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Old 14th September 2007, 09:02 PM   #1
jayme is offline jayme  United States
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Default Aikido circuit question

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Don't the 1M safety resistors on the output stage form a voltage divider, and inject half of the B+ noise into the output coupling capacitor?
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Old 14th September 2007, 09:22 PM   #2
DougL is offline DougL  United States
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It would if you pull the first tube.

Otherwise the Rp of the lower triode + Cathode Resistor of the lower triode will "swamp" the 1 Meg Resistor.

Think Zout vs. 1 meg voltage divider.


I was looking at the other tube. Same principal applies to the output.
Scienta sine ars nihil est - Science without Art is nothing. (Implies the converse as well)
Mater tua criceta fuit, et pater tuo redoluit bacarum sambucus
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Old 14th September 2007, 09:22 PM   #3
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I would say yes and no.

The noise is injected into the output, at 1/2 the magnitude of ripple, and at a 500K source impedance.

The musical content, however, is also injected into the output, at a much higher signal level, and a much lower source impedance.

The end result is that at the source impedances involved, the injected noise is benign to nonexistent.
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Old 14th September 2007, 09:31 PM   #4
jayme is offline jayme  United States
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So...given a "noisy" CLCLC PSU with some B+ fluctuation due to poor wall power, the likelihood that the Low Frequency B+ noise is being injected into the outputs at this point is very low, correct?
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