Timeline for Analog input for ADC with multiple potentiometers noise
Current License: CC BY-SA 4.0
12 events
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| Jan 30, 2022 at 15:17 | comment | added | TonyM | @KellerDev, I'm insisting on nothing and require no comfort. I replied to your further comment questions with an engineering assessment. You have all the information you have asked for in plenty of detail. | |
| Jan 30, 2022 at 14:22 | comment | added | KellerDev | Thanks again Sir/madam for the input. I appreciate your effort, but by insisting on the approach to do it with muxes you are changing the topic. I'm not sure why you are trying to shame me into doing it that way, but if this gives you some comfort, I have used muxes before and know how they operate. They are unfortunately less sutable for my goal due to the other functionality of the device I'm doing. | |
| Jan 30, 2022 at 13:23 | comment | added | TonyM | @KellerDev, I think I see...are you trying to avoid tearing it up and starting again using analogue muxes? I appreciate your honesty about inexperience and admire you putting the work into trying an idea. But the whole pot-to-diode-to-ADC idea is dead in the water for accuracy, especially driven from I/O gates. Don't chase a poor solution and toy with it, it'll only get worse. The mux version is simple, will work and you'll get experience in a good solution, will teach you far more Rip it up and start again, you'll be glad you did :-) | |
| Jan 30, 2022 at 11:11 | comment | added | KellerDev | Thanks again for your input. I am just trying to figure out the correct wayt to bring the high impedance from the diodes output to a low impedance for the ADC. I know the input of the opamp is highimpedance, but the output should be low impedance right? So maybe in this case again the noise comes from the high impedance out of the diodes and then is just amplified by the opamp. What is then the correct way to buffer the output of the diodes and keep the linearity of the potentiometers? | |
| Jan 30, 2022 at 9:36 | comment | added | TonyM | I've extra-clarified the text but it's all explained there. Your op-amp also has a very high impedance, which you'll see when you look at the datasheet. Concentrate on "you're seeing the diodes as 'magic one-way wires' and not designing based on their operating characteristics" because that's the root of all understanding for you. The answer tells you a test resistor will only "demonstrate the point" because it will "ruin the linear behaviour of your pot'" so you're not surprised to see it do so. | |
| Jan 30, 2022 at 9:34 | history | edited | TonyM | CC BY-SA 4.0 |
Clarifications.
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| Jan 30, 2022 at 9:23 | history | edited | TonyM | CC BY-SA 4.0 |
Clarification as OP didn't get it.
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| Jan 30, 2022 at 8:44 | vote | accept | KellerDev | ||
| Jan 30, 2022 at 8:44 | |||||
| Jan 30, 2022 at 7:40 | comment | added | KellerDev | Thanks alot for the input! I was explaining it to myself something along those lines, that the current from the diodes is not sufficient, basically they turn the output of the pots into a very high impedance load. But my concern is why didn't it work with the opamp. Should it work in theory with the circuit on the left, and then buffered through the positive input of an opamp, configured as a voltage follower? This configuration should once again turn it into a low impedance load? The resistor to ground works, but it should be around 10k for good result, and the linearity is ruined | |
| Jan 29, 2022 at 23:51 | history | edited | TonyM | CC BY-SA 4.0 |
Added example circuit.
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| Jan 29, 2022 at 23:23 | history | edited | TonyM | CC BY-SA 4.0 |
Clarifications.
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| Jan 29, 2022 at 22:54 | history | answered | TonyM | CC BY-SA 4.0 |