Electronics > Projects, Designs, and Technical Stuff
Logarithmic pot
akis:
Is there a way to use a simple, passive circuit, with very few parts, to turn a linear, or a badly made "log" pot, into an audio pot? From some charts I have seen it seems we are looking for something like the below:
--- Code: ---potentiometer position experimental log scale
0 0
1 0.2
2 0.6
3 2
4 5
5 10
6 20
7 33
8 54
9 80
10 100
--- End code ---
Prehistoricman:
You can add a resistor from the tap to either end of the pot to affect the curve. See pics.
The graph shows the behaviour of a pot with a resistor 10% of its value to ground. X is the pot setting, Y is the tap output.
Whales:
+1 for "10%" recomendation. I have read suggestions of 50% elsewhere, but some analysis I did a while back showed 10% was a better fit.
IIRC: this gives you about 20dB of db-linear audio range (used as an attenuator, although I think I was also putting a resistor above?). The last few degrees at the far edges of the pot will be distorted and give you a lot more than this.
Prehistoricman:
--- Quote from: Whales on January 05, 2020, 09:15:53 pm ---+1 for "10%" recomendation. I have read suggestions of 50% elsewhere, but some analysis I did a while back showed 10% was a better fit.
IIRC: this gives you about 20dB of db-linear audio range (used as an attenuator, although I think I was also putting a resistor above?). The last few degrees at the far edges of the pot will be distorted and give you a lot more than this.
--- End quote ---
Resistor above would make an anti-log taper.
And yes, you're right about the edge distortion. Here's a quick excel plot of a real log taper against this hack.
But a real log pot will probably contain two linear sections of different resistivity to roughly approximate the ideal log taper (source: internet, no validation has been done).
basinstreetdesign:
Yes I've done a least-squares fit between the ideal exponential curve and what you get from a fixed resistor in parallel with a linear pot as Prehistoricman says. The optimum value (least error) is at resistor = 11.6% of pot. This assumes that there is no other load and the source impedance is zero. It looks like this:
--- End quote ---
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