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Old 7th May 2004, 10:17 PM   #1
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Default Gain in Aleph P 1.7

Hi,
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!
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Old 8th May 2004, 01:20 AM   #2
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You can put in any resistor you like. For 22 dB, you want to
look at the 100 to 200 ohm range.
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Old 8th May 2004, 02:58 AM   #3
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Nelson,

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?
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Old 8th May 2004, 11:39 AM   #4
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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?
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Old 8th May 2004, 03:28 PM   #5
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Quote:
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.
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Old 9th May 2004, 01:27 AM   #6
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Thank you!
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Old 17th May 2004, 02:04 PM   #7
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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?

Thomas
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Old 17th May 2004, 02:26 PM   #8
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I assume that the gain is approximately X = (R53//R56)/(R51+R66+2K pot set value)--i.e. 20 x log(X) dB.
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Old 17th May 2004, 02:35 PM   #9
Klaus is offline Klaus  Germany
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Default R49 missing?

Quote:
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.
Hi,

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

regards

Klaus
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Old 17th May 2004, 03:09 PM   #10
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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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