When setting the voltage input to 2V from potentiometer, why does it increase to 2.55V after the Vout of potentiometer is disconnected for an inverting amplifier? And remains the same voltage of 2V for an non-inverting amplifier?
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\$\begingroup\$ It's just the simulator. Reality doesn't work that way. \$\endgroup\$– JRECommented Oct 30, 2020 at 17:59
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\$\begingroup\$ You have set the pot to 51% of 5V and disconnected it from the 10K resistor. What do you expect the voltage to be? \$\endgroup\$– user16324Commented Oct 30, 2020 at 18:08
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2 Answers
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Find the Thevenin equivalent for the voltage source (the pot) and compare loading for inverting and non-inverting cases.
Note that the Thevenin resistance (as well as the voltage) is a function of pot wiper angle.
If the voltage is set exactly to 2V, the pot voltage when open should be 2.50V not 2.55, by the way.
answered Oct 30, 2020 at 18:01
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\$\begingroup\$ How do you obtained the value of 2.5V when the voltage is set exactly to 2V? Thank you. \$\endgroup\$– HHHCommented Oct 31, 2020 at 2:01
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\$\begingroup\$ If the 10K pot is set in the middle, the equivalent is a 2.50V voltage source with 2.5K resistance. When loaded with 10K from the inverting amplifier the wiper voltage (ideally) is exactly 2V. \$\endgroup\$ Commented Oct 31, 2020 at 7:14
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\$\begingroup\$ Did you use a voltage divider to obtain the 2.5V when pot is disconnect and 2V when pot is connected? Thank you. \$\endgroup\$– HHHCommented Nov 1, 2020 at 2:58
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\$\begingroup\$ And i thought 2.50V voltage source is a 5k resistance at the lower half of the pot when disconnect? \$\endgroup\$– HHHCommented Nov 1, 2020 at 3:07
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\$\begingroup\$ 2.5K source resistance when the 5V source is connected (and the pot wiper is in the center). \$\endgroup\$ Commented Nov 1, 2020 at 6:00
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For the inverting configuration both of the op amp's inputs are at nearly the same potential at ground potential and so the 10k resistor is in parallel with the lower half of the pot drawing extra current through the top half of the pot and lowering its output voltage from that voltage output when the pot is isolated from the 10k resistor.
For the non-inverting configuration the op amp's + input draws only a tiny current and so it doesn't affect the voltage at the pot output.
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\$\begingroup\$ Can i say that it is due to the low input impedance of inverting amplifier and high input impedance of non-inverting amplifier which result to this output voltage? \$\endgroup\$– HHHCommented Oct 31, 2020 at 2:00
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\$\begingroup\$ Yes, that would be fine. The difference in input impedance between the two amplifier's results in different currents being drawn out of the pot which in turn results in a large voltage drop at the pot's output for the inverting amp and a small voltage drop at the pot's output for the non-inverting amp. \$\endgroup\$– user173271Commented Oct 31, 2020 at 5:05
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