Solenoid Question

[Les Jones]

Hi Joe,
Terminal 2 on the timer to C on the small relay (The contacts only need to carry a few milliamps.)
Terminal 6 on the timer to NC on the small relay. (NOT normally open)
When the command from the robot is in the chuck open state the coil on the small relay will be energised. (So C will be connected to the NO contact. (The N state is with no power to the coil.) When the robot sends the command to close the chuck is REMOVES the 24 volts (That would have originaly caused the valve to open and close the chuck) Now this will remove power to the relay coil. This results in C now being connected to NC which will start the timer. On the output of the timer the solenoid valve needs to be powered via the timer's NC contact (Terminal 4 NOT terminal 3 that you said in post #19) so that when the end of the delay is reached the NC contact OPENS removing power from the solenoid valve and closing the chuck. I am assuming we are talking about the MS4SM that strantor suggested using. What is the answer to my question in the last sentence of post #16 ?

Les.

 

[strantor]

Strantor,
As long as you aren't fed up yet, could you please continue to work with me.

Sorry I've had the flu. Now that I've slept 40 out of the last 48 hours I can't sleep anymore.

Les et al,
I get what you are saying about pins 2 and 6 now. The timer is looking for a dry contact closure to start the timer. Right?

right

As far as the relay, because of the way the robot manufacture designed the circuit, normal operation (Chuck Open) outputs 24V. The coil will be energized connecting C to NO during normal operation. When Chuck Closed is activated, the coil will de-energize connecting C to NC. Right?

right

Therefore, I should connect timer pin 2(neutral from the robot PS to C and timer pin 6 to NO. NC on the relay should be left blank.

right

On the contact side of the timer, jumper pin 1 to pin 2. Pin 3 should be wired to the common side of the solenoid. The hot side of the solenoid would be wired in parallel with pin 10(24V from the PS.) Thoughts?

Sure, I drew it (below) the other way around but either way works.

Hi Joe, Terminal 2 on the timer to C on the small relay (The contacts only need to carry a few milliamps.)

Agreed. Although I drew it the other way around, doesn't really matter if you think of it as a SPST switch.

Terminal 6 on the timer to NC on the small relay. (NOT normally open)

I disagree

When the command from the robot is in the chuck open state the coil on the small relay will be energised. (So C will be connected to the NO contact. (The N state is with no power to the coil.) When the robot sends the command to close the chuck is REMOVES the 24 volts (That would have originaly caused the valve to open and close the chuck) Now this will remove power to the relay coil. This results in C now being connected to NC which will start the timer.

Disagree. The timer starts upon LOSS of the "START" signal (chuck OPEN signal).

On the output of the timer the solenoid valve needs to be powered via the timer's NC contact (Terminal 4 NOT terminal 3 that you said in post #19) so that when the end of the delay is reached the NC contact OPENS removing power from the solenoid valve and closing the chuck.

Disagree. Two lefts doesn't make a right in this case. If we follow the timing diagram, we are looking for a LOSS of the "start" signal (chuck OPEN signal), followed by a timed period of no state change, followed by a LOSS of output (chuck CLOSE). The conditions which reflect this are in the timed N.O. line. We see that at the end of the timed period, the NC contact CLOSES.




Here's how I drew it up. "start signal" means "chuck open signal." That's true coming out of the robot controller and coming out of the intermediate relay
"gate signal" and "reset signal" are not needed
I apparently think backwards from everybody else because I drew everything on the opposite sides of the contacts, but as has already been addressed, it doesn't really matter.

 

[strantor]

Oh yeah, and you should put a freewheeling diode in there as well to protect the relay from inductive spikes.
1N4004 is what I normally use.

 

[Les Jones]

Hi strantor,
My comments were for use with the delay-on timing diagram. I agree that your method will work with the signal off delay diagram. I think the original MS4SA timer could be used connected like this.

Les.

 

[strantor]

Hi strantor,
My comments were for use with the delay-on timing diagram. I agree that your method will work with the signal off delay diagram. I think the original MS4SA timer could be used connected like this.Les.

Oh, ok gotcha. I didn't expect that since you said this:

I am assuming we are talking about the MS4SM that strantor suggested using.

But now that we're on the same page, I must tip my hat. Your solution is much better considering he already has the ON-delay timer in-hand.

 

[cowboybob]

The logic is a bit hard to wrap my head around...

chuck closed button

my emphasis)

strantor and Les, I'm thinking that we don't know enough about this "button".

Joe,

Is it a real, manual switch (like a simple, non-spring loaded emergency "ON/OFF" slam switch) or is it a touch screen selection?

If it is a real switch, that if activated (OFF), disconnects ALL signals, from whatever source, to the valve controlling the chuck vise (thereby forcing it's closure), wouldn't that (if left in the OFF position) also continue to interrupt the"robot" (PLC) circuit's 24VDC OPEN command?

Or is it merely a screen selection which forces a program response to a PLC IO port that then turns off the 24VDC controlling the chuck position control valve?

Or is it a combination of the two"

Les, did you mean for the "robot" to be the source for the timer control?

 

[Joe Fricke]

Depressing the chuck close button on the robot pendent removes power from the solenoid.

 

[Les Jones]

Hi Joe,
This does not effect the function of the circuit but my original understanding was the the robot was commanded in some way to move to a position. When the robot reached that position IT sent the signal to the solenoid valve. Are you now saying that a human opperator presses the button when the robot has reached it's target position ? I still do not understand your added requirement in post #15
CBB,
The signal from the robot (Or manual push button ???) that originaly went to the solenoid valve now drives the small relay that effectivly inverts the signal to drive the timer.

Les.

 

[cowboybob]

Depressing the chuck close button on the robot pendent removes power from the solenoid.

I understand that. But how?

Is the button switch, all by itself, directly turning OFF the power to the solenoid (it is between the PLC and the solenoid) or is it signaling the PLC to turn OFF power to the solenoid?

Is the button spring loaded, i.e. when you remove your hand (finger) does the button return to its original position? If spring loaded, do you have to maintain pressure until the chuck is in the position you desire? Or is a single, momentary press all that is required?

Do you know if the button contacts are connected directly to the valve, or the PLC or both? In other words, HOW is the button interrupting the PLC command to OPEN?

These distinctions matter a great deal when trying to determine what that button action is, how that action results in the chuck closing and how to override the PLC command by supplying a timed 24VDC to the solenoid.

And at the same time, NOT wire up a modification that interferes with "normal" operation of the robot arm.

 

[Joe Fricke]

Momentary depressing chuck open button sends signal to robot's PLC, PLC outputs a 24V latching output. When the chuck close button is momentarily depressed the PLC removes the 24V volts.

 

[Joe Fricke]

Les,
When in setup mode the buttons are used to open and close the chucks, this is done through the robot's PLC though. When the robot is in automatic the PLC triggers the action based on where the robot is located in its program.

 

[Joe Fricke]

Les,
I tried your circuit that you had drawn up, but didn't see positive results. I'm guessing it is because I wired it wrong though. What I did was run the 24V in parallel to the positive side of the relay, to pin 7 of the timer and to the positive side of the solenoid valve. This 24V is only present when the chuck open button is depressed. I think this is where I went wrong. Because pressing the chuck close timer kills the power to the timer and the solenoid.

 

[cowboybob]

Momentary depressing chuck open button sends signal to robot's PLC, PLC outputs a 24V latching output. When the chuck close button is momentarily depressed the PLC removes the 24V volts.

OK. I suspected that. The Close button is used solely to momentarily generate a trigger pulse to the PLC. The PLC then responds by changing the latched output to the solenoid from High to Low. The button does not, by itself, control the solenoid voltage changes: it is only telling the PLC what to do. I would expect the Chuck Open button to act in the same fashion, except the output would latch from Low to High.

I do not see a way to use the button as you wish without re-programming the PLC to include the delay you want during the period of time you specified. And if your were to introduce a delay circuit into the line feeding the Close chuck solenoid, it would delay each and every time the PLC executed that command, irrespective of the program's sequence point.

To achieve your goal you'll have to introduce a separate, additional Chuck Close switch, using power from elsewhere in the machine, for the delay circuit. And at that, you may run into problems with appropriate chuck action during other steps in the robot chuck control routines.

<EDIT> And I have to wonder (never having done this before) if a "trained" robot even has a program that can be altered without an actual, physical "re-training", i.e., no lines of code to adjust.

 

[Les Jones]

Hi Joe,
Can you draw the schematic of EXACTLY how the solenoid valve was wired originally. Is the 24 volts DC supply totaly switched off when the chuck is closed. I had the impression that the positive side was connected directly to one end of the solenoid valve coil and the other end of the coil connected to the negative via a "contact". The "contact" could be a relay contact, a bipolar transistor or a mosfet.

Les.

 

[Joe Fricke]

Les,
The robot is pretty old and I could not find a schematic, but every one of the robots we have use the same backwards logic for the grip valves. So I attached a schematic from a newer robot. On the print "PART GRIP OPEN" is the same as what I refer to "CHUCK OPEN" in this thread. I really think there is a simple solution, but I'm just not as smart on these things as you all. I am familiar with reading basic schematics and can build a simple control circuit from scratch. But in this case I need to use the output logic from the robot and just interrupt it at the right time to delay the "closing of the chucks" On a side note, I will have just about every part I need to get this going, I have an On-delay timer here, a SmartRelay here and a Multi-purpose timer coming. If I can't figure it out soon, I will just replace the 2-way, 2-position valve with a 2-way 4-position valve, I'm almost positive that would work as well.

 

[cowboybob]

... Is the 24 volts DC supply totaly switched off when the chuck is closed. I had the impression that the positive side was connected directly to one end of the solenoid valve coil and the other end of the coil connected to the negative via a "contact". ...

Les, the 24VDC would still be present on either connection of the solenoid if the PLC ground leg of the wiring is being switched OFF (and latched). The total disappearance of the 24VDC confirms that it is the leg being latched OFF.

 

[Les Jones]

Hi Joe,
You need to find the negative connection of the 24 volts supply. It is this that you need to connect to the 0 V connection on my diagram. The diagram you have posted only shows the output terminal strip /connector. The negative of the 24 volt supply will be connected to the other side of the contact that originally fed the solenoid valve. (The "contact" may be a relay contact, a transistor or a mosfet.)

Les.

 

[Joe Fricke]

Reading your circuit, this is what I come up with.
1.) In normal condition (chucks open) there will be 24V from the robot control circuit applied to the coil which will pull the relay to NO.
2.) I should wire up a constant 24V direct from the PS to both pin 7 of the timer and the + side of the solenoid.
3.) When robot calls for (chuck close) the 24V from the control circuit will be removed from the coil allowing the 0V side to continue to pin 2 on the timer.
4.) Because there is a constant 24V on pin 7 of the timer once the relay opens it completes the circuit powering up the timer.
5.) Once time expires, timer contact closes which passes 0V from PS to negative side of solenoid
6.) Resetting happens when chuck open signal is called for because it reapplies 24V from the control circuit to the coil which removes the 0V from pin 2 on the timer (breaking the timers power circuit.)

 

[cowboybob]

So, the plan is to remove the negative (Gnd) lead of the solenoid from its current position on the PLC and reattach it to Pin 4 of the delay timer, correct?

That, actually, ought to work. I had not considered that particular change .

There might be a very slight "blip"of solenoid movement during the transition from No power - Power - No power as the various relays shift (x milliseconds). Chuck grip pressure would, hopefully, be unaffected.

 

[Joe Fricke]

Correct