There are two voltage rails on the receiver unit, 5V and 8 volts, is the receiver unit, a separate unit which is in its own enclosure. I have read the comments about the receiver and the voltage is strict, above 5 volts and the unit lets out the magic smoke.
10.9V is the transmitter as 5v on the transmitter by all accounts gives a couple of meters range and I need to have max range of about 20 meters which is 10+ volts.
The PC123 optocouplers (x2) will be needed on the receiver unit as its output will form one of two functions, to help pulse a pin input on a unit that has an 8 volt supply to power and pulse the pin, the reset pins are simply shorted together. So this only needs the optocoupler arranged as a device that bridges the pins when activated.
I don't have any way of showing a wiring diagram, I can however show you via these links the units I am using.
The counter
https://www.amazon.co.uk/dp/B016U4J2RG is a unit that is used as a tachometer or as a counter up and down.
The Tx / Rx units
https://www.amazon.co.uk/gp/product/B08ZMN4PMJ is a set of 433 Mhz units with 4 channels
The optocouplers are needed x2 for the isolation of the higher voltage rail from the lower voltage rail as the counter is operating on its lowest voltage setting.
So what I need is to drop the 5volt rail on the low voltage rail down so that the input voltage is 1.2v.
In this data sheet
https://media.digikey.com/pdf/data%20sheets/sharp%20pdfs/pc123%20series.pdf page 5, it shows a table. If I am reading this right,
Looking up resistor voltage dropping, ( input volts - unit vots ) / milliamps is what I need to use. I think I got the wrong scale earlier as milliamps is 1/1000 making 20mA 0.02amps.
So 5v = 1.2v = 3.8v 3.8v / 20uA or 3.8v / 0.02 = 190 ohms
The closest I can get to that figure is 195 ohm (120 ohm + 75 ohm inline)