Measuring V loss during charging

Measure the resistance of the circuit.

Voltage drop in cables is simply the difference between the voltage measured at the power supply and the voltage measured at the battery.

If you want to measure directly, just put one leg of your voltmeter on the battery and the other on the same wire at the output of the power supply.

Steve

The resistance was .036 ohms. Find a chart with the wire size and resistance per foot. I would say you have no issues. Things must be shiny clean and tight because 12 volts has very little push to overcome resistance.

Quote: "Measure the resistance of the circuit."

To elaborate on this: In this case the resistance is too small to be measured directly with a typical ohm meter because the current generated by the typical ohm meter is so little. Best to measure (an estimate is probably close enough) the current in the wire and the end to end voltage drop in the length of wire (as suggested above). In this way you are creating a "high current" ohm meter which over comes the standard VOM ohm meter by using brute force (15 Amps as opposed to a few milliamps).

You seem to have a lot of time on your hands.

Voltage loss really does not matter much.

Your needs are a charged battery and power to run whatever you feel like at the time you want to run it. Voltage loss will not matter if you have plenty of power. I can run my porch lamp 24/7 because I have a 400 watt solar system, and it takes more effort to shut off the switch and turn it back on later in the day, than to just leave it on all the time. The LED light only consumes 1 amp per hour, and that is more than made up by each of the 120 watt solar panels each day.

So until the solar system needs to run more than 5 hours per day to fully charge the battery, I will leave the porch light on.

Fred.

13.60 less 13.06 is 0.54 voltage drop. That is a lot for 15 amps.

V=IR or .54 = 15 *R; R = .036 ohms.

More realistic is 50 amps. 50 * .036 = 1.8 volt drop with proper voltage to get 50 amps charging. So virtually no converter at 14.4v will supply 50 amps due to the resistance. Need to reduce resistance to get faster charging.

Charging at just 13.6 volts is a disaster

smkettner wrote:
13.60 less 13.06 is 0.54 voltage drop. That is a lot for 15 amps.

V=IR or .54 = 15 *R; R = .036 ohms.

More realistic is 50 amps. 50 * .036 = 1.8 volt drop with proper voltage to get 50 amps charging. So virtually no converter at 14.4v will supply 50 amps due to the resistance. Need to reduce resistance to get faster charging.

Charging at just 13.6 volts is a disaster

X2 And yet to many rigs seem to have inadequate charging. The mfg are only concerned about lower cost and equipment lasting until the warranty runs out.

I used the voltmeter trick (Salvo taught me this) to get the R by adding about 25 ft of speaker wire to one of the voltmeter leads (doesn't matter which one you pick) and alligator clamp ends.(you can't be in two places at once :) ) That long speaker wire did not affect the voltage readings.

I got my amps from the Trimetric for converter-battery paths.

This let me measure between the "battery lugs" at the DC fuse panel inside the trailer out the door and up to the batteries in front.

You get one R for the positive path and another for the neg path. The neg path could be much less than the positive path because it uses the trailer's frame, and the positive has that skinny wire and maybe a switch. (switches and fuse holders are bad for R) It is the total of the two Rs that counts for what amps you will get to the battery from the converter.

First priority is to reduce the R of the positive path which is the big R. You can also make some improvements to the neg path if there is any corrosion at the frame contact points for the short neg path wires.

Progress was measured by the increased amps when R was reduced. There is a limit where there is no point in reducing R anymore once the converter is at its max amps.

It is better not to use the batteries for this testing because they have so much R themselves (more on that below) and it keeps changing. Best to get a big draw on the batts from the inverter running a kettle and see the amps for that, then turn on the converter and see how many amps the Trimetric shows then.

Eg, converter off, kettle on, Tri shows -90amps. Turn on converter, kettle still on, Tri shows -60amps. Means the converter is doing 30 amps. I was able to reduce total R by various means so that the 7355 was doing 56 amps. However on the same paths total R with actual battery charging on low batts, the amps start off at more like 45a and taper quickly to the 30s-mid 20s. That is way better than before though! And at 13.8v too.

http://www.rv.net/forum/index.cfm/fuseaction/thread/tid/26711816/srt/pa/pging/1/page/3.cfm

Quote: "The neg path could be much less than the positive path because it uses the trailer's frame"

It has been done this way in the past but I think current theory connects DC(-) to the Frame/Chassis in only one place to prevent frame/chassis currents. And if so, this one connection is used to service the starter motor on a motorized vehicle. Just saying, I think this is the current theory, could be wrong and if so I am sure we will hear about it ! :)

So in my example, some part of the resistance is coming from the battery, right?

If there was no wiring resistance, would the charger and battery terminals read the same voltage, whatever it may be?