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Old 15th April 2004, 07:33 PM   #1
SI is offline SI  United States
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Default Building Aleph 2

Hi guys, It has been a few weeks and I finally got the power supply done. The problem is: without anything connected to the PS it's almost pushing out +- 50 V for the rails. I think this is to much for my 50 WV caps. So I need to shed 5-6 Volts. I figure I'll lose couple of volts at the thermistors, If I can find some. Both Newark and Digikey are out of them and lead time is 2 months. I'm open to suggestion, I mean please help.


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Old 15th April 2004, 07:42 PM   #2
BrianGT is offline BrianGT  United States
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Default Re: Building Aleph 2

Quote:
Originally posted by SI
Hi guys, It has been a few weeks and I finally got the power supply done. The problem is: without anything connected to the PS it's almost pushing out +- 50 V for the rails. I think this is to much for my 50 WV caps. So I need to shed 5-6 Volts. I figure I'll lose couple of volts at the thermistors, If I can find some. Both Newark and Digikey are out of them and lead time is 2 months. I'm open to suggestion, I mean please help.


Regards.
Mouser has some in stock. I think that this is the same part that you are looking for:
http://www.mouser.com/index.cfm?hand..._pcodeid=52703

Mouser also has no minimum order, and fairly inexpensive shipping. (no handling charges)

You could also try using some big inductors in a CLC setup.

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Brian
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Old 15th April 2004, 07:45 PM   #3
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Hi,

with the Aleph 2 connected, voltage will probably go down 4 or 5% so that will give you some margin.
Opinions about how much margin you need vary on this forum from 20% to 0%.
Personally Ive got no problem connecting 50V caps to 50V DC since the manufacturer also uses some security margin. Some Caps (i.e. PanasonicFC ) can be loaded with 10% more voltage without any problems.

William
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Old 15th April 2004, 08:49 PM   #4
Coulomb is offline Coulomb  England
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I thought the CL-60 was supposed to be on the AC input to prevent Inrush currents, did I miss another application in the DC side of the Power Supply???

Regards

Anthony
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Old 15th April 2004, 09:00 PM   #5
moe29 is offline moe29  United States
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they are used as R's in CRC filter. Basically they turn in to a power
resistor after they heat up.

as shown in the BZLS article you can use them in your grounding
scheme too.

Thermistors are cool ((or hot, i should say))
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Old 15th April 2004, 09:23 PM   #6
Coulomb is offline Coulomb  England
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So they are used as a pi configuration?

Could I have a CRCLC filter setup, I guess I could but at a heck of a VD across the rails.

Anthony
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Old 15th April 2004, 10:03 PM   #7
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Quote:
Originally posted by moe29
[B]they are used as R's in CRC filter. Basically they turn in to a power
resistor after they heat up.
If your current demands mandate the use of more than one thermistor/surge suppressor, you better hope that the additional thermistors that you put in parallel are absolutely identical. Any deviation in response curves can lead to current hogging and the possible failure of the device and damage to the power supply.

If I remember correctly, Mr. Pass addressed this issue with his AX-200 amplifier design. Check out these posts:

Parallel Thermistors
Parallel Thermsitors
And probably a better solution...

I personally like them for breaking ground loops.

Later,
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Old 15th April 2004, 10:38 PM   #8
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Quote:
So I need to shed 5-6 Volts
Go for heavy chokes.They will sure help you drop a couple of volts.

Bartek
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Old 16th April 2004, 12:11 AM   #9
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Inductors will not do much to drop DC voltage in a CLC configuration. They will do some good if you use them in an LC hookup, though.
That said, I'd run the amp for few minutes to see what the actual rail turns out to be. Running a power supply with no load isn't the same as the rail under load.

Grey
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Old 16th April 2004, 12:11 AM   #10
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Default Re: Building Aleph 2

Quote:
Originally posted by SI
The problem is: without anything connected to the PS it's almost pushing out +- 50 V for the rails. I think this is to much for my 50 WV caps
1) 50V is OK for 50V caps, especially unloaded.

2) The moment you load the circuit, the voltage will drop as
much or more than you think you want.

3) Coils and thermistors will not really reduce the voltage of
the unloaded circuit.

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