Aleph X power supply

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I have searched through threads for posts, and posts for threads, and anything I found about the Aleph X I have put into a summary paper. The one part that is still up in the air though is the power supply.

I am hoping for help with designing a power supply capable of delivering 15 volt rails to the transistors into a 4.5 amp load. This will supply roughly 42 watts into an 8 ohm load using hifiZen's board.

I am looking at using a 300VA with dual 120VAC primaries and dual 15VAC secondaries. Another issue here is actual values to use in PSUD2 (such as the resistance of the transformer, and the load to show at the end). I have tried both an 8 ohm load (as suggested if the speakers are 8 ohms), and a 4.5 amp current load. Both yield far different results. Also, changing the resistance of the transformer (as I can't find a value for plitron toroid) makes a large difference. Any help on these values would be appreciated :)

The areas in yellow on my schematic are "options". The 2 0.03 ohm resistors on the mains are thermistors (ala CL-60's). The two smaller caps in parallel on the right along with the RC networks would be placed as close to the transistors as possible.

Thoughts / ideas / help? Thanks :)
 

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PSU

I used PSUD2 for simulating my powersupply. For the regulation number of the transformer I used the values found at www.plitron.com.

But I use a cheap 330VA 2x 15 V trannie so things worked out a little different :). I am using the following setup CRC with for R 0.28 ohm. Both C's are 47.000uF. The voltage I get is ~17 volts. The bias is set to a total of 4,5 A per channel. As bridge I use 4 fast recovery diodes.
 
Re: PSU

Dirt,
Doh... first mistake noticed :) Cut and paste didn't give new numbers on the caps. However, the smaller values should be 100uF then 100nF...


Taco said:
I used PSUD2 for simulating my powersupply. For the regulation number of the transformer I used the values found at www.plitron.com.

But I use a cheap 330VA 2x 15 V trannie so things worked out a little different :). I am using the following setup CRC with for R 0.28 ohm. Both C's are 47.000uF. The voltage I get is ~17 volts. The bias is set to a total of 4,5 A per channel. As bridge I use 4 fast recovery diodes.

I saw that it is 6% regulation, but how does that translate to an ohm rating? since I would be at about half load (it is rated for 10 amps and I'd be pulling just under 5 amps), does it follow that 15 volts at no load / 1.03 = roughly 14.6 volts is what the actual output would be... then ohms law: R=V/I=14.6volts / 4.5amps = 3.2 ohms? So i put 3.2 ohms into the value for the transformer in PSUD?
 
In the transformer dialog of PSUD was the posibility to fill in the VA-rating and the regulation and it calculated itself those other values.

I think the output is higher than 15 volts. So 1.06 * 15 = 15,9 V if no load is connected. But I have to little knowledge about this subject :bawling: .
 
Hi Mach_Y

1. I do not like snubbers on the primaries in the power supply. In my opinion they decrease dynamic. I would leave capacitor 100n – 330n only.
2. I think, that one CL-60 on one supply line is enough. In your case differences in resistance between termistors could disturb work of the transformer.
3. Instead of capacitors C12 and C13 I would give a 1nF or so capacitor parallely to each diode in the bridges.
4. I would add one more R=0.1 ohm and C=20 000uF . Value of 40 000uF per line is too small for me.
5. I would omit R3, C14, R3, C15

This is only my opinion.
Regards
 
Alright... great info, especially on the PSUD [...] button *slaps self* :)

Did a couple of simulations with changes, and came up with the following design. The simulation does not take into account R1 (thermistor) or the final caps (C7 and C8). The numbers in PSUD were for a 15 volt secondary rated at 10 amps (number for 300VA plitron). PSUD filled in values of 15.9 volts at no load with 90milliohms. Using bridge25 for the rectifiers, on a 4.54A load. Mean and RMS voltage across the current load is 15.014 volts with ripple at +/- 0.001 volts :) The numbers for the inductor are from JW Miller model 8113 (digikey part number M9848-ND and are $14.33 each) which is a 5mH inductor rated at 8.9 amps with a DCR of 0.022 ohms. C1 - C6 are all comprised of 10uF caps, C1 and C2 being 3 in parallel, C3 and C4 2 in parallel, and C5 and C6 by themselves. They are all Panasonic ECOS1EP103CA 10000uF caps, 25VDC rating (32VDC surge) with a max ESR of 0.058 ohms, and are $3.71 each at digikey. The resistor is two Mills non-inductive wire-wound resistors in parallel, values 1.0 ohms and 0.33 ohms to yield 0.66 ohms. They are MRA-12 (12 watt rating) and are $3.50 each at www.percyaudio.com. They'll dissipate 9 watts combined.

Any other suggestions? :)
 

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I don't believe those iductors will work... you need a big coil with
fat wire (to paraphrase Mr. Pass).

I use the MCM part number from the SOZ article. I've used them in
a few amps and they work well. Some complain that they're cheap,
but they've been fine for me, no problems.


m.
 
moe29 and jarek: thanks for the suggestion and corrections.

Got ahead of myself on the resistors... :smash: ok, using 1.25 ohm and 1.5 ohm MRA-12 resistors I get 0.682 ohms, which at 4.54A current means 14.1 watts, which should be fine for the Mills.

Not knowing the value of resistance for the 2mH MCM inductor (part number 50-375, at $6.62 each, even better!), I left it at 0.022 ohms. This yields an output of 14.935 volts at 4.54A load, with ripple at 0.003 volts. :)
 

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GRollins said:
A regulated supply, such as the Z5 power supply, is a good idea. Just make sure that you have adequate heatsinking for the regulator MOSFETs. The current draw will be heavy and the Pd will be fairly high.

Grey

Grey, the man who started it all :)

Ok, regulated power supply... I am reading up on this... went through the ZV5 article too.

As far as I understand, I could use a step-down to 15 volt regulator, and decrease the value of R2 and R3 so that the voltage going into the regulator is above 15 volts, correct?

In the ZV5 article, the diodes in parallel with the caps are just for bleeding purposes?
 
Ok, more studying needed by me...

From the ZV5 article, a couple of things would have to change. Primarily, the rating of the transformer would need to increase from 300VA to 500VA. VA rating formula:

VA Rating = 2 * 2.5 * (regulated voltage + 8) * Amp draw
VA Rating = 5 * (15 + 8) * 454
VA Rating = 522

Also, the secondary voltage would need to be changed

Secondary Voltage = (regulated voltage + 11) / sqrt(2)
Secondary Voltage = (15 + 11) / 1.4142
Secondary Voltage = 18.4

That value is under load. The Plitron 500VA transformers have 4% regulation, so 18.4 * 1.04 = 19.14 volts.

A 500VA 19.14vac secondary transformer is not common, and as such would be a custom order. Would it beneficial to go down to 18 volts or up to 20 volts?

Determining the values of R13, R14, R15, R16, Z1, and Z2 I am have trouble understanding though.

Assuming this is built correctly and is capable of producing 15 regulated volts at 4.54 amps, are we left with a better power supply for the Aleph X than the above unregulated?
 
Why the demanding PS?

Why do these amps produce so much ripple? The suggested supplies have big caps, pi filters, even double-pi. Yet an AB amp like the AKSA is supposedly doing well with a single 4700uF cap, for 55W output.

Foolishly I had imagined the PS need not be too fancy, with the Class A balanced design almost self-regulating for constant current draw.

I don't want problems with ripple: I have 105dB/W speakers.

Grant
 
PSUD2

How can I get PSUD2 to simulate the Aleph X power supply pi filter, with R or C on both the plus and minus rails? When I insert and RC element it only puts the R on the plus rail.

Or should I split it into the plus to earth and earth to minus halves?

Grant

Grant
 
Re: PSUD2

nowater said:
How can I get PSUD2 to simulate the Aleph X power supply pi filter, with R or C on both the plus and minus rails? When I insert and RC element it only puts the R on the plus rail.

Or should I split it into the plus to earth and earth to minus halves?

Grant

Grant


Yes, the positive an negative half will behave in the same way.

William
 
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