Battery Monitoring Circuit

The fuse is there to protect the wiring in the event of a fault.

is the fuse expected to protect the load?

For this fuse to blow, your 5.6K resistor must be able to temporarily withstand 1400(!)Watt. And plc must 'witness' about 420V, divided from a 2800V surge. So fuse is totally useless there. If you want to protect your plc from overvoltage you can use an appropriate zener (after first resistor) to ground. A 9v 1W zener there will handle 110ma, enough to produce ~70W of heat on 5.6K resistor, burn it out quickly, and protect the plc. S.M.

Hi Sujinvipin

I think that what you are trying to do is monitor the charge level of the batteries and you are doing this through a resistance bridge, is this correct?

If so the current that is being used is about 0.0085 amps or 8.5ma. So since you are monitoring voltage, the best protection that you could use is a IN4740 Zener Diodes connected in parallel between the PLC and ground that achieve the required current protection required. I would suggest the smallest fuse that you can get, as the load on this circuit is only 8.5ma and the zener that I have suggested is only rated at 1 watt to totally protect the plc you may have to have a several of these connected in Parallel, but don't forget that the plc has a high impedance when in voltage mode and can recover from most problems its self

That is all I have at this stage

Cheers

Joe

Yes, yes, yes.

What about the wiring?