The fact that the DC value across the caps is "very close to" the peak AC voltage IS the point. It doesn't confuse the issue; quite the opposite.
With enough capacitance, the "DC" value across the caps is, exactly, that peak transformer output voltage, minus only the rectifier drops. What happens if the caps are insufficient is not the point, and only confuses the issue being discussed.
And actually, I had no issue to discuss about that, except to try to clarify it. The peak AC voltage at the output of the transformer is the same as the (unloaded) DC voltage across the capacitors, except for the loss of the rectifier voltage drops.
The only "issue" was in how to think about the multiplication by the square root of two, which is only to convert the typically-provided RMS rating to a more-useful peak rating.
The peak is already there, before the rectifiers and we only need to calculate its nominal value, and then subtract the rectifier drops. When I stated that, people said "No!!" and "What!!!".
In reality, the voltage across the caps will often have peaks that are HIGHER than the peak transformer output, due to the peaking in the response that is caused by the leakage inductance of the transformer windings, which forms a second-order system with the reservoir capacitance, when the rectifiers are conducting. But that's a different discussion.
Suffice to say that the DC voltage after rectification and the caps will be approx. 1.4 x the RMS output voltage of the transformer which is the voltage specified on the box that the transformer came in.
I'm all for that. But I am only trying to de-confuse, by responding to posts that attempt to defend, and/or add to, the original confusion.
It looks like the OP kind of got lost in the dust.
Assuming a capacitor-input power supply, and trying to keep math to a minimum...
In electronics, using RMS allows a person to think of things from more of a DC perspective, and the math for DC is easier than for AC.
Your design requires ±45V @ 5A.
Transformers are rated by RMS voltage. The peak voltage is 1.414x (the square root of 2) that value.
We can take the reciprocal of 1.414 (0.707) and multiply it by 45. 0707*45=31.8. Let's call it 32.
The design needs a transformer with two ~32 volt secondaries. You can check that with 32*1.414=45.248.
The capacitor will charge to the peak voltage of the transformer secondary. There will be a diode or two Vdrop loss, but it's a small percentage of the total voltage so there's no need to fuss about it.
The secondary is rated for 32Vrms. Since the DC on the capacitor is at 45V, the continuous DC current must be lower than the AC current rating of the transformer. An example is better than my explanation - for a transformer rated for 32V @ 5A continuous, the power = 160. When that is rectified in our circuit, 45V @ 5A continuous = 225.* That isn't possible with known laws of physics.
So the transformer must have a higher current rating. 150% (32V@7.5A) would be borderline, 175% (32V@8.75A) would be better, and 200% (32V@10A) would be better still.
*I left out power units on purpose. 160 would be VA (volt-amperes), 225 would be W (watts). Suffice to say that VA, like RMS, is an AC term that allows a person to think more from a DC perspective.
Assuming a capacitor-input power supply, and trying to keep math to a minimum...
In electronics, using RMS allows a person to think of things from more of a DC perspective, and the math for DC is easier than for AC.
Your design requires ±45V @ 5A.
Transformers are rated by RMS voltage. The peak voltage is 1.414x (the square root of 2) that value.
We can take the reciprocal of 1.414 (0.707) and multiply it by 45. 0707*45=31.8. Let's call it 32.
The design needs a transformer with two ~32 volt secondaries. You can check that with 32*1.414=45.248.
The capacitor will charge to the peak voltage of the transformer secondary. There will be a diode or two Vdrop loss, but it's a small percentage of the total voltage so there's no need to fuss about it.
The secondary is rated for 32Vrms. Since the DC on the capacitor is at 45V, the continuous DC current must be lower than the AC current rating of the transformer. An example is better than my explanation - for a transformer rated for 32V @ 5A continuous, the power = 160. When that is rectified in our circuit, 45V @ 5A continuous = 225.* That isn't possible with known laws of physics.
So the transformer must have a higher current rating. 150% (32V@7.5A) would be borderline, 175% (32V@8.75A) would be better, and 200% (32V@10A) would be better still.
*I left out power units on purpose. 160 would be VA (volt-amperes), 225 would be W (watts). Suffice to say that VA, like RMS, is an AC term that allows a person to think more from a DC perspective.
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I think that when you did the 32V@nn Amps VA calcs, it should have been RMS current, since the 32V is RMS voltage. So the 200% case would come out to about 200 VA, instead of 320 VA. (But also remember that that's for one channel, only. So if two channels are to be powered by this same PSU, then a transformer with double that VA rating would be needed.)
Leaving out the 2V of diode drops, and the 3V or so of Vclip voltage between the amplifier's DC Power In and Audio Out, means that he would get more like 28V max RMS out, instead of 32V. But that's also assuming ZERO RIPPLE.
Anyway, his spec was for max rated power of 100W into 8 Ohms. So that locks it to 28.28 V RMS max out (40 Volts peak) and 3.54 Amps RMS max out (5 Amps peak).
Adding 40V peak signal out (to get the rated max of 100 W RMS into 8 Ohms) plus 2 V of diode drops plus 3V for amplifier Vclip, plus 5% ripple (= about 2.25V) gives a requirement for 47.25-Volt peak transformer output, i.e. 33.4 V RMS secondary voltage ratings, minimum (under load). Depending on how the transformer is rated, that could mean 36 V RMS rated secondaries, if it was rated unloaded and had 8% regulation. Either way, it should provide +/- 45.25 Volt rails (peak), for the onset of clipping to occur at the 100 W output level, assuming there's enough capacitance to keep the p-p ripple at or below 2.25 V when at that 100W rated max output power.
For a subwoofer, the low-frequency bass peaks could be wide-enough that they might approximate DC, over entire charging periods, occasionally. Therefore the reservoir capacitance should be calculated using peak output figures, not RMS. (The spreadsheet I have attached to this post does both.)
With 45.25-volt rails (which only allows for 2.25 Volts peak-to-peak ripple at the onset of clipping), which equates to 33.4 V RMS transformer secondaries, the required bare-minimum reservoir capacitance would be about 37600 uF per rail, if two channels are being powered from the one supply, with 60 Hz mains (or 44844 uF minimum per rail for 50 Hz mains).
Adding just 0.6 Volts to the transformer RMS output voltages, i.e. using 34 V RMS secondaries, would lower the required capacitance from 37600 uF to 28200 uF for 60 Hz and from 44844 uF to 33633 uF for 50 Hz.
Using 35V RMS secondaries would lower the required bare-minimum capacitance to 19235 uF for 60 Hz and 22941 uF for 50 Hz. And 36V RMS secondaries would lower the required capacitance to 14633 uF for 60 Hz and 17453 uF for 50 Hz.
For powering two channels from the one power supply, at 100 W/ch into 8 Ohms, a transformer rated at 400 VA would probably be safe-enough.
If only one subwoofer channel will be powered from the supply, then cut all of the capacitance values and the VA rating, that were given above, in half.
Note, too, that the decrease in capacitance that was gotten by raising the rail voltage, in the last few paragraphs above, was gotten at the expense of higher peak-to-peak ripple voltage amplitudes. That should be OK, since most audio power amplifiers have good PSRR (power supply rejection ratio, aka "ripple immunity") at low frequencies. But more capacitance could not hurt the sound, and might improve it.
To use the attached Excel spreadsheet, rename the file without the ".txt" at the end of the file name. This is an updated version of the previous spreadsheet, which is friendlier, and starts with the rated RMS voltage of the secondaries, and the diode drops at the rated max current, instead of requiring the user to calculate the rail voltage.
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Leaving out the 2V of diode drops, and the 3V or so of Vclip voltage between the amplifier's DC Power In and Audio Out, means that he would get more like 28V max RMS out, instead of 32V. But that's also assuming ZERO RIPPLE.
Anyway, his spec was for max rated power of 100W into 8 Ohms. So that locks it to 28.28 V RMS max out (40 Volts peak) and 3.54 Amps RMS max out (5 Amps peak).
Adding 40V peak signal out (to get the rated max of 100 W RMS into 8 Ohms) plus 2 V of diode drops plus 3V for amplifier Vclip, plus 5% ripple (= about 2.25V) gives a requirement for 47.25-Volt peak transformer output, i.e. 33.4 V RMS secondary voltage ratings, minimum (under load). Depending on how the transformer is rated, that could mean 36 V RMS rated secondaries, if it was rated unloaded and had 8% regulation. Either way, it should provide +/- 45.25 Volt rails (peak), for the onset of clipping to occur at the 100 W output level, assuming there's enough capacitance to keep the p-p ripple at or below 2.25 V when at that 100W rated max output power.
For a subwoofer, the low-frequency bass peaks could be wide-enough that they might approximate DC, over entire charging periods, occasionally. Therefore the reservoir capacitance should be calculated using peak output figures, not RMS. (The spreadsheet I have attached to this post does both.)
With 45.25-volt rails (which only allows for 2.25 Volts peak-to-peak ripple at the onset of clipping), which equates to 33.4 V RMS transformer secondaries, the required bare-minimum reservoir capacitance would be about 37600 uF per rail, if two channels are being powered from the one supply, with 60 Hz mains (or 44844 uF minimum per rail for 50 Hz mains).
Adding just 0.6 Volts to the transformer RMS output voltages, i.e. using 34 V RMS secondaries, would lower the required capacitance from 37600 uF to 28200 uF for 60 Hz and from 44844 uF to 33633 uF for 50 Hz.
Using 35V RMS secondaries would lower the required bare-minimum capacitance to 19235 uF for 60 Hz and 22941 uF for 50 Hz. And 36V RMS secondaries would lower the required capacitance to 14633 uF for 60 Hz and 17453 uF for 50 Hz.
For powering two channels from the one power supply, at 100 W/ch into 8 Ohms, a transformer rated at 400 VA would probably be safe-enough.
If only one subwoofer channel will be powered from the supply, then cut all of the capacitance values and the VA rating, that were given above, in half.
Note, too, that the decrease in capacitance that was gotten by raising the rail voltage, in the last few paragraphs above, was gotten at the expense of higher peak-to-peak ripple voltage amplitudes. That should be OK, since most audio power amplifiers have good PSRR (power supply rejection ratio, aka "ripple immunity") at low frequencies. But more capacitance could not hurt the sound, and might improve it.
To use the attached Excel spreadsheet, rename the file without the ".txt" at the end of the file name. This is an updated version of the previous spreadsheet, which is friendlier, and starts with the rated RMS voltage of the secondaries, and the diode drops at the rated max current, instead of requiring the user to calculate the rail voltage.
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You're not trying to be irksome, are you?🙂
His very first sentence was, "I am designing a power supply using a center tapped toroidal transformer in order to supply 5Adc and +-45Vdc."
I was confident the OP would know that 32V@nnA meant a 64V center tap transformer.
I do admit that I made an assumption that the OP would sooner find a xformer with dual 34V or 35V secondaries, so I was more concerned with helping to understand some factors used to make a purchase selection.
Or was this intended for me? Or??
His very first sentence was, "I am designing a power supply using a center tapped toroidal transformer in order to supply 5Adc and +-45Vdc."
Maybe. I'm not sure. I assumed that the current referenced to the RMS voltage would also be the RMS value.
I was confident the OP would know that 32V@nnA meant a 64V center tap transformer.
I think know what you're saying here, but others might be confused. Are you saying the Vclip results in an effective 28V out? If so, you should make that more clear.
See above. This was a power supply question, not an amplifier question.
I do admit that I made an assumption that the OP would sooner find a xformer with dual 34V or 35V secondaries, so I was more concerned with helping to understand some factors used to make a purchase selection.
Maybe I don't get out enough. I've never seen a transformer rated unloaded.
The capacitance sees DC with ripple. Is this what you are referring to, peak vs. RMS? Do you think the OP grasps what you're saying?
Or was this intended for me? Or??
Sorry. It's one of my "features". <grin>
I meant that the nn (amps) should have been divided by 1.414, in each case.
I meant that if the 2V of diode drops and the 3V of Vclip had been accounted for, then it would only leave 40 V for the max peak signal output, or 28 V RMS. And if the ripple had also been accounted for, it would have been even less.
Same thing, or at least inextricably linked. Anyway, the "100 W 8 Ohms" came LAST. So it should override everything before it. Right? <grin> (And his schematic says something different than both of the specs we have mentioned.)
Neither have I. But I thought I had read about that possibility, here, somewhere.
The reservoir capacitance is supplying all of the output current for a possibly-very-low-frequency high-power output signal, between charging pulses. Because of the low frequency, possibly as low as 1/3rd of mains frequency or lower, the peaks of the sine wave output can remain near the rated max peak output level for more than one charging period, and possibly for several charging periods in a row. If you imagine a plot of a sine, where one polarity lasts for at least six of the periods between charging pulses, you can see that near the peak of that sine, it's closely approximating a flat line (DC), at the peak rated output level, for probably at least one (or maybe more) charging periods. Therefore, the reservoir capacitance needs to be calculated as if the caps should be able to supply DC current, at the PEAK rated output level (as opposed to RMS), without running out of charge, for at least one whole charging period, so the caps won't run out of charge before the next charging pulse comes along.
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I don't think so. The current values were raised to account for the RMS-to-peak voltage. They don't then get lowered peak-to-RMS. Where's the progress in that?
I guess the answer to the question is in there, but it's no more clear.
100+ words to answer, "Yes."
I've read the comments and am thinking of a 36V@8.34A transformer, giving 300VA and +-50Vdc supply so I have enough voltage after all the drops. I'm in the UK so will be using a 50Hz supply frequency and will be powering a single channel. I think 3x 4700uF caps in parallel per bank will be sufficient. As I'm using 50Hz, it takes 20ms for one cycle and therefore 10ms for each cycle after rectification. So for a small ripple the time constant of the filter should be much in excess to 10ms. Am I correct to say that a 2V ripple is a 4% voltage change, so 0.96*Vdc is the minimum I want to see? Therefore using time const = -10ms / ln(0.96) I get 245ms. Is the filter an LC due to the inductance in the transformer? Please correct me if what I've said is rubbish.
Using a 230:36 dual secondary transformer on a typical UK supply voltage of 240Vac +-6% will normally give much more than +-50Vdc after rectification and smoothing.
Use 63V capacitors.
A 100W amplifier will work well from a 100VA to 200VA transformer.
Two channels of 100W+100W will work well with a 200VA to 400VA transformer.
Do not bother with current definitions when working with the POWER requirements of transformers feeding audio amplifiers.
If you feel inclined to ignore my advice then I'll convert those VA to Amperes for you.
100VA 36+36Vac = ~1.4Aac
200VA 36+36Vac = ~2.8Aac
300VA 36+36Vac = ~4.2Aac
400VA 36+36Vac = ~5.6Aac
BTW,
you do not need an 8Aac transformer to power a pair of 100W audio amplifiers.
And you will find very few stockists of 36+36Vac.
Look for the more available 35+35Vac. 150VA to 200VA is plenty for a single channel 100W Power Amplifier.
Use 63V capacitors.
A 100W amplifier will work well from a 100VA to 200VA transformer.
Two channels of 100W+100W will work well with a 200VA to 400VA transformer.
Do not bother with current definitions when working with the POWER requirements of transformers feeding audio amplifiers.
If you feel inclined to ignore my advice then I'll convert those VA to Amperes for you.
100VA 36+36Vac = ~1.4Aac
200VA 36+36Vac = ~2.8Aac
300VA 36+36Vac = ~4.2Aac
400VA 36+36Vac = ~5.6Aac
BTW,
you do not need an 8Aac transformer to power a pair of 100W audio amplifiers.
And you will find very few stockists of 36+36Vac.
Look for the more available 35+35Vac. 150VA to 200VA is plenty for a single channel 100W Power Amplifier.
It's better to think of the caps as the supply of current for the output signal. As they supply the current, their voltage falls. That creates the ripple. Size them so max output current for the charging period length will only pull their voltage down by the desired ripple amplitude. I gave a link to a post with the relevant equation, and also posted a spreadsheet that calculates it.
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Thanks AndrewT, I actually have a 36-36V 160VA transformer spare, I just didn't think it would be enough.
Ignoring the issue of VA vs. W, doesn't a 100W amp from a 100VA transformer imply 100% efficiency?
This is from Rod Elliott's site, concerning the full wave split supply shown earlier:
No, not efficiency but duty cycle. You could, for a very short while, run a 100W amp from a 50VA transformer.sofaspud said:
Not if you listen to organ music or reggae! Best to design for both common loading and one side loading, as either can happen.
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