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Old December 30th, 2002, 06:58 PM   #1
rbbalaji
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Omron C200h-cp114

Hi again,

I have become a sort of regular visitor here I guess. This question I am asking out of curiosity cos I don't have to deal with it at work but just thought that there is nothing wrong in learning. I had seen a C200H-CP114 PLC and it had like 4 or 5 output blocks connected to it. I was wondering what is the output address you give in the program with so many outputs. If someone could help me it would be great. I am getting well versed with CPM1A PLC. Can we connect output blocks to them too. I guess no.

By the way if anyone of you knows of a very good website like this for robotics please let me know. This is really a great site where people share their experiences.

Bye and wish you all a very happy and prosperous new year.

Balaji

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Old December 31st, 2002, 01:09 AM   #2
BobB
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Talking

That is not a valid Omron CPU number. Would suggest you find out the correct number and go to http://oeiweb.omron.com/oei/ where you can download any manuals you require in PDF format.
Basically, the first rack starts at channel 0 and progresses to channel 9 for a 10 slot rack. However, if an I/O card is a special I/O card (32 I/O or above), reference to the manuals is required to determine which address is to be used.
Addressing for 16 bit I/O cards is 0.00 to 0.15 (left hand end slot), then 1.00 to 1.15 (second slot) and so on. The last standard address on the main rack for 16 bit cards is 9.00 to 9.15. If all slots numbers are not used, then the bits for those slots can be used as internal relays (bits). For example, in the case of an 8 slot rack, channels 8 and 9 are not in use on the rack. These bits, 8.00 to 8.15 and 9.00 to 9.15 are available for use as internal relays.
The PLC will not care if the card is an input or output, it is just a number. No I/Q or X/Y addressing is required.
Having an open sheet can be a bit daunting the first time round but those of us that use Omron most of the time like not being limited as in most "conventional" PLCs to having to use I/Q or X/Y. Takes more time typing them in, depending of the programming package. Even Omron's latest PLCs used with Cx-Programmer will accept the digital address "0" and translate to "0.00" "100" is translated as "1.00".
The other thing you will find with Omron PLCs earlier than the CS1 or CJ1 is that there is a finite number of timer/counter numbers that can be allocated. A long time since I have used a C200H but I believe there are 512 timer/counter allocation numbers. These can be allocated to timers, high speed timers, counters but the same number can only be allocated once. For example, timers 0-510 (a total of 511 timers) counter 511 (a total of 1 counter) using the total of 512 timer/counter memory allocations.
The CS1 and CJ1 have 4096 timers and 4096 counters available in separate memory areas. At last, I will not run out of timers and counters and have to make my own out of registers with @INC and move functions.

Last edited by BobB; December 31st, 2002 at 01:13 AM.
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Old December 31st, 2002, 06:56 PM   #3
Jay Anthony
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Quote:
I had seen a C200H-CP114 PLC and it had like 4 or 5 output blocks connected to it. I was wondering what is the output address you give in the program with so many outputs.
The C200H-CP114 is an Intelligent I/O module that can be mounted to a C200H style I/O rack. The 'CP" in the part number refers to "Cam Positioner." The module is basically a resolver (similar to absolute encoder) input module and has 48 preset outputs surrounding a 720 degree (360 in 1/2 degree steps) rotation. The first 16 presets are available as direct outputs and can directly drive relay blocks. See attachment below. Here is the link to the manual:C200H-CP114 Cam Positioner Module
The CPM1A PLC uses terminal strip I/O blocks but can be terminated to relay blocks with pre-terminated cables. Here is the link:Pre-Terminated Cables
Attached Images
File Type: jpg c200h-cp114.jpg (95.3 KB, 235 views)
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Old December 31st, 2002, 09:06 PM   #4
rbbalaji
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Thanks Jay again. But can you give me some basic introduction to what is an I/O rack. And what is a bus. Cos my automation manager wanted to change the program in the C200H PLC and I think what he did was put a Cam Positioner Unit(that's what i think it was) in the bus(thats what he told me it was) and connected his computer(old one which had LSS) to the cam positioner and programmed the C200H PLC. Can you tell me where I can read more about this bs and I/O rack for C200H PLC. I dont think he connected a power supply to what i thought was a cam positioner so it may not be it cos as I saw in your manual we have to connect power supply to it. And Jay as for the program I asked you by email I know you think I am asking the same question again but it is not true. This program which I currently use has 100 lines but unfortunately the mini SP10 cannot accept a 100 line program so I have to shortedn it(which I did a bit). But I will mail you the100 line program and if you could please help me you can remove all the unwanted lines.
Thanks,
Balaji
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Old December 31st, 2002, 10:01 PM   #5
rsdoran
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Quote:
But can you give me some basic introduction to what is an I/O rack.
I know this will sound sarcastic but I cant believe you are asking that question. If you are attempting to work with plc's of any kind and can not relate to what that means then maybe its time to do alot more studying or find another career.

Jay is very informative and very knowledgeable but some things just take reading the manual(s) involved. Take a few days and do some reading then come back when you need a question answered.
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Old January 2nd, 2003, 01:00 AM   #6
Jay Anthony
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OK, I did the cardinal sin and took it off-line. Here's what happened. I should have known better. BTW rbbalaji, contract programming starts at $125/Hr!

Hi Jay,
The PLC was luckily able to have 120V AC outputss.
Jay I need another favor from you. I need to write a
program for a fixture. It has two touchbuttons, four
sensors for clip presence on clip holders, two
cylinders to lock or clamp the part. and two cyliders
to insert four clips on the part. I have two outputs,
one direct valve connected through tees to the two
locking or clamping cylinders and another direct valve
connected through tees to the two cyliders used for
clip attachment. Now I want the program to be written
in such a way:
1. When you press two touchbuttons, the locking
cylinder locks the part if the four clips are present
on the clipholder.
2. once the part is locked the clip attachment cylider
attaches the clip on the part.
3.If by chance there is any clip left on the
clipholder dont release the locking cylider, so that
the operator cannot take a faulty part. then operator
presses touchbuttons for 3 seconds for reset.
4. If everything is fine release the lock and end
cycle.

I had a similar program and I am using that with some
modifications since I did not have the time or
experience to write another program. But this program
is really very long. Can you help me out. If you want
I can mail you the program I am currently using.
Bye, and wish you a very happy and joyous newy year.
Balaji.

Man, didn't you learn anything from this one!!!!!

September 26th, 2002 09:40 PM

rbbalaji

actually i am using the plc for pneumatics. and i use part sensors and when part is detected i wnat to clamp the part when two touch buttons are pressed. then i check for clips. once the clips are put then i want to release the clamp again when the touch buttons are pressed. but this occurs only once. in the sense that once the clamp is released i want it to continue so that when the next part is put it is clamped again. can anyone help me. the program i have written works only for one cycle. the program is as follows(omron plc)
LD 00006 touchbutton1
TIM 005
#5
LD 00007 touchbutton2
TIM 006
#5
LD 00006
AND 00007
AND 00001 partsensor1
AND 00004 partsensor2
AND NOT TIM 005
AND NOT TIM 006
OR 200
AND NOT 300
OUT 200 clamppart
LD 200
OUT 10.01
OUT 1100
LD 00000 clipsensor1
AND 00001
AND 00002 clipsensor2
AND 00003 clipsensor3
AND 00004
AND 00005 clipsensor4
AND 00006
AND 00007
AND NOT TIM 005
AND NOT TIM 006
OR 300
OUT 300 clippresent
END

September 27th, 2002 05:31 AM

Jay Anthony

This is a classic alternate action or flip-flop circuit where the first press of an input turns an output on and the second press turns it off. Two things are different:
1. There are two inputs instead of one
2. There are permissives in the On side of the circuit and the Off side of the circuit.

Let's see what we can salvage out of the original circuit.

LD 00006 touchbutton1
TIM 005
#5
LD 00007 touchbutton2
TIM 006
#5

This sets up the beginning of an anti-tiedown operator pushbutton circuit.

Now re-arrange your common input conditions to drive a DIFU (one-shot for you AB programmers.)

LD 00006
AND 00007
AND 00001 partsensor1
AND 00004 partsensor2
AND NOT TIM 005
AND NOT TIM 006
DIFU 300

Now use the DIFU 300 to drive the coil to a SET 200.

LD 300
AND NOT 200
SET 200

Finally use the DIFU 300 and all of your release permissives to drive the coil to a RST 200.

LD 300
AND 200
AND 00000 clipsensor1
AND 00002 clipsensor2
AND 00003 clipsensor3
AND 00005 clipsensor4
RST 200

Keep the original circuit that actually controls the clamp.

LD 200
OUT 10.01
OUT 1100


The key to this arrangement is the DIFU 300 (one-shot) which only gives a one scan pulse when PB1 and PB2 and Part1 and Part2 are energized and then released.

September 27th, 2002 09:13 AM

rbbalaji

thanks for your help. i have another doubt. i am using this plc in my fixture wherin i have two sensors. one sensor has a clip detector and when a clip is detected one cylider is fired and it keeps the part. and the part is cut using a cutter on application of a pushbutton. when the cutter cuts the part, the other sensor senses the cut the cylider should go back so that the part can be removed. the sensor that senses the cutter senses it only for a few seconds.
ow should i program the omron plc
LD 00.00 clip sensor
AND NOT 00.01 cutter sensor
OUT 10.00 cylinder

September 30th, 2002 05:15 PM

Jay Anthony

Did the circuit above solve your original problem? Is this the same machine? If it is, then you are not saying the same sequence as above. Make a step by step sequence of operations.
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September 30th, 2002 08:19 PM

rbbalaji


yes that solced the problem. and this is not the same circuit. this is for another fixture. the step by step sequence is as follows:
1. the operator places metal clip.
2. the operator inserts plastic tube into the metal clip and the other end of the tube goes t the cutter.
3. the operator presses pushbutton one. this locks his plastic tube using a cylinder that is directly connected to the puhshbutton(no plc action). but ther is another cylinder lets call it cylinder1 that has to fire when he pushes pushbutton1 depeding on sensor1 that detects the metal clip.
4. the operator presses another pushbutton2 which fires the cutter again without the help of plc(direct switch).
5. when sensor2 detects cutter which is only for a fraction of a second this cylinder1 one should go back.
please help me with this.
Thank You in advance.

October 1st, 2002 05:25 AM

Jay Anthony

(Quote
1. The operator places metal clip.
2. The operator inserts plastic tube into the metal clip and the other end of the tube goes to the cutter.
3. The operator presses Pushbutton one. This locks his plastic tube using a cylinder that is directly connected to the Pushbutton 1(no plc action).


If I understand you correctly, up to this point, none of the above pushbuttons and cylinders are hooked to the PLC.
(Quote
But there is another cylinder (let's call it cylinder 1) that has to fire when he pushes Pushbutton 1 depending on Sensor 1 that detects the metal clip.
4. The operator presses another Pushbutton 2 which fires the cutter again without the help of plc(direct switch).
5. When Sensor 2 detects cutter which is only for a fraction of a second, this Cylinder 1 one should go back.


Try this out:


LD 00.00 ClipSensor1
SET 10.00 Cylinder1
LD 00.01 CutterSensor2
RST 10.00 Cylinder1

Let us know what happens.

October 1st, 2002 07:03 PM

rbbalaji

Hi everyone!

Thanks for everyone who has helped me in this site. I have another doubt. I hope you will be able to guide me as before. I am actually modifying someoneelse's program. Let me explaing the operation.
This is a fixture wherein the steps are as follows:
1. Operator places two metal clips on clip holders. Two sesors are used for this purpose and are inputs to Omron PLC 000 and 001.
2. Operator places two parts on the nest which is on top of clip holder.
3. Operator presses pu8shbutton. Input to PLC 002.
4. Two cylinders press the part from the top. Outputs 10.00 and 10.02.
5. Two cylinders pushes the clip from the bottom. Only one direct valve for both cylinders and output is 10.01.
6. The bottom cylinders for the clip holders is released.
7. The top cylinder is released.

Now this program is working very well. But I altered this program due to new demands. Since the operators were putting parts that already have clips on the nest and as a result the clipholder was damaged we had to find a solution. As a solution this was decided:
1.we put two more sensors to see to it that the part placed on the nest doesnt have clips alreay attached. We used inputs 007 and 008 for thses two sensors.
2. I clip already attached then dont clamp the part,

The program for this is as follows:

LD 007
OR 008
OUT 500

LD 001
AND 002
OUT 200

LD 200
AND NOT 500
OUT 10.00

But this gave rise tom a new problem.
1. Once the part without clip is placed and the clipholder cylinder is activated and the part clamp cylinder is activated it senses the new clip to be attached and then resets the part clamp and clipholder cylinders as a result of which the clip is not attached.

CAN ANYBODY HELP ME WITH THIS PLEASE

October 8th, 2002 08:23 PM

Jay Anthony

(Quote
1. Once the part without clip is placed and the clipholder cylinder is activated and the part clamp cylinder is activated, it senses the new clip to be attached and then resets the part clamp and clipholder cylinders as a result of which the clip is not attached.

This is a case of permissives for a start circuit becoming the early stoppage of the start circuit. The only thing you are concerned about is that the clips are not present before you push the start clamp button.

Let's see what we can salvage from your circuit:

LD 007
OR 008
OUT 500

LD 001 (should this be 000?)
AND 002 (should this be 001?)
OUT 200

This is all OK. Now we need to remember that the clips were not present before you pushed the start clamp pushbutton. Use an open contact of 10.00 to latch the circuit:

LD 200
AND NOT 500
OR 10.00
AND NOT 201
OUT 10.00

Now you need a way to unlatch the circuit. Use the bottom cylinder output to drive a DIFD (trailing edge one-shot for you AB programmers.)

LD 10.01
DIFD 201
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Old January 2nd, 2003, 01:36 AM   #7
rsdoran
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Way and above the call of plc.net or any forum duty Jay, I admire you for your commitment. Then again I am willing to do the same so guess we both aint got good sense, but you are better at programming.

I on the other hand am not that good so can be a little more sarcastic, mean or whatever term is needed to make people think for theirselves. I want some of that buck 25 an hour for myself. It amazes me the people that have these jobs that do not deserve to have them. I may not deserve it myself but I am willing to bet I could do as well if not far better than many that have the job now. Maybe thats my ego talking but it sure would be nice to find out. I would do some of this for free if it offered the opportunity to show I can do it properly.

Oh well, maybe one day.
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Old January 2nd, 2003, 03:36 AM   #8
BobB
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Thumbs down

Fair dinkum , this guy is using the forum to learn about PLCs and how to programme them. I would suggest "rbbalaji" go to technical school and learn what a PLC is. Or better still, buy Phil's book. Trouble is he would then probably annoy Phil for explanations. Sorry Phil, probably a bad idea.
Really, the programme is very short and quite simple to understand. If he thinks that this is a long programme, I would hate to think how he would go on something complex.
Jay, I do not know how you have the patience. If you are being paid by this guy fine, but I think $125.00 per hour is not enough.
bonkhead

Last edited by BobB; January 2nd, 2003 at 03:45 AM.
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