My requirement is to convert a 5v encoder signal to a 24v to be fed to a PLC i/p.
This runs at 100Khz and a marginal current is needed since the i/p of the PLC has a high enough input impedance.Can i just put a MOSFET in between?
Or any other known and proven technique?
I'm puzzled? Are the signal pairs (e.g. A and A~) designed to drive a pair of wires differentially, like RS485? Depending on wire length, how noisy is the environment, I would be inclined to put a RS485 differential receiver near the PLC, followed by a level shifter from 5V to 24V going into a PLC pin, one PLC pin for each for the A, B and Z channels.
I have a very preliminary idea about this and will get the Datasheets for the Encoder and the PLC input card later today.The 3 differential signals comment was made by the engineer who is handling this.So i AM thinking about adding an SN75176 at the input and then convert the TTL to 24V which is where i require all help.
The first thing hit me was using a MOSFET (P channel) then i realised that if i put an optical isolator i will achieve my goal and at the same time get isolation.
Can you comment?
I would look at how much current the encoder can source/sink at each differential pair. If you are planning to drive the photo-emitter in the opto-coupler directly from say the A A~ pins, then the A pin would be sourcing while the A~ pin would be sinking. You would like a minimum of ~ 10mA.
An open-collector photo-transistor on that side of the opto-coupler could be hooked emitter grounded, pullup returned to +24V with the collector also connected to the PLC input.
There are TTL input, high-voltage open-collector output level-shifters (a.k.a. Hex inverters/buffers) out there, too.