I have a non-inverting, single-supply, variable gain op-amp circuit configured in accordance with this diagram:
- Vcc: +16V
- Rf: 800k Ohm Potentiometer
- R2: 100k Ohm
I have the potentiometer set to ~200k Ohm for a gain of:
$$
A=(1+R_f/R_2)=(1+200k/100k)=3
$$
I have confirmed that this works experimentally (A=3.07). However, what puzzles me is that when \$V_{in}\$ is left floating I get an output of 6V instead of approx. 0V. Consequently, any input below approx. 2V yields no change in output.
Per the datasheet for this OpAmp, the maximum \$V_{os}\$ is 5mV, which even when amplified by 3, is nowhere near 6V.
Edit:
Below is a drawing of my circuit.
Best Answer
Extracts from the NTE941M data sheet: -
This means that if one supply of the op-amp is ground (0 volts) then you must have your input signal greater than typically 3 volts above ground or all bets are off.
This means that you cannot expect the output to get any closer to the 0 volt rail than typically +1 volt above it.
Do you understand why now?