Hmmmm...that's an interesting idea...let's see how it would work for a typical preamp/power amp. Most preamps draw less than 100ma. So, for "ultra low ripple", that would be a 1,000uF capacitor. Seems quite low; perhaps that's because most preamp supplies are regulated, and that lowers the requirement??? Most commercial preamps use at least a 2,200uF cap. A 200 watt power amp driving an 8 ohm load would need 5 amps of current; that would a 50,000uF cap. HUGE!! Maybe so big because it's usually UNregulated?? What if it's a regulated high-power supply? Does that lower the requirement?
It depends on the voltage.
I simulated it with 1A CCS when I was building a headphone amp.
if you have a 12V supply peak with 1kuF cap and load it with 1A CCS, you will see 5.9V ripple.
With 10kuF you will see 0,76V ripple.
In bigger amplifiers nobody cares if you have 5-6V ripple as the system have more than enough till clipping as the rail voltage is 10-12V higher than it is needed.
But if you want low ripple, you need a lot of caps.
If you are looking at ripple, then the guide is all that is needed.
If you are looking at signal after a regulator, then you need to understand your regulator and it's characteristics.
If you are looking at signal after a regulator, then you need to understand your regulator and it's characteristics.
The formula comes from the most basic definition of 1 Coulomb. 1Amp for 1s.
1s is 0.01s because of the 100Hz (full rectified 50Hz). So you divide the 1F cap with 100 and you have what I sad: 10kuF for 1A gives 1V ripple.
The linked calculation uses the 0,7 factor. And it actually gets closer to what I also simulated: 0,76V ripple.
So if you can remember the 1Coulomb definition, then you can always get to the equation rather fast.
But these are mentioned at the link. It is good to know why you do things and not just say it works and others do it also like that so it is good for me. Because then you don't really know what you are doing.
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"So you divide the 1F cap with 100 and you have what I sad: 10kuF for 1A gives 1V ripple. The linked calculation uses the 0,7 factor."
Huh?? A 10,000uF cap is .01F. NOT 1F. Ripple is then 0.7v for 1 Amp. NOT 1v. And for a power amp needing 5 Amps (200watts into 8 ohms), this becomes 3.5volts ripple. I guess this is regardless of voltage? 3.5 volts ripple on a 30 volt supply would be ALOT, methinks.
Huh?? A 10,000uF cap is .01F. NOT 1F. Ripple is then 0.7v for 1 Amp. NOT 1v. And for a power amp needing 5 Amps (200watts into 8 ohms), this becomes 3.5volts ripple. I guess this is regardless of voltage? 3.5 volts ripple on a 30 volt supply would be ALOT, methinks.
No, a 200W amplifier does not draw 5A from the smoothing capacitor.
The output is a sinewave current peaking at 7.07aApk with an AC value of 5Arms
This current comes from TWO supply rails, not one.
The supply rail currents are just a bit above 7.071Apk when the peak flow occurs for one half waveform. But both rail currents drop to roughly the quiescent current as the output crosses over the 0A level. The supply rail currents are on alternate half waves and lying virtually idle when the other rail is supplying.
The average current from each rail is roughly 1.8Aavg (2.5Arms) due to the 50% duty cycle.
Now apply the ultra low ripple guidance of 10mF/A and I would get 18mF on each supply rail.
I just happen to use ±20mF for my 8ohms rated Power Amplifiers and recommend this to Members, if they want their amplifiers to pass all of the 20Hz to 20kHz audio band..
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You are not even trying to understand what I wrote and then write silly things.
I never wrote 10kuF is 1F. I wrote "divide the 1F cap with 100" so 1F/100 is 10kuF. Ripple is not 0.7V either. It does differ from the peak voltage and voltage drop as we are talking about shine wave.
My simulations:
12V 10kuF 1A CCS: 0,76V ripple
24V 10kuF 1A CCS: 0,8V ripple
80V 10kuF 1A CCS: 0,86V ripple
So the higher the voltage, the closer you get to 1V.
Where do you get that 30V rail? For 200W? You know you need 40V rails for 100W? And a normal amp would have at leas 50V rails, exatly to have enough headroom, so that 3,5V drop wouldn't matter.
My yamaha 40W amp has 46V rails.
AndrewT tried to explain the rest. Try to understand why what happens and how things are calculated...
Yes, I dig what you're rappin'!! You are correct, and that result makes a lot more sense to me. Thanks for your input.
Well, perhaps it is in semantics--I still don't understand WHICH "1F cap" you are referring to.
The ripple WOULD be 0.7v following the reference formula; but as AndrewT has pointed out, the needed current from each rail would NOT be 5 amps as I had assumed, but 1.8 amps ; thus giving a ripple of 1.26v( Δ V=0.7 I / C F ) for the 10,000uF cap, and half that (0.63v) for AndrewT's recommendation of 20,000uF. THAT makes sense to me!!
I don't need 40 volt rails to produce 200 watts into 8 ohms, as my amplifier is run in bridge mode; I can attain that (just barely) with 30 volt rails.
you're away again.
200W into 8ohms from a bridged amplifier requires two amplifiers, each capable of 100W into 4ohms.
100W into 4ohms is equivalent to 28.3Vpk and 7.07Apk
You can't get that from supply rails
200W into 8ohms from a bridged amplifier requires two amplifiers, each capable of 100W into 4ohms.
100W into 4ohms is equivalent to 28.3Vpk and 7.07Apk
You can't get that from supply rails
And the two amplifier would have a total peak draw of >7.07Apk from BOTH supply rails at the same time.
I mentioned the Coulomb law. That is where I can quickly get back to how much cap is needed. 1A, 1V, 1F over 1s. If I totally simplify it. And as mentioned as time is 0,01s, I only need 1F/100= 10kuF. The equations are still there. That you linked. Check those.
If you need to produce 5A how on earth would you only need to put in only 1.8A? Ok with transformer, but there is none after the main trafo, so you can only require more amp.
I use a wholly different way to figure the power supply reservoir capacitance. First I play with my capacitor collection series to example expected speaker to find out how much, no more, no less, would be required if I had used an output cap for the amp. Double that figure works perfectly for the power supply reservoir figure, per each rail, of a split rail supply.
I've suggested only checking the needs of the load; however, some cases may have needs beyond that.
But, if it was going to be easy...
Average 1 cu-ft ish speaker averages 15000u per each rail of a split rail power supply with tolerable variance plus or minus 15%.
So, 10000u per each rail and a stereo pair of 16 ohm 0.75 cu-ft speakers looks feasible as well.
Otherwise, that 10000u idea, albeit an excellent try, falls short, EXCEPT!!!! The level worst thing to do is increase the load on an overly small transformer by adding too much capacitance, forming a delay timer that sounds even worse than deep sticky clay mud feels on your boots. Problematically, a linear supply is only clean when the caps are brim full; so, if you're considering an upgrade, put the budget towards a big transformer that can charge/refill the caps more quickly.
When the transformer is too small, then have the capacitance smaller than ideal load support by exactly the same percentage. Basically, after the transformer has been milked for all it is worth, then greater capacitance is not wanted.
This consideration is so much easier and a lot more straightforward when the transformer is big enough for the job.
I've suggested only checking the needs of the load; however, some cases may have needs beyond that.
But, if it was going to be easy...
Average 1 cu-ft ish speaker averages 15000u per each rail of a split rail power supply with tolerable variance plus or minus 15%.
So, 10000u per each rail and a stereo pair of 16 ohm 0.75 cu-ft speakers looks feasible as well.
Otherwise, that 10000u idea, albeit an excellent try, falls short, EXCEPT!!!! The level worst thing to do is increase the load on an overly small transformer by adding too much capacitance, forming a delay timer that sounds even worse than deep sticky clay mud feels on your boots. Problematically, a linear supply is only clean when the caps are brim full; so, if you're considering an upgrade, put the budget towards a big transformer that can charge/refill the caps more quickly.
When the transformer is too small, then have the capacitance smaller than ideal load support by exactly the same percentage. Basically, after the transformer has been milked for all it is worth, then greater capacitance is not wanted.
This consideration is so much easier and a lot more straightforward when the transformer is big enough for the job.
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I have proposed the opposite.
When the amplifier is about as big as the slightly small transformer can support, then increasing the smoothing capacitance does make up some of the performance shortfall when the transients get to the "loud" end.
Well, uh, yeah...."bridged" sorta IMPLIES two amps, eh?
Now, according to the T.I. data sheets for a LM4780, a +/- 30 volt supply IS capable of delivering 100 watts into 4 ohms for each amplifier. That IS at clipping, naturally---10% THD. Datasheet Page 13 clearly shows a graph (Clipping Voltage vs. Supply Voltage, R L = 4 ohms) of ~ 4 volts at +/- 30v supply voltage. This leaves 56 volts for AC peak---just barely enough for the required +/-28 voltage swing. This is further illustrated on Datasheet page 11's graph (Output Power vs. Supply Voltage), showing 100 watts into 4 ohms with +/- 30 volt supply rails, with 10 %THD. Certainly one would not operate his amplifier at this level continuously; it just indicates the maximum headroom available at clipping---200 watts from a bridged amplifier.
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Ok so....back to the topic at hand.
I've mounted the power supplies in a recovered case, and I'm off to grab the toroid. Now I plan on running a preamp in this case, with a few filters planned to offload anything <100Hz to the subwoofer and giving me tone controls/etc. This board requires +/-12V and uses 6xOPA2134's.
Is it worthwhile trying to find a +/-25 and +/-12v toroid and provide a filter/regulator board for both voltages, or could you step it down with Zener's/etc to get the correct input voltage to the pre-amp board?
I've mounted the power supplies in a recovered case, and I'm off to grab the toroid. Now I plan on running a preamp in this case, with a few filters planned to offload anything <100Hz to the subwoofer and giving me tone controls/etc. This board requires +/-12V and uses 6xOPA2134's.
Is it worthwhile trying to find a +/-25 and +/-12v toroid and provide a filter/regulator board for both voltages, or could you step it down with Zener's/etc to get the correct input voltage to the pre-amp board?
Well, a Zener diode IS a NOISE generator, and thus not recommended for a high quality preamp. I'd just use the 25v toroid and the adjustable LM3xx-series regulators for your voltages---those regulators are fairly quiet (WAY better than a Zener and better than the LM78xx-series). I'd also adjust the voltage for the OPA2134s to +/- 18 volts to maximize headroom.
I used a Zener to drop off too much supply volts to feed a B1 style crossover filter.
There was no detectable noise.
The CRC after the Zener did it's job of providing a smooth DC supply to the series of B1 in the audio channel. And for reference, the B1 Buffer is prone to low PSRR.
There was no detectable noise.
The CRC after the Zener did it's job of providing a smooth DC supply to the series of B1 in the audio channel. And for reference, the B1 Buffer is prone to low PSRR.
why people say 7xxx regulators are bad, noisy?
how bad solution is use chipamps supply rails dropped with series resistor into cap, then regulator? its not that opamps need much current so resistors should be cold?
how bad solution is use chipamps supply rails dropped with series resistor into cap, then regulator? its not that opamps need much current so resistors should be cold?
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