dear all, i need some help about current conditioning pass the comparator

here's the problem

i want to measure RPM data of AC generator (500W, 60 V) .
from P= V I we get I = 8.3 A
diagram block :
|generator output <sinusoidal wave>|---> |10 A bridge rectifier (fullwave DC)|---->|comparator <squarewave>| ----> |optocoupler<TTL output>|----> |counter|

do not find a comparator can handle that high current. please help

 

If you just need to get the alternator RPM, there are safer ways to do it. If you can get to the rotating shaft, you can use a relectiive photosensor to give you a signal. Paint the end of the shaft black, and then glue a spot of foil near the edge for a reflector.

I would use a Fairchild device, Digi-Key part # QRB1114, but I don't know what parts you may have available. Positioning it close to the shaft end will give a low-going output every time the reflective spot goes by. It's then simple to apply the pulse to a counter circuit to give alternator RPM.

Trying to use the electrical output for the purpose will require something like a transformer to isolate the measuring circuit from the output. It is necessary to reduce the voltage to a level that a comparator can handle.

 

from P= V I we get I = 8.3 A
| do not find a comparator can handle that high current. please help

The generator can provide 8.3A, but a load will only draw current according to Ohm's Law. The comparator is a high input impedance device, and will not draw any appreciable current.