If I correctly understand your question, that unknown resistor can be quite large in value - at least ten times the value of an logarithmic (audio) taper volume control, it's not really critical. It's purpose is to ensure that the op-amp's input never floats by operation of the volume control's wiper.
If you're using a linear taper volume comtrol pot., you could then leverage the input loading resistor to produce a good quasi-logarithmic taper, as per the following linked instructions.
http://sound.westhost.com/project01.htm
If you're using a linear taper volume comtrol pot., you could then leverage the input loading resistor to produce a good quasi-logarithmic taper, as per the following linked instructions.
http://sound.westhost.com/project01.htm
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Sorry, I plan on using a log pot and I usually pick ~20K for the input resistor on a buffer. I just don't know if I should reduce that number because of the added impedance of the voltage divider from the volume pot or if there are things that I have not yet considered.
Do you think there is an advantage using the linear pot plus a parallel resistor like the ESP link shows instead of using a log pot?
Do you think there is an advantage using the linear pot plus a parallel resistor like the ESP link shows instead of using a log pot?
Hi,
in short:
As rule of thumb make it 5 to 10 times the Poti value.
Its value is then minimum 10 times the value as if looking into the poti from the wiper.
in long:
We make following assumptions..
- the sources output impedance equals 0. The sources output looks like a virtual ground.
- let Vs be the voltage at the Source
- let Ru be the upper part of the Poti from the Source to the Wiper
- let Rl be the lower part of the poti from the Wiper to gnd.
- let Ri be the additional Resistor from the OPAmps Input to gnd
- let Rin be the (combined) impedance the OPAmp sees connected from its input to gnd
- let Vin be the voltage at the OPAmps input
and last
- lets assume the OPAmps input impedance be so high it may be perfectly neglected
Now, without Ri connected, the Poti forms a variable voltage divider.
The voltage at the Wiper, Vin calculates to Vin= Vs[Rl/(Ru+Rl)]
The impedance Rin the OPAmp sees looking into the Wiper is then the parallel connection of Ru and Rl ( both connected to gnd or a virtual gnd).
The formula is 1/Rin= 1/Ru + 1/Rl.
For a linear 10k Poti Rin would vary between 0R at the end positions to a maximum of 2k5 with the wiper in middle position.
If You now add a resistor Ri, the resistor will be in parallel to Rl.
Simply change the Rl value in the above formulas to the paralleled value Rli which calculates to 1/Rli= 1/Rl + 1/Ri.
You will notice that the influence of Ri remains small if Ri is chosen 5 to 10 times Rl.
Choosing Ri lower in value You'll notice a reduction in the maximum value of Vin as well as a change in the shape of the Vin over Poti setting.
Adding a low enough in value Ri to a linear Poti also results in a quasi logarithmic impedance curve.
So, if You need Ri only as a dc biasing path, choose Ri large in value.
If You want to additionally shape the impedance curve, resp the Poti character, calculate with a lower Ri.
jauu
Calvin
in short:
As rule of thumb make it 5 to 10 times the Poti value.
Its value is then minimum 10 times the value as if looking into the poti from the wiper.
in long:
We make following assumptions..
- the sources output impedance equals 0. The sources output looks like a virtual ground.
- let Vs be the voltage at the Source
- let Ru be the upper part of the Poti from the Source to the Wiper
- let Rl be the lower part of the poti from the Wiper to gnd.
- let Ri be the additional Resistor from the OPAmps Input to gnd
- let Rin be the (combined) impedance the OPAmp sees connected from its input to gnd
- let Vin be the voltage at the OPAmps input
and last
- lets assume the OPAmps input impedance be so high it may be perfectly neglected
Now, without Ri connected, the Poti forms a variable voltage divider.
The voltage at the Wiper, Vin calculates to Vin= Vs[Rl/(Ru+Rl)]
The impedance Rin the OPAmp sees looking into the Wiper is then the parallel connection of Ru and Rl ( both connected to gnd or a virtual gnd).
The formula is 1/Rin= 1/Ru + 1/Rl.
For a linear 10k Poti Rin would vary between 0R at the end positions to a maximum of 2k5 with the wiper in middle position.
If You now add a resistor Ri, the resistor will be in parallel to Rl.
Simply change the Rl value in the above formulas to the paralleled value Rli which calculates to 1/Rli= 1/Rl + 1/Ri.
You will notice that the influence of Ri remains small if Ri is chosen 5 to 10 times Rl.
Choosing Ri lower in value You'll notice a reduction in the maximum value of Vin as well as a change in the shape of the Vin over Poti setting.
Adding a low enough in value Ri to a linear Poti also results in a quasi logarithmic impedance curve.
So, if You need Ri only as a dc biasing path, choose Ri large in value.
If You want to additionally shape the impedance curve, resp the Poti character, calculate with a lower Ri.
jauu
Calvin
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