Low voltage under load

Jokani

Jokani

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My chartplotter reports that the voltage it is receiving is significantly less than reported by the battery monitor.

If I measure the voltage supplied by the chartplotter power cable it is the same as that measured by the battery monitor.

An example of the voltage difference under load- The BM1 reports 12.4v and a voltmeter on the powercable also reports 12.4, but when I connect the CP, it only reports 12.1v, the BM1 continues to report 12.4v.

This drop in voltage to equipment under load is causing me issues - My Garmin gWind WSI is powered via the NMEA2000 network, and the voltage drop causes the WSI to turn off. Garmin have suggested that I connect a dedicated powers supply to the WSI, whilst I am happy to do this but would also like to get to the bottom of the low voltage under load issue.

If I place a 4.8 ohms diode over the voltmeter + and - inputs, the voltage drops to 12.1, suggesting the CP is correctly reporting the voltage it is receiving.

Is a drop of 0.3v normal/to be expected? The gap widens as the battery voltage lowers.

I have 4 x Trojan T105's new this season, giving around 450 amps, so there should be plenty of power, but it's not getting where it's needed.

A complicated problem that I'm struggling to get my head around, any advice gratefully received.
 
You are measuring the volts on the CP cable without the CP connected and operaing. Thats almost ceratinly going to readthe same as te battery monitor,


However if you have a bad connection somewhere on the supply to the CP . (It could be in the negative or the positive. ) when the CP is on and drwaing current the bad connection will cause a drop in volts.

Confirm by measuring the volts at the CP while it is connected and operating. Easy to say, Maybe not quite so easy to do in practise.
 
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Can you measure the voltage with the chart-plotter connected & powered-up? Measure at the fuse/supply branch point & at the plotter. That may shed some light on the matter.
 
You shouldn't have a 0.3v drop on your plotter/instrument supply. It would suggest either faulty connections or wiring which is too thin for the load. Can you temporarily try a feed using heavier wire and see what difference it makes?
 
Many thanks for the quick replies, that has given me something to gets started with.

I will:

1) Connect CP and test voltage at fuse panel

2) Connect direct to battery with heavier wire and see if the CP report the correct voltage.

3) Connect CP, strip cable wire and test voltage close to CP.

I'll complete these tests this week and report back the results.

Thanks again.
 
Unless it was previously OK, and has now changed, my first port of call for low voltage would be cable thickness and faulty connections. If it was ok, and has changed, faulty connections would be the port of call.

has anything else been added, removed, from the same circuit?
 
Quote If I place a 4.8 ohms diode over the voltmeter + and - inputs, the voltage drops to 12.1, suggesting the CP is correctly reporting the voltage it is receiving.

I presume you added the 4.8 ohm RESISTOR to the volt meter + and - terminals to simulate a load when checking the voltage at the CP plug. That is the correct thing to do and will provide a load of near 3 amps. This might be similar to what the CP draws. I presume it is difficult to open the CP power plug to make connection to pins with it connected to the CP.
Now if you connect this resistor to the CP plug (in lieu of CP) then you can use the volt meter (no load resistor) from your battery pos terminal to the pos of the CP plug. This will measure a low voltage which is the actual drop in the pos wire. Do the same with the negative wire. If the cable is heavy enough for low volt drop you might finds that the drop .3v is mostly lost in the pos cable. If this is the case you work backwards along the pos line to find if the drop is ina fuse (holder) or switch or terminal of some sort. The negative line might be giving the volt drop but oless complex so less likely.
As siad an other option is to add more wire from the pos battery terminal tot he CP plug pos output to see if substituting or bypassing either pos or neg wire will reduce volt drop. (dramatically).
good luck olewill
 
William_H said:
I presume you added the 4.8 ohm RESISTOR to the volt meter + and - terminals to simulate a load when checking the voltage at the CP plug. That is the correct thing to do and will provide a load of near 3 amps. This might be similar to what the CP draws. I presume it is difficult to open the CP power plug to make connection to pins with it connected to the CP.
Click to expand...

You are correct, it was a resistor, it was one of these http://www.ebay.co.uk/itm/171939548548

Will do as you suggest and let you know how I get on.
 
GaryCottam said:
You are correct, it was a resistor, it was one of these http://www.ebay.co.uk/itm/171939548548

Will do as you suggest and let you know how I get on.
Click to expand...

Those resitors are rated at 1/4 watt. That means that if they dissipate more than 1/4 watt they will be overheated and destroyed. Now 4.8 ohm resistor will pass about 3 amps when 12v is applied across it. That comes out as about 36 watts so one would imagine that the resistor would have gone "poof"
You need a resistor with a much higher rating. usually called wire wound resistors but 36 w rating is pretty big. Adding more resitors in parallel will mean you can add the power rating of all of them. However adding ressitors in parallel divides the actual resistance. So 4 resistors 2 in parallel in sewries with another 2 in parallel will give 4X power rating.
A more practical approach is to use a car incandescant globe. About 36 watts or so. However be aware the resistance and current drain of a globe varies with the temperature so it will not always give 4.8 ohm but might be more like 2 ohms when glowing dimly.
 
Before you go changing anything, take some measurements to find the exact location of the problem:
You need access to the power feed points on the instrument with it connected and powered on.
Meter on low DC volts, black to battery neg post, red to neg at instrument - and
" " " " " black to pos at instrument, red to battery pos.
Any voltage measured must be eliminated. Example = if you measured 1volt on the first instance, leave the black lead on the post, and follow the path with the red lead. You may have a shunt, fuse, distribution bus, engine chassis etc to chase.
A shortcut is hinted if the instrument is the ONLY thing that suffers from a drop it has to be in that cable, but have you looked at anything else?
If there's a suspect joint somewhere in the distribution panel try increasing the load by switching on lamps, fridge etc. as it will exacerbate the problem and make it more obvious. I would expect the cable supplied with the instrument should be adequate, so would be more interested in where it is fed from.
 
William_H said:
Those resitors are rated at 1/4 watt. That means that if they dissipate more than 1/4 watt they will be overheated and destroyed. Now 4.8 ohm resistor will pass about 3 amps when 12v is applied across it. That comes out as about 36 watts so one would imagine that the resistor would have gone "poof"
Click to expand...

You are correct, again! The small resistors became too hot to handle, and only worked for a few seconds.

Would a 680 ohm resistor do the same job? If not I'll source a bulb.
 
GaryCottam said:
Thanks for the advice, I'm going to have some fun this weekend!
Click to expand...
Stop playing with lamps and resistors and follow William_H and Billjratt's advice.

What they are describing is how to measure the voltage ALONG a wire on it's journey from the battery positive and negative terminals to your instruments or loads. On its way it might meet a dozen connections or switches or fuses or bus bars. Any one of these 'dirty' connections or faulty wiring could be causing the voltage drop. So what you are doing is measuring that small, less than 1 volt, drop along the wires. So work your way past all these connections until you suddenly find the culprit. You will only get this voltage drop when the CP is turned on.
 
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Billjratt said:
Before you go changing anything, take some measurements to find the exact location of the problem:
You need access to the power feed points on the instrument with it connected and powered on.
Meter on low DC volts, black to battery neg post, red to neg at instrument - and
" " " " " black to pos at instrument, red to battery pos.
Any voltage measured must be eliminated. Example = if you measured 1volt on the first instance, leave the black lead on the post, and follow the path with the red lead. You may have a shunt, fuse, distribution bus, engine chassis etc to chase..
Click to expand...

Just so I am clear on what I am testing and the process.

One or both of the pos and neg cable runs between the post and instrument should display a voltage if there is a fault on that run. If there is, need to work down the cable run until the voltage either goes to zero or reduces which would indicate a fault in that part of the run. Only when the voltage read is zero have I cured/found all faults.

Sorry if I am making hard work of this, just want to be absolutely clear on what needs to be done.
 
GaryCottam said:
Just so I am clear on what I am testing and the process.

One or both of the pos and neg cable runs between the post and instrument should display a voltage if there is a fault on that run. If there is, need to work down the cable run until the voltage either goes to zero or reduces which would indicate a fault in that part of the run. Only when the voltage read is zero have I cured/found all faults.

Sorry if I am making hard work of this, just want to be absolutely clear on what needs to be done.
Click to expand...
You've got it!

it is difficult for people to get their head around this sometimes. How can a cable produce a voltage?

A very simple test is to measure the voltage across your main DC house breaker switch with a high current being drawn by several bits of kit. Put a digital voltmeter directly on the studs on the back of the switch and you will probably see 0.01v or more. If you see 0.1v or more the switch is dirty. Mine showed 0.25v.

Turn it on and off many times to clean the contacts or take it to pieces and check for pitting on the switch contacts. That put my voltage drop back to 0.01v.
 
I have a Fluke meter with max / min.
If you set it to record MAX and connect it across a questionable switch (while it's on obviously!) it will record the voltage drop across the switch as the load changes. Anything over a few mV would suggest a problem.

Our old fashioned 1/2/both switch is a case in point. If I connect the meter across the switch and then pop up on deck and start the engine it will record as much as 1v as the starter motor turns.
 
The usual method of fault-finding along a series circuit is called "halving"
In this instance if you find a significant voltage across a circuit, leave one meter-lead (either) where it is and move the other closer half-way across the circuit ( that can be distance or number of connections) and see if the voltage (fault) stays or goes. Then meter across the failing half (with the voltage on it) and repeat the halving exercise, and so on until the fault is identified.
It's worth noting that the error voltage will diminish as the process proceeds, since a small (acceptable) loss is inherent in most circuits.
A good way to practice is finding the dud lamp in your fairy lights... (using ohms of course)
 
Thanks gain for the advice. Off to the boat tomorrow, can't wait to see what I find out. Will off course report back.
 
GaryCottam said:
You are correct, again! The small resistors became too hot to handle, and only worked for a few seconds.

Would a 680 ohm resistor do the same job? If not I'll source a bulb.
Click to expand...

A 680 ohm resistor will only pass 12 divded by 680 or about 20 milliamps. .02 amp. and about .25 watt. While this small current might help you find significant corroded wire switch or fuse holder faults and is much better than just the volt meter, it will not be much use for your present problem.
I am guessing that you can not or don't want to open up the battery plug to the CP to expose the battery wires +ve and -ve. If you can get to the pins of the plug when conected with the CP running then the CP will be the right load for checking volt drop. But if you can't get to the pins except by removing the plug then you need a resistor or lamp to substitute for the CP load. But to be of best use it will have to dissipate similar power to the CP. ie about 30 watts. Actually the more power taken by the resistor the more obvious will be any fault in the wiring and supply. good luck olewill
 
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