Sample and Hold, input voltage.

Hi,
I am using the Sample and Hold LF198/LF298 in my project.
But there is something that does not make sense.
I am unsure what the diodes in the circuit do?
Are they blocking current or are they maintaining a voltage potential off 1,4 V?

Secondly, when I use -5V +5V as supplyvoltage i can only sample signals in the range -2 to 2V. Is this because of the "The following specifcations apply for −VS + 3.5V ≤ VIN ≤ +VS" on page 3?

Should it not be like: maximum input voltage equal to +Vsupply?
Like written on page 3:
Absolute maximum ratings:
Input Voltage : Equal to Supply Voltage

The datasheet:
https://www.ti.com/lit/ds/symlink/lf198-n.pdf

Best,

The opamp with the diodes in the feedback loop is, I think, the "opamp-assisted peak detector"

Putting a diode in the feedback path is a way of using the (huge!) gain of the op-amp to remove the diode non-linarity offset.
This method is commonly used in "perfect diode" rectifier and "absolute value" analogue computation opamp circuits.

..such as --> HERE

Having two diodes allows sampling of negative values as well. In a simple diode peak detector, the voltage goes through a diode, to charge a "hold" capacitor. Good for high voltages maybe, but when you want accurate sampling of voltages that in normal circumstances would hardly turn on the diode, and needing to get over an offset of several hundred millivolts depending on the type of diode, then using the opamp pumps a voltage onto the "hold" capacitor that will match the input to within whatever is the (tiny) Vos of the opamp.

In the LF198, this is switched into the hold capacitor, and the voltage there drives the extremely high impedance of the output follower, which delivers the sample voltage while only loading the hold capacitor with the very small (nA or pA) current that goes up a opamp input.

if you consider a moment before the sample switch closes and a positive input of 100 micro volts, with no diodes. The switch is open so the first op-amp is running without any feed back so its output stage switches to Vcc and saturates and charges up any parasitic capacitance to + Vcc. When the switch closes this charge then charges up the storage cap, and the output stages take some finite time to come out of saturation. Adding the diodes, stops the output stages saturating because as soon as the output exceeds +- .7 of a volt, the gain of the opamp drops to unity. This means that the output of this amp is curtailed to +- .7V, which also reduces the charge on the parasitic capacitance. This output then goes to +- 0V after some time when the second opamp output follows the storage voltage.
" the range -2 to 2V. Is " no its -2 -> +5V
Frank