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By Mee_n_Mac
#163417
Again look at the specs for an op-amp. You will see both specs. That's because the simple "1'st order" model of how an op-amp works isn't the whole truth. For example the simple assumptions that Ad is infinite and Acm = 0. When you get into demanding applications these assumptions will lead to a circuit that may not work as desired. Ad has some limit and that has effects; on linearity, on frequency response to name two. That Acm != 0 means there will be some output even when the + and - signals are identical but non-zero. If you're designing a differential amp using op-amps, you need to understand that and then do the math to see if the effect of Acm is important (or not) for your design.

A lot of circuitry is so complex that you, as a designer, don't have the time or the resources to understand all the inner workings. Instead, those that do understand their design and it's inner workings, will give you as set of specs that try to describe how the "black box" will work in almost all usages ... assuming you understand the specs and what they mean. That's one reason why EE's get paid the medium $$s.
By 8Volts
#163426
Okay Mee_n_Mac,

I think I get it! I think it's time for me to take two steps back
and clear my head .....

As for the specs, they says

Common-Mode. DC,VCM =0VtoV+ −1.5V,

so I guess this means :

Vcm = V1 + V2/2. where Vcm MAX is (V1) -1.5vdc

okay ... thanks for your help....
very appreciated
By Mee_n_Mac
#163428
8Volts wrote:As for the specs, they says

Common-Mode. DC,VCM =0VtoV+ −1.5V,

so I guess this means :

Vcm = V1 + V2/2. where Vcm MAX is (V1) -1.5vdc
The above looks like a common mode input voltage range (there are various acronyms for this) ... that is that any common mode voltage must be in that range or the op-amp won't work as expected. You might want to read this long thread on that :
https://forum.sparkfun.com/viewtopic.ph ... p&start=15

Op-amps will also have a Acm spec, different from the above, but it's usually expressed as a common mode rejection ratio:
http://en.wikipedia.org/wiki/Common-mod ... tion_ratio
By 8Volts
#163436
Thanks Mee_n_Mac,

I appreciate very much your replies because of their
pertinence and relevance to my questions being
posted. In summary, it's nice to receive straight cut
no beating around the bush replies just as you have done.

I know you have put a significant amount of effort in your
replies and for this I sincerely thank you.

Regards