Thanks a lot !
no ... I was wrong. I checked roughly and more or less there is the same stored energy (higher V but also much less uF in the tube amps)
Thanks again
Regards,
gino
Just for fun, I fixed the method used to calculate the reservoir capacitance required for a sine output from a class AB power amplifier, at the clipping threshold, for the case when the signal frequency is greater than or equal to the AC mains frequency. I will post a corrected spreadsheet, sometime soon.
Still need to also re-work the case for signal frequeny less than the mains frequency, which currently is wrong and gives capacitances that are too low (probably by 10%-25%). I think I know how, but haven't had the time, yet.
Still need to also re-work the case for signal frequeny less than the mains frequency, which currently is wrong and gives capacitances that are too low (probably by 10%-25%). I think I know how, but haven't had the time, yet.
Last edited:
Hi Frank. Well, the sine-only part isn't really important (since the DC@peak vs rated power stuff solves the original problem very well). But it's one of those things I just want to do. It's like a certain type of puzzle that I like to try to solve, just for the love of trying. For me, it's similar to going out in the woods and hunting for Morel mushrooms, or working "Killer Sudoku" puzzles (see krazydad.com). Also, even though the math isn't even close to what I used to be able to do, back in, say, 1980, it still kind-of makes me feel young again, to relearn some small parts of it.
Well, I posted that document too soon. Wish I had more time for this stuff.
It turns out that the sine-only calculations that I said were good for f_signal >= f_mains are really only good up to 2*f_mains, i.e. 2*f_mains >= f_signal >= f_mains.
Back to the drawing board...
Sorry about that.
It turns out that the sine-only calculations that I said were good for f_signal >= f_mains are really only good up to 2*f_mains, i.e. 2*f_mains >= f_signal >= f_mains.
Back to the drawing board...
Sorry about that.
Not really something to worry about, but as a demonstration of what's possible if you absolutely must have the lowest power supply impedance at very high frequencies, the ultimate decoupling strategy so to speak, there's this: http://www.sanmina.com/pdf/solutions/pcbres/buried_capacitance_technical_0106.pdf.
The strategy, and precision required, is very different from that of audio circuits, but it gets results for supply impedance: 10 milliohms at 100Meg, 100 milliohms at 1GHz.
Frank
The strategy, and precision required, is very different from that of audio circuits, but it gets results for supply impedance: 10 milliohms at 100Meg, 100 milliohms at 1GHz.
Frank
For one 120/100 Hz rectified (pulsating DC) peak, the conduction interval, with "reasonable" capacitances, is usually 1/4th to 1/5th of the total time, but can be much less with large capacitances, and can be 1/3rd of the total with less capacitance. At 1/3rd, the capacitance is often too small to prevent clipping at max rated power. For 60 Hz AC mains, with about 8.33 ms per 120 Hz peak, the charging pulses are typically about 2 ms long, for the minimum reservoir capacitances that allow the clipping threshold to be reached just as the rated peak output power is reached.
How can you count two peaks, when there is a conduction period for every peak (for a full-wave rectifier)?
Last edited:
Interesting - I agree with the 'hierarchical capacitance' philosophy. Al electrolytic, tantalum, ceramic. Just I have an aversion to tants taking over the mid-frequency decoupling so instead I use outrageous quantities of ceramics and multiple paralleled lytics to bridge the gap...
The biggest problem with ceramics is how much capacitance they lose when fully biassed up - some are down to only 10% of the sticker value. Tricky to find reliable data on this and it varies between manufacturers quite a lot. TDK look to be one of the best.
Last edited:
A 5% regulated transformer will reach the loaded peak voltage 18 degrees early due to float. It will begin to charge the capacitor at that time with the charging current increasing until the loaded peak current. It's post peak current then drops until the 5% float drops below the now higher voltage charged filter capacitor. Even an infinite value capacitor will not affect that.
OK. Maybe I'm missing your point. I thought we were discussing the typical values of conduction angles.
What does "float" mean?
Here's how I understand it so far:
Charging begins when the rising transformer output voltage minus the rectifier voltage(s) equals the falling capacitor voltage, which will be the minimum loaded voltage (not the peak loaded voltage).
Charging current will typically increase to way above the load's peak current, since the charging current has to supply as much charge, during the short charging interval, as the load uses during the entire 1/(2*fmains) period.
So how does the transformer regulation percentage affect the charging pulse length (or, rather, by how much does it affect it, and how can that be calculated)?
What does "float" mean?
Here's how I understand it so far:
Charging begins when the rising transformer output voltage minus the rectifier voltage(s) equals the falling capacitor voltage, which will be the minimum loaded voltage (not the peak loaded voltage).
Charging current will typically increase to way above the load's peak current, since the charging current has to supply as much charge, during the short charging interval, as the load uses during the entire 1/(2*fmains) period.
So how does the transformer regulation percentage affect the charging pulse length (or, rather, by how much does it affect it, and how can that be calculated)?
Last edited:
So if I understand what is being said here we have a pulsed current flowing to charge the capacitor as the load draws down the capacitors charge. Now is there any way to increase the charging time or is that impossible, because we are trying to keep up with the current demands of the output devices. How would a smps play into this conductance angle, or would that still have the identical pulse charging problems. The last question is is it really a problem just the way it is?
Well for one the first issue was my bad... Long day I didn't rennet correctly my first post.
If you take any power transformer and measure the output voltage without any load it will be greater than when it is loaded to the rated current. A small cheap transformer may go as high as 20%. A well built power transformer suitable for an audio power amplifier is almost always rated at 5%. This is due to the resistance of the windings and other losses that occur under load.
So until the filter capacitor begins to draw current the output voltage will behave as if unloaded and be 5% higher. Now the rectifier also has a lower voltage drop at lower currents but as the voltage drop rise much more quickly with current it really can be considered a constant drop at all times so it doesn't influence the charging time available.
So the result is that current begins to flow as soon as the output voltage goes above the stored voltage. However as the charging current increases so does the power loss from the internal loses. At the peak voltage the transformer is fully loaded. Due to the peak current being higher than the rated current the actual peak voltage is lower than the rated voltage!
After the peak voltage the current begins to drop and the transformer losses decrease so even the post peak voltage begins to rise.
Tomorrow I will try to post some actual scope measurements.
If you take any power transformer and measure the output voltage without any load it will be greater than when it is loaded to the rated current. A small cheap transformer may go as high as 20%. A well built power transformer suitable for an audio power amplifier is almost always rated at 5%. This is due to the resistance of the windings and other losses that occur under load.
So until the filter capacitor begins to draw current the output voltage will behave as if unloaded and be 5% higher. Now the rectifier also has a lower voltage drop at lower currents but as the voltage drop rise much more quickly with current it really can be considered a constant drop at all times so it doesn't influence the charging time available.
So the result is that current begins to flow as soon as the output voltage goes above the stored voltage. However as the charging current increases so does the power loss from the internal loses. At the peak voltage the transformer is fully loaded. Due to the peak current being higher than the rated current the actual peak voltage is lower than the rated voltage!
After the peak voltage the current begins to drop and the transformer losses decrease so even the post peak voltage begins to rise.
Tomorrow I will try to post some actual scope measurements.
But, the actual capacitance is almost irrelevant ... at the high frequencies you're using the overlapping of the impedance minimums at resonance to get the job done - sufficient of around the right value should do it.
Frank
Gootee,
If the charging pulse stops 5 degrees after the halfwave peak, then a 60degree conduction angle requires conduction to begin 55degrees before the peak of that halfwave.
At 55 degrees before the peak, the sinewave has reaching about 57% of the transformer voltage.
If conduction begins at that 57% value then the capacitor voltage must have dropped to the 57% at the end of the previous non-conduction period.
If I had a ripple on my smoothing capacitor that dropped to 57% of the transformer voltage, then I have failed to design and build an adequate PSU.
The conduction angle cannot be 60degrees on any properly designed and built amplifier PSU. Ripple that bad is never tolerated.
Could someone estimate the Amperes per mF of smoothiong that would ensue to give a 60degree conduction angle. I'll guess it is no where near the typical 1A/mF to ½A/mF that we often use as a guide for smoothing capacitance.
If the charging pulse stops 5 degrees after the halfwave peak, then a 60degree conduction angle requires conduction to begin 55degrees before the peak of that halfwave.
At 55 degrees before the peak, the sinewave has reaching about 57% of the transformer voltage.
If conduction begins at that 57% value then the capacitor voltage must have dropped to the 57% at the end of the previous non-conduction period.
If I had a ripple on my smoothing capacitor that dropped to 57% of the transformer voltage, then I have failed to design and build an adequate PSU.
The conduction angle cannot be 60degrees on any properly designed and built amplifier PSU. Ripple that bad is never tolerated.
Could someone estimate the Amperes per mF of smoothiong that would ensue to give a 60degree conduction angle. I'll guess it is no where near the typical 1A/mF to ½A/mF that we often use as a guide for smoothing capacitance.
It is 60 degrees of the total 360. But at maximum current it is a bit longer. To get it shorter you must have a transformer with more capacity than the load requires.
The float doesn't just give you extra charge time on the front end but also on the discharge side.
If you have 5% ripple then you start charging at .9 or 26 degrees before the peak at 5% regulation. After the peak the slope of the sine wave is flatter than the discharge slope for a bit less than 3 degrees then you follow the float for another 18 degrees. That would be about the longest charge time.
Now if your transformer is rated for ten times the capacity required your output voltage would be 5% (almost) higher and your float would be .5%. That is only .3 degrees on the front end for an infinite capacitor. At 5% ripple you would add 18 degrees to that on the front and 3 degrees on the back so the total would be 22 degrees of charging per half cycle.
The charging time is greatly dependent on transformer regulation. It is only with a lot of ripple that the capacitance becomes dominant.
The float doesn't just give you extra charge time on the front end but also on the discharge side.
If you have 5% ripple then you start charging at .9 or 26 degrees before the peak at 5% regulation. After the peak the slope of the sine wave is flatter than the discharge slope for a bit less than 3 degrees then you follow the float for another 18 degrees. That would be about the longest charge time.
Now if your transformer is rated for ten times the capacity required your output voltage would be 5% (almost) higher and your float would be .5%. That is only .3 degrees on the front end for an infinite capacitor. At 5% ripple you would add 18 degrees to that on the front and 3 degrees on the back so the total would be 22 degrees of charging per half cycle.
The charging time is greatly dependent on transformer regulation. It is only with a lot of ripple that the capacitance becomes dominant.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.