Im looking at a comparator that has an absolute maximum on any two pins of 2.5V

but it has a CMVR of 1.5V

Obviously if i go over 2.5 it can be damaged. Question is..... if i go over 1.5V does that mean it will just not respond linearly or will it be damaged?

Im trying to understand if i have (2.5+.6)/2 = 1.6 will that break it and beyond .6

https://www.ti.com/lit/ds/symlink/l...l=https%3A%2F%2Fwww.ti.com%2Fproduct%2FLMV762

Data sheet here

 

The Common Mode Voltage Range is not what you think it is.

It is determined by tying the two inputs together such that the voltage at both inputs is the common mode voltage. But what the spec is actually telling you is what is the range of input voltages that the device is spec'ed for at each input.

So if the CMVR is -0.3 V to 1.5 V, do not let the voltage at either input go outside of those limits. Going outside those limits may not damage the part -- usually won't unless you actually go above or below the supply rails by enough to turn on protection diodes -- but the device functional behavior or, most especially, timing behavior may deviate significantly from the specifications.

 

The absolute min & max is where the part might break. This has nothing to do with function.

CMVR is the area where the part works correctly. In this case -0.3 to 1.5V. If both inputs are at 2V (outside CMVR) the output is undefined. When one input is inside CMVR and one is outside, (voltage comparator), all the parts I have used function but do not meet all the specifications. Example the input current is out of spec. The speed is slow. But the output voltage is right.

I have used J-FET opamps that went crazy if either input was output the CMVR. That is the only example I can think of.

 

The Common Mode Voltage Range is not what you think it is.

It is determined by tying the two inputs together such that the voltage at both inputs is the common mode voltage. But what the spec is actually telling you is what is the range of input voltages that the device is spec'ed for at each input.

So if the CMVR is -0.3 V to 1.5 V, do not let the voltage at either input go outside of those limits. Going outside those limits may not damage the part -- usually won't unless you actually go above or below the supply rails by enough to turn on protection diodes -- but the device functional behavior or, most especially, timing behavior may deviate significantly from the specifications.

What do you mean its not what i think it is?

I think it is this based on a page from AD

https://www.analog.com/en/design-center/glossary/input_cmvr_v.html

 

The comparator can tolerate any input voltage up to the supply voltage without damage.
It will only operate properly when the input is no higher than 1.5V (Edit - 1.2V)below the supply voltage.

 

The 1.5V is only when the operating voltage is 2.7V. If you look down further in the datasheet you will see that it is 3.8V when the operating voltage is 5V.

You need to use a supply voltage that is at least 1.2V higher than any voltage you will put on the inputs.

 

The comparator can tolerate any input voltage up to the supply voltage without damage.
It will only operate properly when the input is no higher than 1.5V below the supply voltage.

1.2 V for that part.

 

What do you mean its not what i think it is?

I think it is this based on a page from AD

https://www.analog.com/en/design-center/glossary/input_cmvr_v.html

For opamps operating in the linear range -- which is the goal for this spec -- there is no appreciable difference between the voltage of each input and the common mode voltage of the two since the differential voltage is nearly zero. But for a comparator, this is typically not the case.

 

But I really don’t think they mean the average of the two inputs, as in the definition linked by the TS.

With a 5V supply and 3.8V common mode limit, does that really mean that if one input is at 0, the other can be at 7.8V?

 

1.2 V for that part.

Correct.
A typo in my post.

 

But I really don’t think they mean the average of the two inputs, as in the definition linked by the TS.

With a 5V supply and 3.8V common mode limit, does that really mean that if one input is at 0, the other can be at 7.8V?

As I said, that definition is for opamps operating in the linear region. If an opamp has one input at 0 V and the other at 7.8 V, it is nowhere near operating in the linear region. Plus, the 7.8 V would exceed the absolute maximum voltage on that pin, which is a separate limit that has to also be honored.