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Old 19th August 2012, 01:25 PM   #1
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Join Date: Dec 2009
Default RF-blocking???


I was looking at the schematic of a guitar pedal's input buffer and there was something I couldn't quite understand. Click the image to open in full size.
How should the 1K resistor block very high frequencies? The input cap and the 510K resistor form a high-pass filter and block DC, right? But what's the role of that 1K resistor? What sort of filter does it form and how? Please, explain.

Thanks in advance!
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Old 19th August 2012, 01:55 PM   #2
DF96 is offline DF96  England
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Join Date: May 2007
The 1K works with stray capacitance in the wiring and parasitic capacitance in the transistor. Assume 5pF, then the rolloff will be from around 300MHz. A larger resistor would be better, say 10k. 300MHz rolloff means no protection from broadcast radio and VHF TV, little protection from UHF TV and only some protection from cell phones.

A 1K would be fine if the first stage is a common emitter amplifier as then Miller effect would amplify the collector-base capacitance.
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Old 19th August 2012, 09:56 PM   #3
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Thank you for the explanation. I'll try with 10k. Two more questions: since this is the input buffer of a guitar pedal, will the pickups' internal resistance help roll of very high frequencies and in effect add up to the resistor and will the increased resistance (10k instead of 1k) introduce more noise?

Last edited by ribolovec2; 19th August 2012 at 10:05 PM.
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Old 19th August 2012, 09:59 PM   #4
DF96 is offline DF96  England
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Yes, if noise could be a problem then stick with 1k and keep cell phones well away from it!
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