Me thinks you are totally confusing LW ripple cancellation with low power supply Z and ultra fast PEAK instantaneous current delivery my friend. Ripple cancellation does not provide the latter.
There IS no free lunch, the basic supply has to have the later "in spades" to perform at a high level !!
Jeff Medwin
Ultrapath DC current: power supply through primary, tube, cathode resistor, return to PS
Standard design current: Same (equivalent).
Ultra path ripple current: PS cap to ground, AC to ground in series with cathode resistor return to PS
Standard: PS cap to ground + PS through primary, tube, and cathode resistor in parallel with cathode cap.
Ultrapath ripple current: PS cap to ground + PS through primary, tube, and ultrapath cap, in series with cathode resistor.
Standard signal: Tube through cathode cap, in parallel with cathode resistor to ground.
Ultrapath signal: Tube through cathode resistor, and tube through ultrapath cap to B+.
The AC current can travel in some proportion through each of those paths. And each of those distinct paths have different impedances, depending on the elements that they are in series or in parallel with. So clearly they are not equivalent.
Sheldon
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Obviously not.
The L-W circuit provides the latter, and simply does not require the former.
Topology is correct. There is an error however.
You are correct that the supply is split after the first LC. The two following LC's are in parallel, so must be represented as 1/2 the value for the L's and 2x the value for the C's.
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Hi Jeff
For the ultra fast current delivery, I simply used a choke loaded supply... The choke is an excellent current store as you know. I don't know if this can be illustrated by any simulation in PSU II.
If you do simulate my supply (or something similar to it), you will probably be shocked at what you see. According to my quick simulations, I have something like 4 AC Volts of ripple.
Yes, that's 4 AC VOLTS peak to peak of ripple in my supply. But I don't hear any hum. Scout's honor. 🙂
Ian
For the ultra fast current delivery, I simply used a choke loaded supply... The choke is an excellent current store as you know. I don't know if this can be illustrated by any simulation in PSU II.
If you do simulate my supply (or something similar to it), you will probably be shocked at what you see. According to my quick simulations, I have something like 4 AC Volts of ripple.
Yes, that's 4 AC VOLTS peak to peak of ripple in my supply. But I don't hear any hum. Scout's honor. 🙂
Ian
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For lowest output impedance PSUs, why not simply use a regulator? Even a simple, $10 reg with two MOSFETs (one as pass, one as error amp) will provide a very good noise reduction and a very low output impedance.
Even a simple regulator will allow to use cheaper parts in the raw supply preceding it, with absolutely no sonic penalties.
Furthermore it will provide higher fidelity than pretty much any non regulated PSU will.
Then of course there are better regulator schemes as well to further experiment. But even the simplest 2 MOSFET reg will make a huge difference.
Even a simple regulator will allow to use cheaper parts in the raw supply preceding it, with absolutely no sonic penalties.
Furthermore it will provide higher fidelity than pretty much any non regulated PSU will.
Then of course there are better regulator schemes as well to further experiment. But even the simplest 2 MOSFET reg will make a huge difference.
Sheldon !!
I say this paragraph by you is incorrect, certainly not accurate. Why the words " must be " below ????
"You are correct that the supply is split after the first LC. The two following LC's are in parallel, so must be represented as 1/2 the value for the L's and 2x the value for the C's."
The two L/Cs following the Y split are "NOT in parallel at all", each 400 Ohm / 8 uF combo feeds only ONE Channel ....... individually.
I would agree with you ONLY if you connected the output of the left C2 to the output of the right channel C2, hard wired them to drive both channels simultaneously. Only THEN WOULD they really be in parallel !!!
But that is certainly not the case here. Each 400 Ohm L / 8 uF combo feeds ONE Channel individually.
Actually, IF the left and right C2s were hard wired, you MIGHT like it better, the ringing of the 400 Ohm Ls goes away and you get the fastest settle, 100 mS. Ripple would be 530.9 mVAC. Try hearing that !!!
Unfortunately, the computed Z of the L2/C2s hard wired in parallel becomes the poorest we have obtained to date. (Its 561.19 VDC minus 553.52 VDC divided by a 15% step, or .0081 A. which is 947 Ohms computed Z.) My favorite - ALT C - was 611 Ohms, while "truly" paralleled C2s are 54% higher in Z than 611 Ohms.
Peace in the Valley.
Jeff Medwin
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Hi Jeff
Do you know why I can't do a split load after only a single choke in PSU II ? is it a bug in the software?
Ian
Do you know why I can't do a split load after only a single choke in PSU II ? is it a bug in the software?
Ian
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This is the only way it can be modeled in PSUD.
Learn spice you can model the full amp split after the first LC. Then you can apply different component values and different signals to the two amp sections and see what you get. It won't be much different than what you see here.
I don't know what you are looking at but there is no ringing. There is a small overshoot, which settles immediately to a steady ripple. You should really try to understand definitions properly.
BTW the ripple cancellation circuit will also remove most of the influence of supply impedance. Model it in spice and see.
Sheldon
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Not a bug. It will only model a series string of sections. Of course, spice does not have this limitation.
Sheldon
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I agree, although I've seen many argue that these caps are somehow outside of the signal path!
Anyhow, I appreciate the help in understanding this but I'm still of the opinion that a non-LW topology does not force the two current loops to coincide in a single cap since you can use two caps, separated by an impedance (which is what I usually do). I think its also an approach that many follow when designing the grounding - layered star earths and all that.
I agree, these are two different topologies, all I'm trying to understand is why the L-W topology is described as unique because the signal current flows in a separate loop from the power supply, when it seems possible for a non L-W design to separate the loops too.
Both topologies have to create a loop for the signal current which must pass through the output tube and the OPT and the power supply must also provide current through that path.
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One of the FIRST things I learned through DIY experiments was that EVERYTHING I did before the regulators, was heard THROUGH the regulators !!
How does this jive with your words
" Even a simple regulator will allow to use cheaper parts in the raw supply preceding it, with absolutely no sonic penalties. "
Sorry, 100% disagree from experience direct.
The next thing I learned was, there is no substitute I know of, for a great supply. Maybe LW is an exception, we shall see, or hear, over time !! I am open to this as a possibility.
I LOVE two stage DC triode amps !!
Jeff Medwin
The key is how much current passes through each path, which is a function of their relative impedance (which differs with frequency also). That differs from one design to the other. Virtually any cap is in some sense part of the signal path. But since not all paths are created equal, the influence is not equal. For instance any cap distortion in a cathode bypass cap is multiplied by the mu of the tube. A coupling caps distortion is added at unity gain.
Sheldon
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Although I don't have any direct experience myself (never used a regulator) I have read many reports from others with observations that agree with yours , i.e. the power supply will be heard and felt because we have not yet a perfect way to isolate it from the amp with any practical regulator design and real world components.
The L-W is not perfect either of course. In order to isolate the output tube from B+ it 'float's it on the B+ noise. To do this it has to ensure the B+ noise is present at the grid of the output tube and it can't do that perfectly. It feeds the B+ noise via capacitors and resistors to the cathode of the input tube which then amplifies it the anode which must therefore introduce some non-linearity to the B+ signal it's trying to replicate at the grid of the Final.
And it means that even if the L-W topology separates the output signal current loop from the power supply, the input signal and the B+ noise are both flowing through the high gain input triode.
Perhaps it's possible that a very good shunt regulator on B+ is technically superior to the L-W but it will accomplish that only at the expense of more complexity and power dissipation.
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Of course you can kill all questions about PSU ripple and Zout with the use of modern regulators. But that was not the point of the discussion.
About the sonic penalties, if your audio circuit's performance depends on low ripple and/or low Zout, the regulator will become part of the audio circuit. Any detrimental sonic impact may be subject to discussion and depend on the individual case, but you are certainly introducing new active components into the audio circuit.
About the cost, L-W will do with 2 small caps in the psu and a 5H choke from the junk box. Not exactly what I would call pricey, although the SS regulator parts may be even cheaper.
A cathode to cathode resistance string question, Sheldon or ??
Sheldon,
I am presently contemplating building two different two-stage DC amps, a LW and a conventional Rk bypassed one, and compare the two amps for myself, to learn from direct experience.
A LOT of work, and expense, but ....both should be really nice. I will be able to test them full range on ALTEC A7-800s.
I have a question, when calculating the "cathode to cathode resistance string". Do we use the mu of the input tube "as if its Rk is bypassed", (as I THINK it is shown in the "average characteristic graph "), OR, do we reduce the mu displayed with some formula to represent an "unbypassed Rk tube mu" ?
Sheldon, your original advice was as so, shown next:
"Look at soulmerchant's schema for how to set up the PS noise cancelation. This calculation is simple: The cathode to cathode resistance string, divided by the input cathode value, should be equal to the u of the input tube. The actual optimal value is determined by adjusting by ear, around the calculated value, and will vary some from tube to tube."
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What is the best approach to define mu, as a bypassed Rk on the data sheet, ( IF I am interpreting that correctly ), or otherwise ???
TIA,
Jeff Medwin
PS :
As I begin to develop the DC LW designs, I will probably post some of my choices, to have folks confirm I am doing it right.
Initially I am THINKING of using 1/2 of a 6SL7 at about 0.7 mA and 195 VDC Ea, as the plate resistance there is only 52,000 Ohms. For the TAC or Triode Amp Characteristic, Rp only needs to be 208 K Ohms or higher, to get a factor of 4 times happening. I prefer no cathode follower / Darius middle stage added, and I think its easy to get a TAC of 5.5 or so.
Output tubes will be either a 6AH4GT at 25 mA. or a Type 45 at 27 mA. Power supply B+ filter, four Hammond 159ZBs in either amp.
JM
Sheldon,
I am presently contemplating building two different two-stage DC amps, a LW and a conventional Rk bypassed one, and compare the two amps for myself, to learn from direct experience.
A LOT of work, and expense, but ....both should be really nice. I will be able to test them full range on ALTEC A7-800s.
I have a question, when calculating the "cathode to cathode resistance string". Do we use the mu of the input tube "as if its Rk is bypassed", (as I THINK it is shown in the "average characteristic graph "), OR, do we reduce the mu displayed with some formula to represent an "unbypassed Rk tube mu" ?
Sheldon, your original advice was as so, shown next:
"Look at soulmerchant's schema for how to set up the PS noise cancelation. This calculation is simple: The cathode to cathode resistance string, divided by the input cathode value, should be equal to the u of the input tube. The actual optimal value is determined by adjusting by ear, around the calculated value, and will vary some from tube to tube."
-------------------------------------------
What is the best approach to define mu, as a bypassed Rk on the data sheet, ( IF I am interpreting that correctly ), or otherwise ???
TIA,
Jeff Medwin
PS :
As I begin to develop the DC LW designs, I will probably post some of my choices, to have folks confirm I am doing it right.
Initially I am THINKING of using 1/2 of a 6SL7 at about 0.7 mA and 195 VDC Ea, as the plate resistance there is only 52,000 Ohms. For the TAC or Triode Amp Characteristic, Rp only needs to be 208 K Ohms or higher, to get a factor of 4 times happening. I prefer no cathode follower / Darius middle stage added, and I think its easy to get a TAC of 5.5 or so.
Output tubes will be either a 6AH4GT at 25 mA. or a Type 45 at 27 mA. Power supply B+ filter, four Hammond 159ZBs in either amp.
JM
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