# Electronic – How to properly use an op-amp

operational-amplifier

ever since posting here I never been so lost of using op-amps before, hearing new things that I have never heard before (Vom, Vcm etc). I always thought OP AMPS is just plug it in and it will work everytime… Very wrong.

I have a couple of questions which would be most appreciated if anyone of you could answer them, before I asked them, yes I have been looking for the past 2 hours in this forums for previous questions that were asked. Still a little confused but it did clear some things up.

To keep things consistent I would be using This OP AMP throughout this whole example. MCP601

VCM: Common Mode Input Range

Here is what I understand -Its the range of which the MCP601 can happily accept with nothing going wrong, if one were to go over or under these ranges the op you will see unexpected error.

Example: Input = Audio Signal (1.2V pk-pk) VDD = 4.8V VSS = GND

VCM – Upper limit = 4.8-1.2 = 3.6

VCM – Lower limit = 0-0.3 = -0.3

VCM – \\$V_{CM_{PP}}\\$ = 3.6-(-0.3) = 3.9V

\\$V_{IN}\\$ – Positive Cycle of input = 600mV + (VDD/2) = 3

\\$V_{IN}\\$ – Negative Cycle of input = -600mV + (VDD/2) = 1.8

\\$V_{IN}\\$ = 1.2Vpk-pk

Meaning the input Vpk-pk is suitable?

VOM: Output Voltage Swing

Here is what I understand – Its the range of which the MCP601 is capable of outputting before clipping.

Example: Input = Audio Signal (1.2V pk-pk) VDD = 4.8V VSS = GND GAIN = 3.2

Input Bias = VDD/2 RL = 5k

VOM – Upper limit = 0+100mV = 100mV

VOM – Lower limit = 4.8-100mV = 4.7V

VOM – \\$V_{OM_{PP}}\\$ = 4.7-100mV = 4.6V

\\$V_o\\$ – Positive Cycle of input = (3.2*600mV)+(VDD/2) = 4.32V

\\$V_o\\$ – Negative Cycle of input = (3.2*-600mV)+(VDD/2) = 0.48V

\\$V_o\\$ – \\$V_{o_{PP}}\\$ = (4.32-0.48) = 3.84V (Before decoupling Cap).

This is how I understood to calculate for both \\$V_{CM}\\$ and \\$V_{OM}\\$. To me this OP-AMP shouldn't have a problem with the Vin as well as happily amplifying the Vin as well, however the opposite happened as it clips at 2.84Vpp. This doesn't make much sense to me from the calculation above. The VCM should be satisfied as well as the VOM. As the VOM has a Vpp of 4.6V which is > then my Vo of ideally 3.84Vpp and my VDD being 4.8V it should amplify to 3.84Vpp no problem?

If anyone can show me how to actually calculate VCM and VOM that would be amazing, I believe this method is missing something or I am not understanding some fundamental logic. I would like to gain the ability to understand input and output limitations through this method.

This configuration works if I increase VDD to ~6.1V if anyone can explain why through the VCM and VOM calculations I can probably correlate the two and probably will clear up any confusions I had.

simulate this circuit – Schematic created using CircuitLab