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bbakota2000 7th May 2004 10:17 PM

Gain in Aleph P 1.7
I have a question regarding Aleph P gain. In the position of 2K pot on the original schematics I want to put a resistor, so I can achieve max gain (22dB balanced/ 16 dB unbalanced). Which resistor value should I use? I will use Aleph P together with an Aleph 5. I will put stepped attenuator on the output. Thanks for advise!

Nelson Pass 8th May 2004 01:20 AM

You can put in any resistor you like. For 22 dB, you want to
look at the 100 to 200 ohm range.

kilowattski 8th May 2004 02:58 AM


My Aleph P 1.7 is coming along nicely also. It seems there has been a rash of Aleph P 1.7's being built lately. I would like to include a pair of Gain Controls on the front panel just like the genuine article. Can I use a linear taper pot? Would you recommend a high quality pot like an Alps? I have seen alot of press about conductive plastic pots. They are claimed to be very long lasting. Would that be suitable, not that it would get changed often?

bbakota2000 8th May 2004 11:39 AM

I remember reading somewhere that I shouldn't go below 800ohms, and the max is around 1.5K. I think it was related to Aleph P.
Is that feedback resistor?

jh6you 8th May 2004 03:28 PM


Originally posted by kilowattski

Can I use a linear taper pot?

I would try to use 6-step rotary switch of good guality with five 400 ohm registers--i.e. to get 0, 400, 800, 1.2K, 1.6K and 2K at each step. I approximate that these register values will give you 22dB, 14dB, 10dB, 7 dB, 5dB and 3dB gains. Or, you could use different register values.

Good luck.

kilowattski 9th May 2004 01:27 AM

Thank you!

tomchaoda 17th May 2004 02:04 PM

How do you achieve the maximum 22dB (balanced) from Aleph P1.7? by lowering the 2K VR or by close the sw 1.1 & 1.2?

What's the formula of this gain?


jh6you 17th May 2004 02:26 PM

I assume that the gain is approximately X = (R53//R56)/(R51+R66+2K pot set value)--i.e. 20 x log(X) dB.

Klaus 17th May 2004 02:35 PM

R49 missing?

Originally posted by jh6you
I assume that the gain is approximately X = (R53//R56)/(R51+R66+2K pot set value)--i.e. 20 x log(X) dB.

I think, we also need to consider R49 - it lays in parallel to R53//R56 for AC.



bbakota2000 17th May 2004 03:09 PM

I tried to use this formula with 820 R value for a fixed resistor instead of a pot, but I got a value of 0.00206 for X. When I calculate the 20 x log(X) it turns out to be -53.72dB. What does this value mean?

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