ADL5201 Common Mode Input Voltage

Hi,

I still don't get the full picture.

With a common mode voltage of +1.5 V, for example,  the negative is allowed to swing down to -1.2 V (presumably). resulting in a  2.7 V peak swing. At the same time, for example, the positive input can swing up to +3.6 V, resulting in a 2.1 V peak for a total of 4.8 V p-p (not 10.8V as per the data sheet).

Assuming that the positive input could also swing by +2.7 V, the total would now be 5.4 V p-p. And, assuming the 10.8 V specification quoted is with s source impedance of 150 Ohm (the impedance of the device) then the 10.8 V figure comes into the picture.

But, there are a lot of assumptions in the above.

Is there a simpler explanation for how the 10.8V p-p is derived?

Regards Anthony

Hi,

I still don't get the full picture.

With a common mode voltage of +1.5 V, for example,  the negative is allowed to swing down to -1.2 V (presumably). resulting in a  2.7 V peak swing. At the same time, for example, the positive input can swing up to +3.6 V, resulting in a 2.1 V peak for a total of 4.8 V p-p (not 10.8V as per the data sheet).

Assuming that the positive input could also swing by +2.7 V, the total would now be 5.4 V p-p. And, assuming the 10.8 V specification quoted is with s source impedance of 150 Ohm (the impedance of the device) then the 10.8 V figure comes into the picture.

But, there are a lot of assumptions in the above.

Is there a simpler explanation for how the 10.8V p-p is derived?

Regards Anthony