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#1 |
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diyAudio Member
Join Date: Feb 2003
Location: Los Angeles, CA
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I've read all the posts and Nuuk's site for info on law-faking linear pots, but there is one question I haven't been able to answer:
Is the final resistence of the pot equal to it's initial value divided by the law-faking resistor value? e.g . a 100k pot with a 5k law-faking resistor produce a 20k result? Also, most posts around here(and Elliot Sounds article) seem to indicate that that the faking rsistor should be between 1/10th and 1/20th the value of pot. Is this true? |
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#2 |
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diyAudio Member
Join Date: May 2004
Location: Melbourne
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>Is the final resistence of the pot equal to it's initial value divided by the law-faking resistor value?
No, the input resistance is the pot resistance at min volume setting. At max volume, it is the faking resistor, pot resistance and the output load impedance in parallel. 10% is a good compromise for Rfake / Rpot. I've even used 15 - 20% and its seems OK. The output resistance (ie. as seen looking back by the load) is zero at min volume setting. At max volume it is the faking resistor in parallel with the pot resistance (at max volume) and the source impedance. HTH, Glenn.
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Glenn. |
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#3 |
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diyAudio Member
Join Date: Jul 2003
Location: Newcastle, Australia
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Rod's info shows that a ratio of 6.67:1 should be maintained. For 100K pot the resistor should be 15K (100/15=6.67).
As far as I know, the pot remains the same, 100K pot remains a 100K pot when adding the resistor from wiper to ground. From info on the Leach Amp site, to alter the resistance of the pot you add the resistor from the input to the wiper. If I'm right, I'm right, if I'm wrong I'm sure someone will tell me. |
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#4 | ||
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diyAudio Member
Join Date: Feb 2003
Location: Los Angeles, CA
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Great explanation, Glennb. I think I've got the hang of it now.
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