[SOLVED] questions on common mode input voltage

Hello, I have two questions related to input common mode voltage.

I understand that for an fully differential op amp, the two input signals are usually V1=Vcm+Vid/2 and V2=Vcm-Vid/2. In this case, Vcm (the input common mode voltage) is defined as (V1+V2)/2, which equals to Vcm. However, when there is only one single-ended input V1 and the other input V2 is grounded, assume no feedback, then if the Vcm is still defined as (V1+V2)/2 then it becomes V1/2. Does it make sense? For example, when ground V2, and V1=sin(wt). By intuition, I think the common mode input should be 0 volt (the DC offset of V1). Is that correct? But then it doesn't equal to V1/2.

Another question is: in some component datasheet (I am looking at LT6044-15, whose common mode input range is from 0 to 1.5 V at 3V supply voltage), it usually specifies the common mode input range. I have been thinking: does this spec make sense? If the DC offset for V1 is super high (100V) and V2 is super low (-100V), the op amp surely won't work, but by definition, Vcm=(V1+V2)/2=0V, which is still within the allowed range. Isn't this weird? Unless this spec refers to one terminal, i.e. the DC offset of one terminal. How should I understand the dilemma here?

Thanks a lot!

Common mode input range is the average DC value of the input. When V1=sin(ωt). and V2 = 0, the CM voltage is still 0 since the DC of the sine wave is still zero.
The common mode input can be taken as the DC around which the voltages are varying.

grit_fire said:

If the DC offset for V1 is super high (100V) and V2 is super low (-100V), the op amp surely won't work, but by definition, Vcm=(V1+V2)/2=0V, which is still within the allowed range. Isn't this weird?

Click to expand...

While the input common mode is still within the required range, the actual inputs are not within the maximum and minimum range. This would be a separate specification.

nitishn5 said:

Common mode input range is the average DC value of the input. When V1=sin(ωt). and V2 = 0, the CM voltage is still 0 since the DC of the sine wave is still zero.
The common mode input can be taken as the DC around which the voltages are varying.




While the input common mode is still within the required range, the actual inputs are not within the maximum and minimum range. This would be a separate specification.

Click to expand...

Thank you.

Would you say rather than writing as Vcm=(V1+V2)/2, it is more accurate to say Vcm=[avg(V1)+avg(V2)]/2? Afterall, it's weird that textbooks never imply this but only use the former equation or may be it is just me who missed it.

I think it would b more accurate to say " Vcm=[avg(V1)+avg(V2)]/2? "
I think the textbooks hand-wave it away by mentioning somewhere that it is the DC value of the signal in question.

thanks a lot. That makes much sense.

Now, let's consider a close loop case. I was looking at the examples given in this link https://www.planetanalog.com/document.asp?doc_id=528175&site=planetanalog. It sounds to me that, when it is close-loop connected (figure 3), the Vcm=Vin. Is that correct?

grit_fire said:

However, when there is only one single-ended input V1 and the other input V2 is grounded, assume no feedback, then if the Vcm is still defined as (V1+V2)/2 then it becomes V1/2. Does it make sense?

Click to expand...

Yes

grit_fire said:

For example, when ground V2, and V1=sin(wt). By intuition, I think the common mode input should be 0 volt (the DC offset of V1). Is that correct?

Click to expand...

No

grit_fire said:

Another question is: in some component datasheet (I am looking at LT6044-15, whose common mode input range is from 0 to 1.5 V at 3V supply voltage), it usually specifies the common mode input range. I have been thinking: does this spec make sense? If the DC offset for V1 is super high (100V) and V2 is super low (-100V), the op amp surely won't work, but by definition, Vcm=(V1+V2)/2=0V, which is still within the allowed range. Isn't this weird? Unless this spec refers to one terminal, i.e. the DC offset of one terminal. How should I understand the dilemma here?

Click to expand...

The common mode input range isn't an "allowed" range as such; it's the range of input voltages over which the opamp behaves in the expected linear manner i.e. Vout = A0 * (Vin1 - Vin2), where A0 is some very large number.

However the opamp can only operate in that linear way for very small diferential input voltages. For example, if you put 100mV between the two input pins, the output will be stuck at either it's maximum or minimum value, so it's not operating in a linear manner anyway whether or not the input voltages are within the common mode input range.

In other words you only care about common mode input range when the differential input voltage is very small.

grit_fire said:

Would you say rather than writing as Vcm=(V1+V2)/2, it is more accurate to say Vcm=[avg(V1)+avg(V2)]/2?

Click to expand...

I think I prefer the original for 2 reasons:

• Averages ignore peaks, so the input voltage may be outside the common mode input range half the time even though the average is inside that range.
• Averages ignore AC, and it's often the AC component of the common mode voltage that we're interested in. Hence why darasheets often give a graph of common mode rejection ratio vs frequency, for example.

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grit_fire said:

It sounds to me that, when it is close-loop connected (figure 3), the Vcm=Vin. Is that correct?

Click to expand...

Not exactly equal, but very close.

Thanks a lot for your input.

Now we have two voices with opposite standings..

alright, I think the right way to put it is:

The DC common mode input is 0V and AC common mode input is V1/2. Common mode input range should consider the combination of DC and AC common mode input.

This link that you gave earlier seems to explain things quite well. In particular, example 2 illustrates the point I was trying to make nicely. The DC (or average) value of the input voltage is well within the common mode input range. However Figures 5 and 6 clearly show that whenever the instantaneous input voltage exceeds that range, the opamp becomes completely non-linear.

Sorry, I forgot about the AC part of the common mode...

Thanks you so much! to both godfreyl and nitishn5