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#26 pre amp
#26 pre amp
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Old 1st June 2011, 07:27 PM   #1031
Rod Coleman is offline Rod Coleman  United Kingdom
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#26 pre amp
Quote:
Originally Posted by E.T. View Post
Hi Rod,

I don't understand why the first cap should have such a low value. Is this to protect the rectifier tube?

With the PT I have, a choke input would drop the B+ voltage too far. I was assuming 40uF / choke / 40 / choke / 40 plus various bypass caps. How would I integrate a 0.47uF into that? Does the ground lug of the small-value cap still function as the star ground for high-current stuff?
Hi, The reason for choosing choke-input is that the rectifier conducts for a longer part of the mains waveform.

This is kinder to the rectifiers, regulates better, and reduces electromagnetic emissions.

It is relative though, so if you need to have cap-input, just use a 40uF in place of the 0.47uF.
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Old 1st June 2011, 08:08 PM   #1032
TheGimp is offline TheGimp  United States
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You can t"Tune" the output voltage of a CLC filter by adjusting the value of the first cap.

The lower the value of the first cap, the higher the output voltage and the more efficient the system becomes as the inductor actually stores energy temporarly and returns it to the system.
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Old 1st June 2011, 09:39 PM   #1033
andyjevans is offline andyjevans  United Kingdom
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Quote:
Originally Posted by TheGimp View Post
You can t"Tune" the output voltage of a CLC filter by adjusting the value of the first cap.

The lower the value of the first cap, the higher the output voltage and the more efficient the system becomes as the inductor actually stores energy temporarly and returns it to the system.
I thought the lower the value of the first cap, the lower the output voltage since you are going towards choke input which is near 1:1 rather than 1:1.4 with cap input. Am I not seeing something here?

Andy
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Old 1st June 2011, 09:42 PM   #1034
massimo is offline massimo  Italy
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Quote:
Originally Posted by andyjevans View Post
I thought the lower the value of the first cap, the lower the output voltage....
it goes without saying.....
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Old 2nd June 2011, 01:22 AM   #1035
kevinkr is offline kevinkr  United States
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#26 pre amp
Quote:
Originally Posted by TheGimp View Post
You can t"Tune" the output voltage of a CLC filter by adjusting the value of the first cap.

The lower the value of the first cap, the higher the output voltage and the more efficient the system becomes as the inductor actually stores energy temporarly and returns it to the system.
I think they were discussing choke input filters with a small input capacitor to tweak the output voltage. In any event the inductor is storing energy in its magnetic field whenever load current flows. From direct experience this works and it is possible to adjust the output voltage over a range of about 0.9 to 1.4 times the rms value (not that I would over that range) depending on the capacitor used - but only with a supply intended to operate as a choke input. Note the above comments are predicated on the fact that the load current is above the inductor's critical current for regulation if not it behaves like a standard CLC filter.

Some old timers actually considered it to be bad form to use an input cap to trim the output voltage on a choke input filter.
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Last edited by kevinkr; 2nd June 2011 at 01:27 AM.
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Old 4th June 2011, 05:38 AM   #1036
airegin is offline airegin  Italy
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Default Telefunken RE 034

Hello, i have a pair of tfk re 034 (and a pair of equivalent Philips A 425) that would like to use in a linestage like this. any suggestion on operating point ? thanks!
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Old 4th June 2011, 10:20 AM   #1037
merlin el mago is offline merlin el mago
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I'm happy, arrived today Yamamoto teflon UX sockets for 26, socket for AZ1 & tube Siemens AZ1 NOS
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Old 4th June 2011, 10:29 AM   #1038
merlin el mago is offline merlin el mago
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I want to use my 6.3 output but I need to reduce it to 4VAC for AZ1, how can calculate the resistor value because Ducan PSUD2 can't have AZ1 mixed with diodes as hybrid rectifier?
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Old 4th June 2011, 11:56 AM   #1039
massimo is offline massimo  Italy
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Ley de Ohm: Ley de Ohm - Wikipedia, la enciclopedia libre
The AZ1 has a 4 V filament rated at 1.1 A
6.3 - 4 = 2.3 Vdrop
R = V/I thus R = 2.3/1.1 then R = 2.09 Ohm
The resistor should dissipate P = I2 x R thus 1.21 x 2.09 then P = 2.53 W
I would use 2-2.2 Ohm/10 W minimum, or, better, split the resistor and use 2 x 1-1.1 Ohm/5 W, one resistor on each cathode.

Quite easy

Edit: http://www.stud.feec.vutbr.cz/~xvape...ty/ohm_zak.php

Last edited by massimo; 4th June 2011 at 12:01 PM.
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Old 4th June 2011, 01:26 PM   #1040
merlin el mago is offline merlin el mago
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Grazie per l'aiuto ed il link mille caro sei un sole

Thank you very much for help & the link friend
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Last edited by merlin el mago; 4th June 2011 at 01:42 PM.
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