Interval Time Delay Relay

[camservo]

I'm hoping to make something similar to this http://www.ssac.com/animation/interspo.htm . (described in the "Interval On" relay type at this page.. http://www.macromatic.com/support/kb/kb.php?kba=kba116tdr ). Anyone have a link to a schematic like this one? I'm relatively new to electronics, but I get the terms pretty well. Thanks!

 

[KrisBlueNZ]

Yes, probably. You need to give a LOT more information. From the first link I can't even tell if you're switching low or high voltage, AC or DC, and whether you want to use an electromechanical relay or something else. You can probably make a fairly simple circuit using a 555 timer driving a SPCO relay but PLEASE give as much detail as you can about your application.

 

[camservo]

Yes, probably. You need to give a LOT more information. From the first link I can't even tell if you're switching low or high voltage, AC or DC, and whether you want to use an electromechanical relay or something else. You can probably make a fairly simple circuit using a 555 timer driving a SPCO relay but PLEASE give as much detail as you can about your application.

I actually just need to create a dry contact, so I'm really just hoping to make a switch that operates in this fashion. It's connecting to another control circuit that has isolated power for the lift I'm controlling. Is relay the wrong term for what I want to do?

 

[camservo]

I suppose the control system would be pulling a very small amount of current, but it's negligible.

 

[KrisBlueNZ]

You're right, you want a relay. If you use one with SPDT contacts, your output will have a common contact, a normally open contact, and a normally closed contact, like in the diagram you linked to. One of these contacts will close DURING the delay, and one will close AFTER the delay, so you can use whichever you want.
Now you need a circuit to drive the relay coil and provide the time delay. This will need to be powered somehow.
What voltage is being switched to provide the trigger for the time delay? If it's mains voltage, you'll need a power supply for the circuit.
The simplest circuit would be something built around a 555 timer. These can drive a relay coil directly, and they use a resistor and capacitor to control the delay. The resistor can be a potentiometer, so you can adjust the delay, and it's fairly stable and repeatable. The 555 needs a supply voltage round 5~15VDC. I would use 12V and a 12V relay coil.
Can you tell me about the switch that starts the timer process? What voltage is it switching? How long will it be closed for? Anything else?

 

[KrisBlueNZ]

Another thing. What behaviour do you want when the circuit is not powered? Do you want your output contact to be open or closed? Also, do you want it closed during the delay, or closed after the delay? These factors affect which relay contact you use (normally closed or normally open) and whether the relay coil is activated during the delay, or after the delay.

 

[CDRIVE]

How long does the delay have to be? For relatively short delays there are time delay relays available.

 

[camservo]

Thanks for the responses. The entire idea is that i have a relay as part of an automated system that's controlling a lift control circuit. The lift currently exists in two states "Up" or "Down", and I can't modify the original relay or the lift controls, so I want to put something in between. The motors are burning out because they're constantly in one state or the other. My idea is that when the original relay turns to either the "up" or "down" state, I want to detect a signal, close the corresponding switch for about 5-6 seconds while the lift runs, and then ignore the "up" input signal, until it switches states.

Does that make sense? Think of someone at a hospital bed. They want to watch TV, so they lift the bed, but they don't know to stop holding the button once the lift is extended until they want to lower it again. Then they hold the lower button while they sleep until they want to lift it again. I'm using a modified hospital bed left, so it was designed with the idea a human would be controlling it. =).

Thanks for any help you might offer!

 

[KrisBlueNZ]

This is normally done with limit switches, that detect when the moving object has reached its upper and lower limit. In this case, the limit switches would be normally closed, and would open when the moving part reaches the limit position.
You can use latching relays to control the motor. The UP latching relay is closed by an UP request, and opened when the lift reaches the top; the DOWN latching relay is closed by a DOWN request, and opened when the lift reaches the bottom.
In your case, since the control signal is permanently either UP or DOWN, all you need is to disconnect the power to the motor when the appropriate limit switch opens.
Do you have any kind of diagram of what you currently have there?
Is it possible to just connect the limit switches in series with the UP and DOWN controls?
UP control ------------ top limit switch ---------- motor run forwards
DOWN control ------ bottom limit switch ----- motor run backwards

 

[camservo]

I don't have a diagram unfortunately. I had originally thought a limit switch was the way to go, since it would be very easy but I can't figure out a way to physically fit a switch in the lift. I also am reluctant to add something like a switch triggered by current spike only because I'm trying to avoid working with high voltage (since I'm obviously not a qualified electrician ).

 

[camservo]

currently though it does seem POSSIBLE. It's relay --> isolated lift control and then the lift control provides power to the lift. I'm really hoping to work in between the relay and the lift control, but if it seems out of reach given my knowledge/ability I'm willing to look into other routes without wasting any of your time. =)

 

[KrisBlueNZ]

OK, you really need to describe the whole setup in a lot more detail.
Otherwise we're just going back and forth suggesting things that aren't going to work.
Can you describe the control signals please. I assume you have an UP signal and a DOWN signal, and these latch on permanently. What form are they? Are they switched mains voltage, dry contacts, or what? Is one always ON when the other is OFF? If so, do you need both of them, or would one control signal be enough?
Can the control signal change state while the motor is running? If so, do you want it to reverse immediately?
Please give a thorough description of what you have, and what you want to do. It may seem like you're wasting our time by making us read all that detail, but really it wastes a lot more time when we make assumptions that turn out to be wrong, and suggest things that are not workable. That will help us (me) to best help you. Thanks.

 

[KrisBlueNZ]

Sorry, I didn't see your most recent post.
So you have two dry contacts, one for UP and one for DOWN, that latch on permanently?
And you need to provide two dry contact outputs to control the motor through the lift interface?
I guess it's allowable for the motor to run on slightly once the lift reaches its limit, as long as the motor doesn't run continuously, right?
I can design you a circuit using a couple of ICs and two relays. It would need a power source of say 12V.
Give me some more details so I know I've got some chance of coming up with something you can use.

 

[CDRIVE]

Odd to see a hospital bed without limit contol. It's not a Hillrom is it?

 

[KrisBlueNZ]

I've designed a simple circuit that might do what you want, if you're still interested.
It accepts two dry contact inputs and provides two dry contact outputs. It needs a supply voltage of 12VDC.
The two relays are SPST (Form A) relays with 12VDC coils.
The run timeouts are set by the two trimpots in conjunction with C1 and C2. If you're working close to either end of the trimpot, you can adjust C1 and C2.
Each part of the circuit ("UP" and "DOWN") is independent and there is no interlock between them; if the UP and DOWN inputs are activated together, the outputs will be activated together too. Here is a circuit description for the "UP" circuit.
While the contact opens, C1 charges quickly via R1 and D1, and C1 is held charged while the contact is open. Q1's emitter is at the positive supply rail so Q1 is not biased, so Q3 does not conduct, so K1 is not activated.
When the contact closes, Q1's emitter is pulled to ground, and its base receives bias from C1 via R3. Q1 is a Darlington transistor so it requires very little base current. It conducts and energises Q3, activating K1 and closing the output contact.
C1 now discharges through R7. When C1's voltage drops below about 1.4V it is no longer enough to keep Q1 forward-biased, so Q1, Q3 and the relay turn off. The output will not activate again until the input contact has gone open, then closes again.