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Voltage Drop - front to back


nb maggie

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The 12v wiring in NB Maggie is not as heavy duty as it could be, so I have voltage drop problems at the Front of the boat from the rear batteries. This only causes a minor problem with the CD/Radio which is very sensitive. So I plan to fit a smaller battery at the front directly by the radio connection. Would this suffice on its own just connected in parallel with the existing 12 circuit or (a) should I fit a diode so the front batery can be charged but not discharged; (:) will the front battery tricke charge OK. My assumption is that a front remote battery just connected to the 12v circuit will be fine.

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The 12v wiring in NB Maggie is not as heavy duty as it could be, so I have voltage drop problems at the Front of the boat from the rear batteries. This only causes a minor problem with the CD/Radio which is very sensitive. So I plan to fit a smaller battery at the front directly by the radio connection. Would this suffice on its own just connected in parallel with the existing 12 circuit or (a) should I fit a diode so the front batery can be charged but not discharged; ( :) will the front battery tricke charge OK. My assumption is that a front remote battery just connected to the 12v circuit will be fine.

 

 

1. If you fit a diode to isolate the new battery from discharging into anything but the radio then the diode will need to be fitted downstream of the radio.

 

2. Unless the new battery is the same type as the rest of the batteries you may have problems in charging the new battery from the existing system.

 

eg: if your domestic batteries are wet lead-acid (normally charged at 14.8v) and your new battery is a sealed gel battery (as it may be because you said it was small) which should not be charged above 14.4v then you will have an issue with gassing.

 

3. Further, (but this will circumvent point 2) if you do insert a diode, you will lose close to one volt across it and so even if you are charging your main batteries at 14.8v there will only be about 13.8v at the new battery. If you are charging from the alternator, without an alternator controller, it is likely the new battery will only see around 13.3v. This may be insufficient over time to charge the new battery correctly.

 

What type of battery do you intend to buy and what type are your existing batteries?

 

Chris

Edited by chris w
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On the subject of Diodes:- are their sole purpose to allow current to pass one way only through them or do they have other uses?

I know Alternators have them but havnt a clue what they do.

 

 

Have a look at

 

http://www.google.co.uk/search?hl=en&d...on&ct=title

 

DIODE - A two-terminal semiconductor device that will allow current to flow through it in only one direction. With the proper voltage polarity across the device, it will act as a conductor. When the voltage polarity is reversed, the device will act as a nonconductor, allowing no current to flow.

Edited by stuart
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The 12v wiring in NB Maggie is not as heavy duty as it could be, so I have voltage drop problems at the Front of the boat from the rear batteries. This only causes a minor problem with the CD/Radio which is very sensitive. So I plan to fit a smaller battery at the front directly by the radio connection. Would this suffice on its own just connected in parallel with the existing 12 circuit or (a) should I fit a diode so the front batery can be charged but not discharged; (:) will the front battery tricke charge OK. My assumption is that a front remote battery just connected to the 12v circuit will be fine.

Go to www.bluesea.com . Look at battery isolators and automatic charging relays. We install 2 or 3 ACR's a month and don't see the 0.7 voltage drop that comes with isolator use.

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They are also used in Galvanic isolators and rectifiers

They are blindingly useful things and crop up all over.

 

I would suggest that the correct, and probably easiest way to fix the volt drop problem would be to double up the cable. By the time you sort out a BSC friendly battery carrier to hold a battery that won't charge 'cos of your volt drop, not to mention more volt drop from a diode, you may as well bite the bullet and throw a couple of lengths of wire up the boat.

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They are blindingly useful things and crop up all over.

 

I would suggest that the correct, and probably easiest way to fix the volt drop problem would be to double up the cable. By the time you sort out a BSC friendly battery carrier to hold a battery that won't charge 'cos of your volt drop, not to mention more volt drop from a diode, you may as well bite the bullet and throw a couple of lengths of wire up the boat.

Or, :) put the wireless at the back of the boat with long speaker leads.

 

wheresmecoat

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They are blindingly useful things and crop up all over.

 

I would suggest that the correct, and probably easiest way to fix the volt drop problem would be to double up the cable. By the time you sort out a BSC friendly battery carrier to hold a battery that won't charge 'cos of your volt drop, not to mention more volt drop from a diode, you may as well bite the bullet and throw a couple of lengths of wire up the boat.

I tend to agree. My mates first boat had insufficient wiring to the horn, and he tried a similar thing using a motorbike battery and eventually a relay involved. It got rather messy and he ended up drawing a larger wire in. Problem solved with less complication.

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On the subject of Diodes:- are their sole purpose to allow current to pass one way only through them or do they have other uses?

I know Alternators have them but havnt a clue what they do.

 

Diodes have a myriad of uses in electronics other than just being a simple one-way valve (which is still a very imporatnt use of them).

 

Chris

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I would suggest that the correct, and probably easiest way to fix the volt drop problem would be to double up the cable. By the time you sort out a BSC friendly battery carrier to hold a battery that won't charge 'cos of your volt drop, not to mention more volt drop from a diode, you may as well bite the bullet and throw a couple of lengths of wire up the boat.

Tbh, i was going to write the same thing.

- Either that or get a electrnic voltage regulator that can take in your 10/11volts and bump it upto 12v (as well as reducing 12/14v to 12)

 

 

Daniel

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- Either that or get a electrnic voltage regulator that can take in your 10/11volts and bump it upto 12v

 

Daniel

 

I'm not sure that would work. The regulator can increase the voltage but not the power (that would be perpetual motion). So if the correct current is supplied at say 12v on the output, then a higher current will need to be supplied on the input at say 10v to balance the power equation. Additionally there will be some power lost in the regulator itself.

 

Therefore this increased current at the input will cause an increased volts drop which will need an even larger current on the input to supply the current output current at 12v and so on ad nauseum......

 

Chris

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Well, obvously if the cables are masivly undersized as to make the transfer of the required power unfeasable, indeed.

- However if the issue is mearly that voltage drops a little bit (say a volt) and the radio is just very sensotive to undervolting when the batterys s low, then having the regulator raise the voltage that volt or so to 12 would probebrly solve you problems. And the cost of the regulator be of simular magnitude as that of the preposed batterys and diode setup.

 

 

 

Daniel

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Well, obvously if the cables are masivly undersized as to make the transfer of the required power unfeasable, indeed.

- However if the issue is mearly that voltage drops a little bit (say a volt) and the radio is just very sensotive to undervolting when the batterys s low, then having the regulator raise the voltage that volt or so to 12 would probebrly solve you problems. And the cost of the regulator be of simular magnitude as that of the preposed batterys and diode setup.

 

 

 

Daniel

 

Actually if you run the numbers you'll find that's not the case. Say it's a 12W, 12V radio and the input voltage is 10V. Work backwards to calculate the source resistance and therefore the change in current at the input to the regulator and you'll get into the loop to which I alluded.

 

Chris

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