Gain in Aleph P 1.7

[bbakota2000]

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!

 

[Nelson Pass]

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

 

[kilowattski]

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?

 

[bbakota2000]

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]

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]

Thank you!

 

[tomchaoda]

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

 

[jh6you]

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

 

[Klaus]

R49 missing?

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

 

[bbakota2000]

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?

 

[tomchaoda]

Klaus:

If I put R49 into perspective, the gain is about 5 dB less than without it. That is, if my VR set to 150 ohms, then the gain:

without R49 is ~15dB
with R49 in the picture is ~10dB

In my current setting, I think without it is more correct when I test it with my ear... (but maybe I am wrong!)

Thomas

 

[jh6you]

Hi, Klaus

Yes, we have R49. But at the same time we have R64, too, on the opposite side. I think that they have a +/- zero effect. Therefore, I did not consider them in my gain consideration.

Mr Pass might give me a strong punch to .

Regards

 

[jh6you]

... this formula ...

dB ~ 20 x log((R53xR56/(R53+R56))/(R51+R66+2K pot set value))

JH

 

[Klaus]

dB ~ 20 x log((R53xR56/(R53+R56))/(R51+R66+2K pot set value))

JH

Hi,

I just asked again my colleague, my personal chief for any analog schematics…

He explained my that the story is a bit of more complicated. The P1.7 has two overlaid gain settings. One given, like described, by the parallel value of R53/56/49 + the input R of the power amp in relation to R51+poti+R66. BUT we also have a voltage feedback given by R48 to the input. Here the gain is also limited. So the total gain to be calculated with both – but I did not understand how…

I am stranded, I go and listen music over my P1.7, however its gain is controlled…

Regards

Klaus

 

[jh6you]

I agree that the x-feedback registers, R48 and R63, take part of the overall amp gain control.

Regards

JH