Power Supply Resevoir Size

[Kindhornman]

Daniel,
If you do use the Schottky diodes instead of a simple diode bridge would a snubber circuit take care of the rest of the filtering to keep the noise out of the circuit. I am talking about the resistor/capacitor snubbing, not just the capacitor across the diodes. I assume that you would want to run two separate bridge rectifiers one for each channel with the snubber circuits.

Steven

[OnAudio]

Originally Posted by Terry Given

OnAudio #799 - feel free to try this; you will attempt to do so only once. regardless of how you choose to pick Zsec for termination, you'll end up dumping metric truckloads of power into your termination - anywhere from 100% to 2000% of rated VA.

an unused winding generates no H field - there is no loop within which current can flow. there is of course an E field, but its pretty well behaved with very low dV/dt = 2*pi*Fac*Vpk. this can be minimised by chopping the leads off, or just folding them back and forth a few times and securing with a cable tie (handy if you want to re-use the xfmr later).

Any other hairy mess (eg nastiness from ac line) that might be present on an unused winding will also appear on the other (presumably in use) windings.

Thanks for your contribution.

Originally Posted by gootee

"Tonight" might have been a WEE bit optimistic.

I haven't had much time to work on anything in the last few days. I did have a nice set of distortion vs reservoir capacitance plots and a table of the actual minimum C_reservoir for SQUARE-wave signals, with various transformer and load combinations, similar to the stuff I posted for sine signals, but decided I needed to re-run everything, before posting them, which will take a few more hours.

Tonight is also good .

[OnAudio]

Originally Posted by Kindhornman

Daniel,
If you do use the Schottky diodes instead of a simple diode bridge would a snubber circuit take care of the rest of the filtering to keep the noise out of the circuit. I am talking about the resistor/capacitor snubbing, not just the capacitor across the diodes. I assume that you would want to run two separate bridge rectifiers one for each channel with the snubber circuits.

Steven

This power supply is deceptively simple. Try it first If you no like then examine alternatives



Also examine the power supply here: To reach such a design requires hours and hours of comparisons. This is vintage wine

http://passdiy.com/pdf/BA2%20r1.pdf

[danielwritesbac]

Originally Posted by Kindhornman

Daniel,
If you do use the Schottky diodes instead of a simple diode bridge would a snubber circuit take care of the rest of the filtering to keep the noise out of the circuit. I am talking about the resistor/capacitor snubbing, not just the capacitor across the diodes. I assume that you would want to run two separate bridge rectifiers one for each channel with the snubber circuits.
Steven

I don't know anything about running two bridge rectifiers from a single transformer. Perhaps it would complicate something?
P.S.
At post 840, I wasn't talking about bridge rectifiers at all.
P.P.S.
What I was actually talking about is, instead of regulators for v+ and v- private per left and right channels, schottky diode drops instead of regulators for accomplishing virtual dual mono more cheaply.

[gootee]

Originally Posted by seppstefano

Hi Steven,
yes exactly, I was wondering if plainly move the capacitance where it's needed might be a simple solution to the issues given by parasitic inductance of the rails. I might be completely wrong but I understand the CRC as in Resistor-Capacitor (RC) Filters. In case of a power amp (like mine) with 2 caps/rail I was wondering if it might be the case to get rid one of one of them and move more capacitance (at least 1/2 capacitance for the +rail and 1/2 on the negative rail, with film bypass and a snubber) as close as possible to the output stage. This way we wouldn't have to add a R as the rails "offer" a resistive component "for free".

Hope it makes some sense,

Stefano

Exactly, yes!! You have seen the light!! (All on your own. I'm impressed.)

We definitely need local "decoupling" capacitance for each active device, usually between the power rail input and the load ground return point, usually impossibly-close to the device, if done well-enough.

The local capacitors act almost like small point-of-load power supplies. They are usually meant to supply the fast or large transient current demands, which can't accurately (in time) get through the rail inductance, and which would also induce a relatively-large voltage across the rail inductance. That large voltage is why most people want to "decouple" the load from the rest of the power distribution circuit.

(You also need small-sized local "bypass" caps from power rail to load ground, for each device. But those are for HF stability, to short out (at HF) the hidden HF positive feedback loop through the power rail that almost all transistor amp circuits have. An extra couple of mm in connection length might make them almost useless, by the way.)

With reasonably-long power/gnd rail conductors from the PSU, the lower frequencies can mostly come directly from the PSU, so I don't think that the decoupling caps need to be as large as the power supply reservoir caps, although it couldn't hurt if they were. But the decoupling caps are absolutely essential for the higher-frequency response, which includes both the closed-loop "internal" response as well as fast edges etc in the signal. Depending on the amplifier's transient frequency response and max slew rate, they will probably need to be able to support frequencies up to 100kHz to 300 kHz, and for some equivalent full-scale rise-time, where frequency, f = 1 / ( π ∙ trise ) .

NOTE: You will have to calculate BOTH the required capacitance AND the inductance (impedance, actually) that can be tolerated in the connection length of the capacitance to the decoupling points. (In practice, you might have to just get as much capacitance as close as you can with the lowest total inductance and then calculate (or simulate) backwards to see how you did, and how much rail-voltage disturbance there will be, worst-case.)

I figured out how to do many of those calculations (and also came up with a way to make the PSU impedance, as seen by the active device, as low as you want). See Post 27, at:

http://www.diyaudio.com/forums/solid...ml#post3097232

and especially the links near the bottom of that post, which go to the actual calculations, et al.

NOTE that in many of those calculations, I was using the WRONG value for the self-inductance of a conductor. There are on-line calculators for when you need to be accurate but an accepted rule-of-thumb is 25 nH per inch, or 1 nH per mm. I think I was using 15 nH per inch, at some of those links. (25 nH per inch will make it even more difficult to implement it well, physically.)

Cheers,

Tom

[gootee]

Attached are a few results for the Cmin reservoir capacitance, with square waves.

Not enough data for comparisons etc, yet, but:

The 360 VA-per-secondary case for 4 Ohms and 150 Watts looks reasonable.

It looks like 240 VA per secondary is either working hard to get 150 Watts into 4 Ohms, or, the 360VA secondaries finds it to be very easy.

And even 480 VA per secondary looks very close to not being enough for 200 Watts into 4 Ohms, with only a 36 Volt transformer output.

Cheers,

Tom

[Kindhornman]

Tom,
I know that it isn't a 1 to 1 ratio but could you approximately double the values for an 8ohm load and have reasonable values? I was surprised how much higher the Cmin value was when you increased the power output from 150 to about 200 watts at 4 ohms. It looks like a steep gain in capacitance value for a 30% increase in power output.

[gootee]

Originally Posted by Kindhornman

Tom,
I know that it isn't a 1 to 1 ratio but could you approximately double the values for an 8ohm load and have reasonable values? I was surprised how much higher the Cmin value was when you increased the power output from 150 to about 200 watts at 4 ohms. It looks like a steep gain in capacitance value for a 30% increase in power output.

Yeah, but the transformer VOLTAGE is then apparently way too LOW, for that (200 Watts), so it needs a whole lot more C just to keep the voltage and current ripple in the right zones, otherwise the signal gets bashed. (Or, maybe the crappy transformer model has something to do with it. I haven't done the calculations for that case.)

[gootee]

Originally Posted by Kindhornman

Tom,
I know that it isn't a 1 to 1 ratio but could you approximately double the values for an 8ohm load and have reasonable values? I was surprised how much higher the Cmin value was when you increased the power output from 150 to about 200 watts at 4 ohms. It looks like a steep gain in capacitance value for a 30% increase in power output.

For an 8 Ohm load instead of 4 Ohms, you would divide them by two, rather than double them.

See the results for sine signals, which also include some 8-Ohm cases, in Post 740, at:

http://www.diyaudio.com/forums/solid...ml#post3134179

Cheers,

Tom

[Kindhornman]

So something like a 600VA transformer would handle the ripple problem with half the capacitor value then?