PLC to solenoids - Another Senior Project help

Hello,
My name is Khoa and I, like another whom posted a similar thread, am a senior in college working on my final design project (though nowhere near as complicated) and was wondering if I could get some help or input from the professionals here at PLCs.net on how to go about creating a switch/solenoid system controlled by a PLC.

First off, please excuse me if I don't use the proper terminology or am missing information or anything like that, wiring and electronics are a bit of a weak spot for me.

What my group and I's project entails is developing a filling system with a reservoir and a spool valve below and above the reservoir to control draining and filling. What we are hoping to do with our project is to use solenoids to move a pair of spool valves so that when one opens, the other closes simultaneously. They would only have to move perhaps .2 inches max. We would also like to be able to control that action using a manual switch.

We were looking to use a PLC that interacts with the solenoids to control their actions as well as a manual switch to turn on our device.

Our adviser directed to automationdirect.com and specifically their CLICK series, specifically the C0-00DD1-D 24VDC combined with the C0-00AC 24VDC .5A power supply (essentially their cheaper models in the interest of saving money as we are on a $400 total budget). In regards to the solenoids I was thinking of 2 Ledex Low Profile Push/Pulls, maybe part# 173921-0XX. By the way, does Push/Pull mean they can do both? If not, then it'll mean I'd need 4 rather than 2 so the spool valves can be moved in both directions correct?

I read on another thread here that we can't connect the solenoid directly to the PLC. If so, then how do we go about connecting it so it can be controlled by the PLC? Also, is the PLC's power supply sufficient to power everything or would the solenoids require a separate power source?

Would the system we are trying to do cost more than $150-$200? Are there cheaper but still suitable alternatives? If it takes up too much of our $400 budget, we may have to scrap the idea since we still have other parts of the project to build.

Sorry for the long post but wanted to include what detail I could. Please feel free to ask for any other details if you need them and I can try to find them.

Thank you.

That post was NOT too long it had enough information to get our interest and enough to get basic understanding of what you are doing.

Yes you can connect solenoids directly to PLC outputs if and
IF you
1. Match the PLC output voltage to that of the solenoid
2. Ensure solenoid current (dont forget inrush) is less than the currnt rating of the PLC output
3. Ensure total watts and or current of switched loads especially if energized simultaneously do not exceed teh power supply rating. Easy if all are ON or OFF at any one time - bit tougher to figure if some are on and some are off at differing times and for differing intervals.
4. It is best although NOT mandatory to put on MOV s to prevent damage to PLC outputs when coil is turned off (these result from collapsing field). If you keep your solenoids small say 15 watt you should be OK. MOVs are so cheap they should be used in my opinion.

IF you are doing just one fill station with just one reservoir and one each fill and drain valve it should be not too difficult. If you use common ASCO solenoid valves and you tell us what the voltage of your PLC outputs are
AND
If you ask then there may be a generous soul here who may sell at discount or even donate.
DO NOT BE CHEAP
offering to pay the freight and throw in a few bucks for a bottle of their favorite booze works quite well - it did for me.

As alternative you do NOT have to build the whole machine. I was allowed to simulate an automated drill press. I did not have a drill press but used bits and pieces out of junk box to simulate drive motors, drill bit drive and other driven loads.

You can also do a lot of simulation with just light bulbs. Level can be simulated with a momentary switch.

The harder part of PLC work is the actual machine design and setup from concept ie we want it to do this, then decide what the operation sequences are, how to sense errors ie jammed drill bit etc etc, to selection of components, safety considerations add a bit of complexity to it also.

The hardest part really is when it does not operate in accordance with plan, figuring out how did I make that mistake and what am I going to do about it?

Dan Bentler

Our adviser directed to automationdirect.com and specifically their CLICK series, specifically the C0-00DD1-D 24VDC combined with the C0-00AC 24VDC .5A power supply

Click to expand...

Khoa, I would use instead the Click C0-00DR-D PLC with 6 relay outputs. The reason is that the sinking transistor outputs that your instructor picked are very easy for students to burn out. Once a transistor output is shorted or overloaded (common events on student work), that output is useless. You only have 6 Click outputs to pick from, and you will need at least two to operate each Push-Pull solenoid.

In regards to the solenoids I was thinking of 2 Ledex Low Profile Push/Pulls, maybe part# 173921-0XX. By the way, does Push/Pull mean they can do both?

Click to expand...

Yes, I think the Ledex part number 173921-025 would be suitable for your project. From the Ledex® Low Profile Size 6EC — Push or Pull data sheet, it has 25 AWG wire, resistance of 12.9 ohms, will work on supply voltage of 20 to 28 volts.

http://www.ledex.com/solenoid-part-numbers/part-numbers-imperial-1.html

Yes, that one will push and pull. I assume that you have to switch the direction of the power to make it reverse. Therefore you will need two DPDT (double-pole, double-throw) relays with 10 Amp contacts. You should be able to buy two cheap relays with mounting bases for about $15 each.

I read on another thread here that we can't connect the solenoid directly to the PLC. If so, then how do we go about connecting it so it can be controlled by the PLC?

Click to expand...

Thanks for searching before asking! There are two reasons that you need the extra relays. (1) Your PLC relay outputs are only rated for 1.0 Amps, and you need to operate these Ledex solenoids that may pull 3 or 4 Amps inrush current, and (2) you need two sets of contacts to connect each solenoid for Forward and another two sets of contacts to connect each solenoid for Reverse. Two DPDT 10 Amp relays will provide that for both of your spool-valve operating solenoids.

A possible alternative would be to parallel two relay outputs on your Click (another reason to go with the CO-OOR-D version) for each solenoid direction, so that the current capacity of each Click output does not exceed 1 Amp. That means that you would need two forward and two reverse outputs for each solenoid, or a total of 8 outputs, but you only have 6. So forget that and go with the two additional relays.

Also, is the PLC's power supply sufficient to power everything or would the solenoids require a separate power source?

Click to expand...

Answer that yourself. "C0-00AC 24VDC .5A power supply" powering the Click PLC AND also two Ledex solenoids that each require 24/12.9 (1.86) Amps operating current? Add it up and you will be somewhat short. You need an additional 24 VDC power supply, or one large single power supply. Instead of buying the C0-00AC 0.5 Amp power supply for $29, consider the Automation Direct PSB24-060-P 60 watt, 2.5 Amps supply for $27.75. It should have enough power to run your Click PLC and also operate one solenoid valve at a time. If you need to operate both at the exact same time, then you will need about 4 amps or 96 watts. Note that with a PLC, you can set up a timer to operate your solenoids sequentially one after the other, so that you do not overload your power supply. You probably only need to pulse the Ledex for about 0.5 seconds to operate it, so if you do that sequentially for both valves, for all practical purposes they will function as if they operate at the same time.

It is all about the engineering to make things work with the least problems and at the least cost.

use a relais output PLC it can direct drive the valvesolenoids. in any voltage.
You can use the smallest PLC you can find, you can even use one small control relais but he if you can afford a PLC please do.

Lancie1 said:

Yes, that one will push and pull. I assume that you have to switch the direction of the power to make it reverse.

Click to expand...

Lancie,

The push-pulls I have worked with were not polarity dependent. The push or pull feature is dependent on how the mechanical connection was made.

khoa152 said:

They would only have to move perhaps .2 inches max.

Click to expand...

khoa,

Be sure to check the force charts, because the force is not linear over the length of the stroke. Notice the 6EC is listed at 49 Lbs holding force but the available force at .2" is around 5Lbs.

khoa152 said:

By the way, does Push/Pull mean they can do both? If not, then it'll mean I'd need 4 rather than 2 so the spool valves can be moved in both directions correct?

Click to expand...

Yes you will need 4 solenoids.

Do the spool valves already exist?
If yes, how much solenoid power (force) is required to operate the spool valves?

If no, you may want to purchase single solenoid two position valves suitable for your liquid (including pressure) and any existing tank piping connections and/or if more power is needed, check into motorized valve actuators as opposed to solenoids.

Also, be sure to include vessel safety (pop-off) protection if the supply valve pressure warrants.

Paul

Do the spool valves already exist?
If yes, how much solenoid power (force) is required to operate the spool valves?

If no, you may want to purchase single solenoid two position valves suitable for your liquid (including pressure) and any existing tank piping connections and/or if more power is needed, check into motorized valve actuators as opposed to solenoids.

Also, be sure to include vessel safety (pop-off) protection if the supply valve pressure warrants.

Paul

The push-pulls I have worked with were not polarity dependent. The push or pull feature is dependent on how the mechanical connection was made.

Click to expand...

That makes sense, but Ledex makes a point of calling these Push OR Pull, where as all others in the series are called either Push-type or Pull-type. So a closer study of the product manual will be needed. (I am available to perform that service for a small honorarium - or Khoa can do that and save the cost!)

Also Ledex has a linear magnetic-latching type that DOES reverse when the voltage is reversed. Using this type might mean only needing 2 instead of 4 solenoids. A cost comparison will be needed.
http://www.ledex.com/linear-solenoids/magnetic-latching-solenoid.html

Don't forget V =-L*di/dt. It WILL kill your relay.

Great little student project that gives you an opportunity to learn many useful lessons. Regarding your question about driving the solenoids, I hope your instruction has taught you about V = -L * di/dt. Your solenoid is an excellent example of the effects caused by collapsing current.

Every inductive device (L) will generate an opposing voltage spike when the inductor (your solenoid) is turned off. Any time you drive a DC solenoid, your driving device will see an appreciable voltage spike when the driving device (your relay) is opened. And it absolutely, positively WILL damage your relay over time since it will arc across the contacts. You will find many PLC engineers that claim you can skip protection of your DC relays on inductive loads but that simply violates physics. With an AC coil, the inductor never reaches a steady state current, hence the di/dt is small: with your DC coil, it (slowly) charges to a contant steady state current and then you turn off the coil, hence the induced voltage can EASILY go into the hundreds of volts.

For an AC solenoid, it will have some specified inrush current that you need to overcome. For your DC coil, you look up the resistance (for the -025 version it is 12.9 ohms @ 24VDC) and calculate you will need 24/12.9 = 1.86 amps steady state DC current per solenoid. That will help you figure your DC power supply load and the minimum ratings of your relay. An inexpensive interposing ice-cube style relay will do the trick (around $10 for a 7 amp relay and base).

But you must also be concerned what happens when you turn off your relay. For that, you can put an inexpensive (30 cents) reverse biased diode across that solenoid output: it will clamp the (large) negative voltage spike and save your relay contacts from failure.

As a student, I hope your instructor encourages you to put an oscilloscope across those leads of your solenoid without the diode so you can see the dramatic effects of L*di/dt and just how effective your diode is at clamping those effects.

Best of luck

nwboson said:

For an AC solenoid, it will have some specified inrush current that you need to overcome. For your DC coil, you look up the resistance (for the -025 version it is 12.9 ohms @ 24VDC) and calculate you will need 24/12.9 = 1.86 amps steady state DC current per solenoid. That will help you figure your DC power supply load and the minimum ratings of your relay. An inexpensive interposing ice-cube style relay will do the trick (around $10 for a 7 amp relay and base).

Click to expand...

This implies that DC solenoids do not require any "starting current". I used to think that until I had to redesign and rewire a large DC Brake solenoid for a 10-ton overhead crane. My brake kept dragging, did not completely release until about 3 seconds. Finally I determined it was because the pull-in current of the solenoid was more than my 100 volt DC power supply could supply in a short time. I added another 120 VAC rectifier and doubled the current - still had the dragging problem, only maybe for 2 seconds instead of 3. I finally had to install a timer and a 110 VDC power supply. The timer kept the 110 volts on for 3 seconds, then switched to a lower 48 volts DC that was enough to hold the solenoid on once the brake was released.

Bottom line: A DC solenoid does have "starting current", commonly called Pull-In current that is higher than the normal operating (hold in) current. The only thing that might save Khoa is that for the size under discussion, the pull-in current will not be very much higher than the operating current. If it were me, I still would want a nice comfortable margin between Calculated Total Current Needed and the Power Supply Rated Output Current.

PS: I doubt that the long-term wear on the relay contacts is going to be a large problem for this particular project. The diodes are normal for a normal project, but probably just another distraction for this low-budget, run-it-one-time-and-throw-away student project.