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On/Off switch with a difference

Hi all,

I have a machine that I'd like it to turn the coolant system off when it has finished a job as it sprays crap everywhere on a stopped cutter.

The cutting head always returns to a "home" position at the end of the work so I put a cherry switch there. It sort of works but it does not go cleanly "home"

It hits a limit switch then backs off a little then advances slowly to the home position. During this little dance, it seems to encourage the coolant system to spray crap even further as the coolant turns off and on rapidly.

The homing tales about 3 seconds so I'd like a switch that will delay until it has reached the home position or a few seconds after that then shuts down the coolant.

I would like it to turn the coolant on when it leaves the "home" position to begin a new job. Fortunately that is a straight turn on with no delay required.

Can anyone suggest something that will do this?

Thanks.

Dave

Reply to
Dave, I can't do that
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A retriggerable monostable can do the job. Set the time delay for 3 seconds. As long as the home switch is held active, the time out continues. As the home switch is released, your delay interval begins, the end of which, the coolant will start. I'm assuming your cutting head doesn't instantly engage as home is released.

If the cutter needs coolant instantly, you'll need another switch to force a reset to the monostable.

Reply to
Lord Garth

Does your CNC system support auxiliary outputs? I've never, ever programmed for CNC so I could be blowing smoke, but I understand that the language supports coolant on-off commands.

Perhaps a quick go-through of the manual is in order, to make sure you aren't reinventing the wheel.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply to
Tim Wescott

How about a micro switch with a long flexible arm.??

Switch xxxxxxx \\ \\ arm\\

This would depress switch to shut off but allow arm pressure to release somewhat with out turning on.

Reply to
WW

Get your self a Time Delay relay for the coolant. Time Delay on..

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I suppose you could make one but it would cost you more than that relay.. Using a 555 timer would do it.

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Reply to
Jamie

Or your flexible arm, but with a magnet on it (assuming it works against something ferrous). Then the "make" will happen when the thing is quite close to home, but it'll have to pull well away from home before it "breaks".

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply to
Tim Wescott

You can use this general circuit:

+12 ---+-------------+ | | Ry1 N/O +---- N/O [Ry1] |
Reply to
ehsjr

Hi Jamie and others,

Thank you all for the replies.

Jamie: That is exactly what I am looking for and under 20-bucks delivered it will do fine as I can get it to control the existing coolant pump relay.

WW: I had tried the long arm but it worked only some of the time as the first home bounce is quite fast.

Tim: I didn't think of the magnet idea but that is quite clever.

Tim: It is an older machine and does not have M-codes for coolant, but thanks for thinking of that.

Ed: That looks exactly what I need. Thanks for going to the trouble to lay that out. I do appreciate you taking the time to do the design and ASCII art though.

I will go with the time delay relay as it is a plug 'n' play thing. You guys are gooood!

Dave

Reply to
Dave, I can't do that

You might need a resistor in series with the base as indicated by the "R" I added above. But even better would be to use an n-channel MOSFET. The problem with a BJT is that the B-E junction is like a diode, and it does not take much voltage change to go from full on to full off. A MOSFET will turn on fully at the 12 volts, and will stay on until the voltage drops to about 3 volts. Your time constant will be about 2RC, so 10 uF and 100k will give you about 2 seconds max, and adjustable down to about 100 mSec when the pot is at 5k.

Also don't forget to add a commutating diode across the relay coil.

Good luck,

Paul

Reply to
Paul E. Schoen

Good points. Right on the base resistor - it needs to be there. And it has to be a fairly high value. Also, the transistor needs to be high gain - a darlington would be good. I should have said that. So I'll redo as a specific rather than a general schematic, add the missing base resistor, and use a darlington:

+12 ---+---------------------+ | | Ry1 N/O +---- N/O [Ry1] |
Reply to
ehsjr

On Nov 13, 9:02 am,

You guys are good, thanks to both of you.

Paul shouldn't you be hammering nails instead of swanning around on a newsgroup?? LOL

Dave

Reply to
Dave, I can't do that

Hi Paul,

I am vacillating about using the delay timer Jamie suggested as I will still need to add in another relay. On checking, the coolant is controlled by the motor on relay and I want the spindle motor to stay running. I want the coolant off so I can get in and change parts without getting wet.

With 2 seconds it will be marginal and 5 or 10 seconds would be better . Would I just slip a MOSFET in place of the 2N5308 or would there be other changes as well?

I don't care about the coolant running for 10 seconds or so, but I do need to get past the 3-second homing sequence and you are suggesting Ed's design will only give me around 2 seconds max.

Dave

Reply to
Dave, I can't do that

Ed's improved design will probably give about 10 seconds with 100k and 100 uF. It all depends on the RC time constant and the point at which the transistor will turn off. It will also depend on the relay drop-out characteristic. The MOSFET design takes less parts and should be reliable. Best bet is to make a quick prototype and test it.

If the relay coil draws less than 100 mA or so, a 555 timer might be better. It has well-defined turn-on and turn-off points of 1/3 and 2/3 Vcc. It will also give a sharp output transition, which means you need a diode across the coil.

Paul

Reply to
Paul E. Schoen

No - you should get at least 5 seconds with the darlington circuit. (As I recall, Paul computed based on 100K and 10uF, for the mosfet design.) The bipolar circuit uses 100 uF, and the added R reduces the resistance to ~50K for RC computation. So in rough terms, the base of the transistor will see about

4 volts after 5 seconds, and should still be able to conduct plenty of current for the relay.

If you need a longer time, you can change the cap from 100 uF to something bigger. For ~10 seconds, you can use a 220 uF, for 15 seconds, a 330 uF, and so on. The times are approximate, but the pot lets you refine that within a range. The exact time depends a lot on the relay used in addition to the cap and pot values. Also, I used an extremely conservative computation (1 RC instead of 2 RC) to be sure to get at least the minimum time. It is closer to use 2RC, but the darlington can't be counted on to conduct for the full 2 RC, so I chose

1 RC for the minimum computation.

Ed

Reply to
ehsjr

Thanks Ed and Paul,

I will breadboard Ed's layout. The delay is not critical as long as it is more than about 3 seconds. I timed it with a stop watch so my tired old reaction time would have had some effect too.

I will guess a fixed value in place of the the pot as I really don't need that degree of accuracy. It takes me about 20 seconds to remove the splash shield and brush the crap away before I need to get hands in there.

Dave

Reply to
Dave, I can't do that

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