Voltage Loss = Amperage Loss
By valterra, Section Diaries Posted on Mon Apr 28, 2008 at 12:29:17 AM MST
If you're not losing voltage in a "long" cable run, could you be losing amperage?

My Ametek Mill, and my HF solar panels are both running to the battery back via a "standard" sized extension cord.  Simply cut off the end plugs and used the Black for negative and the White for positive.  Easy enough.  50' Wires if I remember right.  Maybe 100'. I've read a lot about voltage loss, so I did a little experiment.  When I first cut the wire for my solar panels, I first tested the voltage directly from the actual panels.  It was something like 25 volts.  Then I hooked up the extension cord to the panels and tested the wire from the other end.  Pretty much the same reading.  So in this example, we'll say it was 24.9 volts.  I don't think the output of the panels is going to change much within the 30 seconds it took me to do that experiment. But all this talk of upwards of 60% voltage loss and how 12v sucks because you can't do long cable runs has me worried that I'm losing something somewhere.   Regardless on the ROI on the cost of larger cable, would much thicker cable greatly increase the performance?  Is this just a math anomaly because my Amperage is so low?  For example a 10% loss at 3A is only 300mA, whereas a 10% loss at 30A is 3A? Or is power loss over these runs one of those things that maybe isn't such a big deal in the "real" world (just like some people seem to do okay with car batteries even though they're not "real" RE batteries)?

Voltage Loss = Amperage Loss | 20 comments (20 topical)

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#1) by ghurd on Sun Apr 27, 2008 at 06:53:12 PM MST

There was no voltage loss, because there was no amperage. Check the wire voltage at the battery and at the PV, with current flowing. PVs are designed to deal with some loss.  That's why the peak power is at 17V, so a couple can be lost in wire, diodes, etc, and still charge OK. G- Ghurd.info

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#2) by valterra on Sun Apr 27, 2008 at 07:30:52 PM MST

(Now just to give more info, with the solar, I also did do a shorted Current test, and I believe had the same results) Okay, so I may, indeed, be experiencing a significant amount of loss and now realising it.   With my Ametek motor (a 40 vdc one) I get just over 3A peak when it is crazy windy.  At 14.5 volts that is about 44 watts.  That motor is spinning like a mofo, and sometimes I think there is more to be had. [ Parent ]

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#3) by valterra on Sun Apr 27, 2008 at 07:38:24 PM MST

If I measured the resistance of the motor whilst not spinning, could I use Ohm's Law to figure out what I 'could' be getting at specific voltages? [ Parent ]

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#5) by ghurd on Sun Apr 27, 2008 at 08:17:24 PM MST

On paper, yes. It never works out for me.  Inductance, brushes, maybe other things. (it did work out 100.00% perfect a time or 2, with PMAs working well under what they were capable of making) Speo posted some nice numbers on Ameteks (and Indianas) he had.  Might check his files. G- Ghurd.info [ Parent ]

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#4) by ghurd on Sun Apr 27, 2008 at 08:04:04 PM MST

I'm confused.  If the battery end of the cable is shorted, there is no voltage at that end.  All the PV voltage is lost in the wire. I don't think your losses are very significant, for your system. I figure a 50 or 100 foot extension cord is #14?  Maybe #12. A #14 cord, 50' long, and 3A... Or 100' of #12... is only about a 7% loss.  Quick mental math. The PVs rarely make 3A for long, and if the windmill does then the battery voltage is probably up. G- Ghurd.info [ Parent ]

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#6) by valterra on Sun Apr 27, 2008 at 08:44:57 PM MST

eh, confusion is my fault account bad verbiage. Both tests were done the same way: Meter set to 200vdc, test solar panel leads.  Hook up extension cord to those leads and test the opposite end. Amperage test same scenario, with meter set to 10A.  "Shorted" through the meter. [ Parent ]

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#7) by wooferhound on Mon Apr 28, 2008 at 12:41:57 AM MST

This page has a lot of information on it but Look at the bottom of the page It's a Voltage drop Calculator tell it the wire size, distance in Feet, Amperage, and system voltage then click calculate. You will get lots of information about what will come out of the wire at the other end. http://www.powerstream.com/Wire_Size.htm W o o f -={(

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#8) by tecker on Mon Apr 28, 2008 at 04:38:43 AM MST

The resistance of 14 with good connections is .002 ohms per 100 feet that should be doubled to include both wires .The resistance of the batteries is changing with the charging process . Checkit with an accurate resistor when the battery has settled .  [voltage drop accross the resistor * the resistor used / the voltage ] that gives a rough look at the impeadance at that moment.

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#9) by valterra on Mon Apr 28, 2008 at 08:12:07 AM MST

That sounds about right, because my dmm barely showed anything (0.0, 0.1) when I directly tested the wire itself. [ Parent ]

Just double it over? (3.00 / 0) (#10) by valterra on Mon Apr 28, 2008 at 07:51:53 PM MST

Bear with me here, and feel free to tell me if I am wrong. If 14 gauge wire isn't  available readily (inexpensively) but, say, 18 is, then running the wire twice should yield good results, right? I am assuming, using smaller wire versus larger: Multiple runs of smaller wire from the battery to individual loads will increase the voltage drop as opposed to the larger wire. Multiple runs of smallerwire from the battery to a BUS would be more or less equal to running a single, larger wire.  The bus creates an electrically common connection on the far end, basically making your multiple smaller wires the same as a larger wire (the same principle as stranded versus solid)   In the first example, the complete circuit of batt-load-batt on each load guarantees a loss in voltage according to the size of the wire. In the second example, loads would connect to the bus, therefore drawing power from all of the wires more or less equally. I ask because it looks like I can run 18 GA wire for much cheaper than 14, and also that I might be being foolish in not using the third wire of the extension cord.  After all, 2 runs gets you 6 conductors instead of 4.

Re: Just double it over? (3.00 / 0) (#11) by DamonHD on Tue Apr 29, 2008 at 01:22:56 AM MST

I just rewired my off-grid RE system since I moved my desk from the back of the house to the front, and I've had to bring the power up through a floor.  I used normal 2x 3-core mains cable for various reasons (eg not having to make one huge 15mm hole in the plasterboard and floorboards) and thus left myself some spare conductors. I found that I have already had to double up on one side of the run (ie knock about 25% off my impedance) else my whole system is unstable with the LVD cutting in and out.  With the doubling up all is fine again.  Clearly, if load increases I may double up on the other side of the run too to knock a little more (33%) off the impedance (and I may add another smoothing cap). So yes, practice matches theory here.  Paralleling the smaller wires really does work. Rgds Damon "Once you have licked the windows of freedom your tongue gets stuck." [ Parent ]

Re: Just double it over? (3.00 / 0) (#12) by valterra on Tue Apr 29, 2008 at 06:34:53 AM MST

Thanks a lot, Damon!  When you said 'double up on one side,' do you mean you doubled up the positive or negative side only? [ Parent ]

Re: Just double it over? (3.00 / 0) (#13) by DamonHD on Tue Apr 29, 2008 at 09:10:37 AM MST

As it happens I doubled up the negative/ground side, but it does not matter of course for the impedance purposes which I did. Rgds Damon "Once you have licked the windows of freedom your tongue gets stuck." [ Parent ]

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#14) by Sparky01 on Tue Apr 29, 2008 at 11:43:16 AM MST

Yes you can run paralell wires to cut down the total resistance of both conductors.  But if this is to be a location that could be inspected or you want to be within code regulations. NEC states that you can not run paralell conductors less than a #2 (if my memory serves me correct).

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#15) by valterra on Tue Apr 29, 2008 at 07:07:21 PM MST

Roger that.  It'd be okay for my particular application. Would there be some kind of formula for what the net result of paralleling would be?  If you ran two 2mm wires, could you figure it's nearly the same as a single 4mm one? [ Parent ]

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#17) by DamonHD on Wed Apr 30, 2008 at 01:33:58 AM MST

By cross-section (mmsq), not diameter (mm), yes, it would be right for DC.  (For high frequencies all sorts of odd transmission line and skin effects get involved, which I regard as black magic.) Rgds Damon "Once you have licked the windows of freedom your tongue gets stuck." [ Parent ]

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#19) by wooferhound on Wed Apr 30, 2008 at 05:23:27 PM MST

Like "Monster Cable" for speaker wireing. W o o f -={( [ Parent ]

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#20) by valterra on Thu May 01, 2008 at 08:06:25 AM MST

That's an easy one.  Monster Cable is OBVIOUSLY better.  They couldn't charge extra just because of the name.  I think it's actually copper-colored silver, but the salesman told me that it actually acts as a "superconductor" he said. [ Parent ]

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#16) by stephent on Tue Apr 29, 2008 at 09:30:29 PM MST

Simple formula Voltage drop= 2 times one way length times amps times "R" factor for the wire ________________________ product of above divided by 1000 Or 2xLxI(amps)xR ------------------      1000 The "R" can be found in the back of the National Electrical Code book Just use the one for either the single conductor type (the 1 in the leftish column) or the stranded wire with a normally a 7 in the column. But use the "bare" header column instead of the "coated"--even if it has insulation....ie...the wire inside is indeed "bare" and doesn't have an enamel individual coating under the insulation. The formula is good for DC or low frequency AC (like maybe up to 50khz er so) The "R" factor in the book seems to be close,,,,but lately I think with the copper or aluminum prices rising...it's looking a tad low. They must be adding more cheaper alloying stuff to the mix ??? So yes... voltage loss (drop) does equal Amperage loss at the other end---either one indicates a resistance (the "R" thing) in the circuit.

Re: Voltage Loss = Amperage Loss (3.00 / 0) (#18) by valterra on Wed Apr 30, 2008 at 06:43:35 AM MST

Okay then.  Thanks!  :-)  With these permanent magnet type gennies (variable voltage output) I think the biggest concern is loss of Amperage or charging capability.  But whether you call it Voltage Drop or Amperage Loss, your point is well taken.  Resistance is resistance, which means you are losing POWER along the way, and that needs to be minimized as much as possible. [ Parent ]

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