consequently
do not have anything like the same v.l.f. content as AV in general.

rgds, sreten.
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Old 2nd January 2013, 01:41 AM   #23
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Originally Posted by realflow100 View Post
If you see your speaker cone moving around 20hz 10hz or less. you got a good subwoofer and amplifier..
I have a soundcard modified to be DC coupled, the small signal opamp output will drive a small 3" speaker with a 1hz sine tone, with movement I can feel with my fingers. There is certain audible output but does not produce a discernible tone. This is a case when audio efficiency is so low that the distortion dominates over any meaningful signal, you will see it, but not necessarily hear it, even if you had an adequate HIFI setup. You won't get anywhere close to actually producing a meaningful 1hz sound unless you have built your house as part of the speaker system.
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Old 2nd January 2013, 04:56 AM   #24
gootee is offline gootee  United States
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Quote:
Originally Posted by realflow100 View Post
it wouldn't have as much power would it though right? and wouldn't that introduce ultrasonic pulses into the audio signal instead of normally just using a capacitor to block DC?
Oh yikes I just tried putting 19 volts to the amplifier chip for a few seconds
And Omg the 10 watt speaker bottomed out from a 5 watt IC chip amplifier
And there was no amplifier distortion.... only the bottoming out of the 10 watt speaker
Yikes I just got over 10 watts out of a 5 watt amplifier chip O.O that's a little scary..
But I hooked it back up to the 12 volt power supply instead so I think the chip is okay... It didn't get warm when I unplugged the power cord and checked the chip..
Maybe the massive heatsink helped save the chip?
[Edit: Wrote this much earlier but walked away without hitting "Post Reply".]

I would also be worried about possibly killing the speaker, in that situation.

If everything still works, count yourself lucky. You might have degraded the performance of the chipamp, somewhat, or shortened its lifespan. But maybe not.

Your high-pass input filter is probably not restricting "ultrasonic pulses" from getting into the amp. I don't know why there would be any ultrasonic signal at your input. But there WILL be RF (Radio Frequency) junk, there. So you should, ideally, also use a low-pass filter, RIGHT before the input pin.

Unless you have a dual-polarity supply and the chip has both plus and minus input pins, you should be able to just insert a series resistor, very close to the input pin, and insert a capacitor to ground, connected between the resistor and the input pin.

You'll want the cutoff frequency of that low-pass RC filter to be around 350 kHz, or maybe lower. You could try a 1K series resistor, which would mean a capacitor to ground of 1 / (2π x 350000 x 1000) = 450 pF. You could try two 220 pF caps in parallel, for that, or use a 470 pF cap, or just drop it back to 330 pF, which would make the cutoff frequency 1 / (2π x .00000000033 x 1000) = 482 kHz. That's getting into the AM radio band so it might be a little too high. A 470 pF cap would make it 339 kHz, which should be fine. Or you could even go lower, with a slightly larger cap.

Switching gears somewhat:

If your power leads are longer than an inch or so, you should have some capacitance from RIGHT AT the power pin to ground. I would rcommend a 0.1 uF X7R ceramic cap in parallel with an electrolytic of at least 10 uF. I would also try 100 uF or 220 uF or more, there, to see if you can hear a difference.

I calculated that 7.9 uF wold be the minimum, at the power pin, if you wanted to be able to go from 0 Watts to 5 Watts (into 8 Ohms) in one microsecond, while disturbing the power supply voltage by no more than 0.1 Volt. But your amp might very well be able to do it faster than that, and its feedback loop might need to be able to respond to frequencies of up to a few hundred kHz. So larger might be better. Also, that figure doesn't account for your bass-driving needs. So the bass response might also benefit from a much larger capacitor between the power pin and ground, especially if your power wiring is very long, or there is much inductance anywhere upstream from the power pin.

You should also tightly twist together the signal and signal ground reference input wiring, ALL the way from end to end. And do the same for your positive and negative DC power wiring, and your speaker wiring. That will help keep RF and other interference out. Everything is an input, for RF. And it might cause subtle degradation of the chip's performance (or it might sometimes cause not-so-subtle problems).

Also, if you run a separate wire all the way back to the power source's ground or negative output, for the grounds that are from components that go to the input pin, it might help the sound quality significantly. i.e. Don't run them together with the speaker ground, or the ground for the caps at the power pin.
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Old 2nd January 2013, 05:32 AM   #25
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Originally Posted by gootee View Post
I calculated that 7.9 uF wold be the minimum, at the power pin, if you wanted to be able to go from 0 Watts to 5 Watts (into 8 Ohms) in one microsecond, while disturbing the power supply voltage by no more than 0.1 Volt. But your amp might very well be able to do it faster than that
I would be very interested to know how you performed that calculation, if you'd care to explain it.

Also, under what circumstances would an amp ever shift from zero volts to the rail voltage in one microsecond (assuming that is the case you're considering)? Wouldn't it take something like a 250kHz sine wave to do that?

EDIT: also, how exactly does a larger power supply capacitance improve bass response? To my knowledge all the capacitance does is decrease ripple voltage, for a given current. Is it just that heavy bass draws higher currents for longer (due to the longer lasting peaks of the waveform) and hence causes more ripple? And if the lowest points of the rippling rail voltage are low enough your maximum output power drops?

Last edited by lmaurits; 2nd January 2013 at 05:53 AM.
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Old 2nd January 2013, 05:59 AM   #26
gootee is offline gootee  United States
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I noticed some duscussion about whether or not sub-1-Hz energy couold exist in a recording. I noticed some very low frequency activity in a song that I was using to test a DC servo circuit.

This was while I was using a WAV file of a portion of a song (less than one minute of music, if I recall correctly), as an input to an LT-Spice simulation of an amplifier with a DC Servo.

It was a bit of an eye-opener, for me, because I had always reasoned that every music signal's average would only stray from zero for as long as a half-cycle of the frequency of the lowest note that was played.

Apparently I was completely wrong about that. And it made my servo's output (before summing with the signal path) wander around, as it "corrected" the average to zero. (I think that I realized then that the servo's time-constant needed to be much longer.)

Anyway, attached is a screen image that I just made, from LT-Spice. It shows the amplifier's INPUT signal, from a WAV file containing the beginning of an AC/DC song called "Highway to Hell". I chose that song for the servo test precisely because it is very "bursty".

As can be seen from the time-domain plot in the attached image, the song starts with drum strikes only, for about 9.5 seconds, and then short electric guitar "chord-bursts" are added, until a little after 18 seconds, when vocals are added (if I am recalling it correctly, just from looking at the plots).

The DC offset appears to be near zero, judging by the level during the gaps between drum strikes. But there is significant asymmetry evident. For example, examine the plot between 10 and 12 seconds. This is the input voltage so assuming an amplifier gain of 20 there could easily be output excursions of more than 1.5 volts away from zero, during the passage shown in the plot, at rates of roughly 0.5 Hz.

There is also some tabulated dynamic DC offset data that I posted during the original servo simulation experimentation that was performed and reported here back in 2007, which can be seen at:

DC Servo question...

That post at that link also includes a screen image with a much longer duration plot of the same song (60 seconds of it).

---

There is another possibility, though: It could be a form of "motorboating" that is causing the OP's speaker to deflect at a slow rate. I have not read all of the details of the OP's configuration from this thread, yet (my apologies). But, perhaps there are mismatched RC time-constants in the system. I can't remember the details of the mechanism, right now, but perhaps something like having a lower RC time-constant (faster response) for something upstream (e.g. input filter or feedback loop), compared to the RC time constant of something downstream (e.g. feedback loop or output filter), might cause a low-frequency instability. I believe that the response speed of subcircuits should never decrease too drastically, as the signal makes its way from input to output.

Cheers,

Tom
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Old 2nd January 2013, 08:30 AM   #27
gootee is offline gootee  United States
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Quote:
Originally Posted by lmaurits View Post
I would be very interested to know how you performed that calculation, if you'd care to explain it.

Also, under what circumstances would an amp ever shift from zero volts to the rail voltage in one microsecond (assuming that is the case you're considering)? Wouldn't it take something like a 250kHz sine wave to do that?

EDIT: also, how exactly does a larger power supply capacitance improve bass response? To my knowledge all the capacitance does is decrease ripple voltage, for a given current. Is it just that heavy bass draws higher currents for longer (due to the longer lasting peaks of the waveform) and hence causes more ripple? And if the lowest points of the rippling rail voltage are low enough your maximum output power drops?
Hi Imaurits,

[I do not consider myself to be an expert. But I am slowly re-learning a lot of things.]

Sure. A while back, I was wondering how to choose the values for decoupling caps, and ended up opening a fairly-large but quite-appetizing "can of worms". <grin>

First, you have to understand that the signal that we hear, from an audio power amplifier, is, EXACTLY, the CURRENT from the power supply reservoir capacitors and the decoupling capacitors. Almost all of that current comes from either the PSU's reservoir caps or the power output devics' decoupling caps. (A small fraction comes directly from the rectifiers, but only when a charging pulse is occurring.)

The proof that makes that completely obvious is in an image that I posted:

Power Supply Resevoir Size

where you can see that the amplifier output signal is coming directly from the PSU reservoir caps (there are no decoupling caps in that example).

So that means that the power supply VOLTAGE is relatively uninteresting. After all, it's not even supposed to change! And the rectifier-induced ripple voltage is easy to make as small as we want, and is only important when an amp's PSRR (power supply rejection ratio) is poor, or craps out at higher frequencies, OR, if we ask too much of the supply and drag its voltage too low. The CURRENT is where the action is. [The _REAL_ "signal path", the one that we actually HEAR, is current that goes straight from the PSU and decoupling caps through the power output devices and directly to the speakers.]

Second, let me explain that while the terms "bypass capacitor" and "decoupling capacitor" are often used interchangeably, I look at them in one particular way, which might also motivate you to appreciate the importance of, especially, decoupling capacitors:

Bypass caps are usually small-size (and usually also small-value) caps that are placed and connected very near to the power supply input for an active device, in order to try to prevent high-frequency instability, because most transistor-based amplifiers have a hidden positive feedback path for high-frequencies, through the power rail. So the bypass capacitance is there to try to short-circuit any HF to ground, to prevent instability (e.g. oscillation).

Decoupling caps, which are much more interesting, to me, are placed at the power supply inputs for active devices so that when a device suddenly demands (or, actually, "allows") a fast change in its DC input current (i.e. a "transient"), that fast-changing current doesn't have to try to come through the self-inductance (and resistance) of the power rail conductors.

The decoupling cap acts as a small point-of-load power supply, with very low impedance because it's only millimeters away, which means that there isn't a lot of inductance or resistance due to the lengths of the power and ground conductors.

The decoupling cap accomplishes at least three very important tasks:

1) [Voltage Spikes:] If the fast transient currents had to come all the way from a more-distant capacitor, e.g. from the power supply reservoir caps, then a large voltage spike would be induced on the power and ground rail conductors, mainly because of their inductance, because, for an inductor, V = L di/dt, which points out that the voltage developed across the rail conductor's inductance is proportional to the time-rate-of-change of the current.

So the voltage spikes can be quite large, even if the current is tiny, as long as the current's amplitude is changing rapidly. The name "decoupling cap" comes from the fact that they are used, in part, to try to "decouple" a local subcircuit's or device's potential for creating power rail voltage disturbances from the rest of the power distribution system.

2) [Transient Distortions:] a) If the fast transient currents had to come all the way from a more-distant capacitor, e.g. from the power supply reservoir caps, then they would often not be rendered accurately-enough, as reflected in the devices output current or voltage.

The inductance of the power rail conductor will always cause a delay in the arrival of the current and a change in the shape of the arriving current's amplitude vs time waveform, because it will delay different frequency components by differing lengths of time.

When an analog signal is being reproduced by the power output stage of an audio amplifier, for example, trying to pull transient currents through too much conductor length results in transient distortion. The audio waveforms' rise-times are altered. Basically, the phase angles and amplitudes of the many Fourier frequency components, that are what gives the correct shape to anything more-complex than a single-tone sine wave, would all be altered in frequency-dependent ways.

That would be "a BAD thing". For something like a square wave, it would result in distortion of the rising and falling edges, and the appearance of what looks like ringing (decaying-amplitude oscillations) just after each edge. But, in general, it could alter or blur the timing of everything in the sound waveform, possibly harming the information that our brains use to perceive the soundstage image.

b) There is also an important "need for speed" because of the feedback loop that is usually used for amplifiers. The term "harmonic distortion" implies the existence of harmonics. The feedback system has to try to react to, and cancel, any harmonics, to prevent them from going through to the output and becoming part of a higher level of harmonic distortion. Harmonics are undesired multiples of some desired frequency. That means they are at higher frequencies. In a fast-responding active device, it is necessary for the feedback system to be able to deal with distortion signals that contain frequencies up to the limit of the device's ability to respond. Chipamps are usually very slow, compared to most discrete transistor circuits. But an LM3886 chipamp, for example, has a typical maximum slew rate spec of 19 Volts per microsecond! If we use Henry W. Ott's equation for the maximum frequency content in the rising edge of a pulse waveform, f = 1 / (Pi x trise), then a 1 microsecond edge is equivalent to 318.31 kHz. But if it can slew 19 Volts in 1 us, maybe it can slew 1 Volt in 1/19th of a us.

At any rate, in order to not leave a bottleneck that degrades a device's ability to perform at its rated specifications, we must use decoupling capacitors with low-inductance connections that can support the supply of high-frequency and rfast-changing-transient currents. (How low the inductance needs to be, and thus how much LENGTH can be tolerated in the decoupling cap round-trip connections, can also be calculated. See farther below.)

3) [Hum and Interference:] Faraday's Law tells us that a time-varying electromagnetic field will cause a corresponding time-varying current to flow in any conductive loop in its presence, and, all else being equal, the current's amplitude will be proportional to the geometric area enclosed by the loop. It also tells us that a time-varying current in a conductive loop will cause a time-varying electromagnetic field in the air, with amplitude proportional to the geometric area enclosed by the loop.

That means that "loops are bad, and big loops are worse", because they can act as either receiving antennas, or transmitting antennas, or both. Unfortunately, current ONLY flows in loops. So we need to always minimize their enclosed geometric area.

That's why I constantly harp on the benefits of tightly twisting together the input signal and signal ground wire pairs, and the AC input and transformer secondary wire pairs. Any space left between the AC wires makes them a tansmitting antenna for hum. And any space left between the input conductors makes them a hum-receiving antenna that's connected to a high-gain high-power amplifier! (Yikes!) (Receiving RF is als oa big concern. Also, all other conductor pairs should get the same treatment. On PCBs, close ground planes under or over input circuitry are your friend.)

Decoupling capacitors are meant to be connected as physically close as possible to an active device's DC power input, and connect to (typically) the device's load ground, via the shortest possible path. That results in the least inductance that the transient currents have to contend with.

Decoupling capacitors are a way to use the (geometrically) smallest-possible loop, for the transient currents that need to flow. That, in turn, creates the smallest-possible radiated EM field, and therefore the least amount of interference with other loops in the circuitry (or in outside systems). It also allows the least amount of interference to be induced in the decoupling connections.

As our very-astute member Terry Given has suggested should be a mantra: "Current flows in loops. Minimize them."

Quote:
Also, under what circumstances would an amp ever shift from zero volts to the rail voltage in one microsecond (assuming that is the case you're considering)? Wouldn't it take something like a 250kHz sine wave to do that?
I think I covered that in 2b, above. Basically, if an amplifier is capable of a certain response speed, then we need to make sure that it can respond that fast, ACCURATELY, because it's going to be trying to cancel high-frequency distortions either way. And it can't do it accurately-enough if it has to try to pull the required currents through too much inductance (or, in general, impedance, since conductors also have resistance).

Quote:
EDIT: also, how exactly does a larger power supply capacitance improve bass response? To my knowledge all the capacitance does is decrease ripple voltage, for a given current. Is it just that heavy bass draws higher currents for longer (due to the longer lasting peaks of the waveform) and hence causes more ripple? And if the lowest points of the rippling rail voltage are low enough your maximum output power drops?
I think that you are correctly thinking about at least the main gist of that. If we used a total capacitance that was below some required amount of reservoir capacitance, the bass tones (maybe especially those that have cycle periods that last as long or longer than the time between rectifier charging pulses) could make the capacitor voltage, and thus the power rail voltage, drop by too much, which could make the minima of what is then "super ripple" bite into the voltage range that is occupied by the amplifier itself, which is usually a 3 or 4 Volt range (Vceo plus whatever emitter resistor voltage) that sits above the maximum signal peak level, and (hopefully) below the ripple voltage minimum. It's almost exactly like a regulator's dropout voltage. If the rail's ripple dips into that area, it gouges ripple-shaped chunks out of the output voltage waveform!

Another aspect of bass vs reservoir capacitance size is that whenever bass drags down the rail voltage, the linearity of the amplifier itself is degraded, because no amplifier has perfect PSRR. Think of the power output transistors as controllable current valves. Transistors are controllable resistances. If the small-signal control input (current to BJT base or voltage for FET gate) changes by a certain amount, then the device's channel resistance changes by a certain amount. If the rail voltage is always exactly the same unchanging voltage, then small-signal changes will always produce a perfectly-proportional change in the current that the device's channel resistance allows to whoosh down from the power rail and into the speaker. But if the rail voltage changes, then the SAME small-signal change would allow a DIFFERENT amount of current to flow. That's the DEFINITION of non-linearity! Luckily, the built-in PSRR of the amplifier counteracts the changes in rail voltage, for the most part. But not completely. That's one main reason why less ripple is usually better, and is also why having more capacitance so bass doesn't change the rail voltage as much is usually better.

This post is getting so long that I'll post what I have so far and will get to the calculation of decoupling capacitance values and their allowable maximum inductance (connection length!) in another post.

Sorry to have blathered-on for so long, about all of that!

Cheers,

Tom

Last edited by gootee; 2nd January 2013 at 08:42 AM.
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Old 2nd January 2013, 09:22 AM   #28
gootee is offline gootee  United States
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I still haven't ever been able to go back and finish the development of the decoupling cap calculations. I will give links to most of what I did do so far, below. The basis for the calculations came mainly from writings by Henry W. Ott and Bruce Archambeault. But their stuff was all concerned with high-speed digital circuits. So I decided to try to apply and extend it for audio circuits.

The first basic equation is a simple-enough one, arrived at by re-arranging the ideal capacitor's differential equation. Start with

i = C dv/dt (the basic capacitor equation)

Rearranging and slightly modifying that equation gives:

C = Δi Δt / Δv

The first trick is to realize that WE get to pick Δv, which is the amplitude of the rail voltage disturbance that we are willing and able to tolerate, in our circuit.

Δi is a change in current that is allowed/demanded by the active device being decoupled.

Δt is the time duration of the change in curent.

I have so far mostly only looked at finding the C required for the minimum Δt and maximum Δi, with small choices of Δv. That approach gives a minimum decoupling capacitance needed in order to supply a particular maximum Δi in a specified minumum time Δt. (Note that for larger Δt, the required capacitance gets larger, as does the maximum tolerable inductance (connection length). This should be a clue for calculating the C needed for accurate lower-frequency, e.g. bass, response.)

The minimum Δt can also help us calculate the maximum tolerable inductance in the decoupling network:

V = L di/dt (the standard ideal inductor equation) gives

L = Δv Δt / Δi

For large Δi and small Δv and Δt, L will tend to be small, giving the maximum tolerable inductance in the decoupling cap and its connections that will still enable responding fast-enough, with enough current.

From the inductance, we can calculate or estimate the maximum total round-trip connection length that can be used for decoupling (cap inductance is usually only the equivalent self-inductance of a conductor with length equal to the cap's lead-length).

A rule-of-thumb estimate for a typical conductor's self-inductance is 1 nH per mm.

It turns out (for the few examples I have worked out) that even for audio power amplifiers, e.g. a chipamp, it is usually not quite possible to fit one capacitor close-enough to the power pin to satisfy both the capacitance and maximum tolerable connection length requirements. So we will probably usually end up wanting to use several smaller caps in parallel, with separate parallel connections for each, in order to divide the inductance and resistance of the connection by the number of caps used in parallel. Smaller caps should also be easier to mount in closer proximity to the device pins, producing even more benefit.

Below are links to most of the posts where I tried to perform detailed example calculations. Also included is a very simple method for converting the requirements into the calculation of the maximum impedance that would need to be seen by the device pins, from DC up to a maximum frequency, which is also calculated at one of the links.

paralleling film caps with electrolytic caps

paralleling film caps with electrolytic caps

paralleling film caps with electrolytic caps

+/-30vDC @ 10A PSU

(Note, too, that at some of those links, I was using too low a value for a typical conductor's self-inductance, instead of using the accepted value of 1 nH per mm. There are also on-line inductance calculators, if you need better accuracy for a particular situation.)

Cheers,

Tom

Last edited by gootee; 2nd January 2013 at 09:32 AM.
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Old 2nd January 2013, 01:38 PM   #29
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Originally Posted by realflow100 View Post
is it possible to get 1 watt rms or even 5 watts rms?from a 5 watt chip amp?
how much capacitance should I need for the output? 2200uF? 10,000uF? I want the amp to nail the low frequencies down to 0.5hz as best as it can without distorting..
BTW, I LOVE old TV's for a source of electronic parts and also coil wire for making transformers.

The best setup IMO for any chipamp is BTL (Bridged Tied Load) Not only is it easier to use, it's MUCH louder than a regular SE amp chip on a lower voltage. (only 9-12V makes it very loud)

1st, you do NOT need a capacitor for the speaker, because a bridged amp connects both the + and - each to an output, which effectively cancels out the DC. The only part that needs a capacitor for the signal is the input leads.

Here's an example. I have removed several AN7522N stereo chips from several old stereo TV's.
http://www.lemona.lt/LIUSE/Pdf/AN7522N.pdf
Click the image to open in full size.

You directly connect the speaker to the outputs, and use 1uf plastic capacitors for the input. Even at only 1uf, I get amazing low frequency response. If you are wanting lower bass frequencies, then parallel a 10 or 22uf cap across the 1uf plastic cap.
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Old 2nd January 2013, 09:54 PM   #30
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what's the maximum input capacitor that I could use for the little IC chip amp that I made?
Is 4.7uF already too much? Because it seems to work just fine for me... And the sound quality is spot on..
And the input capacitors maximum voltage rating. Should I use a higher voltage input capacitor? like 50 volt or more? or is 25 just fine?
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