Ready Steady Eddy

[cncJim]

I'm thinking this is going to be my Charge Pump relay circuit. My BOB ENA signal is active low so the relay will be in the 'healthy' state when de-energised, this is not fail safe and I could have inverted the signal but it's going to do. The BOB is supplied by 24vdc and so GND is common to the 5v and the 24V.

Attachment 10822

Irving, you are right, I just threw the drawing together on Paint and I have indeed got e and c crossed over. What would your plan be to make it failsafe ?

Maybe this ? (sorry for poor drawing, no tools at hand on this pc)

Attachment 10832

that'll do it. i'd put a series resistor between BOB and base of transistor, value depends on pull-up on BOB. If there is no pull-up (i.e. its an o/c output) use 10k to 5v rail.

R needs to be something like 2k7 0.5W rating if connected to 24v, or 470R 0.125W if connected to 5v rail.

BD679 is fine as output transistor if a bit of an overkill. Any small signal NPN with a gain >50, Vce >40v and Ic >200mA will do for the input transistor e.g 2N3904, BC337 or similar

R needse 2k7 0.5W rating if connected to 24v, or 470R 0.125W if connected to 5v rail.

BD679 is fine as output transistor if a bit of an overkill. Any small signal NPN with a gain >50, Vce >40v and Ic >200mA will do for the input transistor e.g 2N3904, BC337 or similar

Please forgive my ignorance but could someone explain how this will work in basic terms? Why is Eddy's first circuit not fail safe and how does the next circuit achieve that?

Sorry, I've not had a lot of sleep over the past few nights (babies all have colds....) and I think my brain may be turning to mush.

[Jonathan]

Please forgive my ignorance but could someone explain how this will work in basic terms? Why is Eddy's first circuit not fail safe and how does the next circuit achieve that?

The transistor is used as a switch. When positive current is applied to the base of an NPN it switches on. In the original circuit, the transistor is simply connected in series with the relay, so when the current is applied the transistor switches on and so does the relay. The problem is, if the signal to the base is broken (e.g. wire accidently cut), the transistor and thus the relay will switch on. We want the relay to be off in this situation, so another transistor is used to invert the signal. When the current is applied to 'new' transistor, it switches on so connects the base of the second transistor to ground, which in turn switches it off - hence the signal is inverted.

[cncJim]

The transistor is used as a switch. When positive current is applied to the base of an NPN it switches on. In the original circuit, the transistor is simply connected in series with the relay, so when the current is applied the transistor switches on and so does the relay. The problem is, if the signal to the base is broken (e.g. wire accidently cut), the transistor and thus the relay will switch on. We want the relay to be off in this situation, so another transistor is used to invert the signal. When the current is applied to 'new' transistor, it switches on so connects the base of the second transistor to ground, which in turn switches it off - hence the signal is inverted.

Thanks for taking the time to reply Jonathan. Very helpful. I was also having hard time understanding/visualising the "active low" part of the puzzle.

This also helped me understand:-

"If the input signal is high there will flow current through R2 and the transistor's base-emitter junction (base, not gate). This current will be amplified, and the collector current through R1 will cause a voltage drop so that the output will be low. Input high, output low.
If the input signal is low there won't be any base current, and no collector current. No current through R1 means no voltage drop, so that the output will be at +V. Input low, output high."

[EddyCurrent]

Good stuff EddyCurrent, I'm still working my way through your diagrams trying to fully understand :) - Do you have any provision for activating your estop circuit from the controlling software? I suppose the argument could be made that its not needed unless the pc a distance from the machine?

Yes that's what the Charge Pump relay is doing. The BOB is turning a 12khz signal from Mach3 into a 0 or 5vdc output, on my particular BOB the output of ENA is 0v when Mach3 is healthy and +5v dc when it stops.

I have just acquired 10 of these from a control panel I stripped down (7x 24v and 3x 230v) which seem to be a similar spec to yours:-
Attachment 10833
Finder TYPE 55.34
The 5A rating, is that for the entire relay or for each contact?

Those look just right, I looked on Farnell and they appear to be 4 pole Change Over which is what you need if using my diagram, the 5A rating is for each contact.

[EddyCurrent]

BD679 is fine as output transistor if a bit of an overkill.

Thank's for info on component sizes.
That is correct, it's just that I have some and they have a high gain because I need to keep the load on ENA as low as possible as it's also supplying 4 x stepper drivers.

[cncJim]

Yes that's what the Charge Pump relay is doing. The BOB is turning a 12khz signal from Mach3 into a 0 or 5vdc output, on my particular BOB the output of ENA is 0v when Mach3 is healthy and +5v dc when it stops.

Of course, what a spanner! sorry that makes perfect sense now.

[EddyCurrent]

If the breakout board output is open collector, why not reduce the component count by using a PNP transistor to invert the signal?

e.g:
Attachment 10834

Edit: Also don't forget the back-emf protection diode!

That would be a sensible solution Jonathan and one that I shall test, thank you

[EddyCurrent]

The breakout board I'm using is this one SYSTEM1
I emailed Roy regarding the charge pump relay and he obviously suggested a transistor buffer but more interestingly the board has 'spare' 12v and 5v outputs and his suggestion was to connect a small 5v reed type relay between +5v and ENA. This has the advantage that when Mach3 is healthy ENA is at 0v therefore a potential of 5v exists between +5v and ENA making it fail safe plus it would share the load better with the 4 connected AM882's. This might be the way to go other than the reed relay does not sound too appealing although in practice one of those encapsulated type that look like a chip with about 8 legs have proved to be reliable in industrial situations.

[EddyCurrent]

There are also 4 contactors which I haven't taken out yet. Are there any practical/safety differences between a relay and a contactor?

With electrical systems there are generally two main circuits, the 'control circuit' and the 'power circuit'. Contactors are regarded as components of the power circuit whereas relays belong to the control circuit, it's mostly to do with the current carrying capacity and the ability to break that circuit without arcing or at least controlling the arcing. There are relays with positive guided contacts and these are for use in safety circuits, I have not come across equivalent contactors but that's not to say they don't exist. I'm not 100% on this but I think relays originated in communications areas such as telephone exchanges. Contactors can have auxilliary contacts fitted which in essence are relay equivalents and are used in the control circuits for those contactors.

[irving2008]

Just to clarify Jonathan's answer, the BOB ENable signal is Active Low i.e. enabled = 0v, disabled = 5v. In Eddy's original circuit the relay would be OFF when the BOB says Enabled and ON when disabled. But a BOB failure, a supply voltage failure, or burn out of the transistor would all turn the relay OFF, an enabled state, so not fail safe. As J says, the addition of the second transistor inverts the logic so relay is ON only when system is enabled and voltages are present.

J's later solution achieves the same result by using a PNP transistor to invert the logic.

[edit] typed this ages ago but forgot to hit submit.... :roll: and now its out of date lol

The transistor is used as a switch. When positive current is applied to the base of an NPN it switches on. In the original circuit, the transistor is simply connected in series with the relay, so when the current is applied the transistor switches on and so does the relay. The problem is, if the signal to the base is broken (e.g. wire accidently cut), the transistor and thus the relay will switch on. We want the relay to be off in this situation, so another transistor is used to invert the signal. When the current is applied to 'new' transistor, it switches on so connects the base of the second transistor to ground, which in turn switches it off - hence the signal is inverted.