QI generate a sinusoidal waveform with a I/V converter using an operational
amplifier. (simply using a feedback resistor) The resultant output waveform is
2 Vpp and its mean value is zero. I read that the maximun input voltage of the
ad835 is +\- 1 Volt but it can resist to an overvoltage of at least 20%. My
question is : i have to protect in some way the input of my ad835?
AThe different ranges of differential input voltage are as follows
+/-1V - the device will operate to datasheet specifications
+/-1.2Vmin, +/-1.4Vtyp - the device will operate without clipping the input
+/-VS (nominally +/-5V) - the device will not be damaged
one caveat is that if the one of the inputs goes beyond the positive supply
rail and the other input goes beyond the negative rail very quickly, the device
may be damaged. There is current limiting built in to the IC to protect against
this but you should try not to swing the inputs beyond the rails too quickly.
You should keep the common-mode input voltage within the -2.5V to +3V range.
The part will not break if you bring the common-mode up to the rails, but you
will turn off the front-end and multiplier core as you leave the -2.5V to +3V
range. We do not foresee phase reversal. The device will "misbehave" only so
far as the multiplier core will turn off at which time the output stage will
peg to the rail.