transformer specs
according to #98, the transformer will saturate at startup, however what we need to be concerned about is the current drawn during normal operations, if you get the capacitance right then the transformer will only need to be capable of supplying the loads current plus a little more to keep up the charge.
Thus too little capacitance makes the transformer work harder during normal operation and rails vary alot
Too much capacitance, blows your mains, the transformer does nothing after that
according to #98, the transformer will saturate at startup, however what we need to be concerned about is the current drawn during normal operations, if you get the capacitance right then the transformer will only need to be capable of supplying the loads current plus a little more to keep up the charge.
Thus too little capacitance makes the transformer work harder during normal operation and rails vary alot
Too much capacitance, blows your mains, the transformer does nothing after that
just a thought: a bank of smaller caps is better than two XBOX HUGE caps, because when they fail it will be cheaper to replace one 10 mF instead of one 100 mF
when i said there can't be too much capacitance it is implied that the rest of the circuit is built to withstand the current they will demand.
now i have to think what happens when you start up a psu (tranny-rect-caps).
the transformer will provide as much current as the capacitors demand. which is, how much? is it limited? i think not. So it will peak. Perhaps then a circuit should delay turning on the psu until the AC crosses zero so at least that way it won't be BAM max current to the caps
but isn't the transformer also a set of coils? doesn't it have inductance? perhaps it is too small to delay rising of the current... or it already does that so it doesn't matter
hm. it's time i put the oscilloscope to my amp's psu. now, where did i stash that fire extinguisher...
when i said there can't be too much capacitance it is implied that the rest of the circuit is built to withstand the current they will demand.
now i have to think what happens when you start up a psu (tranny-rect-caps).
the transformer will provide as much current as the capacitors demand. which is, how much? is it limited? i think not. So it will peak. Perhaps then a circuit should delay turning on the psu until the AC crosses zero so at least that way it won't be BAM max current to the caps
but isn't the transformer also a set of coils? doesn't it have inductance? perhaps it is too small to delay rising of the current... or it already does that so it doesn't matter
hm. it's time i put the oscilloscope to my amp's psu. now, where did i stash that fire extinguisher...
Wow, sorry Nico. I just now saw this post. I've been mucking about for hours on other things. Now it's almost 11pm here. I'll see what I can come up with.
(I am still working on the psu design that Nico suggested. This is basically unrelated to that.)
You might find that the phase angle at the transformer, when the AC was last switched off, if it leaves the core magnetized in a particular way, can affect how bad the next startup transient can be. I've forgoten the rest of the details but it's on diyaudio somewhere.
In any case, although the capacitor currents will have very large peaks, they are only that large for a relatively short time. So certain types of components' limits that are exceeded will not necessarily result in automatic immediate destruction of anything, basically because the duty cycle is so low. But the bottom line here is that you should check the specs for the affected components (especially the rectifier bridge), to see how they are rated for such currents. Some types of components might fry at the very first pulse while others could operate fine with them, for decades.
I remember designing a small (22 Volts, 4 Amp) regulated power supply, once, where I had to use an inrush-current limiter for the caps, because otherwise the adjustable regulator couldn't have a large-enough cap to bypass its adjust pin, because that would make the regulator's input-output differential voltage possibly go beyond the maximum spec for the regulator, during startup. Anyway, for the first 170 ms or so, a one-Ohm resistor had to carry the full inrush current, which peaked at about 20 Amps (down from 38 Amps without the limiter). At the peaks, the resistor had to dissipate 400 Watts. After looking at resistor datasheets, it turned out that I could use a 5-Watt resistor (of a particular type) and still have so much safety margin that the resistor never got warm at all.
I attached a couple of screen shots from LTSpice simulations of the startup, with and without the inrush limiter. These might give people an idea of what happens at startup.
This is for a 22-Volt regulated supply, with 4 amps being drawn by the load after the output voltage stabilizes.
(Note that one potential advantage of using multiple parallel caps is illustrated: The total ripple current is divided among the caps, making each cap's ripple current requirement much lower.)
The power transformer was also modeled, with more than the usual amount of detail. It was a 120 VA Hammond toroidal type, with paralleled dual secondaries, which made the rating 25V@4.8A.
Upstream from the transformers, I even added a model for a portion of the AC Mains wiring, that someone had given to me. (But it looks a little wrong, now.)
The schematic is actually still on line, at
Spice Component and Circuit Modeling and Simulation .
The transformer modeling information (and downloadable spice model and transformer measurement instructions) is also there, at
Spice Component and Circuit Modeling and Simulation .
(Note that nothing is for sale, on any of my webpages. Some of them still look like it but there should be something saying nothing is for sale, any more. I left most of them on line, at first, because people liked having them for references. It's been long-enough now that I could take some of them down but I haven't taken the time for that. Anyway, nothing is for sale there.)
- Tom
Attachments
Harrison, I don't even grasp the question.
However regarding 101, firstly one must remember that the charge voltage is not a DC pulse, but a sinusoid that rises from zero to max and that the capacitor will store this max voltage it reached. The next charge pulse would rise from the previous max to a new max in time RC. Also the voltage lags the current by 90 degrees.
During the first charge impulse the voltage across the capacitor would rise to some value determined by the duration of the charge pulse, the envelope of the charge pulse, the value of the capacitor, its internal equivalent series resistance and the series resistance of the transformer secondary (and of course the inductance of the windings and magnetic properties of the core).
When you say the transformer would saturate while attempting to charge the capacitors, what exactly do you understand by the term saturating a transformer - does the secondary become zero volts and zero amps due to the magnetic field flat lining?
However regarding 101, firstly one must remember that the charge voltage is not a DC pulse, but a sinusoid that rises from zero to max and that the capacitor will store this max voltage it reached. The next charge pulse would rise from the previous max to a new max in time RC. Also the voltage lags the current by 90 degrees.
During the first charge impulse the voltage across the capacitor would rise to some value determined by the duration of the charge pulse, the envelope of the charge pulse, the value of the capacitor, its internal equivalent series resistance and the series resistance of the transformer secondary (and of course the inductance of the windings and magnetic properties of the core).
When you say the transformer would saturate while attempting to charge the capacitors, what exactly do you understand by the term saturating a transformer - does the secondary become zero volts and zero amps due to the magnetic field flat lining?
OK. Here is a nice webpage about unregulated power supply design, with just the bare basics:
Unregulated Power Supply Design
Their equation for the peak-to-peak amplitude of the ripple voltage is equivalent to the equation I gave for calculating capacitance based on Δi, Δt, and a desired Δv:
(1) Δv = i / (2fC)
where f is the supply frequency (60 Hz, in our case).
Solving for C, we get:
(2) C = i / (2fΔv)
which is equivalent to the equation I gave in post # 61:
(3) C ≥ (Δi Δt) / Δv
if we realize they are using 1 / 2f in place of Δt, because the rectified AC mains (and ripple) frequency is twice the mains supply frequency, f, and the time for one cycle at frequency f is 1/f, AND we realize that in (3), the Δi is usually PEAK current, not RMS, since it's usually applied for transients or short time periods.
First we can note that the "3300 uF per Amp" rule would give, using equation (1), with C = 3300uF x 3.54A rms = 11682uF, a p-p ripple amplitude of Δv = 2.52 Volts. (Or was that rule for Amps PEAK? That would be 16500 uF, and 1.79 Volts p-p ripple.)
But the ripple voltage is not the only rail disturbance. And the ripple calculation in (1) only took into account a constant load current. Our load will pull current from the caps that will not be constant. It will be sinusoidal, probably a half-cycle sine at a time, in the simplest case of one pure tone. It might be worse than that but that means that the capacitance requirement for that can be a lower bound, i.e. a minimum.
So I started out thinking that the lowest frequency would need the longest current draw and would be the worst case. I looked at 15 Hz as an example. The period for one cycle of 15 Hz can be found using
(4) T (sec) = 1 / f (Hz)
as 1 / 15 = .0667 sec.
So we know that just to supply the current for the rising half of the positive part of a 15 Hz sine wave, we would have to go from 0 Amps to 5 Amps Peak (worst case) in one-fourth of a cycle, or 0.0167 sec.
Using equation (4), we can get C in terms of the Δv we decide we can tolerate: C = (5A)(0.0167 sec) / Δv, or C = .083500 / Δv , for 15 Hz, which can also be rearranged to give Δv = .083500 / C for 15 Hz.
BUT THEN I realized that since the charging pulses are faster than 15 Hz, and would actually re-charge the caps up to four times during a half-cycle of 15 Hz, that the equations in the previous paragraph were not valid because of that, and that 60 Hz would probably be a better "worst case".
60 Hz is 0.00417 sec per 1/4th cycle. So, C ≥ (5A)(0.004170 sec) / Δv, or
(5) C ≥ 0.020850 / Δv , for 60 Hz (note that it's 20850 uF for 1V Δv), and
(6) Δv = 0.020850 / C , for 60 Hz.
Note that the Δv in (5) and (6) is in addition to the mains ripple.
If we used the 11682 uF given by the "3300 uF pr Amp" rule, that would give Δv = 1.78 Volt of ADDITIONAL rail-voltage disturbance, which is 71% more (1/√2 more) than what we thought the 3300 uF rule would give us. (Or if that rule was supposed to be for PEAK amps, it would have given 16500 uF and 1.79 V p-p 120 Hz ripple, which would give an additional 1.26 Volts of rail disturbance due to the positive half of our 60 Hz tone, or 42% more.)
That might be especially inconvenient if we were also using a linear regulator, since the current humps pulled by the load would cause voltage sags that would essentially pull the troughs of the ripple voltage downward farther than planned, by up to 71% of its originally-calculated amplitude, which might be likely to cause the regulator's input minus output voltage difference to become less than it's dropout voltage spec, during parts of the troughs, which would get really ugly.
It would only possibly be 71% [or 42%, if the rule was for peak amps] worse when we were driving the amp all the way to the rail. But the bottom of the ripple waveform would come down lower than we thought, by some percentage, at every volume setting.
If we were happy with the mains ripple that the "3300 uF per Amp" rule was going to give us, then it looks like we should ADD 71% (or 42% if the rule used peak current instead of RMS) to the capacitance value it gave us, AND to the rule, in order to keep the worst-case rail disturbance the same.
In any case, we might as well come up with a general equation that will take into account BOTH the worst-case mains ripple AND the transient voltage sags that our signals might induce. There might be some "overlap" between those causes and effects but this should give a safe minimum capacitance:
For the transient component, from (3) we can substitute 1/4f for Δt, assuming that 1/4 cycle at the AC Mains frequency is the worst case, and get
(7) C = ipeak / (4fΔv)
and in order to take into account both the DC and transient components of Δv we can combine (7) with (2) to get the total C needed:
(8) C = [ iRMS / (2fΔv) ] + [ ipeak / (4fΔv) ]
Then we can solve for Δv:
(9) Δv = [ iRMS / (2fC) ] + [ ipeak / (4fC) ]
If we convert iRMS to ipeak, pull out the common denominator factor, then clean up the remaining fractions, we get:
(10) Δv = (0.6036)(ipeak)/fC
or
(11) C = (0.6036)(ipeak) / (f Δv)
where
f is the AC Mains frequency,
ipeak is the power supply's peak (rail) voltage (not RMS), i.e. ipeak = Vrail/Rload, and
Δv is your choice of an acceptable worst-case voltage-rail variation.
(So it looks like the new rule should be "≥ 5600 uF per Amp (peak)".)
So, assuming that the "3300 uF per amp" rule meant PEAK amps, and without considering whether or not the 1.79 Volts worst-case peak-to-peak ripple amplitude that it gave is "good enough", the minimum total reservoir capacitance needed in order to REALLY get a worst-case of less than or equal to 1.79 Volts Δv should be, using (11):
C = [(0.6036)(5)] / [(60)(1.79)] = 28100 uF
We might also want to add 15% or more to that, to account for variations in mains voltage and transformer regulation.
So maybe 33000 uF would be a better number.
(I hope that someone will check my math and logic.)
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Tom, short and sweet... What capacitor value would you arrive at for the reservoir having to dump a worse case square wave at 3.54A into an 8 ohm load while the transformer secondary is moving through a minimum, i.e. the diodes are turned off.
Thus this cap will supplying the energy to dissipate 100 watt into an 8 ohm load for probably a period of 20mS.
In other words the Vc cannot be less than 28V and Ic cannot be less than 3.6A. This should be a straight forward calculation.
Thus this cap will supplying the energy to dissipate 100 watt into an 8 ohm load for probably a period of 20mS.
In other words the Vc cannot be less than 28V and Ic cannot be less than 3.6A. This should be a straight forward calculation.
Hugh,
Thanks.
Interesting? Yes. Accurate? Not quite sure yet. It might simply be about twice as much as is needed.
I'm not really an expert at this stuff. I'm just sort-of making it up as I go along.
I havn't even simulated this one, yet.
I'm not completely comfortable with using the 1.79 V ripple figure, since it came from the 3300uF/amp rule, which tends to imply that 3300uF/amp is already good enough. But in one sense, I am comfortable with it, because it might mean that 3300 uF/amp is USUALLY enough.
Maybe one good way to think better about it would be to look at the two contributions to the capacitance, separately. In this case, the "standard" equation for reservoir capacitance, based on a max load current of 5 Amps and maintaining less than a 1.79 V p-p mains ripple amplitude gives 16480 uF and the one for suddenly providing the current for driving the beginning of a 60 Hz sine from 0 to the 40V rail (or almost to the rail) into 8 Ohms contributes 11638 uF.
Could the 16.5 mF do that job, instead, making the 11.6mF unneeded? It should at least be able to supply a CONSTANT 5 Amps with the ripple not getting bigger than 1.79 V. And there would be nothing for the 11.6 mF to even do, in terms of transient currents, if the current was already at max.
On the other hand, if nothing is happening and then the 11.6 mF is suddenly required to blow ALL of its charge out, to change the current from 0 to 5 amps in order to make the load go from 0 to 40 V, then it can't help at all with the ripple, and its voltage is going to have to come down by 1.79 Volts, just to make that amount of current come out of it and change it at that speed. Of course there's more than 11.6mF around, since the other 16.5mF is in parallel with it. So the rail should come down less than 1.79 V and there should still be the equivalent of 16.5mF that's not blowing its whole wad in order to perform the worst-case transient event, and it should be able to keep the ripple in check as usual, since it's able to do that even when the current goes to max.
Thinking again about the 16.5 mF being there alone, if an 11.6mF portion of it needed to blow ALL of its charge, to change the load current from 0 to 5 amps, then there would only be 4.9 mF left with any charge, to do the job we think would require the whole 16.5 mF.
I guess maybe I'm now convinced that there could be cases where the 16.5 mF by itself would not be enough. And I think I'm also convincd that there could be cases where the equivalent of 11.6 mF is just finishing a worst-cae transient, and the current is at almost at 5A, and the whole remaining 16.6 mF would need to also be available to keep the ripple within spec.
But I'm not sure how rare or common such cases might be, or even how significant the effects might be if there wasn't enough capacitance available, much less how it might affect the sound. But it would either not be noticeable or it would be bad to some degree.
I guess that a 60 Hz bass tone at max output would create one of the situations I just mentioned. Any very large transient could also do it, or at least partially create those conditions.
The main question that's left, then, seems to be whether it would require an extreme event, near the rail, to show any difference, or how much difference it might make under more-average output conditions.
Because of parasitics like rail inductance, it might be a little more beneficial than the calculations imply, too.
Regards,
Tom
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Nicely put Tom.
Hugh maybe music does not represent even close to a worst case situation, however if one is aware that a certain value of cap would be required for a worst case, then one can decide whether you want to go that far or not.
Think of it, how often here are we are adamant in placing X number of output devices in parallel to cater for the occasional low impedance worst case current, but do not have the energy available from the power supply to actually perform the function of driving such load. Then what is the point.
If one insist on the best of the best then obviously one must consider all the factors - only then would you have the "best "sounding amplifier, regardless of topology or does topology make up for other basic shortcomings.
I cannot say that I have encountered a fully designed amp in any thread on DIY so our colleagues presenting us with a schematic insinuating the best of the best while little consideration was given to what the power supply requirements are would be misleading. This probably applies equally to the commercial and most high-end amplifiers out there.
Hugh maybe music does not represent even close to a worst case situation, however if one is aware that a certain value of cap would be required for a worst case, then one can decide whether you want to go that far or not.
Think of it, how often here are we are adamant in placing X number of output devices in parallel to cater for the occasional low impedance worst case current, but do not have the energy available from the power supply to actually perform the function of driving such load. Then what is the point.
If one insist on the best of the best then obviously one must consider all the factors - only then would you have the "best "sounding amplifier, regardless of topology or does topology make up for other basic shortcomings.
I cannot say that I have encountered a fully designed amp in any thread on DIY so our colleagues presenting us with a schematic insinuating the best of the best while little consideration was given to what the power supply requirements are would be misleading. This probably applies equally to the commercial and most high-end amplifiers out there.
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Finally consider this. If we at DIY have a working rule of thumb or easily calculable power supply that can full-fill the best of the best requirement for a design then there exists a standard to measure or hear the differences between topologies or certain designs.
We can then expect that a design would perform in practice that could come close to simulation and that an FFT indicating amplitudes of H2 and H3, etc actually become meaningful quantities that can be discussed and optimized and may have a real impact on amplifier performance. This gentlemen and ladies is where this thread is leading that is why it requires as much input from all hands as it may become one of the most important threads in the community. At the end of the day we arrive at a standard and that takes the guess work out of it
Mooly, where the heck did you old experienced hands go!
We can then expect that a design would perform in practice that could come close to simulation and that an FFT indicating amplitudes of H2 and H3, etc actually become meaningful quantities that can be discussed and optimized and may have a real impact on amplifier performance. This gentlemen and ladies is where this thread is leading that is why it requires as much input from all hands as it may become one of the most important threads in the community. At the end of the day we arrive at a standard and that takes the guess work out of it
Mooly, where the heck did you old experienced hands go!
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All good points Nico, it's why I use separated supplies for each amp stage as to be able to optimize and make most efficient the required supplies for those stages. The output stage supply can be dedicated and designed for the output stage completely without having to take the other stages into consideration.
The harmonics distortion in my design are odd harmonics prominently due to the J-FET differential input pair, but they are quite consistent in the simulations..
The harmonics distortion in my design are odd harmonics prominently due to the J-FET differential input pair, but they are quite consistent in the simulations..
Tom,
My real concern is the level of the charge pulse coming off the rectifier to charge the filter cap.
Go too high, and this current is huge, impacting the earth return currents from the speaker voice coil.
With around 10mF, I have noted that a BYV32E UFSR rectifier is needed; a BYC28E with surge rating around 70A will blow from time to time.
There comes a point where large filter caps need large rectification and drive large charge currents, which seem to have deleterious effect on sound quality. I see similar compromises here to say, the cam timing on an IC engine. It is vexing indeed......
I have built 75W amps with just 2000uF of filter capacitance. Bass and impression of speed has been exceptional. I have even used just 1000uF with little perceived loss of quality or noise performance. Make the caps 4,700uF and again, there is little perceived improvement in sound quality. Make it 10,000uF and rectifiers begin to break, at least the 35nS rectifiers I like to use. And the perception of speed seems somehow to slow.
These are all subjective observations, and as such are raised because they are interesting, but the explanation does not seem clear to me.
Cheers,
Hugh
My real concern is the level of the charge pulse coming off the rectifier to charge the filter cap.
Go too high, and this current is huge, impacting the earth return currents from the speaker voice coil.
With around 10mF, I have noted that a BYV32E UFSR rectifier is needed; a BYC28E with surge rating around 70A will blow from time to time.
There comes a point where large filter caps need large rectification and drive large charge currents, which seem to have deleterious effect on sound quality. I see similar compromises here to say, the cam timing on an IC engine. It is vexing indeed......
I have built 75W amps with just 2000uF of filter capacitance. Bass and impression of speed has been exceptional. I have even used just 1000uF with little perceived loss of quality or noise performance. Make the caps 4,700uF and again, there is little perceived improvement in sound quality. Make it 10,000uF and rectifiers begin to break, at least the 35nS rectifiers I like to use. And the perception of speed seems somehow to slow.
These are all subjective observations, and as such are raised because they are interesting, but the explanation does not seem clear to me.
Cheers,
Hugh
Well.. for the amp I'll be using 2 toroids of 55V/7.5Ax2 / 850VA for a total of 1700VA AC power, one for each channel. It's followed by a 25A fast rectifier with smooting caps and a 4x10mF/100V capacitor bank, 2 for each power rail. This leaves a practical DC ripple bottom of 70-75V depending on load, leaving about 5V Vds minimal over the output transistors with a full power signal. It's really not going to be special other than the mandatory supply and return wiring (star et all). Optionally I might try Hugh's intermediate filter, but I wasn't initially planning on it.
The power rails will be decoupled by 220uF/100V upon entering the board. Each power transistor will get its own personal 100nF MKT reservoir, fed from a common 100uF/100V cap as close as possible as well. The decoupling ground will have its own return, as well as the snubber network ground which is also current that's supposed to go back to the output power supply.
The input stage and the voltage converter stage are fed by regulated DC supplies using a stacked DC regulator/zener diode tower to obtain the intermediate voltages. Each stackable voltage is fed by its own transformer winding. 3 relatively low power transformers totalling the 6 needed windings for +100/+30/+15/0/-15/-30/-100. I've employed this stacked supply technique in MF80, my 80W/4Ohm symmetric MOSamp design which is currently playing my music.
Hope that helps?
Edit: I'm going even further though. There will be one output supply ground return from the mono board, using a 2.5mm2 wire. On the board itself I use a separated ground plane (not in the small signal area of the amp) for the output power GND where the decoupling returns to as well as the snubber, a physically straight line from each component ground pin to the bolt pad for the 2.5mm2 wire in such way that no return path will intersect with another, when you would imagine all physical return lines in the plane surface. Remember, current wants to go the shortest way. A straight line is. The other ground plane is the regulated input/vas stage ground plane which serves as a split-half plane between supply and signal, also physically laid out so no return path will ever cross another physically.
Yes, it helps. MF80 does not hum an you won't hear any noise at max gain with your ear against the tweater
(Visaton DSM25 FFL)
The power rails will be decoupled by 220uF/100V upon entering the board. Each power transistor will get its own personal 100nF MKT reservoir, fed from a common 100uF/100V cap as close as possible as well. The decoupling ground will have its own return, as well as the snubber network ground which is also current that's supposed to go back to the output power supply.
The input stage and the voltage converter stage are fed by regulated DC supplies using a stacked DC regulator/zener diode tower to obtain the intermediate voltages. Each stackable voltage is fed by its own transformer winding. 3 relatively low power transformers totalling the 6 needed windings for +100/+30/+15/0/-15/-30/-100. I've employed this stacked supply technique in MF80, my 80W/4Ohm symmetric MOSamp design which is currently playing my music.
Hope that helps?
Edit: I'm going even further though. There will be one output supply ground return from the mono board, using a 2.5mm2 wire. On the board itself I use a separated ground plane (not in the small signal area of the amp) for the output power GND where the decoupling returns to as well as the snubber, a physically straight line from each component ground pin to the bolt pad for the 2.5mm2 wire in such way that no return path will intersect with another, when you would imagine all physical return lines in the plane surface. Remember, current wants to go the shortest way. A straight line is. The other ground plane is the regulated input/vas stage ground plane which serves as a split-half plane between supply and signal, also physically laid out so no return path will ever cross another physically.
Yes, it helps. MF80 does not hum an you won't hear any noise at max gain with your ear against the tweater
(Visaton DSM25 FFL)
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They do, huge toroids can have excessive startup currents depending on how flux was left at time of turn-off. It adds up with empty cap banks behind the transformer. There are basic ways, fancy ways and state of the art ways to deal with this.
The basic way is using a relay that inserts a a big 100ish ohms power resister in one of the primary leads of the transformer. It's engaged for a second to a few but then the relais shortcuts the resistor after which the transformer is directly connected. A power resistor can deal with the initial powerdraw and makes a bad case startup significantly less noisy. But that very hard initial pulse is still considerable.
The fancy way is to PWM one of the primary leads using a bipolar mosfet switch. the transformer can be started more silently and gradual, but triacs still have no real control on their on-off transition and due to a transformer being an inductor essentially, there's phase shift between voltage and current, further troubling triac off-timing.
The state of the art way is to do the same thing, though now using a process that precharges magnetic flux most favourable for startup using a DC pulsed signal into the primary, before power is enganged at the optimal point in time which is now a constant due to the flux being primed. This method is patented and I'm not sure if you can use it for non-commercial purposes.
I haven't decided yet on which method I'l use. The power mosfet solid state switches still pose a problem with their reverse bias diodes. At every half period, both a diode and a FET are conducting, causing a drop across the diode for each half period which causes additional harmonics and a constant powerloss across the FETs' reverse bias diodes.
The basic way is using a relay that inserts a a big 100ish ohms power resister in one of the primary leads of the transformer. It's engaged for a second to a few but then the relais shortcuts the resistor after which the transformer is directly connected. A power resistor can deal with the initial powerdraw and makes a bad case startup significantly less noisy. But that very hard initial pulse is still considerable.
The fancy way is to PWM one of the primary leads using a bipolar mosfet switch. the transformer can be started more silently and gradual, but triacs still have no real control on their on-off transition and due to a transformer being an inductor essentially, there's phase shift between voltage and current, further troubling triac off-timing.
The state of the art way is to do the same thing, though now using a process that precharges magnetic flux most favourable for startup using a DC pulsed signal into the primary, before power is enganged at the optimal point in time which is now a constant due to the flux being primed. This method is patented and I'm not sure if you can use it for non-commercial purposes.
I haven't decided yet on which method I'l use. The power mosfet solid state switches still pose a problem with their reverse bias diodes. At every half period, both a diode and a FET are conducting, causing a drop across the diode for each half period which causes additional harmonics and a constant powerloss across the FETs' reverse bias diodes.
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Is there a reason for it standing on oven mittens?
Only to stop it scratching the table as it didn't have its feet fitted at the time.
The PSU in Post #89 uses 2 x 600VA transformers at 240V into 12 x 33000uF. There is no soft start or anything else except a 15A DPDT switch. There are 4 x 50A rectifiers which work well in this application.
It is in CLCC CLCC CLCC CLCC configuration with each of the four supplies feeding +48V(L) +48V(R) -48V(L) and -48V(R)
Incidentally the metal bolts through the inductors and the transformers have all been replaced with nylon threaded bar in the final iteration.
It is in CLCC CLCC CLCC CLCC configuration with each of the four supplies feeding +48V(L) +48V(R) -48V(L) and -48V(R)
Incidentally the metal bolts through the inductors and the transformers have all been replaced with nylon threaded bar in the final iteration.
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Nico Ras - One factor that has not been mentioned yet is COST and / or Physical Size.
The best solution might well be 1000 x 100uF Black Gate caps in parallel. But at £5 each, £5000 for capacitors might be prohibitive.
We often have to take the ideal solution and then compare that with the bank balance or the Real Estate available.
The Aleph 4 is a Class A amplifier so it always uses 2.5A regardless of the volume that it is producing.
Allowing for 200mV ripple C calculates at 2.5/(100 x 0.2) = 125 000uF. After searching E-Bay I went for 3 x 33 000uF which is 100 000uF. In this case it was a mixture of cost against real estate.
The Aleph 4 has a good PSRR so the 200mV ripple is inaudible and the result is beautiful.
The best solution might well be 1000 x 100uF Black Gate caps in parallel. But at £5 each, £5000 for capacitors might be prohibitive.
We often have to take the ideal solution and then compare that with the bank balance or the Real Estate available.
The Aleph 4 is a Class A amplifier so it always uses 2.5A regardless of the volume that it is producing.
Allowing for 200mV ripple C calculates at 2.5/(100 x 0.2) = 125 000uF. After searching E-Bay I went for 3 x 33 000uF which is 100 000uF. In this case it was a mixture of cost against real estate.
The Aleph 4 has a good PSRR so the 200mV ripple is inaudible and the result is beautiful.
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