Is regulated PSU for phono preamp a MUST?

[chris ma]

The one difference which a transformer can make (particularly for a phono-preamp) is the energy it radiates itself -- EI cores and torroids have different patterns.

The other difference is that attibutable to from the tank circuit formed by the secondary inductance, the diode's capacitance AND the capacitance between primary and secondary windings. The ringing can be dealt with by placing a RC across each diode, or across the diode bridge. If you don't have the equipment to measure the transformer inductance, try using 10nF and 220R in series.

I am not sure how to measure this inductance and what all this really mean. I measure the secondary between V and center tap and it display 29.1mH when the multimeter set to measure L. See pic...The primary measured 1.22H if this mean anything?

 

[jackinnj]

I am not sure how to measure this inductance and what all this really mean. I measure the secondary between V and center tap and it display 29.1mH when the multimeter set to measure L. See pic...The primary measured 1.22H if this mean anything?

Short the two primary leads together, then measure the secondary inductance. Then measure the capacitance between the primary and secondary windings.

The manufacturer's datasheet for the diodes will specify a capacity at a specific voltage. for the MUR860 it should be 100pF at 30V.

the ringing frequency can be found from this formula:

F = 1/ [2* pi * sqrt (LC)]

where C is the sum of the diode capacitance and the interwinding capacitance.

the snubber resistor is then:

Rs = sqrt (L/C)

the snubber capacitor is then:

Cs = [2*pi*sqrt(LC)] /Rs

all of the above is clearly discussed on Jim Hagerman's paper "Calculating Optimum Snubbers". http://www.hagtech.com/pdf/snubber.pdf
This in turn is based upon a white paper which Cornell Dubilier put out.

 

[chris ma]

I shorted the primary and measured L between the secondary I read 2.91mH
I also shorted the primary as +ve and shorted teh secondary
as -ve and measured 68.2pf

so for the F = 1/ [2* pi * sqrt (LC)]
F = 1/[2* pi * sqrt(2.91*168.2)

 

[chris ma]

I used R=sqrt(L/C)

R=sqrt(2.91*100/168.2)
R=1.315

I guessed 1.315*1000 = 1k3 ohm


+++++++++++

C=2pi * sqrt[LC] / 1300

I am having trouble with the mH and pF, conversion and number of zeros is driving me nuts!

 

[jackinnj]

don't forget that the units of millihenries are 10^-3, and picofarads are 10^-12.

(I am guessing you are using +/- 15 V rails, the diode capacitance is around 180pF at this voltage (capacitance varies inversely with voltage).

Rs = 3.3K

Cs = 2.2 nF

you can try a "before and after" to see if it makes a listenable difference.

 

[chris ma]

chris ma,
if you encounter problems how to stuff the regulator boards i could help you out with the circuit for that specific boards.

and the AD825: is there any dip8 replacement that provides similar performance?

Rüdiger

Do you mind sharing how do you connect to the regulators. I am yet to make those two jumper wires that I can see from picture on the AOS site. These jumper wires are not mentioned in the schematic. Is the +ve load means the output +ve to the phono stage?

 

[Onvinyl]

Its best to connect the ve load and the vreg_sense direclty at the load. if you for what reasons ever choose not to do so, jumperwire it on the board.
hth,
Rüdiger

 

[chris ma]

I am not quite sure how to connect to the load if not jumper like the photo.

 

[chris ma]

updated picture

 

[Mikett]

Please post a picture of a positive board that you have. Post both sides so I can follow the traces.
BTW, have you read the description of the original article from AudioXpress/Audio Amateur as yet? It will provide much insight to what is going on.

Also see this
http://home.comcast.net/~walt-jung/wsb/PDFs/Improved_PN_Regs.pdf

I'm busy this week, otherwise I could always head out your direction one evening next week and get it started.

 

[chris ma]

Is the red dots that I am not sure how to connect.

Positive reg

 

[chris ma]

I got confused with the remote sense and local sense that it does not make any sense to me 🙁
Call me dumb🙂

Negative reg

 

[Mikett]

OK, it is confusing to someone who has not really understood what the heck the super reg is trying to do as yet since this regulator is unlike others you might have done before.

http://www.alw.audio.dsl.pipex.com/Manuals/ALWSR_rev2.9_Iss005s.pdf

Take some time tonite and read the assembly manual on connections.

When you connect everything up in a "local" sense fashion, you are turning the super reg configuration into a more conventional power regulator. The voltage is maintained and regulated UP TO the output of the board itself.

When you connect everything up in the REMOTE sense configuration, you are regulating UP TO WHERE the remote sense is connected. This way, any impedances along the connections to the powered circuit is compensated for by the feedback. In essence you are EXTENDING the actual regulator power circuit RIGHT UP TO WHERE the IC actually is.

Since you are going dual mono P2P on the phono, this will likely be right up to the IC power connection. (Don't forget decoupling caps for the IC)

If you read the link supplied you'll get a better understanding of this down around page 24 or so. Due to the high bandwidth, the remote sense is best done using shielded wire.

I can't be specific on the tracking connection since I don't know exactly how AOS has done this on the board. In other words I don't know if the tracking feedback is connected remotely of from the return on the regulator itself.

Figure 4 should shed some light

 

[chris ma]

I think I guessed it now after reading the pdf.

The aos pcb is already set up to join the OP1, OP2 and TRS1 as 1 point. The TROUT and SRIN already connected. So as remote sense connection please verify the attached

 

[Mikett]

Good, you're on the right track. Keep leads as short as possible and remember the gnd sense screen is there for a reason. Hope it turns out well.

 

[chris ma]

don't forget that the units of millihenries are 10^-3, and picofarads are 10^-12.

(I am guessing you are using +/- 15 V rails, the diode capacitance is around 180pF at this voltage (capacitance varies inversely with voltage).

Rs = 3.3K

Cs = 2.2 nF

you can try a "before and after" to see if it makes a listenable difference.

Hi,
Do you mind show me how you get the Rs=3.3k with the formula from the optimum snubber pdf ? I am still confused with the number of Zeros😕

 

[chris ma]

Help with maths

Please see if I have any mistake here? I am not trying to prove you wrong I just want to make sure I understand how to use the formulas so that when I use a different transformer I can calculate new parts. That's why I am not sure how Carlos can recommend all those snubber values without knowing all different transformers that people are using in the chip amps.

See my calculation here:

L=2.91mH
C=68.2pF + 180pF=248.2pF



L=2.91mH=0.0291H
C=248.2pF=0.0000000002482F

Rs=sqrt(L/C)
Rs=sqrt(0.0291/0.0000000002482)
Rs=sqrt(117244157.937147461)
Rs=10827.9341


Cs=[2*pi*sqrt(LC)]/Rs
Cs=[2*(22/7)*sqrt(0.0291*0.0000000002482)]/10827.9341
Cs=0.000000000156010404798

If 10^-9 = n
Then Cs=1.56nF

So I get
Rs=10k8
Cs=1.56nF


Sorry that I have only O level maths and that was grade B I got many many many years ago!!

 

[AndrewT]

Hi Chris,
almost there, only lost one decimal place.
3k42 & 1n56F therefore use 3k3 & 1n5F.

Try using a calculator with Exp function then it's much easier.
e.g. 2.91mH enter 2.91 Exp 3 +- => .00291H
The exp multiplies 2.91 by 10^3 (or 1000) +- changes the sign from + to - which gives 2.91 * 10^-3
and 68.2pF is 68.2 Exp 12+- => 68.2 *10^-12

This way you don't have to keep count of all the zeros
I used the formulae from your last post, check them again to be sure. I am very surprised that the resistor is so large but maybe this is due to the smallness of the transformer.

Congratulations on taking on board such a complex piece of research with limited maths abilities. As a result of you sharing your work I have learned something new as well.

 

[chris ma]

Hi Chris,
snip.... I am very surprised that the resistor is so large but maybe this is due to the smallness of the transformer.

snip..l.

Can I assume that the bigger the transformer gets the value of the resistor would tend to be smaller?

 

[AndrewT]

Hi,
I hope some expert can tell us.
Maybe a table of component values related to transformer type/size would be ideal.