Check your battery terminals!

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Bru

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It's pretty much a routine check now when I get on a customer's boat, even if I'm not there to work on the batteries

About three quarters of the boats I work on have loose battery terminals. So loose that I can remove the terminals without using any tools in many cases

And a fair proportion have screw terminal cable connections that are also loose.

Loose battery connections are bad news. The high resistance causes voltage drop, increased current (and therefore more rapid battery drain) and poor charging

So get down there and check those terminals folks!
 
Bru said:
It's pretty much a routine check now when I get on a customer's boat, even if I'm not there to work on the batteries

About three quarters of the boats I work on have loose battery terminals. So loose that I can remove the terminals without using any tools in many cases

And a fair proportion have screw terminal cable connections that are also loose.

Loose battery connections are bad news. The high resistance causes voltage drop, increased current (and therefore more rapid battery drain) and poor charging

So get down there and check those terminals folks!
Click to expand...

I appreciate that poor connections will introduce resistance into the circuit which will cause voltage drop but I don't understand how that will lead to increased current. Can you please explain as my non electrical engineering logic says increased resistance will mean less current.
 
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VicS said:
I appreciate that poor connections will introduce resistance into the circuit which will cause voltage drop but I don't understand how that will lead to increased current. Can you please explain.
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Are you doubting the word of a marine electrical engineer?:eek:
 
VicS said:
I appreciate that poor connections will introduce resistance into the circuit which will cause voltage drop but I don't understand how that will lead to increased current. Can you please explain.
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I'm intrigued by the loose connections causing rapid battery drain ....... although I guess that might be the same as the increased current? :confused:

In extremis, if the connection is so loose that it is not connected at all, does the battery go flat instantly?

Richard
 
VicS said:
I appreciate that poor connections will introduce resistance into the circuit which will cause voltage drop but I don't understand how that will lead to increased current. Can you please explain as my non electrical engineering logic says increased resistance will mean less current.
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This one had me jump at the keyboard to quote Ohm's law, but then the lil grey cells kicked in... what if...
What if not all, or even the majority, of your loads arent resistive? What if, say you have an inverter that needs to fullfill a power demand, if the input volts drop and it dosent barf at the low voltage, then to get the same power it has to draw more juice. A fridge on low volts may not chill as fast, leading to a higher duty cycle... more Ah if not more A per se. Then we come to electronikery.. if the kit has a bad old shunt psu then no change in the consumption (maybe a little less, but not as for a pure resistor) if, it is up to date (post Ark) then a switch mode PSU will draw more A at less V to make up for it.

So, it may seem weird, but then life is like that innit?
 
Dougal Tolan said:
This one had me jump at the keyboard to quote Ohm's law, but then the lil grey cells kicked in... what if...
What if not all, or even the majority, of your loads arent resistive? What if, say you have an inverter that needs to fullfill a power demand, if the input volts drop and it dosent barf at the low voltage, then to get the same power it has to draw more juice. A fridge on low volts may not chill as fast, leading to a higher duty cycle... more Ah if not more A per se. Then we come to electronikery.. if the kit has a bad old shunt psu then no change in the consumption (maybe a little less, but not as for a pure resistor) if, it is up to date (post Ark) then a switch mode PSU will draw more A at less V to make up for it.

So, it may seem weird, but then life is like that innit?
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Ah, right, so as the connection deteriorates, and the voltage drops further, the connection is magically capable of delivering ever-increasing amounts of current? That's a novel concept.
 
pvb said:
Ah, right, so as the connection deteriorates, and the voltage drops further, the connection is magically capable of delivering ever-increasing amounts of current? That's a novel concept.
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One of the terminals will be better connected than the other and the (increased) current will naturally go out of that one:encouragement:
 
pvb said:
Ah, right, so as the connection deteriorates, and the voltage drops further, the connection is magically capable of delivering ever-increasing amounts of current? That's a novel concept.
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Good old forum, jumping to the end before enjoying the middle bits.
Ferinstance:
Terminal adds 0.01 ohm on a 480W load.
In a good condition the current drawn will be 40A but with an R in the way:
let volts at load be v
let current in circuit be i
let resistance of contact be r
let the battery be 12V
just for the sake of a frictionless billiard ball we will assume superconducting wires (infinitely thin ones too)
so:
v=12-ir (battery less drop across terminal)
using p=iv we get
p=12i-i2r
formalise this as a quadratic we get
0=ri2-12i+480
use: http://www.math.com/students/calculators/source/quadratic.htm (because I've forgotten that bit)
i = 1158A or 41.43A
dump the daft one (which is an alternate reality) and you have a stable circuit drawing 41.43A with a volt drop of 0.4143V at the battery terminal.
Nothing novel in it.
What I think you are imagining is the soon to be result of further deterioration. In the example the terminal will dissipate 17W, should the resistance creep up to 0.02 ohm then it will dissipate 37W. Pretty soon we enter the realm of the escaping magic smoke when it all stops working. But not quite the run away situation that would seem to be presented. Well, never fear that can happen. Should the equipment not burn up, things get funky at 0.075 ohms. The load will demand 80A and the terminal will be glowing at 480W. Should the resistance increase any more then there is no solution to the equation, meaning that the load will not be able to function.
 
Bru said:
It's too late in the evening, I've been on the beer and I want to go to bed. JUST TIGHTEN THE DAMN TERMINALS UP! :D
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Tomorrow will be soon enough. It was a serious request for an explanation . Not a "dig" or sarcasm like some of the other responses
 
The statement that current will increase with a bad battery terminal is a bit misleading and perhaps only slightly true in a few cases. Mostly not.
A starter motor will turn more slowly at reduced voltage (volts being lost in bad connection) which will result in it trying to draw more current which could conceivably end up with a total more current out of the battery but not much. (or maybe not at all)
Like wise as said some electronic devices with switch mode power converters will draw more current try to provide the same stable power. Fridges and inverters come to mind. Hence lower voltage more current. Some power lost in the bad connection and so yes more current from the battery. (a little more)
Big problem with bad connection is that heat dissipated and possible arcing produces even worse contact so rapidly failing connection. olewill
 
When I fitted my smart gauge a few weeks back, I noticed that the battery terminals I was working with were loose. I hadn't checked them previously as it was a new (to me) boat and I had the receipt showing the entire battery bank being professionally replaced in the summer of the previous Year.
Out of curiosity, I checked all the other connections and not a single one had been tightened up.
It was a beautifully neat install, and looked perfect. but I think what probably happened is that the engineer got everything fitted and finger-tight, got distracted, and never clamped everything down.
So, yes, check your terminals!
 
Correct me if I am wrong, but the missing power disappears as heat at the high resistance, loose battery terminal. If there is no heat then there is very low resistance and little is wrong. No missing amps or volts and no flat battery. The poor battery connection would go un-noticed most of the time. It would be noticed at very high loads when the poor battery connection was starting an engine, say 80 amps, but since this is usually only a split second, the loose terminal may not get too warm and the fault would go un-noticed. A one bar electric fire uses 1000w. Sitting in front of that electric fire you really feel the heat. A very bad electrical terminal would have to be glowing red to provide any heat you can feel. The surface area just doesnt exist to put out any major heat in the short time it is likely to be glowing unless you are churning the engine for several minutes. This is where the problem lies. One problem such as a problematic engine not starting shows up another problem such as loose battery terminals. I suspect most of us would never notice a problem of loose connections and it would not really be a problem until we have that sustained high battery load such as a faulty engine.
 
Dougal nailed it nicely, not all loads are resistive or simply on or off.

Increased resistance on a purely resistive load will, of course, result in reduced current in the circuit and that too is bad - your nav lights will be dimmer for example
 
Bru said:
Dougal nailed it nicely, not all loads are resistive or simply on or off.

Increased resistance on a purely resistive load will, of course, result in reduced current in the circuit and that too is bad - your nav lights will be dimmer for example
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Ever wished you never bothered ?
 
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