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#11 |
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diyAudio Member
Join Date: Sep 2006
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Hi Salas,
Thanks for the reply. What about CLC after the reg ? Would it cause oscillation & drag down the voltage ? Also if higher current is require how can it be achieve ? Many thanks |
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#12 |
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diyAudio Chief Moderator
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Not after. It is totally noiseless as it is and you will ruin its transient advantages that way. It is already correctly terminated.
Before it is the right thing, so to pre filter noise passively up high. Actually you can see a 10H frame choke in the left side of the PSU box pictured above. Part of an input CLC. If you have read my previous directions, you will see that the constant current is defined by the total Led string voltage drop minus the Q1 Vgs divided by R1. For the IRFP9240, Vgs at such voltage conditions is about 3.1-3.2V. Measure the string Led in series with a 10k resistor using a 9V battery. Note their total voltage drop. Then I=(LedsVdrop-3.1)/R1. In practice it will deviate around that, but you are in the ballpark. |
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#13 |
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diyAudio Member
Join Date: Nov 2002
Location: Viña del Mar, Torreon
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Hi Salas,
This shuntreg looks pretty interesting and simple design; I’ll have a closer look to it later. It also looks as a very good candidate to the application we were talking about earlier, the 6V6 lineamp. Cheers |
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#14 |
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diyAudio Chief Moderator
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I already look for a nice looking big sink good for 30W so to make monoblock shunts at 100mA/350V for that. Should look funky on that shiny copper chassis.
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#15 |
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diyAudio Member
Join Date: Jun 2007
Location: UK
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http://www.diyaudio.com/forums/attac...amp=1229193896
Hi Salas, Very nice work. Exactly what I was looking for my tube preamp (on the works) I have a question about the input, How does one get 400V rectified and filtered? can you give an example? I mean what transformer did you use? Thanks
__________________
"I don't remember fighting Godzilla, but that's probably what I would've done" |
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#17 | |
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diyAudio Chief Moderator
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Quote:
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#18 |
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diyAudio Member
Join Date: Sep 2006
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Hi Salas,
Just wanted to confirm. Is V out taken at C3 ? If so what is the wattage of resistor R6 assuming if I built a 50ma version ? Thanks |
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#19 | |
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diyAudio Member
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Quote:
R6 and C3 are a Zobel element when I am reading this right. |
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#20 | |
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diyAudio Chief Moderator
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Quote:
Also if there isn't any special Wattage need for R6, you can use 1/4W. C3 is a dead end for DC, so virtually no energy consumption on R6. Together they form a Zobel terminal block that helps transient overshoot performance. See how it overshoots at IRF840's gate for current without the Zobel. |
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