Digital Output parallel with power supply

[Karipain99]

Hello!
I would like to ask a question about connecting a digital output parallel to the positive of the power supply!
My application consist in keep the microcontroller turned on untill a certain message via CAN BUS is received even after turning off the switch.
so the idea was to put a digital output (with a diode in series) to high and connect it in parallel to the positive which energize the relay, thus when turning off the switch the digital output keeps the relay energized untill the specific data is received and then put the output pin to ground and auto disconnecting.

In attached a picture that maybe could help understand better what I've said!
my question :
Is it raccomanded to do like this?
P.Sigital output and battery have the same votlage 24V and the ground is common.
Thanks in advance!

 

[betwixt]

You are not as confused as you think I am.:-?

Please explain what this output signal from the ECM does. You show it with an arrow pointing in toward the ECM but describe it as an output. Are you trying to sense whether the switch is closed or use it to energize the relay? In other words, it is an input that detects if the power is supposed to be turned on and if that is correct, what powers the ECM to be able to use the signal?

From your diagram it looks like the ECM powers the battery, it is normally the other way around.

Brian.

 

[Karipain99]

Hello Brian!
I will explain better what i wanted to do in this application by steps!
Turning on the switch (latch switch) - energize the coil - power up the ECM - after a certain CAN msg - put the output status to high (this output is connected just to the coil, thus we will have two source voltage which energize the coil of the relay, one is from the battery and the second is from the output voltage).
Now when turning off the latch switch, the coil will remain energized from the ouput voltage of the ECM and the ECM will not turn off, Is this Right?
Finally when The ECM receive another msg via CAN it puts the output voltage to false and so cut the power supply from the ECM!
Hope this could help better!
Thank you again.

 

[betwixt]

OK, so what you are asking for is a latched system, one that has to be started manually but afterwards keeps it's power switch on by itself.

Firstly - you must have another connection to maintain an uninterrupted supply to the ECM or it will disconnect it's own power source when the relay contacts open. That will obviously stop all outputs and at best make it cyclically reset itself.

I would suggest the way to make the relay power up correctly would be to use three diodes, each rated to carry at least the relay coil current + a 50% safety margin. As most relays only consume low power you can probably use 1N4001 or similar which are very inexpensive.

Connect one diode across the relay coil (ideally at the relay itself), with it's cathode toward the switched end. It will not normally conduct but at the moment the relay de-energizes it will trap any nasty voltages that could damage the switch or ECM.

Diode 2 goes in series with the switch, before it joins the junction of 'digital output' and the relay. This is to allow current flow through the switch when it closes so the relay operates but it blocks current from the 'digital output' going back through the switch to the battery.

Diode 3 goes in series with the 'digital output' with it's cathode toward the relay coil. This one is to prevent battery voltage feeding back in to the 'digital output' when the switch is closed.

If wired crrectly, the relay will energize with current through one or both diodes but the two sources cannot feed into each other.

If the 'power supply input' is a sense signal you may have to scale it to a lower voltage with a potential divider made from two resistors. The 'digital output' must also be capable of providing enough current to operate the relay. Without information on the internal circuit of the ECM I can't say if these points are valid.

Brian.

 

[Karipain99]

Hi Brian!
When you say: Diode 2 goes in series with the switch, before it joins the junction of 'digital output' and the relay. This is to allow current flow through the switch when it closes so the relay operates but it blocks current from the 'digital output' going back through the switch to the battery.
Why should the current goes from the output pin to the battery if they have the same voltage?
Thanks again!

Karim

 

[betwixt]

No current will flow if they ARE the same voltage can you guarantee that? A battery is an excellent sink of current, if the digital output carries a higher voltage it will be dragged down to battery voltage, possibly many Amps will flow.

Brian.

 

[Karipain99]

Hi Brian!
I think that If the battery is powering the ECM the output voltage can not be higher than the voltage of the battery!
is it right?
Thanks again.

Karim

 

[betwixt]

That is probably true but it is possible the ECM has circuitry that boosts or reduces the voltage, that's why I suggested the cautious route of using isolating diodes. You should also consider what might happen if the engine is cranked and the battery voltage falls under the load, is there anything in the ECM that stores energy and may be forced to release it back to the battery?

Although it can mean anything with logic states, most of us would consider a 'digital output' to be a control signal from a digital device rather than a power source so check it can produce enough current under load to operate the relay coil.

Brian.

 

[Karipain99]

Hi Brian!
you are right.
It would be better to put also this diode.
Thanks again.

Karim