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Old 29th October 2003, 08:20 PM   #1
00940 is offline 00940  Belgium
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Default Tube amp, output capacitor and headphones impedance

On my old RKV, there are two big output caps. Electrolytic, 220uf, 350V. Fact is, they need to be changed, they made their time.

The question is : are such big caps useful for 300ohms headphones (I don't plan to use anything below) ?

I found this formula on headwize, giving the frequency corner (3db down) in function of the size of the output cap and the impedance of the cans:

C (uF) = 1,000,000 / [2pi (corner frequency) Rheadphones]

So :

47 = 1,000,000 / [2 x 3.1416 x corner frequency x 300]
=> 47 x 2 x 3.1416 x 300 = 1,000,000 / corner frequency
==> corner frequency = 1,000,000 / 88,593 = 11.30Hz

Which is fine.

The good thing now is that a 47uf, 400V polyprop audyn cap is only 10.25 a piece. And a high quality electrolytic caps, with this voltage rating would cost arms and legs. And I'm not even speaking of 220uf/385V blackgate (around 130$ a piece).

Is everything here allright or is the output capacitor also dependant of the circuit as a whole ?
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Old 30th October 2003, 01:03 AM   #2
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Hi,

Quote:
Is everything here allright or is the output capacitor also dependant of the circuit as a whole ?
No, it's load dependent.

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Old 30th October 2003, 09:50 AM   #3
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Quote:

No, it's load dependent.
Oh, really?
Actually it depends on both load and Rout of the circuit...

C (uF) = 1,000,000 / [2pi (corner frequency) (Rheadphones-Rout)]
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Old 30th October 2003, 09:59 AM   #4
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Hi,

Quote:
Actually it depends on both load and Rout of the circuit...
Sure but I considered that as a known factor not a variable.

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Old 30th October 2003, 10:22 AM   #5
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Quote:

Sure but I considered that as a known factor not a variable.
Well it don't have to be a variable to cause troubles
With Rout=200 Ohm (not so uncommon with tube amps) the output cap has to be 3 times bigger to keep the same corner frequency...

Good luck
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Old 30th October 2003, 11:19 AM   #6
00940 is offline 00940  Belgium
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how easy (difficult) is it to determine Rout ?

I don't know a lot about tube amps, it's just maintenance here

I should perhaps just get a pana FC of the same value and bypass it with a high quality polyprop
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Old 30th October 2003, 11:24 AM   #7
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Hi,

An easy to read paper explaining basics can be found here:

AIKENAMPS

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Old 30th October 2003, 11:51 AM   #8
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hum, the design of the RKV is awfully complicated and i can't associate it with any of those schematics.

Is there a practical way to measure the Rout, i mean with a multimeter or such ?

http://www.audiovalve.de/rkv/rkvpatent.jpg here's a bad pic of the schematic; if you feel you could help, i can provide you the actual schematic but i've been asked not to make it public.
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Old 30th October 2003, 12:11 PM   #9
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Quote:

Is there a practical way to measure the Rout, i mean with a multimeter or such ?
Yes, there is. First apply some signal to the input and measure output (say U0) unloaded. Then load output with resistor Rl(e.g. 0.5-1 kOhm) and measure output (U1) retaining the same input.

Rout = Rl*(U0/U1 - 1)

Good luck!
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Old 30th October 2003, 12:15 PM   #10
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Hi,

The outputstage on the patent looks like some sort of SEPP circuit.

It should have very low Zout if the tubes are run in triode mode.
I can't tell from the patent drawing...

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