Instruments turn off when starting engine having added extra battery,

PaulRainbow

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Yes, and Boater Sam's post 13 may well have a general bearing on matters.


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See post #19, Boater Sams post is nonsense, you cannot get a voltage drop from one battery to another without a circuit, having the negatives all in parallel has no bearing on the issue. Almost every boat on the planet has the engine battery and domestic battery/bank negatives in parallel.
 

Refueler

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A poor negative connection somewhere will cause the neg to rise when you are cranking, resulting in an effective voltage drop on the cabin batteries.

A poor connection will just create a high resistance .... neg has no actual figure to rise ....

To have a voltage drop on any bank - requires that bank to be supplying + and - ... not just -
 

Refueler

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Yes, and Boater Sam's post 13 may well have a general bearing on matters.

I don't know the size of the engine or the electronics involved but I sometimes start my 20hp engine on the domestic ,100Ah, deep cycle battery and the Plotter, VHF, AIS and sounder all function without a flicker. So if the above + battery, connections wiring size all come up good, you should be ok?

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Post 13 is nonsense sorry.

Your domestic bank not causing plotter etc to drop out is lucky and particularly as you say its a Deep Cycle Battery - not best to use for engine starting ... says to me - you engine starts very easy !

I know my Perkins 4-107 needs quite a bit more grunt than your 20hp - but if I have plotter on when I press starter - if engine is cold - my plotter will switch off with the voltage drop ... (I have the old 1-0-2-B switch)

My other boat with 20hp Yanmar - I can get a flicker on plotter when I hit 'tit' - but will stay on if I use domestic bank (needs a bridging cable to connect to starter circuit).
 

sailor211

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Thanks for all the input, it has been useful in solififying my thoughts. I believed that the only change I made was the increased battery bank. I have obviously made another change to allow the +ve to be connected. The start battery is good, being new and bigger than proviously. (I has to buy in chandlers in a marina with no other source.) However I wll make a list of possible causes and try to be logical in my testing . If / when I solve the problem i will try to remember to report back.
 

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The old solution was to fit a capacitor of suitable size to supplement and cover the short voltage drop. Basically it charges up while circuit is OK .. when voltage drops - it discharges to prop up the voltage ...

Fitting a regulator as you suggest ? is relying on the supply voltage remaining at 12v - which it is not - its why the gear is dropping out - 12v is not maintained... infact voltage is dropping significantly lower than 12v to cause this drop out.
 

PaulRainbow

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Thanks for all the input, it has been useful in solififying my thoughts. I believed that the only change I made was the increased battery bank. I have obviously made another change to allow the +ve to be connected. The start battery is good, being new and bigger than proviously. (I has to buy in chandlers in a marina with no other source.) However I wll make a list of possible causes and try to be logical in my testing . If / when I solve the problem i will try to remember to report back.
First step, disconnect the VSR negative.

If that doesn't fix it, you'll need to look into how the engine battery and domestic batteries are sharing a positive connection.

Please don't waste any time checking negatives for bad connections, totally irrelevant in your case.
 

PaulRainbow

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Sorry, pasted the wrong link!

Ment to link to one of these:

8-40V to 12V 10A DC Voltage Stabilizer Car Power Supply Regulator Waterproof | eBay

There regulators will take an input down to 8v and kick out 12v.

With my old (pre split charge relay) system I had exactly the same issue with just 1 instrument. One of these solved the issue...
If the OP had a system where the engine and domestics shared a battery that could work Andrew. But, he has separate engine and domestic systems (or he should have), so the problem should not exist.
 

Refueler

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Sorry, pasted the wrong link!

Ment to link to one of these:

8-40V to 12V 10A DC Voltage Stabilizer Car Power Supply Regulator Waterproof | eBay

There regulators will take an input down to 8v and kick out 12v.

With my old (pre split charge relay) system I had exactly the same issue with just 1 instrument. One of these solved the issue...

Ok .. thats' better than the previous link ... but still a capacitor will do the job ....

Each to their own !

But at end of day - OP has to find out why he has the problem ... not put a sticky plaster on it ...
 
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bedouin

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To get a voltage drop in the domestic systems by starting the engine, you need a circuit between the two systems. You cannot pull the voltage of the domestic systems down by just having a connection between the negatives.
Read what I wrote again - it is simple physics. There is voltage drop in the shared element of the negative circuit. Whenever current flows along any wire there is a voltage drop and all circuits including that section of wire will experience the same voltage drop. It is quite feasible for the inrush current of a starter motor to cause a momentary drop of 1 or 2 V in the negative circuit and the plotter would see the same drop and could trip itself off

Can be corrected by providing a negative supply to the domestic side that is NOT shared by the starting circuit.
 

PaulRainbow

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Read what I wrote again - it is simple physics.
No need to read it again, it's not physics, it's wrong.

There is voltage drop in the shared element of the negative circuit. Whenever current flows along any wire
Current has to have a circuit to flow, only having the negative connected isn't a circuit.
there is a voltage drop and all circuits including that section of wire will experience the same voltage drop. It is quite feasible for the inrush current of a starter motor to cause a momentary drop of 1 or 2 V in the negative circuit and the plotter would see the same drop and could trip itself off

Can be corrected by providing a negative supply to the domestic side that is NOT shared by the starting circuit.
As i have said before, almost every boat on the planet has the negatives from the domestic batteries and the engine battery in parallel. Providing that the positives are kept separate there cannot be a voltage drop on the domestic systems as a result of the engine being started.

This is my day job.
 
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Buck Turgidson

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Read what I wrote again - it is simple physics. There is voltage drop in the shared element of the negative circuit. Whenever current flows along any wire there is a voltage drop and all circuits including that section of wire will experience the same voltage drop. It is quite feasible for the inrush current of a starter motor to cause a momentary drop of 1 or 2 V in the negative circuit and the plotter would see the same drop and could trip itself off

Can be corrected by providing a negative supply to the domestic side that is NOT shared by the starting circuit.
this makes no sense at all. Are you saying the negative side of the start circuit will go above zero and this will drive the house negative above zero thus lowering the domestic circuit potential? how else do you lower the voltage in the domestic circuit?
 

bedouin

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this makes no sense at all. Are you saying the negative side of the start circuit will go above zero and this will drive the house negative above zero thus lowering the domestic circuit potential? how else do you lower the voltage in the domestic circuit?

If the domestic negative shares wiring with the starting circuit....

The inrush current of the starter will cause a voltage drop in all the wiring - any circuit that shares that wiring will experience the same voltage drop. So that means that the point at which the domestic circuit takes it's negative from the common supply is at a higher voltage than the battery negative. That can easily be 1 or 2 volts - the positive remains at 12V so the instrument actually only sees a supply of 10 or 11V. A bad connection in the negative would just make things worse.
 

Andrew_Trayfoot

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If the domestic negative shares wiring with the starting circuit....

The inrush current of the starter will cause a voltage drop in all the wiring - any circuit that shares that wiring will experience the same voltage drop. So that means that the point at which the domestic circuit takes it's negative from the common supply is at a higher voltage than the battery negative. That can easily be 1 or 2 volts - the positive remains at 12V so the instrument actually only sees a supply of 10 or 11V. A bad connection in the negative would just make things worse.
I just can't see how that can happen...

I suppose it is possible with a shared -ve that if more overall current is being drawn though it by the starter and domestic load than it was rated for so it got hot and its resistance went up, but I can't see this happening in the time it takes to start a small diesel.

As others have said, sounds more like an issue with the VSR...
 

bedouin

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I just can't see how that can happen...

I suppose it is possible with a shared -ve that if more overall current is being drawn though it by the starter and domestic load than it was rated for so it got hot and its resistance went up, but I can't see this happening in the time it takes to start a small diesel.

As others have said, sounds more like an issue with the VSR...
Doesn't need to warm up. Any current in any wire will cause a voltage drop (it is called Ohm's Law). In rush current in a starter motor is said to be about 8-10x the normal current so can easily be above 500A for a couple of seconds. That will cause a voltage drop of more than 1V for every metre of cable (with 16mm cable that is undersized).

It is always amusing people telling me it can't happen when I have seen it for myself :) In my case my GPS has reported a voltage of 9.6V when the engine is starting.
 

Buck Turgidson

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If the domestic negative shares wiring with the starting circuit....

The inrush current of the starter will cause a voltage drop in all the wiring - any circuit that shares that wiring will experience the same voltage drop. So that means that the point at which the domestic circuit takes it's negative from the common supply is at a higher voltage than the battery negative. That can easily be 1 or 2 volts - the positive remains at 12V so the instrument actually only sees a supply of 10 or 11V. A bad connection in the negative would just make things worse.
where is this "inrush" current that is increasing the voltage on the negative side of the starter motor(the thing doing work) coming from? I'm afraid you need to draw a circuit diagram and show me where the positive voltage that is lifting the battery neg is being generated.
 
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