Battery Isolator with Battery Drain

[DeepFrz]

Congratulations to getting it all back together and working. I hope you didn't have to pay your consultant.

[Albro359]

Maybe not directly in topic...but connected (sic)
Seems like none of you have heard of the FET isolators made by both Victron and Vetus...they use high current FETs instead of diodes which limits the voltage drop to less than 0.1 volt.
Even with direct battery sensing you still lose with a diode isolator as energy that should be going to your batteries is burned off a heat. The FETs get only mildly warm in my system witha 110A alternator charging 5 x 105Ah batteries.

[bill good]

I never saw you report on the made /model of the isolator you used. Could you provide the details so the matter can have a correct technical response. I am only assuming that the inbuilt reverse protection of the (assumed) FET? sees the faulty battery battery with lower voltage & attempts to lift the faulty battery to that of the house battery. i.e. the isolator only works with good batteries. You can prove this by disconnecting the starter battery & checking with a volt meter that connection & you will see the voltage from the house battery. Have had this type of result with solid state relays with FETS & DC voltage on both sides. Albro might be in the same boat!!(sic)

Regards Bill

[DeborahLee]

Check your ground. Try seperating the fridge and see what happens then the lights.

[Bensigler]

Thank you Depfrz, but yes, I had to pay the "consultant". That was a bummer b/c he took my money and 3 weekends of my time.

In regards to the make and Model (Bill Good), the Isolator is Newmar Model 2-3-70 rated at 70amps per post. I tested the odolator myself after my consultant said it was brokem and it was working.

Thank you Deboralee, but that woulnt have been my issue as the whole thing was resolved when I "isolated" the isolator from the system. If that was the problem I would have still been affected by that issue after re-installing the isolator. Also, ground was fine.

It was also not due to coroded wiring and/or contacts. I assert this b/c I did not clean anything when I took it off and I even noted that the contacts looked ok when I was removing everything.

Thank you all again. Although I am happy everything is working, I just don't have confidence in the system. With the same seemingly randomness that it began to work with, it can also cease to work.

-Ben

[sailorboy1]

Originally Posted by Bensigler

Thank you all again. Although I am happy everything is working, I just don't have confidence in the system. With the same seemingly randomness that it began to work with, it can also cease to work.

-Ben

Hey if it starts acting up again; take it all apart and put it back together like you did to "fix" it this time. Sometimes you just have to go with PFM when you don't know why, but know what to do about it.

[DeepFrz]

They are reliable systems. Chalk it up to something you missed and leave it alone and you should be fine. Possibly not as "good" as something like an "echo charge" but that is subjective. I have used diode isolators and they work fine. If it breaks down again, get an "echo charge", but I wouldn't worry to much about it if it were my boat.

[Bensigler]

@ Don Lucas -> LOL!!

[Bensigler]

Thank you Deepfrz Amen to that!

[nnyerges]

Ben (Bensigler),

I carefully read all the posts on this topic you have started. Nice debate on how a diode works.. but, I’m left with a doubt about the alternator sense terminal. Noted that most participants emphasis on where to connect the sense, as if there were only one alternator in your entire case.

I read you have a twin engines and a 2ALT-IN x 3BATT-OUT ISOLATOR (terminals A1, A2 and B1, B2 & B3 I suppose), so I assume that each of your engine has its own alternator with its own sense terminal (not only one), so I do not understand why all they where talking about. You have two sense terminal i guess, so I would like to verify with you if the sense of each alternator, are connected to the Isolator output side B1 and B3, or failing that, more effective, on each of the batteries connected to terminals B1 and B3. Is that correct?

Thank you in advance,
Regards
Nicolas

[nigelmercier]

Originally Posted by nnyerges

Nice debate on how a diode works...

Unfortunately there was rather too much debate, as some folks didn't comprehend what I was trying to explain. Against my better judgement, I'm going to try one more time...

Diodes are often thought of as being "one way" only, but once they are conducting, they provide a low resistance in both directions: in other words either terminal can "see" the other terminal via a low resistance. Because of this, what happens at one terminal (more or less) happens at the other.

[Andina Marie]

Originally Posted by nigelmercier

Unfortunately there was rather too much debate, as some folks didn't comprehend what I was trying to explain. Against my better judgement, I'm going to try one more time...

Diodes are often thought of as being "one way" only, but once they are conducting, they provide a low resistance in both directions: in other words either terminal can "see" the other terminal via a low resistance. Because of this, what happens at one terminal (more or less) happens at the other.

Correct so long as they continue to be conducting but if the output voltage gets higher than the input voltage they will no longer be conducting and they will no longer be tied together.
Also due to the internal drop, the output when conducting will be approximately 0.6 volts less.

[cal40john]

Originally Posted by nigelmercier

Unfortunately there was rather too much debate, as some folks didn't comprehend what I was trying to explain. Against my better judgement, I'm going to try one more time...

Diodes are often thought of as being "one way" only, but once they are conducting, they provide a low resistance in both directions: in other words either terminal can "see" the other terminal via a low resistance. Because of this, what happens at one terminal (more or less) happens at the other.

Yes while forward biased the voltage at one terminal is forced to be approx. 0.7 volts higher than the other for a silicon diode, but you have been stating that a higher charged battery will discharge into a lower charged battery via a diode battery isolator.

Since I and everyone else has disagreed with you, maybe if you supplied us with a reference it would help your case.

John

[nigelmercier]

Originally Posted by Andina Marie

Correct so long as they continue to be conducting ...

Of course, as I said above.

Originally Posted by cal40john

you have been stating that a higher charged battery will discharge into a lower charged battery via a diode battery isolator.

No I haven't. What I said was that the diodes caused "some current from the higher voltage battery to flow back round the loop to the other battery: subtracting itself from the incoming current to the higher voltage battery, and adding to the current to the lower."

This negative current is only a vector, it could not of course be measured as travelling in the oposite direction. Perhaps sailing terms would help. There is a tidal flow of 2 knots coming from the direction you are going in. Your speed through the water is 5 knots. The result is 3 knots COG. From inside the boat, you could not measure the tidal flow, it is a vector. But it is still there.

[cal40john]

I thought I had asked if you were only looking at net current flow in a specific loop, but going back through my posts I see that I didn't.

Anyway I feel sorry for your students. You have described this so poorly that everyone thought you were saying that current was going backwards through the diodes, in particular look at your posts #33, and #44. While your explanation can make analysis of the circuit easier when trying to look at something specific, it is actually harder to understand in terms of the entire circuit, at least in my opinion, and you're telling everyone that if they don't look at it your way they're wrong. If you confuse a bunch of people who are obviously in the electronics field, what do you think you're doing to your students?

John

Originally Posted by nigelmercier

Of course, as I said above.



No I haven't. What I said was that the diodes caused "some current from the higher voltage battery to flow back round the loop to the other battery: subtracting itself from the incoming current to the higher voltage battery, and adding to the current to the lower."

This negative current is only a vector, it could not of course be measured as travelling in the oposite direction. Perhaps sailing terms would help. There is a tidal flow of 2 knots coming from the direction you are going in. Your speed through the water is 5 knots. The result is 3 knots COG. From inside the boat, you could not measure the tidal flow, it is a vector. But it is still there.