Set/reset

userxyz · Jan 19, 2007

Code:

	  A	 #S_Spoelen_In_A
	  BLD   100
	  FP	#Membit_1
	  =	 L	  0.0
	  A	 L	  0.0
	  AN	#Spo_Stat_1
	  AN	#Spoelen_A
	  S	 #Spoelen_A
	  =	 #Spo_Stat_2
	  A	 L	  0.0
	  AN	#Spo_Stat_2
	  A	 #Spoelen_A
	  R	 #Spoelen_A
	  =	 #Spo_Stat_1

convert to LAD possible

Is there an other way also to do this (maybe a standard function from siemens or something?) ?

krk · Jan 19, 2007

A bit toggle eh?, no standard function AFAIK but you could simplify it a bit...

Code:

	 A	 #S_Spoelen_In_A
	 FP	#Membit_1
	 =	 L	 0.0
 
	 A	 L	 0.0
	 AN	#Spoelen_A
	 S	 #Spoelen_A
	 JC	_001
 
	 A	 L	 0.0
	 A	 #Spoelen_A
	 R	 #Spoelen_A
_001:	 NOP 0

I'd replace the absolute adressed temp bit with a symbolic one though...

PeterW · Jan 19, 2007

It looks like it is ladder displayed in STL. the BLD instruction and the use of L bits gives it away.

Try changing the block view from STL to ladder and it should display in ladder.

L D[AR2P#0.0] · Jan 19, 2007

Here's another version:

Code:

	  A	 #S_Spoelen_In_A
	  FP	#Membit_1
	  JCN   skip
	  AN	#Spoelen_A
	  =	 #Spoelen_A
skip: NOP   0

krk · Jan 19, 2007

very elegant Simon....

S7Guy · Jan 19, 2007

My all time favorite is still this one:

A M10.0
FP M10.1
X M10.2
= M10.2

It will display in FBD, but not ladder. It's a shame that Siemens didn't include XOR instructions in their ladder conversions.

krk · Jan 19, 2007

....and even more elegant S7_guy!

...that's gotta be as compact as you can get a toggle...

userxyz · Jan 20, 2007

Great

Great pieces of code !!

Really thanks, learned again ...

userxyz · Jan 20, 2007

Hey

A M10.0
FP M10.1
X M10.2
= M10.2

This is great, It's much better then my own code

this one I don't understand, is this a real toggle ? :::

A #S_Spoelen_In_A
FP #Membit_1
JCN skip
AN #Spoelen_A
= #Spoelen_A
skip: NOP 0

L D[AR2P#0.0] · Jan 20, 2007

If there is no rising edge, JCN skip will execute and #Spoelen_A will not be modified. If there is a rising edge the jump will not execute and then if #Spoelen_A=0 it will become a 1 and if #Spoelen_A=1 it will become a 0 due to:

Code:

AN #Spoelen_A
= #Spoelen_A

userxyz · Jan 21, 2007

Hey

Now I understand I think, because AN spoelen wil make the RLO 0 (when spelen is true), = Spoelen wil become zero ...

And when spoelen zero, AN Spoelen makes RLO 1...

?

userxyz · Jan 22, 2007

Hey

It's confusing me:

A #Input
FP membit
X #OUT
= #OUT

I tought I understand this code but noooooo...

Okay, we got an XOR function in this short code, an XOR on the positive edge from #Input and the #OUT bit.

So we get something like this:

A positive edge
AN #OUT
O
AN positive edge
A #OUT

That´s what I understood of it.

But the XOR is only executed on a positive edge and is only executed for 1 cycle...

So when #OUT is low...
A positive edge
AN #OUT

this in the XOR function wil make #OUT high, and in remains high because there isn´t written to it after that cycle.

When #OUT was 1, then
AN positive edge
A #OUT

so the positive edge wil is one in that cycle, the AN positive edge wil prevent to write 1 to #OUT, so, because #OUT is called in that cycle it will become 0.

Is this correct, or can anyone explain it correctly

Thanks In forward

regards,

Combo

L D[AR2P#0.0] · Jan 22, 2007

The XOR is always executed, the FP merely conditions the RLO prior to the XOR.

S7Guy · Jan 22, 2007

I think you understand it ok. I'll put it in my words too:

Ok, assuming all bits are false, we will be waiting for the input to go true. So....

1. #Input goes true. The RLO after the FP Membit is true for one scan.

2. The #Out bit is XORed with the RLO. The conditions are true, so "= #Out" is true.

3. On the next scan, we can assume that #Input is still true (probably a Push Button), but it doesn't matter because the RLO after the FP Membit is now false #Out is true, so the XOR condition is still true.

4. If the button is released, nothing changes.

5. Someone pushes the button again, the RLO after FP Membit is true, and the XOR condition is no longer true. #Out is set to false, and needs another rising edge at the input to set it again.

userxyz · Jan 23, 2007

S7Guy said:
I think you understand it ok. I'll put it in my words too:

Ok, assuming all bits are false, we will be waiting for the input to go true. So....

1. #Input goes true. The RLO after the FP Membit is true for one scan.

2. The #Out bit is XORed with the RLO. The conditions are true, so "= #Out" is true.

3. On the next scan, we can assume that #Input is still true (probably a Push Button), but it doesn't matter because the RLO after the FP Membit is now false #Out is true, so the XOR condition is still true.

4. If the button is released, nothing changes.

5. Someone pushes the button again, the RLO after FP Membit is true, and the XOR condition is no longer true. #Out is set to false, and needs another rising edge at the input to set it again.

L D[AR2,P#0.0] said that the XOR allways functions, And I understand why, I was thinking in Lad, XOR after FP does something else in ladder, the XOR would only work one cycle in lad.
And then I understand S7Guy also, it's the RLO of the positive edge that I must XOR with the Output.

XOR is basically a START/STOP circuit

Am I right ?

Set/reset

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