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Old 1st September 2009, 12:03 AM   #1
jmillerdoc is offline jmillerdoc  United States
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Default PSUD2 questions

I know this probably isn't an easy one, two question to answer but....can anybody give me a fairly straight forward approach to calcuating the "load" resistance you use to run the calculations. For instance, I am buiding a fairly simple preamp using a single 5687 tube and I wanted to use this program to design a PSU....but I would't know the load resiator value to inpt into the program. I iagine if I am using 2 sides foe stereo then part of the value will be 1/2 the plate resistance at my voltage, but what about any other resistances in the circuit....do you just add them up? Th 5687 tube has a plate resistance of about 3000 at 250v as I recall, so two of these in parallel would be a resistance of 1500. WHAT ABOUT THE PLATE RESISTORS TOO? DO YOU JUST ADD THOSE IN TOO? these are 10k, so would it be the same...add 1/2 of this to the 1500 which equals 7500. Is there any thing else I am missing.
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Old 1st September 2009, 12:11 AM   #2
aardvarkash10 is offline aardvarkash10  New Zealand
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Default set it up as constant current

don't put a resistance in - put the total circuit current. Since your preamp will be running Class A the current is the same at all times, regardless of where its going.
"It may not be easy for some to not hear differences, even if they are not there." - Vacuphile,
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