chongmagic Well-known member Jan 6, 2020 #1 I am working on a pedal that lists a 5kC pot for fuzz level, but all I have is 1kC and 10kC pots. Here are the build docs. https://www.dropbox.com/s/67g2txnnvkg7ord/penumbra_documentation.pdf?dl=1 Last edited: Jan 6, 2020
I am working on a pedal that lists a 5kC pot for fuzz level, but all I have is 1kC and 10kC pots. Here are the build docs. https://www.dropbox.com/s/67g2txnnvkg7ord/penumbra_documentation.pdf?dl=1
B benny_profane Well-known member Jan 6, 2020 #2 Depends on the purpose and load required, but you can put a 10k resistor in parallel across the relevant lugs and change it to 5k. The sweep will be affected, however. You can’t change a 1k to 5k and retain full sweep.
Depends on the purpose and load required, but you can put a 10k resistor in parallel across the relevant lugs and change it to 5k. The sweep will be affected, however. You can’t change a 1k to 5k and retain full sweep.
B Boba7 Active member Jan 7, 2020 #4 In that configuration a 10kC with a 10k in parallel will work fine I think
B benny_profane Well-known member Jan 7, 2020 #5 Do you have a 10kB pot? The affected sweep might actually be similar to the 5kC if you use the parallel resistor. I can do a simulation later today.
Do you have a 10kB pot? The affected sweep might actually be similar to the 5kC if you use the parallel resistor. I can do a simulation later today.
chongmagic Well-known member Jan 7, 2020 #6 benny_profane said: Do you have a 10kB pot? The affected sweep might actually be similar to the 5kC if you use the parallel resistor. I can do a simulation later today. Click to expand... Yes I believe I have plenty of 10kb pots.
benny_profane said: Do you have a 10kB pot? The affected sweep might actually be similar to the 5kC if you use the parallel resistor. I can do a simulation later today. Click to expand... Yes I believe I have plenty of 10kb pots.
B benny_profane Well-known member Jan 7, 2020 #7 Found this graph someone already made. I believe the plotting is correct. It’s not as extreme as a true inverse log, but it’s getting there. plot y=1/(1/R+100/Rx) and y=Rx/200 where R=10000 from x=0 to x=100 - Wolfram|Alpha Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels. www.wolframalpha.com
Found this graph someone already made. I believe the plotting is correct. It’s not as extreme as a true inverse log, but it’s getting there. plot y=1/(1/R+100/Rx) and y=Rx/200 where R=10000 from x=0 to x=100 - Wolfram|Alpha Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels. www.wolframalpha.com
Chuck D. Bones Circuit Wizard Jan 8, 2020 #8 Since there is already a 2.2K in parallel (R7), there's no need to add another resistor in parallel with a 10K pot. If you tend to use the top half of the FUZZ pot's rotation, then C1K would be fine.
Since there is already a 2.2K in parallel (R7), there's no need to add another resistor in parallel with a 10K pot. If you tend to use the top half of the FUZZ pot's rotation, then C1K would be fine.