BCD help

Bilbo1161 · Sep 15, 2015

Only just learning ladder so please bear with me.

If I have 4 bits to show the following states of a motor, stopped, starting, running and fault. How do I get these out of the PLC as a BCD to read 1 = stopped, 2 = starting, 3 = running and 4 = fault.
I am using a GE Fanuc 90-30 PLC

Thanks

widelto · Sep 15, 2015

You can go from 0000 up to 1001, that is 10 different states.
in this case:
0001 as stopped
0010 as starting
0011 as running
0100 as fault.

danatomega · Sep 15, 2015

Four bits:
bit value (weight) 8 4 2 1
bit number ------- 3 2 1 0

A value of one = 0 0 0 1
A value of two = 0 0 1 0
A value of three = 0 0 1 1
A value of four = 0 1 0 0

You can see that it is possible to go clear up to 1 1 1 1 (15 decimal) BUT since it's BCD (Binary Coded Decimal) it's not legal to go above 9 (bit pattern 1 0 0 1) then you have to move on to the next digit.

chavak · Sep 15, 2015

As Widelto pointed out you need only 3 bits to cater for 4 states, [Maybe can shrink it to 2 bits, if taken zero as another state]. If your PLC had a 'BCD compare' instruction, you may use it to monitor the states. Take note not to use the balance of the bits of that word in your logic if you plan to use a compare instruction.

You may watch this clock to see BCD in action

http://www.plcs.net/downloads/index...vnpok18m6o2&direction=0&order=&directory=Misc

Regards

Bilbo1161 · Sep 16, 2015

I'm now wondering if i need to use bcd at all, or if i just use the MOV function to move the constant of either 1, 2, 3 or 4 into a register on the appropriate state being true.
What i need is the decimal value out as a register to use in a Scada package

the_msp · Sep 16, 2015

Yes I would do that (MOV). What is your SCADA package?

Bilbo1161 · Sep 16, 2015

Wonderware Intouch Standalone

gibbom · Sep 16, 2015

Ladder option

bit 0 = stopped
bit 1 = starting
bit 2 = running
bit 3 = fault

then

XIC(bit0), XIO(bit3) MOV(0,Result);
XIC(bit1), XIO(bit3) MOV(1,Result);
XIC(bit2), XIO(bit3), MOV(2,Result);
XIC(bit3), MOV(3,Result);

Bilbo1161 · Sep 16, 2015

gibbom said:
bit 0 = stopped
bit 1 = starting
bit 2 = running
bit 3 = fault

then

XIC(bit0), XIO(bit3) MOV(0,Result);
XIC(bit1), XIO(bit3) MOV(1,Result);
XIC(bit2), XIO(bit3), MOV(2,Result);
XIC(bit3), MOV(3,Result);

Thanks Gibbom.
That's probably what I am looking at.

Mickey · Sep 16, 2015

What if bit 1 and bit 2 are both on? etc...

It can't happen? Are you sure?

BCD help

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