Power Supply Resevoir Size

[AndrewT]

Connect the HF decoupling together.
Connect the MF decoupling together.
Connect these two and call them Power Ground.
Connect the output Zobel to Power Ground.
Connect Power Ground to Main Audio Ground.
Connect Speaker Return to Main Audio Ground.

This, as far as I can see, is the best way. I don't have H.Ott's book, nor articles so can't compare my suggestion to his bible, but I suspect there is little, if anything, that is fundamentally different.

Fas,
an opamp does not have internal local decoupling. Instead we fit external local decoupling. They get connected just as described above. I consider opamps and chipamps and Power Amplifiers as the same thing when I try to layout my wiring/traces. They all work to the same physical laws.

Does that help?

I have downloaded the link to Ott and will read it.

Can we have a subsection somewhere in the Forum to store a compendium of really useful articles. eg a link to Jung's archive, a list of Ott's openly published works/articles. There are many other links, or pdfs that can be included.

[AndrewT]

Originally Posted by costis_n

............ zero-voltage switching relay.

but, such a device does not by itself solve the peak current demand problem.
The emf driving the current and the capacitor demanding the current in response to the emf will try to charge the capacitor in one quarter cycle. The SS relay timed to close at zero crossing will still suffer an enormous current pulse.

One must consider the components in the charging circuit and decide what the one shot peak current can be to allow all components to have a long lifetime. That peak current may be limited by some slow charging circuit.

If one decides that a slow charging circuit is required, it can be as simple as an added resistance in the charging circuit.

[seppstefano]

Originally Posted by AndrewT

Connect the HF decoupling together.
Connect the MF decoupling together.
Connect these two and call them Power Ground.
Connect the output Zobel to Power Ground.
Connect Power Ground to Main Audio Ground.
Connect Speaker Return to Main Audio Ground.

This, as far as I can see, is the best way. I don't have H.Ott's book, nor articles so can't compare my suggestion to his bible, but I suspect there is little, if anything, that is fundamentally different.

Does that help?

Almost... have you eventually got any drawing around?

HF/MF decoupling?

Thanks for your patience,

Stefano

[AndrewT]

HF decoupling = High Frequency = ceramics or similar with trace/lead lengths <20mm round trip route.
MF decoupling = Medium Frequency = electrolytics or similar with trace lengths >>20mm round trip route.

Do you really need a drawing to implement that set of suggested instructions?

[seppstefano]

Thank you Andrew,
I asked because I seem to remember a drawing by mike on another thread sharing the opportunity of placing a resistor between Power and Main Audio grounds, but I'm not so sure, and guessed you probably were referring to something similar.

I'm too lazy a fellow. No, no need for you to draw something for me.

Thanks a lot

[fas42]

Originally Posted by AndrewT

Connect Power Ground to Main Audio Ground.
Connect Speaker Return to

So what do you see as defining "Main Audio Ground"; I'm coming from the position that every piece of wire or trace connecting one place to another is an impedance to some degree, so where precisely should be that reference point be?

Fas,
an opamp does not have internal local decoupling. Instead we fit external local decoupling. They get connected just as described above. I consider opamps and chipamps and Power Amplifiers as the same thing when I try to layout my wiring/traces. They all work to the same physical laws. .

Sorry, I'm not quite sure what you're saying here with "internal" and "external" ...

Frank

[AndrewT]

Chipamps and Power amps have an on PCB set of decoupling capacitors. Since these are on board I think of them as "internal" to the amplifier. They are there by design and result in a Power Ground Pin/Tab/solder pad.

But an opamp does not have an internal set of decoupling caps.
Instead we add them externally. The junction of those decoupling caps is the Power Ground (PG) and that PG is outside the opamp. We have created a PG just like a Power Amplifier, but in the opamp it is "external" whereas for the PA it is "internal".
The two systems are equivalent, all that varies is the actual location of the decoupling and PG pin.

The Main Audio Ground (MAG) is the "point" in space where all the circuits pick up their reference voltage.

I choose to locate it at the centroid of: Speaker terminal, Input terminal, PG & PSU Zero Volts.
NOTE !!!!
the MAG is never coincident with the PSU Zero Volts.

By placing the MAG at the centroid, all the wires coming from various parts of the amplifier are kept to reasonably short lengths. No single wire is excessively long, except maybe the PSU Zero Volts link. The PSU may be the furthest from the centroid if the decision is taken to maximise the transformer to amplifier distance within the constraints of the chassis.

[AndrewT]

Originally Posted by AndrewT

Connect the HF decoupling together.
Connect the MF decoupling together.
Connect these two and call them Power Ground.
Connect the output Zobel to Power Ground.
Connect Power Ground to Main Audio Ground.
Connect Speaker Return to Main Audio Ground.

This, as far as I can see, is the best way. I don't have H.Ott's book, nor articles so can't compare my suggestion to his bible, but I suspect there is little, if anything, that is fundamentally different.
....................I have downloaded the link to Ott and will read it...........................................

Originally Posted by fas42

So what do you see as defining "Main Audio Ground"; I'm coming from the position that every piece of wire or trace connecting one place to another is an impedance to some degree, so where precisely should be that reference point be?
..................

Originally Posted by AndrewT

..................................The junction of those decoupling caps is the Power Ground (PG) and that PG is outside the opamp. We have created a PG just like a Power Amplifier, but in the opamp it is "external" whereas for the PA it is "internal".
The two systems are equivalent, all that varies is the actual location of the decoupling and PG pin.

The Main Audio Ground (MAG) is the "point" in space where all the circuits pick up their reference voltage.

I choose to locate it at the centroid of: Speaker terminal, Input terminal, PG & PSU Zero Volts.
NOTE !!!!
the MAG is never coincident with the PSU Zero Volts..............................

I have now read H.Ott's 1983 paper.
As far as I can see my suggested connections follow Ott's recommendations in all details.
There are certainly no fundamental differences between Ott's and mine. If there were I would look at how to modify my method.

All my suggestions have been developed without sight of Ott's papers and virtually all the information came from posts on this Forum. It would appear that there are many Members posting here that know what to do, I am simply the messenger, you Members are the Source. It is just that you get outnumbered by the ill-informed Members who post rubbish.

Before I close, Ott makes numerous mentions of ground planes in that paper. Nowhere in that paper does he state, nor show that a Flow and Return pair cannot perform as well as a ground plane. I would suggest that his paper although based on Ground Plane implementation is actually an argument for not using a Ground Plane. Maybe his other papers show otherwise.
As an example look at ex1 and fig5.
Here a pair of wires or a pair of traces would perform the same as a coax and none of those require or benefit from a connection to the ground plane at the load end. Except where he considers the case of parasitics that become significant enough to bring the VHF effect down into HF. That will effect the HF performance. I'm thinking here as HF = 100kHz to 10MHz.

Comments?

[OnAudio]

Originally Posted by gootee

Progress Update:

Just FYI, I am still working on a power supply spreadsheet (transformer, rectifier bridge, capacitor, and active load). I know that I said that it was almost done, way back when. But I found some problems and started a "deep dive" into it. Wow this sh!t is complicated and difficult. I think that I have well over half of it figured out, now. But implementing it has been a bit of a nightmare. However, if I can do it successfully, then it will probably be quite accurate. And it won't be only for finding Cmin.

It would have been far easier to just have people use the LT-Spice model that I have been using, especially now that we have the scalable transformer model, thanks to Terry Given, and AndrewT. But a spreadsheet will be far easier to use, for most people.

However, had I realized how many times I would think that I was done and then realize that there was still a major discrepancy somewhere, that I didn't understand, I might have given up, long ago. (I'm trying to not think about the possibility that I am at a similar point, yet again.)

One big revelation came, somewhat recently, when I saw that everything seemed to be looking very good EXCEPT where I was trying to account for the effects of transformer losses on the maximum rail voltage, due to the secondary resistance and the secondary and primary leakage inductances, during charging pulses.

At that point I was mainly working with peak values of everything, including the peak charging pulse current (because the peak values of things were relatively easy to calculate and were mostly all that was needed). I apparently hadn't thought about it enough and was thinking that I could multiply the peak rectifier current by the sum of the transformer impedances to determine the voltage drop and then subtract that from the peak input voltage divided by the turns ratio to get the secondary's peak output voltage.

Wow that was so wrong! Consider the charging pulse. A large swing of fast-rising current and then a large swing of fast-falling current. It goes through a resistance and... an inductance, in the transformer model. This is not like AC analysis. Think "time domain", with a one-shot pulse through an RL series network. The voltage across the resistor follows the shape of the current pulse exactly, with a peak voltage determined by its resistance. So far so good. But the voltage across the inductor is V = L di/dt. And di/dt, the slope of the current waveform vs time, is first a large positive number and then a large negative number. So the inductance basically gets two opposite-polarity voltage pulses across it, during each charging pulse. And it doesn't end at zero, at the end of the pulse. And the total series voltage across the inductance plus the resistance is needed, to calculate what happens to the secondary output voltage.

Actually, it's a little harder than that. The secondary "INPUT" voltage, at the output of the ideal transformer in the model (since the parasitics are separated out), during a charging pulse, is the theoretical secondary input voltage ("ideal" sinusoidal primary voltage divided by turns ratio) MINUS the voltage across the secondary leakage inductance. And, of course, VSEC_IN - VSEC_OUT = V_Ls + V_Rs, also.

I noticed, while studying time-domain plots of the various voltages and currents from simulations, that at the beginning and end of a charging pulse, at least, VSEC_OUT = V_primary/turns_ratio -2 * V_Ls - V_Rs. But V_Rs is zero at the beginning and end of the pulse, since the current goes to zero there (but the slope of the current vs time doesn't go to zero there). So the delta_VSEC due to a pulse is Vprimary/turns_ratio (at pulse end) - 2 * V_Ls (at pulse end) - Vprimary/turns_ratio (at pulse start). That's nice, except I didn't know how to calculate where the pulse start and end times were, in order to know where to pick the Vin values off of the input sine wave, so I'd be able to calculate the secondary's output voltage after a charging pulse (Hint: It can be well-above the "ideal" theoretical maximum peak secondary voltage!).

I had originally thought that VSEC_MAX was the theoretical peak secondary input voltage minus the voltage across the secondary leakage inductance at rectifier turn-off (pulse end), which would only require multiplying the final down-slope rate of the pulse by the leakage inductance. That turned out to not always be correct. But I did learn to use Excel's curve-fitting apparatus, and found a polynomial for the terminal slope of the charging pulses, under various conditions. But that equation changed for different VA and output power combinations. (So I started looking into converting sets of equations that give parallel plots in a plane into a more-general 2-D "field" equation. But it turned out that "parameterization" is probably the usual best way to go, for that.) But then I found a better way to account for different VA and output power ratings, for that equation (mentioned farther below).

I was able to plot all of the (simulation) voltages and currents that happen during a charging pulse, and had studied them so much that I finally deduced the simple equations showing which things add and subtract to produce the plots, which for some reason were just not obvious to me, at the beginning.

But there seemed to be no easy way to calculate the actual values needed, in a spreadsheet, without an closed-form mathematical equation for the charging pulse itself, i.e. some algebraic expression that would give current in amps if I plugged in a time value.

The charging pulses somewhat resemble half of a rectified sine, in shape. But unlike a rectified sinusoid the beginning has a gradual starting slope at first, and the whole thing is asymmetrical and "slanted" and stretched toward the right, while the trailing edge's slope usually just gets steeper until it hits zero. (It turns out that, in my simulations at least, those right-triangle-like pulse shapes that some scientific papers' authors have used as an approximate model for the pulses are only produced when the parasitics are removed from the transformer model.)

The charging pulses appeared to have the same basic shape and features, almost no matter what their peak values were. So I decided to "take the plunge" and measured about twenty or thirty data points from one, off of the LT-Spice plot window screen, with my mouse cursor, and put the "measured" times and currents into Excel columns. Then I used the Excel curve-fitting feature on that data and got a very good match by using a fourth-order polynomial. I measured two other ones, with different peak current values, too.

I can get the maximum peak rectifier current, in my speadsheet, already, quite accurately, through other means. So I thought that it would probably be helpful (to say the least) if I could somehow "scale" the pulse-shape equation from the one particular (16.36 Amp peak) rectifier pulse that I had an equation for, in order to be able to get an equation for a pulse that had any other peak value.

So first I plotted, in Excel, the widths of the bottoms of several charging pulses, in milliseconds, versus their peak amps. Then I used the Excel curve fitter again and got a second-order polynomial with zero error that will give the width of the bottom of a charging pulse (in milliseconds), given only its peak value.

I then shifted the pulse data so that time=0 was at the peak and re-fitted the original equation, for that situation. I also normalized both the pulse widths and the peak values and re-did that polynomial. That way, I could take a new peak value, calculate the width with the polynomial equation for that, and then use the original equation, but with the input time values simply scaled for the new width, and then scale the resulting current (amps) values by the new peak current divided by the one used for the original polynomial.

It works! And it's trivial to differentiate and integrate polynomials. So now I also have equations, as functions of time relative to the peak's time, for the slope at any point in time, ih Amps per millisecond, and for the area under the curve between any two points in time, for almost any size of rectifier charging pulse (for the particular transformer model parameters supplied by AndrewT, at least). And note that the slope, for which I now have an equation, is the same "di/dt" that is needed to calculate the voltage across the secondary leakage inductance at any time.

And the capacitor current's equations are now also known, since it's just the rectifier current minus the load current and my load current is a simple square wave.

The rectifier pulse width also determines the exact time interval (or, equivalently, the phase angle range) for where the input sine wave gets "hacked into", during the charging pulses.

So now, that simple equation that I gave earlier, for the "delta Vsecondary due to a charging pulse", will become very useful, as soon as I get one more step completed, which is calculating the exact time when the decaying capacitor voltage runs into the next rectified sine peak (or maybe the one after that, in some cases). It shouldn't be too difficult. It just happens to be the step I'm on right now, thanks to the fact that I now finally have a way to use that information to continue on and calculate the actual solution.

It turns out that finding that intersection point, between the decaying exponential capacitor voltage and the rectified sinusoidal input voltage, can ONLY be done numerically (or graphically), since there is no closed-form mathematical solution for a transcendental equation such as that. So it's a good application for a computer.

I will (almost certainly) also have to go back one more time, at least, and re-do all of the charging-pulse-related equations. Earlier, when I was playing around with curve-fitting an equation for the maximum downslope at the end of a charging pulse, I eventually figured out that I could make it work for more than one output power and VA rating if I first multiplied the independent variable (the peak current value of a pulse for which the ending slope was desired) by the output power divided by the VA rating, and then curve-fitted to find the equation.

Then I could just first multiply any desired new peak charging pulse current value by Watts/VA, and then plug the result into the polynomial that had been found by Excel's curve fitter, and get the correct final slope for a charging pulse of any height, for any output power and transformer VA rating that was being used.

So I assume that I will need to do something similar for the pulse_amps(t) polynomial and the related equations for its derivative (slope) and integral (area under curve). "So many fun things to do. But so little time."

At any rate (at some rate?), I think that I will eventually get it done.

Cheers,

Tom

Hi Tom,

Thanks for the nice job. Looking forward to your findings

[OnAudio]

There are multiple solutions to the problem. This is just one of them