After all the discussion, I believe that it may be as simple as this...

Instead of tying grounds together, there is no reason that we cannot tie the +Vs together.
The microprocessors only see one supply, as the grounds are separated.
At worst, the batteries are paralled in the event there is a sneak path through the transistors,
but there's no harm in that.

 

Because there is only one burn wire, there is only one return for the multiple sources of current. This is good; it makes things much more simple, and reduces or eliminates the need for optocouplers or some other method of isolated control.

If you replace Q1 (the 2N4401; always use reference designators) with a medium-power n-channel, logic-level MOSFET then you can connect the three circuits in direct parallel. No OR-ing diodes. The FETs that are on will sink current, the ones that are off will not, and that's about all there is to it. Open-collector or open-drain drivers are nice that way. With MOSFETs, you can eliminate R1 (the 620 ohm resistor).

Separate from that, you *will* need isolation diodes between the batteries and the burn wire to prevent one battery trying to back-charge another. If efficiency is important, these can be replaced with p-channel MOSFETs, but that is an added layer of complexity.

ak

dj beat me with his drawing. You can replace the three transistors and resistors with three MOSFETs, and add diodes between the battery + terminals and a common power buss to all uCs and the burn wire. This gives you redundant power to the system.

 

So are you now saying that the "burn wire" (does not burn or melt) but is eroded away by electrolysis ? My original image was of some resistance wire wrapped around the anchor rope that burned through the rope. My new image is now a metal wire that forms part of the anchor cable. This metal is not normally eroded by sea water but it eroded when the wire is the anode of an electrolytic cell. Is this assumption correct ?

Les.

Yes, that's correct! I apologize for the lack of clarity, I'm only just understanding how this functions. The term "burn wire" is being thrown around by some other folks I've been talking to so that's what I've been calling it.

 

Just seeing these more recent posts now - very exciting, thanks everyone! Makes a heck of a lot of sense. I'll try this out on Tuesday and report back (long weekend up here in Canada).

 

So my final question (for now) is about the diode placement. I'm pretty sure I've done this correctly, keeping the + terminals separated, but still wired to a common bus. I've also replaced the transistors w/Mosfets, and added the correct voltage to the power supplies. (Apologies if it got a little pixelated in the process).

 

Not there yet. If you look at your sch, you will see that the common power bus (the three diode cathodes tied together) goes nowhere. Try again. Working left to right:

Batteries

Three battery anodes (negative terminals) form the common ground point.

Three battery cathodes (positive terminals) go to diode anodes

Common power bus (three diode cathodes)

Connection to burn wire and all three uCs

ak

Note: If you think of electronic components as having a "positive end" and "negative end", batteries are backwards. When supplying power to a circuit, the positive terminal is the cathode, whereas the "plus side" of a diode is the anode.

 

Okay, second attempt... though I read the previous post many times, somehow I feel like this is still wrong. Basically what user:djfantasi drew on Friday except with the diodes properly oriented. Please forgive me if I'm not getting this...

 

Not there yet. If you look at your sch, you will see that the common power bus (the three diode cathodes tied together) goes nowhere. Try again. Working left to right:

Batteries

Three battery anodes (negative terminals) form the common ground point.

Three battery cathodes (positive terminals) go to diode anodes

Common power bus (three diode cathodes)

Connection to burn wire and all three uCs

ak

Note: If you think of electronic components as having a "positive end" and "negative end", batteries are backwards. When supplying power to a circuit, the positive terminal is the cathode, whereas the "plus side" of a diode is the anode.

The diodes were drawn backwards. Looking at the rest of the circuit, the three MOSFETs are grounding the load. So the batteries polarity are correct. With the diode cathodes pointing to the battery supplies, they are protected from reverse voltage and the load polarity is correct.

 

The diodes were drawn backwards. Looking at the rest of the circuit, the three MOSFETs are grounding the load. So the batteries polarity are correct. With the diode cathodes pointing to the battery supplies, they are protected from reverse voltage and the load polarity is correct.

Forgive me but I'm a little confused now. Yr just saying flip the diodes on my last drawing and it should be theoretically correct?

 

Schematic #27:

1. Reverse all three diodes: Anodes to the batteries, cathodes to the common power bus.

2. Connect the three battery negatives (and uC grounds and FET sources) to form a common ground. The common power bus works for redundancy only if there is a common return.

ak

 

Schematic #27:

1. Reverse all three diodes: Anodes to the batteries, cathodes to the common power bus.

2. Connect the three battery negatives (and uC grounds and FET sources) to form a common ground. The common power bus works for redundancy only if there is a common return.

ak

Like so, eh?

 

Looks good to me.

ak

 

Update: this was working fine for a while (added "thanks" to everyone who helped design it), until, from what I can tell, some of the MOSFETs began to fail. I've come to this conclusion based on the fact that the load is always on (just driving an LED at the moment) regardless of the state of the MOSFET gates, and that two of the three MOSFETs failed a basic DMM test.

Can anyone suggest what may have caused my MOSFETs to short out? Is there some protective circuitry I could add to prevent this in the future?

Just to clarify, my circuit is looking like this at the moment (and again, for simplicity's sake, consider all the sea water/anchor/burn wire part as one load resistor).

 

A couple of questions, have I might!*

1. Are the MOSFETs pins protected from seawater?
2. Since the load is always on, I suggest adding a 10K resistor between the MOSFET gate pins and ground (1 per MOSFET)

dj

* for Star Wars fans.

 

3 uC, 1 load ?!

 

3 uC, 1 load ?!

Yep! Have you read the entire thread? It makes sense after you do.

 

The MOSFETs aren't exposed to the seawater in any way. They're sealed up behind very thick glass, which the wire runs through via a special connector. Is that what you mean by "protected"? There's no components between them and the load, however.

Yes, I sensed something like a 10K resistor was in order. That will likely solve the primary issue of things being on all the time.

But as far as further protection, I've read some things about incorporating zener diodes into MOSFET circuits, though I don't feel confident that it's applicable or necessary for this circuit.

3 uC, 1 load ?!

Ya I know it's weird. It's an intentionally redundant, mission critical sort of circuit.

 

Since the load is always on, I suggest adding a 10K resistor between the MOSFET gate pins and ground (1 per MOSFET).

Since each uC is redundantly powered, each output pin is continuously in a digital I/O mode, a pretty low output impedance in either logic state. I doubt that an external resistor in parallel will improve things.

Decoupling caps?

ak

 

Decoupling caps?

I actually have some decoupling caps on there already from my voltage regulators, which aren't pictured in my circuit (must get a decent cad - if anyone knows of some decent open source circuit diagram software, please let me know!). They're two 0.22 uF caps across the power supply bus. Would love to hear any further suggestions.

As for the gate-ground resistors, they've solved the issue of the arbitrary load activation - primarily because of my own ignorance (is it ever anything other than that?). One of my uC's was simply disconnected, leaving its respective MOSFET gate floating and consequentially active. Lesson learned!

 

One more question for the community:

Many posts ago it was mentioned that there could be a problem if more than one of my uC's went high at the same time. As far as I can tell, this was a potential issue only before the circuit was designed for power redundancy with each of the individual supplies diode protected.

So as it stands now, with each power supply protected by diodes, having more than one uC go high at once shouldn't be a problem, correct? I've also considered the maximum current for the MOSFETS, which is 17A, much higher than anything I could imagine my circuit producing.