I have a simple op amp buffer circuit and was wondering whether there would be any downsides to replacing the feedback resistor with a pot wired as a variable resistor in order to control op amp gain. As far as I can tell the below circuit should give me roughly 20 dB of gain with the pot fully CW (2K feedback resistor, 220R to VREF) and unity gain with the pot fully CCW.
Non-inverting op amp buffer with gain control
- Thread starter lowpitch
- Start date
benny_profane
Well-known member
What are you biasing the input to? Vref from a voltage divider? I’d probably want to add a cap to ground in series with R1 to avoid problems with DC offset, but, otherwise, that’s fine.
JTEX
Well-known member
I have a feeling lowpitch wants to use an OPA205, so I wouldn't worry too much about DC offset, even with gain. I'd recommend higher values for R1 and the pot. Maybe around 1k and 10k. 220R is a pretty heavy load (at unity gain) for this application.
True 
See jtex's thread here for the full circuit: https://forum.pedalpcb.com/threads/an-idea-for-a-universal-input-buffer-booster-onboard.21235/ Although I'm keeping the feedback path as above, not with the switching.
What would be the result of the 220R load? And in general what is the philosophy behind choosing the resistor values for the feedback loop?
See jtex's thread here for the full circuit: https://forum.pedalpcb.com/threads/an-idea-for-a-universal-input-buffer-booster-onboard.21235/ Although I'm keeping the feedback path as above, not with the switching.What would be the result of the 220R load? And in general what is the philosophy behind choosing the resistor values for the feedback loop?
JTEX
Well-known member
Generally you want to choose feedback resistors low enough so they don't add significant noise to the op amp's self noise, but not so low that they load its output too heavily. This could limit the output voltage swing before clipping occurs, and also needlessly increases power consumption when signal is present. OPA205 has a self noise equivalent to that of a 3k resistor, so going much lower than 3k for your R1 won't noticeably lower noise, while loading the output for no good reason.
There's another little subtlety. R1's AC (signal) path to true ground (the power supply negative in a typical single supply pedal) goes through the bypass capacitor of the VREF divider. On my schematic, that's C2, 10uF. 220R and 10uF would form a high pass filter that rolls off the lows below about 70Hz. You can increase C2 to maybe 47uF to fix this, but then another funny thing happens. With 100k in the Vref divider, the circuit will take several seconds after powerup before a 47uF C2 finishes charging and Vref stabilizes to mid-supply. You might get no output at all for a second or two while this charging happens. To fix it, you'll have to reduce the 2x100k divider to something like 2x 20k. A 2x20k divider across a 9V supply draws 0.22mA. As much as the op amp itself, in case you care about battery life. Details!
TL;DR: go with R1= 1k or so
There's another little subtlety. R1's AC (signal) path to true ground (the power supply negative in a typical single supply pedal) goes through the bypass capacitor of the VREF divider. On my schematic, that's C2, 10uF. 220R and 10uF would form a high pass filter that rolls off the lows below about 70Hz. You can increase C2 to maybe 47uF to fix this, but then another funny thing happens. With 100k in the Vref divider, the circuit will take several seconds after powerup before a 47uF C2 finishes charging and Vref stabilizes to mid-supply. You might get no output at all for a second or two while this charging happens. To fix it, you'll have to reduce the 2x100k divider to something like 2x 20k. A 2x20k divider across a 9V supply draws 0.22mA. As much as the op amp itself, in case you care about battery life. Details!
TL;DR: go with R1= 1k or so
Sturdag Lagernathy
Active member
College? Night school? Barring those 2, it just takes time, soak up everything you can!! Oh, and learn to love math!
