bufferimpedanceoperational-amplifier
So before adding an op-amp buffer between sensor and the non-inverting input of a schmitt trigger, if the output of the sensor was for example 0.4V, the moment I connected it to the non-inverting input somehow the output of the sensor would suddenly become about 2V. Ater I added the buffer the problem was solved and the output of the sensor didn’t change anymore. So I have done some reading on buffers applications but can’t really understand what’s going on here and why buffer was the solution.
As TEMLIB pointed out, a comparator is not an op-amp, even if it happens to be drawn as a triangle with with 2 inputs and one output.
A typical design for a comparator is an open drain comparator, which can be modeled like this:
simulate this circuit – Schematic created using CircuitLab
Here, OA1 is an ideal op-amp.
If you connect Vout to V-, you won't get a voltage follower.
Additionally, even if you stick to the world of "true op-amps", there are some restrictions on what op-amps you can use and how you use it. For example, some op-amps are not unity gain stable. Since a voltage follower has by definition a gain of one, trying to build a voltage follower with one of these will always be unstable, and you won't get the desired voltage follower behavior.
You could start with an advertised unity gain stable op-amp and drive it unstable by what you connect the output to and what input waveform you give to the op-amp. It is also possible to do the reverse and add components to stabilize an unstable configuration. There's a wide variety of literature available online on op amp stability.
LM358 is not a rail-to-rail output opamp. So it means it is not able to drive its outputs hard enough to reach the supply voltage. There will always be a voltage drop even when it saturates. This exists both on the high side and, to a lesser extent, low side.
See the datasheet to know the magnitude of this drop (which depends on the current you need on the output - the more you draw, the bigger the drop):
For the high side:
For the low side:
Best Answer
Your sensor has a high output impedance.
Imagine the sensor as a voltage source in series with a resistor Rout. If the output of the comparator is high, you get the following equivalent circuit:
simulate this circuit – Schematic created using CircuitLab
Here, V_OpAmp is the output voltage of the op amp. I've assumed it's railed high at 5V. The open-circuit voltage V_sensor of the sensor is 0.4V, but because of it's got a high output impedance, the op amp output voltage pulls it high - what you actually measure at the output of the sensor is V_sensor_out = 2V.
I have picked some resistor values that replicate the measured results you mention. Knowing Rin and Rf, you could easily calculate the actual value of Rout.
Adding the buffer presents the sensor with a high-impedance load, and the Schmidt trigger with a low impedance driver, effectively eliminating the effect.