Gain of the Thorsten/Peter Daniels gainclone? - diyAudio
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Old 20th July 2003, 09:52 PM   #1
JoeBob is offline JoeBob  Canada
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Default Gain of the Thorsten/Peter Daniels gainclone?

What would the gain of the regular Thorsten/Peter Daniels gainclone be? 220k/10k, so a gain of 22, right?

Also, how high a gain could this design be pushed to? 40, 80, 100?
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Old 20th July 2003, 10:06 PM   #2
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Actually, with 10k and 220k, the gain is 23... I don't know howhigh I would push the gain, though. (personally, for me, g=23 was higher than I like... but it worked out ok)
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Old 20th July 2003, 11:04 PM   #3
moamps is offline moamps  Croatia
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Quote:
Originally posted by needtubes
Actually, with 10k and 220k, the gain is 23... I don't know howhigh I would push the gain, though. (personally, for me, g=23 was higher than I like... but it worked out ok)
Hi,

Actually, gain is 22. For inverting topology gain is R2/R1. Maximum gain is limited with bandwith you need.

Regards
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Old 20th July 2003, 11:12 PM   #4
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Oops... I was thinking non-inverting... forgot that their circuits were inverting... Sorry.
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Old 20th July 2003, 11:12 PM   #5
BrianGT is offline BrianGT  United States
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Quote:
Originally posted by moamps


Hi,

Actually, gain is 22. For inverting topology gain is R2/R1. Maximum gain is limited with bandwith you need.

Regards
If you account for the potentiometer in the gainclone, which is in series with R1, then it is not always 22, and changes with the movement of the wiper on the potentiometer.

I am wondering if the actual perfect gain for the gainclone is quite a bit lower then 22, since the average listening level isn't full volume. It might be interesting to figure out what exactly the gain is at the setting that you use on the gainclone in inverting configuration, and setting the same volume level and same gain in non-inverting configuration to compare.

Assuming a 100k potentiometer (thorsten's schematic), and the desired listening level is in the middle (linearly in the middle), then the gain would be 220k/(10k+50k) = a gain of 3.667.

--
Brian
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Old 21st July 2003, 12:53 AM   #6
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Konnichiwa,

Quote:
Originally posted by BrianGT


Assuming a 100k potentiometer (thorsten's schematic), and the desired listening level is in the middle (linearly in the middle), then the gain would be 220k/(10k+50k) = a gain of 3.667.

Your calculation is based on assumption (middle of travel) and incomplete math. The reality looks a lot more complex due to a number of interactions.

Sayonara
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Old 21st July 2003, 01:35 AM   #7
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my sound card kept making my amp clip so i changed feedback resistors. i have 20k and 1.3k or 1.6k, i don't recall. doesn't oscillate though.
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Old 21st July 2003, 01:59 AM   #8
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I'm using 250k for feedback, 10k series resistor and 50K log. pot. I don't know about other people, but my source outputs only so much signal that I usually have the pot set at around 28th pos. on 31 detented potentiometer. It sounds pretty good this way and the gain is close to 30.
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Old 21st July 2003, 02:10 PM   #9
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Quote:
Originally posted by Kuei Yang Wang
Konnichiwa,



Your calculation is based on assumption (middle of travel) and incomplete math. The reality looks a lot more complex due to a number of interactions.

Sayonara
Incomplete maths indeed - at mid position of pot... a quick check on the ol' sliderule... trumpets blaring... 6.286 .

The 100K pot must be regarded as 25K as the assumption must be that the source impedance, say from your CD Player, is low. So:

(220)/(10+25) = 6.286 .

When wiper is at the top (hold on to your ears) the gain = 22. So likewise when at the bottom, except gain of 22 times zero (no signal to amplify).

Now the math are complete!

Seriously though, anyone got a diff result?

Isn't it interesting that the calculation is different from shunt feedback, where gain is (R1+R2)/R2, whereas series (inverted) feedback is R1/R2.

Joe R.

PS: Oh yes, my preferred gain for LM3875 is +33dB.
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Old 21st July 2003, 02:20 PM   #10
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Konnichiwa,

Quote:
Originally posted by Joe Rasmussen


Incomplete maths indeed - at mid position of pot... a quick check on the ol' sliderule... trumpets blaring... 6.286 .

Seriously though, anyone got a diff result?

Well, if you calculate the actual gain from the input of the Pot with the pot in the center....

Then you get 50K : 50K//10k = 1:7 which in turn is amplified by 22, giving a net of 3.1428, yet the noisegain is as you have pointed out 6.286.

As said, it is a complex sort of circuit (simplicity is deceptive) but it works okay, especially if the feedback resistor is increased to 330K, which should avoid the sound degradation observed with a (22k) resistor from negative input to ground.

There are a great many ways to skin this particular Dog (I never skin cat's, I love cat's - can't stand dogs though).

Sayonara
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