This circuit takes a 0-50V, 0-50Khz pulse signal at the input and outputs the buffered signal to a microcontroller pin. The zener at the output is 3.3v. Anyone know how this circuit works and what the diodes are for?

 

Follow up question. Is the voltage at the positive op-amp terminal always ~.7V because of the diode drop?..

 

The diodes at the input limit the pulse in to an appropriate voltage for the circuit. The diodes on the output limit the output to voltages the are acceptable to the microcontroller. They probably used a microcontroller that is power off of a 3.3V supply.

The amplifier provides gain for the for the limited pulse input.

 

Follow up question. Is the voltage at the positive op-amp terminal always ~.7V because of the diode drop?..

It would be about 0.7V for a positive pulse and about 0.7V for a negative pulse.

 

It would be about 0.7V for a positive pulse and about 0.7V for a negative pulse.

Thanks for your response. It makes sense. The pulse signal is limited to 0-50VDC so it doesnt go below zero. Does this mean D5 and D6 would always be off and D2 and D3 will switch from ON/OFF? If this is true, then what is the purpose of D5 and D6, can they serve as protection?

 

Many cases they are there just in case. Things like ESD and other possible surge sources justify them. Figure a diode is around 2¢, the rest of the circuit can be $20. Cheap at the price.

 

Does this mean D5 and D6 would always be off and D2 and D3 will switch from ON/OFF?

In the ideal case, this is true. As both you and Bill have suggested, the other diodes probably serve as protection against the non-ideal pulse at the input. Pulses have a tendancy to overshoot their desired voltage before they settle where you want them. In this case, the positive pulse might go slightly negative as it returns to 0V. The other diodes make sure it doesn't go too far into the electronics and have an affect. Depends on the pulse, so you can put the diodes in "just in case."