Of course.Does voltage follower has common mode voltage?
So because of the fact that "-" > "+" the op-amp output will be at "negative saturation".Try again because your calculation is wrong. Also, do not forget that all voltage is measured with respect to ground.
And notice that your example does not show a stable situation. We have 2.5V at "-" input and 0V at "+" input.
So because of the fact that "-" > "+" the op-amp output will be at "negative saturation".
Are we looking at the same diagram ??"-" is the reference voltage (3.3V) of the inverting schmitt trigger while the voltage at "+" is 2.5V, it is lower than the reference voltage.
Therefore, the output of the inverting schmitt trigger should be 5V. (Supply is 5V)
Here is my calculationStep 1: Calculate the UTP & LTP of the inverting schmitt trigger to determine its output when the input voltage is 2.5V.
Wrong again. First notice that the voltage at "-" input is 2.5V not 0.0115V.Step 2: Determine the common mode voltage.
Common mode voltage = ((V+)+(V-))/2; (V-) = 2.5V while (V+) is unknown due to positive feedback circuit.
For easier visualisation, i have simplify the feedback circuit as shown below and through the calculation, the voltage at the (V+) pin is 0.015V.
Therefore, Vcm = (2.5V+0.0115V)/2 = 1.256V
Appreciate and thanks for the example shown. I understand and realize my mistake.The correct way of finding the voltage at "+" input is :
Because of the fact that Vo > Vref the current will flow in this direction Vo --->1MΩ--->6.8kΩ---->Vref.
And from KVL we can find V+ voltage
V+ = Vo - I1*1MΩ or
V+ = Vref + I1*6.8kΩ