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Hoffman Amps Forum image Author Topic: power supply help  (Read 3571 times)

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Offline ernie_jr

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power supply help
« on: August 17, 2011, 01:51:43 pm »
Schematic calls for a 750 volt CT P/T with 3 and 4th nodes opn their own 22K resistor. I have a P/T that gives 650 volt C/T
Would I need to increase or decrease the value of the resistors to bring the voltage at each node to the proper level.
Thanks,
Ernie

Offline sluckey

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Re: power supply help
« Reply #1 on: August 17, 2011, 02:00:56 pm »
Still talking about the B15? If so, decrease.
A schematic, layout, and hi-rez pics are very useful for troubleshooting your amp. Don't wait to be asked. JUST DO IT!

Offline ernie_jr

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Re: power supply help
« Reply #2 on: August 18, 2011, 07:42:07 am »
Thanks Slukey
will keep forum posted on progress
Ernie

Offline ernie_jr

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Re: power supply help
« Reply #3 on: August 22, 2011, 07:23:23 am »
Ok, If I have a p/t  325 volt split, I should get 455 volt d/c (325 x 1.4) if I run a choke for the 2nd node, voltage should be  about 450. For my 3rd and 4th nodes, if I use a 1K to the 3rd and a 12K to the 4th node, will I get approx the voltage I need at 425 for the 3rd node and 325 for the 4th? tubes are a 6an8 and 2 x 12ax7.
Thanks,
Ernie

Offline jjasilli

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Re: power supply help
« Reply #4 on: August 22, 2011, 10:40:56 am »
Ohm's law: Vdrop = R X I;  Go to tube charts:  http://www.mif.pg.gda.pl/homepages/frank/sheets/106/6/6AN8.pdf

6AN8 draws: 15 + 12 + 3.1mA = 30.1mA.  You want to drop 25V (I think?) to the 6AN8.  25 = R X 30.1mA; R = 25/.031 = 806Ω.  (Note that your choke may not drop any voltage at the screen node, so yo may need to re-compute.)

Do same for 12ax7.  Assume it draws 2mA (or plot a load line for more accuracy); tweak from there.

stratele52

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Re: power supply help
« Reply #5 on: August 22, 2011, 02:23:33 pm »
3.1 ma or 3.8 ma ?

I try to follow your project

Offline jjasilli

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Re: power supply help
« Reply #6 on: August 22, 2011, 03:48:26 pm »
Whoops: should be 3.8mA, not 3.1 for the screen draw.  25/.038 = about 660Ω.  If a 30V drop is needed, then 30/.038 = abut 790Ω.

Offline ernie_jr

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Re: power supply help
« Reply #7 on: August 23, 2011, 10:21:19 am »
looks like i can try a 680 and an 820 to see which gets me closest, I have both in stock
thanks
ernie

Offline jjasilli

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Re: power supply help
« Reply #8 on: August 24, 2011, 02:59:42 pm »
You can also put whatever you have on hand in series or parallel (just pay attention to wattage ratings).  Close enough is good enough.  I was trying to be precise to illustrate calculations; but there's lots of wiggle room in the actual circuitry.

Offline PRR

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Re: power supply help
« Reply #9 on: August 24, 2011, 11:55:31 pm »
> 6AN8 draws: 15 + 12 + 3.1mA = 30.1mA.

Only as an extreme tube-book show-off example!

In the Ampeg, in audio generally, we have 100K-200K resistors dropping about half the supply voltage. Worst-case, 400V supply, 200V across 100K, is 2mA. (Pentode screen may draw another 25%, 2.5mA.)

After you get past the power-tube screen-grids, it is often simplest to just drop in a coupled 10K resistors. The preamp stuff WILL work. Then read the actual voltages, compare to "desired" voltages, and adjust the 10K up/down to shift B+ down/up.

 


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