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So I have a LM358 Dual opamp. I only care about the first op amp, the second I just have set to put out 12V to keep it from floating. The the first is reading in feedback from a 1k potentiometer from 10-0 volts. The voltage divider knocks it down to the uC analogue voltage level. it all works fine, but when the pot is disconnected the output of the op amp rails to 24 volts. Why is that? if you have 0 volts on the non inverting input it should be 0 for the output correct?

EDIT: Maybe this picture is easier to follow

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  • \$\begingroup\$ This would be a lot easier to read and understand if you used the op-amp symbol rather than a box. \$\endgroup\$ – Transistor 19 hours ago
  • \$\begingroup\$ ...and explain what "receptor X" is and where R11 is connected to... \$\endgroup\$ – Unimportant 19 hours ago
  • \$\begingroup\$ "when the pot is disconnected.." this is not a 0 Volt input, it is an open circuit. Add a pull down resistor across C10, and see what you get. \$\endgroup\$ – Chris Knudsen 19 hours ago
  • \$\begingroup\$ Sorry, just used what TI provided. Receptor X is just the name of a signal going to the microcontroller (uC). \$\endgroup\$ – Ih8th3c0ld 19 hours ago
  • \$\begingroup\$ Chris, ok I will. \$\endgroup\$ – Ih8th3c0ld 19 hours ago
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This kind of op-amp has a bias current that flows out of the input, so opening it will cause the output to go to about +22V.

If you connect a resistor (say 100K) from the pot wiper connection to ground you will affect the linearity slightly, but an open wiper will result in an output voltage of only a few mV typically (worst case could be maybe 50mV).

Linearity will be affected by 0.25% (slight droop at 50.0%).

C62 is a really bad idea and can cause oscillation. Put it across the power supply instead.

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If you disconnect a pot and leave the top end of R11 floating then the bias current from the non-inverting input will charge up C10 until it reaches one of the power rails.

Since you have a voltage following buffer circuit the output will just follow the input.

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  • \$\begingroup\$ So if I measured across C10 I should see the max supply voltage? \$\endgroup\$ – Ih8th3c0ld 19 hours ago
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    \$\begingroup\$ You might. Your voltmeter will discharge C10. You can estimate the time constant from \$ \tau = RC \$ where R is the resistance of your meter - probably 10 M\$ \Omega\$. \$\endgroup\$ – Transistor 18 hours ago

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