To answer your questions I recall the circuit here:




The actual value of reference voltage is rather irrelevant. Important is only to find a suited compromise:

1. In order to provide a symmetrical hystersis, being equal for positive and negative half waves, reference voltage should be at half the supply voltage.

2. Reference voltage at non-inverting input should be small, because otherwise the performance of the bipolar to unipolar transformation becomes too asymmetrical. So, a small reference voltage is needed to keep the difference in performance for positive and negative half waves small.

3. On the other hand, if the reference voltage is too small, then the circuit becomes highly susceptible against offset voltage changes (drift). So, the reference voltage should be much higher than the offset voltage. From this point of view choosing a reference voltage of about 1V, which is a third of supply voltage is a good compromise here.



In order to transform a bipolar signal ranging from -2.5V to 2.5V (as Prahlad stated) into a unipolar signal ranging from 0V to 3.3V-0.3V (or 3.0V-0.3V as I read from Prahalds first schematic, but which actually doesn't matter), the original circuit contains a zener diode, which in our case unfortunately allows a current to flow far below its turn-on voltage. This current is called leakage current, because the diode seems to leak: A 1N4148 silicon diode does allow a current to flow of about 1µA at 0.25V, which dramatically rises, when increasing the voltage. This leakage current must be taken into consideration, when designing the circuit, because this leakage current is additionally flowing to the regular current of voltage divider and can very badly influence the fabricated voltage levels. Even more problematic is the very high temperature dependence of this leakage current.
Why do I talk here about the leakage current of 1N4148, when we should use a zener diode as in Prahlads first schematic?
Because we can't use a standard zener diode here, due to the low supply voltage of micro. In order to protect the micro (also when the micro's supply is switched-off, but the transformer still delivers power...) we need a zener diode providing a threshold of 2V maximum, but at the same time showing ultra low leakage current. Normally one would put two silicon diodes in series and connect them in forward direction to form such a zener, because a 2V zener diode isn't fabricated. But the leakage current would be tremendeously high, not at all suited to desgin a low drift circuit. That's the reason why here a red LED is used, which shows a forward voltage drop of 1.6...1.8V at 20mA.

All we talked about above had to do with leakage current, not reverse current. What you are talking about is reverse current, which is the current that flows, when a reverse voltage is applied.
The 1N6263 wasn't chosen because of its small leakage current or small reverse current, but because of its small forward voltage drop (it's a small signal Schottky diode)! Even if input voltage is 10V, forward voltage drop of first 1N6263 will be no more than about 0.4V. So, when supply voltage of micro is off, then still no unsane voltage is present at its inverting input.
The same for the non-inverting input of micro if the supply voltage of micro is off: The red LED limits the potential to about 1.6V...1.8V. The second 1N6263 limits the input voltage to about Vcc + 0.3V.
A standard 1N4148 would violate the maximum specifications of micro, because then the potentials at inputs can exceed the 0.5V limits!

Also, only a small signal Schottky diode like 1N6263 or similar can be used here. A Schottky rectifier like 1N5819 would cause a too high reverse current, would badly influence the potentials and the whole circuit would suffer from the reverse current's temperature drift.



No spoecial care. 5k1 is ok. Are you sure that it's needed?



You could. But the 1M...10M feedback resistor must still be connected from output to non-inverting input!! As I stated in an earlier post, to provide a hysteresis you need a POSITIVE feedback, which is only fullfilled when connecting the feedback resistor to the non-inverting input.

Kai