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Old 21st December 2002, 08:43 AM   #1
lykkedk is offline lykkedk  Denmark
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Default Would this work

OK, I give up, I just dissasambled my entire PCB, for the balanced line stage, planning to make a complete new design, unbalanced only.

I have found this schematics (Thank's to the author),

Would this work ???? ( The one I had redrawn ? )
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Old 21st December 2002, 08:45 AM   #2
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Default Re: Would this work

And the Schematics I had found is this :
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Old 21st December 2002, 09:00 AM   #3
lykkedk is offline lykkedk  Denmark
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Default Re: Re: Would this work

And connected like this to RCA :
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Old 21st December 2002, 09:07 AM   #4
2Bak is offline 2Bak  Denmark
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this will work fine
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Old 21st December 2002, 09:14 AM   #5
lykkedk is offline lykkedk  Denmark
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Quote:
Originally posted by 2Bak
this will work fine
OK THANX
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Old 21st December 2002, 09:22 AM   #6
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This will work but you'll get better results if you retain the gate stopper and connect to ground through a similar resistance as the i/p mosfet. Differential amps achieve best CMRR when both inputs are driven by equal impedances. Also may be a good idea to retain the trimpot.

cheers

peter
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Old 21st December 2002, 09:33 AM   #7
lykkedk is offline lykkedk  Denmark
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Thanks, but could you perhaps draw that in on my schematics, cause I don't understand all of this !!!
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Old 21st December 2002, 10:13 AM   #8
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Only a few more modifications and you would have a Bride of Zen ?
Why all that trouble, i would go for a BOZ right away if only unbalanced is needed.

Gr,
Nick
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Old 21st December 2002, 10:25 AM   #9
lykkedk is offline lykkedk  Denmark
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Quote:
Originally posted by Electro-Nick
Only a few more modifications and you would have a Bride of Zen ?
Why all that trouble, i would go for a BOZ right away if only unbalanced is needed.

Gr,
Nick
Cause I think, the balanced, is better as unbalanced, than the BOZ
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Old 21st December 2002, 10:51 AM   #10
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Referring to the schematic in the first posting, seeing the Q2 drain signal is just an inverted but equal amlitude version of Q1 drain signal, it doesn't appear that Q2 is really doing anything besides making the schematic look pretty. You could probably just take the signal from Q1 drain and throw Q2 and it's extra pieces away! The total source resistance may need to be altered because there is now only half the current in it.
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