Indirect addressing in S7-300

mohitkataria012 · Feb 9, 2020

Hi All,

As i am new to STL programming I am facing a problem with the code below

OPN DB 103
A M [DBD 140]
= M 206.0
A M [DBD 384]
= M 206.2
OPN DB 104
A M [DBD 20]
= M 206.1

what i understood from the code above is that

1) open db 103

2) Load the value of db103.dbd140 in the memory bit. Let say value of dbd140 is 10.0

3) Now if M10.0 is high then it will make M 206.0 high.

Am i right??

Best Regards
Mohit_kataria

parky · Feb 9, 2020

Yes correct, The value in DB103 DBD140 will contain a M bit pointer i.e. 10.0 so it equates to M10.0
ignoring the indirect it is the same as
A M 10.0 // A M[DBD140] contains 10.0
= M206.0

mohitkataria012 · Feb 9, 2020

Thanks Parky for the quick response!!!

Could you please explain what does indirect mean??
Please see the attached Picture.
If i make M 250.0 high then M 206.1 is still LOW

Please Explain......

L D[AR2P#0.0] · Feb 9, 2020

The address in DB14.dbd20 is P#31.2 - the address is bit based so

250 decimal = (byte 31 * 8) + 2

If you want to address M250.0 then the number required in DB14.DBD20 is (250*8) + 0 = 2000

mohitkataria012 · Feb 9, 2020

Thanks for the reply!! @L D[AR2,P#0.0]

Bit confusing pointer in general is P#byte.bit as in Memory indirect addressing
Here Let say i want to point M 31.2 for that i have to make it in pointer format that P#byte.bit. but what is the reason behind multiplying 31 with eight

Is it because of 1byte =8 bits.

Lare · Feb 9, 2020

mohitkataria012 said:
Thanks for the reply!! @L D[AR2,P#0.0]

Bit confusing pointer in general is P#byte.bit as in Memory indirect addressing
Here Let say i want to point M 31.2 for that i have to make it in pointer format that P#byte.bit. but what is the reason behind multiplying 31 with eight
Is it because of 1byte =8 bits.

Exactly.
You can also shift 3, as it is same than multiply by 8

mohitkataria012 · Feb 9, 2020

Thanks Lare:beerchug:

But in my above attachment under the standard tab it is showing 250 for M 31.1,M 31.2 & M 225.1. However I know the address are right as db's contain different value so why only 250 is shown under the Standard tab not the other different values as per addresses?

L D[AR2P#0.0] · Feb 9, 2020

The standard tab shows the values loaded into accumulator 1

mohitkataria012 · Feb 9, 2020

ohhh

Also db103.dbd384 contains 1815 (DW16#00000717) if i divide it by 8 address comes out to be M 226.7.
1815=226*8 +7. why it is pointing to address 225.1???

please see the attachment.

L D[AR2P#0.0] · Feb 9, 2020

Put L DB103.DBD384 immediately before the A M[DBD384] instruction to verify

mohitkataria012 · Feb 10, 2020

@ L D[AR2,P#0.0]

You are right

the value is 1801 in decimal format and which gives an address of M 225.1. But in my db value is 1815 then how it is loading 1801??

Lare · Feb 10, 2020

Is it writed on program several times?
Last writed value is showed on db block

mohitkataria012 · Feb 10, 2020

Hi Lare,

You were right the address is written two times.I have attached screenshot of both the places.Please have a look. And according to Program cycle it will first execute network 2 & 4(SNIP2) after that it will execute Network 33 & 34(SNIP3) that is why DB is showing value 1815(db103.dbd384).

Thanks

:site:

parky · Feb 10, 2020

Personally I like the old S5 Indirect addressing but then again I'm old school so cut my teeth on S5.

mohitkataria012 · Feb 10, 2020

Thanks Everyone !!!

for your support @ Parky i have no idea of S5 indirect addressing not even S7 indirect addressing. Being a newbee in STL programming i am learning new things.

Indirect addressing in S7-300

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