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Old 31st July 2008, 02:31 PM   #1
anbello is offline anbello  Italy
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Default Headphone amplifier with buffer

I would like to build an headphone amplifier based on the well known opa + buf634 pair configured as seen on the datasheet, but i would also add an input buffer before the pot to have a better impedance interface with various sources.
My questions are:
1) is this correct or i have to add series and/or parallel resistor between the first opamp and the pot and/or between the pot and the second opamp?
2) which is the correct value for the pot? (i have one 20K and one 100K ALPS)

Ciao
Andrea
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Old 31st July 2008, 06:02 PM   #2
paulb is offline paulb  Canada
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The basic circuit already has a very high input impedance; you don't need the buffer, unless you need a higher input impedance than the pot value. If you use the 100K pot, that will be the input impedance. You haven't gained anything by adding the buffer with its 100K input impedance.
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Old 1st August 2008, 06:57 AM   #3
Dxvideo is offline Dxvideo  Turkey
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I am agree with Paul..
You dont need the first opamp. Just you might put a 100K pot and a 100K fixed resistor as input load (to prevent oscillation in case of pot failures) that makes at least 50K input impedance which is enough for a sound system.
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Old 1st August 2008, 08:16 AM   #4
anbello is offline anbello  Italy
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Thanks Dx and paul, but without the input buffer i have a variable input impedance or am i wrong?

Ciao
Andrea
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Old 1st August 2008, 08:18 AM   #5
Dxvideo is offline Dxvideo  Turkey
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Right,
As I mentioned, you will have AT LEAST 50K input impedance.. Is it a problem for you?
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Old 1st August 2008, 08:29 AM   #6
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Quote:
Originally posted by anbello
Thanks Dx and paul, but without the input buffer i have a variable input impedance or am i wrong?

Ciao
Andrea

For a 100 k pot or lower, the input impedance will vary less than 0.000001 %. Hardly anything to worry about.
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Old 1st August 2008, 08:53 AM   #7
Dxvideo is offline Dxvideo  Turkey
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Wihout parallel resistor, thats right. But in this case your only input load will be the potentiometer. Thats a risk I guess..
However with parallel resistor to pot, then input impedance will vary between 50K to ~100K ...
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Old 1st August 2008, 08:57 AM   #8
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Quote:
Originally posted by Dxvideo
Wihout parallel resistor, thats right. But in this case your only input load will be the potentiometer. Thats a risk I guess..
However with parallel resistor to pot, then input impedance will vary between 50K to ~100K ...

Why? The pot itself has a fixed resistance between input and ground, and the wiper is connected to a 10 GOhm op amp input, which can be neglected. Hence the very low figure I posted.
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Old 1st August 2008, 09:04 AM   #9
Dxvideo is offline Dxvideo  Turkey
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May be there is a misunderstanding.
I meant,
Yes youre right, there will be no input impedance swing if you use just a potentiometer on input.
However, I adviced a parallel resistor to middle pin of the pot to gnd. So in this case you will have a variable input impedance. But this is a need I think. Because as my experiences, if an opamp works without an input load resistor then it becomes unstable situation. And if your potentiometer fails while running without a parallel resistor then your headamp may oscillate and you may say goodbye forever to your precious headphone..
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Old 1st August 2008, 09:05 AM   #10
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OK, I see. I was commenting on the original schematic.
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