Triggering a 555 timer on opening of a line

[zookopf]

Hi guys,

Firstly, I am by no means experienced with electronics, so if I say something stupid, please forgive me.

I am working on an interlock system for a project I'm assigned to as part of an industrial placement, and I am not sure how to overcome a couple of issues.

The way the interlock works is as follows;

In normal operation of the system, two separate sets of 24V lines are shorted to each other, i.e. pins 1&4 and pins 2&3 on the serial interface are connected.

When one of these lines is opened the system shuts down, and in order to get it powered up again the un-opened line needs to be opened and then both lines need to be closed.

One of these lines is operated on a switch, and as such poses no problem, but unfortunately that is not an option for the second line.

So, my goal is to somehow automatically trigger the opening and closing of the second line when the first line has been broken.

I was thinking along the lines of a monostable 555 timer circuit with its output connected to a relay, which would open the second line for a period of time defined by the 555 circuit, then close it again, but my problem is how to go about triggering the 555 timer circuit on the breaking of the first line?

Any help with accomplishing the above, or any suggestions on alternative methods would be greatly appreciated.


Thanks

 

[KrisBlueNZ]

Hi zookopf and welcome to Electronics Point

I'm sure we can help you, but you need to explain your problem more clearly.

If you add the timing components to the 555, your circuit will generate a pulse when the contacts at the input go open. You could use a relay, with its coil connected across one 24V source, to provide the contacts. There are probably simpler ways but without more information I can't really be sure.

The 555's output can drive a relay coil directly, and this could break your second 24VDC line for a short time then re-connect it, if you use the normally closed terminal on a relay.

Relays are a simple and generic way to switch circuits. They are available in single switch (SPST) configurations, changeover (SPDT) configurations where one terminal is switched between two contacts, and other larger configurations such as DPDT (two independent SPDT switch circuits). For an example, see http://www.digikey.com/product-detail/en/G5LE-1-VDDC24/Z3624-ND/1815686. There's a link to the manufacturer's data sheet there. Also do a Google image search on relay; you'll see lots of circuit diagrams that show how they can be used.

Can you describe your application more generally? Also, a diagram is worth a thousand words.

 

[zookopf]

The image attached to my first post is a transistor based triggering system suggested by someone on another forum, but I am not sure how it works and the post has gone dry. If anyone could explain this circuit to me, I would be grateful.

 

[zookopf]

Hi Kris, thanks for the reply. I will do up a simple diagram on what I want to accomplish and get back to you ASAP.

 

[zookopf]

Hi Kris, sorry about the slow reply, got very busy in work.

Here is a diagram which may explain what I am trying to do more clearly.

 

[KrisBlueNZ]

I'm afraid that doesn't really help me understand what you have there.

In normal operation of the system, two separate sets of 24V lines are shorted to each other, i.e. pins 1&4 and pins 2&3 on the serial interface are connected.

So you have two independent 24V DC sources? And they are joined together, positive to positive and negative to negative, on a "serial interface" connector? Does this connector carry power and serial data? What does it connect to? Are these 24V DC sources used for power or as some kind of signal?

When one of these lines is opened the system shuts down, and in order to get it powered up again the un-opened line needs to be opened and then both lines need to be closed.

If both 24V DC sources are connected together, how can you detect when one of them has failed? There will still be 24V DC across both of them (unless one of them fails in such a way that it shorts out the other one, in which case there will be 0V across both of them).

One of these lines is operated on a switch, and as such poses no problem, but unfortunately that is not an option for the second line.

So there is a switch between the first power supply and the serial connector? I really can't picture this. You need to draw a diagram.

So, my goal is to somehow automatically trigger the opening and closing of the second line when the first line has been broken.

OK, that's possible, if you can interrupt the second line and insert a relay into it. You need some way to detect when the first power supply is switched OFF. When this happens, you want the relay to open, so both power supplies are OFF. Then a short time later, you want the relay to close again, so the second power supply is connected again. Then you can turn the switch back ON again. Is that what you mean? It doesn't sound right to me.

I was thinking along the lines of a monostable 555 timer circuit with its output connected to a relay, which would open the second line for a period of time defined by the 555 circuit, then close it again, but my problem is how to go about triggering the 555 timer circuit on the breaking of the first line?

Right, and that's what I'm trying to figure out. What changes, from an electrical point of view, when the first line is broken? Does any voltage go to zero? That would be the easiest thing to detect.

 

[zookopf]

Hi Kris, I apologise for my lack of clarity and thanks for hanging in there.

I will try to answer a few of your questions, but first some background info;

The system being used is an industrial laser. The interlocks act as a safety mechanism so that if, for example, a door is opened in the lab the system shuts down. Normally these types of laser have only one interlock channel but this one has two, which makes it awkward. As said previously, when these two channels are in a contact closed (i.e. pins 1&4 making up channel 1 and pins 2&3 making up channel 2) state the system operates. pins 1 & 4 of channel 1 are connected via a circuit breaker box, which breaks the connection between pins 1 & 4 when the lab door has been opened, thereby shutting off the lasers power supply. The problem is that there are not enough available ports on the circuit breaker box accommodate channel two pins 2 & 3, so I must find a way to open and close the connection between the channel two pins automatically, when the first channel has been opened by the circuit breaker; otherwise the laser's power supply cannot be restarted.

"So you have two independent 24V DC sources?"

Four sources, each of the four pins output 24VDC RSE

"And they are joined together, positive to positive and negative to negative, on a "serial interface" connector?"

pins 1 & 4, and pins 2 & 3 are tied into each other in normal operation

Are these 24V DC sources used for power or as some kind of signal?

All four pins carry no serial data, they remain at 24VDC RSE at all times. Whatever way the internal circuitry of the device is set up, tying 1&4 and 2&3 into each other allows for activation of the power supply

"Does this connector carry power and serial data?"

Other pins on the serial interface carry variable data, but not these four, they just act as an enable for the power supply when tied into each other.

"You need some way to detect when the first power supply is switched OFF. When this happens, you want the relay to open, so both power supplies are OFF. Then a short time later, you want the relay to close again, so the second power supply is connected again. Then you can turn the switch back ON again."

That's pretty much it.

"What changes, from an electrical point of view, when the first line is broken? Does any voltage go to zero? That would be the easiest thing to detect."

Nothing really changes, all four pins still exhibit 24VDC

 

[KrisBlueNZ]

OK, I think I'm getting the hang of it now. Correct me where I go wrong.

The laser uses four pins on its serial connector to detect external events that require the laser to shut down, for safety. For the laser to run, an external circuit must provide continuity between pin 1 and pin 4, and another external circuit must provide continuity between pin 2 and pin 3.

If either of those circuits is broken, the laser turns off, and it can't be turned on until the other circuit has been broken as well, then both circuits have been restored to continuity.

You currently have pins 1 and 4 connected to a contact in the circuit breaker box that goes open circuit when the lab door is opened, so that opening the lab door will shut down the laser. But you don't have another contact in the circuit breaker box to drive the second circuit on pins 2 and 3.

Connecting pins 2 and 3 together permanently doesn't work because when the first circuit goes open and the laser shuts down, the laser expects to see the second circuit go open then closed as well, before it can be turned back on.

So I guess you've tried to find a setting in the laser's configuration to disable one of the safety circuits? And you've looked in the manual to see whether you can just connect the two circuits together (pin 2 to pin 1, and pin 3 to pin 4) so they can use the same safety contact?

You say there's +24V DC on those pins. Is that measured relative to an earth or 0V connection on that connector?

When the lab door is actually open, and continuity is broken between pins 1 and 4, what are the voltages on those pins? Presumably there will still be +24V on one of them, because that's a +24V output from the laser, but the voltage on the other pin, the detect pin, will jump to 0V so the laser can sense that the circuit is broken, right?

If you break the connection between pin 2 and pin 3, one of those pins will also go to 0V for the same reason, right? If so, it should be possible to jumper the two sense pins together, so that when the circuit goes open, both sense pins drop to 0V. Depending on how the laser expects those pins to be used, this might be enough to allow it to be turned back on.

If the laser actually requires a delay between the first and second input changes, then you will need a circuit. But it may not.

 

[zookopf]

"The laser uses four pins on its serial connector to detect external events that require the laser to shut down, for safety. For the laser to run, an external circuit must provide continuity between pin 1 and pin 4, and another external circuit must provide continuity between pin 2 and pin 3.

If either of those circuits is broken, the laser turns off, and it can't be turned on until the other circuit has been broken as well, then both circuits have been restored to continuity.

You currently have pins 1 and 4 connected to a contact in the circuit breaker box that goes open circuit when the lab door is opened, so that opening the lab door will shut down the laser. But you don't have another contact in the circuit breaker box to drive the second circuit on pins 2 and 3.

Connecting pins 2 and 3 together permanently doesn't work because when the first circuit goes open and the laser shuts down, the laser expects to see the second circuit go open then closed as well, before it can be turned back on."

All exactly correct

"So I guess you've tried to find a setting in the laser's configuration to disable one of the safety circuits? And you've looked in the manual to see whether you can just connect the two circuits together (pin 2 to pin 1, and pin 3 to pin 4) so they can use the same safety contact?"

The manual for the laser is poor at best, the company has switched location twice I believe, from Japan to France, then from France to the UK, so many of the manuals have been translated, at least once. The manual seems to suggest that you cannot connect pins 1&2 and pins 3&4, but I have not tried it. Here is all the manual says on the interlock channels (I have confirmed this experimentally);

'To have a possibility of the internal main power supply activation it is necessary that pin1 is connected with
pin4 and pin2 is connected with pin3. Otherwise, the internal main power supply is switched off and the
emission cannot be turned on. Once any of the pairs of the mentioned above contacts has been opened it is
impossible to switch the lasers power supply on till the second pair is opened and then both pairs are
shortened.'

"You say there's +24V DC on those pins. Is that measured relative to an earth or 0V connection on that connector?"

Relative to earth (actually around 22.9V, just quoted in the manual as 24V)

"When the lab door is actually open, and continuity is broken between pins 1 and 4, what are the voltages on those pins? Presumably there will still be +24V on one of them, because that's a +24V output from the laser, but the voltage on the other pin, the detect pin, will jump to 0V so the laser can sense that the circuit is broken, right?"

Yes that is correct, I hadn't thought so but I just went down to the lab to double check and you are correct, one of them is a 0V line. The Voltages on the lines are as follows (relative to earth);

Pin 1 (when channel 1 is open) - '24V'
Pin 4 (when channel 1 is open) - 0V

Pins 1 & 4 (when channel 1 is closed) - '24V'

Pin 2 (when channel's 1 & 2 are open) - 0V
Pin 3 (when channel's 1 & 2 are open) - 0V

Pin 2 (when channel 1 is closed and channel 2 is open) - 0V
Pin 3 (when channel 1 is closed and channel 2 is open) - '24V'

Pins 2 & 3 (when channel's 1 & 2 are closed) - '24V'

"If you break the connection between pin 2 and pin 3, one of those pins will also go to 0V for the same reason, right? If so, it should be possible to jumper the two sense pins together, so that when the circuit goes open, both sense pins drop to 0V. Depending on how the laser expects those pins to be used, this might be enough to allow it to be turned back on.

If the laser actually requires a delay between the first and second input changes, then you will need a circuit. But it may not."

That might just work, I won't be able test it until next week though, as I am required to book lab time.

Thanks again for the feedback.

 

[KrisBlueNZ]

I hadn't thought so but I just went down to the lab to double check and you are correct, one of them is a 0V line.

It's not a "0V line"; it just reads 0V when there is no connection from that pin to the other pin that provides the +24V. It's the input to the laser.

Pin 1 (when channel 1 is open) - '24V'
Pin 4 (when channel 1 is open) - 0V

Right. So the laser provides +24V on pin 1 and monitors the voltage on pin 4. There will be a pulldown resistor or something similar inside the laser that pulls pin 4 down to 0V unless it's actively being driven to +24V by the external loop.

When the external circuit is present, the +24V from pin 1 is fed back into pin 4, where it overrides the pulldown. The laser detects this +24V on pin 4 and knows that the external circuit is closed, so it knows it's safe to run.

When the external circuit is broken, the pulldown resistor inside the laser pulls pin 4 down to 0V. The laser detects this voltage and shuts down.

Pin 2 (when channel's 1 & 2 are open) - 0V
Pin 3 (when channel's 1 & 2 are open) - 0V
Pin 2 (when channel 1 is closed and channel 2 is open) - 0V
Pin 3 (when channel 1 is closed and channel 2 is open) - '24V'

If that's all correct, the situation is a bit more complicated.

1. Are you sure that the external loops are supposed to be connected between pins 1 and 4, and between pins 2 and 3? Wouldn't it make more sense for them to be connected between pins 1 and 2, and between pins 3 and 4?

2. Disconnect everything from the pins and measure the voltages on all four pins.

3. Link pins 1 and 4 and repeat the measurements.

4. Unlink pins 1 and 4, and link pins 2 and 3, and repeat the measurements.

 

[zookopf]

"1. Are you sure that the external loops are supposed to be connected between pins 1 and 4, and between pins 2 and 3? Wouldn't it make more sense for them to be connected between pins 1 and 2, and between pins 3 and 4?"

I agree that 1 to 4 and 2 to 3 seems a little strange but that's what the manual says, and I have confirmed that the system functions correctly in this configuration using two manual switches (not using the circuit breaker at all), one of which was connected between pins 1 & 4, and the other between pins 2 & 3.

"2. Disconnect everything from the pins and measure the voltages on all four pins."

Pin 1 - 24V
Pin 2 - 0V
Pin 3 - 0V
Pin 4 - 0V

"3. Link pins 1 and 4 and repeat the measurements."

Pin 1 - 24V
Pin 2 - 0V
Pin 3 - 24V
Pin 4 - 24V

"4. Unlink pins 1 and 4, and link pins 2 and 3, and repeat the measurements."

Pin 1 - 24V
Pin 2 - 0V
Pin 3 - 0V
Pin 4 - 0V

 

[KrisBlueNZ]

It looks like pin 1 is the +24V output to the external loop, pin 2 is the detection input, and pins 3 and 4 are just connected together internally. But it can't be that simple. That wouldn't allow the laser to detect the two circuits independently and require the second one to be broken independently of the first one.

If I understand correctly, the steps are:
1. The lab door is opened. The first circuit (pins 1 and 4) goes open. The laser turns itself off.
2. The second circuit (pins 2 and 3) goes open.
3. The second circuit (pins 2 and 3) goes closed.
4. The first circuit (pins 1 and 4) goes closed. The laser can now be turned back on.

Do you know whether the laser requires any delays between the different steps of the interlock process? In other words, would it work if both circuits went open at the same time, and both circuits went closed at the same time? You could test this using a double-pole switch to switch both circuits simultaneously.

If no delays are required, a double pole relay would be enough. If delays are required, I can design a circuit. It would help to know whether steps 3 and 4 above can be done in the opposite order. In other words, would it be OK for the second circuit to just follow the first circuit with a delay? This would simplify the circuit. It would give this sequence:
1. The lab door is opened. Circuit 1-4 goes open. Laser turns itself off.
2. After a short delay, circuit 2-3 goes open.
3. The lab door is closed. Circuit 1-4 goes closed.
4. After a short delay, circuit 2-3 goes closed. The laser can now be turned back on...

Can you test that sequence using two switches to see whether the laser will accept it? And if so, what is the shortest delay between steps 1 and 2, and between steps 3 and 4, that it will accept?

 

[zookopf]

Sorry about the delayed reply, I have limited internet access outside work hours.

The steps you stated above are correct.

I will go into work early tomorrow and check out whether time delays are nessesary, and if so I will determine the shortest possible period, as well as the reversibility of steps 3 and 4.

Once again, thanks for the assistance.

 

[zookopf]

Hi, just checked the system and it does not appear to require a delay.

That being said, I did not have a double pole switch available so I had to use two single pole switches actuated simultaneously. Because of this, there was probably a very small delay inherent in the switching, but I did repeat the process >20 times, with no problems, so I'm fairly sure no delay is required.

 

[zookopf]

Do you think something like this work would work if the relay was very low power?

 

[KrisBlueNZ]

Do you think something like this work would work if the relay was very low power?

Like what?

Here's my suggestion.



The connector on the laser is shown on the right. The circuit uses the +24V rail from pin 1 (but it draws less than 10 mA from it, so it's very unlikely to cause any problem). It also needs a connection to the 0V rail, which I assume is available on another pin on that connector.

K1 is a DPDT (double pole, double throw) relay. It has two sets of contacts which change over together in response to 24V DC being applied across the coil. The contacts are shown in the default (not energised) position. I've listed three suitable parts, all available for a few dollars from Digi-Key. The numbers on the left are the manufacturers' part numbers; numbers on the right are Digi-Key catalogue numbers.

While the circuit (connected on the left) has continuity, i.e. the lab door is closed, current flows from the +24V supply on pin 1 of the laser's connector, through the door switch, through the relay coil and back to 0V on the laser. This energises the coil and pulls the contacts to the opposite position from that shown.

This completes the circuit from pin 1 to pin 4, and the circuit from pin 2 to pin 3.

If the lab door is opened, the normally closed circuit is broken. The relay de-energises. Diode D1 is needed to absorb a voltage pulse called "back EMF" that is generated by the relay coil as its magnetic field collapses, due to its inductance. The paths from pin 1 to pin 4 and from pin 2 to pin 3 are broken. When the door is closed again, those paths are completed again.

Those relays come in a small through-hole package that needs to be mounted onto a circuit board of some kind. I suggest using a small piece of stripboard. You can mount the diode on the stripboard as well, and you could use sockets and plugs for the connections to the off-board parts (the normally closed loop, two connections, and the laser, five connections).

 

[zookopf]

Perfect, just what what I needed, and pretty simple to boot.

Many thanks Kris, you have been a tremendous help.

 

[KrisBlueNZ]

You're welcome. I hope it works for you!