<font size=2 face="sans-serif">Hi:</font>
<br>
<br><font size=2 face="sans-serif">From time to time over the years I've
come across systems where the routing of DC cables between the batteries
and the inverter has been the cause of issues.</font>
<br>
<br><font size=2 face="sans-serif">I'm not referring to wire thickness
or quality of terminations. For the purposes of this discussion,
just assume that wire thickness and terminations are perfect.</font>
<br>
<br><font size=2 face="sans-serif">What I am referring to is the routing
of the positive and negative battery cables. In particular, the loop
area within the + and - cables as shown in the image below,</font>
<br>
<br><img src=cid:_1_0D3137480D312EB400618A9F88257E70>
<br><font size=2 face="sans-serif">The problem I've seen in systems with
a large loop in the setup is that the inverter does not provide good surge
power and can even go into low voltage shutdown during large surges.</font>
<br>
<br><font size=2 face="sans-serif">Recently this happened again and I wanted
to get a better feel for it, so I did some math. </font>
<br>
<br><font size=2 face="sans-serif">For a cable length of about 12', the
loop is an inductor which has a value of inductance of about 1 uH for side
by side cables and as much as 6 uH for cables about 1 foot apart. </font>
<br>
<br><font size=2 face="sans-serif">This inductance is greatly multiplied
by any ferrous metal in the loop and can easily be in the range of 10's
to 100's of uH. Examples being cables which run in steel conduits
or along the steel frame of a motor home.</font>
<br>
<br><font size=2 face="sans-serif">Inductance causes a voltage drop proportional
to the rate at which the current is changing. To get an idea of how
large that rate can be for typical inverters, I did surge tests with a
5kW inverter and found that the rate of change of current can be as high
as 100A per milli-second or 100,000 Amps/second.</font>
<br>
<br><font size=2 face="sans-serif">Given that, the voltage drop of the
wire inductance is then , Vdrop = (rate of change of current) x (inductance),
</font>
<br>
<br><font size=2 face="sans-serif">Vdrop for 1 uH = (100,000 A/s) x (0.000001
H) = 0.1V</font>
<br><font size=2 face="sans-serif">Vdrop for 10 uH = (100,000 A/s) x (0.000001
H) = 1.0V</font>
<br><font size=2 face="sans-serif">Vdrop for 100 uH = (100,000 A/s) x (0.000001
H) = 10.0V clearly this is a problem.</font>
<br>
<br><font size=2 face="sans-serif">Have any of the wrenches had systems
with this issue? If so, how often.</font>
<br>
<br>
<br><font size=2 face="sans-serif">JARMO</font>
<br>
<br>
<br><font size=2 face="sans-serif"> <br>
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