Solar panel is hooked up and working! - Page 5 - Fiberglass RV


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Old 05-06-2013, 08:31 AM   #57
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Name: Paul
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Something I am not understanding on Mike's design is the charging process. Maybe I am missing something. The Sunguard controller can function as a stand alone charger? Is that correct?

If we are putting the solar power through the trailer side connector are we not using the charger on or in the trailer to do the charging? My Escape uses a WFCO Ultra III distribution center. One of the comments I have read about the charging portion of this device is it is slow to charge. Or, becasue we are using the power line normally used for charging from the tow vehicle, is the trailer charger bypassed.

On a related note, is there a method to measure what charge you are getting and what charge you have after finishing. I am aware that taking a measurement just after completeing a charge will give higher readings due to the topping off feature. I have an inexpensive Equus as shown below and a multi tester but not the knowledge of where and when to test.
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Old 05-06-2013, 09:39 AM   #58
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Originally Posted by Paul Braun View Post
Something I am not understanding on Mike's design is the charging process. Maybe I am missing something. The Sunguard controller can function as a stand alone charger? Is that correct?

If we are putting the solar power through the trailer side connector are we not using the charger on or in the trailer to do the charging? My Escape uses a WFCO Ultra III distribution center. One of the comments I have read about the charging portion of this device is it is slow to charge. Or, because we are using the power line normally used for charging from the tow vehicle, is the trailer charger bypassed.

On a related note, is there a method to measure what charge you are getting and what charge you have after finishing. I am aware that taking a measurement just after completing a charge will give higher readings due to the topping off feature. I have an inexpensive Equus as shown below and a multi tester but not the knowledge of where and when to test.
You can (and should) wire the trailer so that the on-board converter, the tow vehicle and the solar panel all connect to the battery. Whichever source has the highest voltage will charge the battery.

To measure the charging you need to use an ammeter. I wired in one when I made a control panel, but you can use a multimeter to measure current flow. I have a disconnect switch on my battery and it is possible to put the multimeter leads across the the terminals when the switch is in the off position, and measure current flow.

One of many on Youtube: How To Use A Multimeter - YouTube
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Old 05-06-2013, 10:41 AM   #59
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Mike, I didn't catch where you bought your solar set up. Anyway you could turn me on to where you got yours? I am not that far to get one now (still working on ac) but My baby will be sitting in the sun everyday and I think it might be a good ideal to have a solar setup and run my vent during the hottest part of the day. Thanks Mike!
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Old 07-13-2013, 04:42 PM   #60
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A couple of things here.
1) As Thomas mentioned, a 50' cord is 100 feet of wire, 50 foot there and 50 feet back.
2) "The current" is NOT being delivered. The current is being wasted as heat (power loss).
To make the math easy to understand assume that the battery is a 1 Ohm load and the solar panel is delivering 12 V. Put 12 volts across a 1 ohm load = 12 amps. I = E/R. Now add .1 ohm resistance in the wire as loss (just an example). The wire is placed in series with the original load and so the resulting load is 1.1 ohms. 12V / 1.1 Ohm = 10.909 amps. You have just "lost" .9 amps. Where did it go? To heat loss.
3) And finally solder is an excellent conductor, assuming you are not using plumbing solder.

http://et.nmsu.edu/~etti/fall97/electronics/solder.html

Solder is a tin (or silver) / lead mix.
The flux in electrical solders help the wire surfaces to "wet" with the solder and form an immaculate electrical connection. Solder also absolutely 100% prevents poor connections in the crimp since there is no crimp. Poor connections lose power in... high resistance... which causes... heat.

I had an electrician come in to look at my power panel in my home. The lights would dim when stuff turned on around the house. It turns out that every screw in the panel (to each individual breaker) had loosened over the years. The main power cable from the power company is a massive thing which came into a massive screw. It too was loose. The black plastic block around the screw was MELTED from the heat generated from the poor connection.

Obviously this is an extreme example, but it amply demonstrates the issue behind the scenes. Crimps will never be as good as a soldered connection. They will ALWAYS corrode over time. Heat causes them to flex. Flexing causes the connection to slowly deteriorate. Poor connection causes resistance which causes heat (which is power loss), which causes more flexing causing worse connection and so a vicious cycle ensues. A poor connection also allows oxidation of the wire surface (corrosion). Corrosion causes resistance, which causes heat, which causes (is) power loss which feeds back into that vicious cycle.
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Old 07-13-2013, 04:53 PM   #61
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Interesting. I have an old charger that uses PWM. I've always assumed the off time was to provide the sensing circuitry with a "disconnect" to check the battery voltage and the lowering of the duty cycle was to slow the charging down to avoid overheating the electrolyte. I'm not sure what you mean by throttle.
RAZ, PWM is used because the current applied to the circuit is directly proportional to the on time if the wave. Assume that the pulse width is zero (always off) - no current delivered. At 50% the current delivered is 50% of the available current. At 100% (always on) the current delivered is 100% of the available current.

Usually the width is varied based on some "algorithm". The current may be applied across a very small resistance (a shunt). The resistance causes a very small voltage across the shunt which is amplified by an op amp and then that voltage is measured. Probably the "wave" is averaged (using a capacitor or inductor) in the op amp circuit so that you get a filtered voltage which can then be measured. That voltage can be used to feed back into the PWM circuit to vary the width as required.
That control circuit is not going to be extraordinarily precise but "good enough". Turn the thing on most of the way for a bulk charge (very low battery charge), cut it back when the current into the battery starts to drop below a certain threshold, and then at some point cut it almost off for the float charge.
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Old 07-13-2013, 05:20 PM   #62
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Originally Posted by jwcolby54 View Post
2) "The current" is NOT being delivered. The current is being wasted as heat (power loss).
To make the math easy to understand assume that the battery is a 1 Ohm load and the solar panel is delivering 12 V. Put 12 volts across a 1 ohm load = 12 amps. I = E/R. Now add .1 ohm resistance in the wire as loss (just an example). The wire is placed in series with the original load and so the resulting load is 1.1 ohms. 12V / 1.1 Ohm = 10.909 amps. You have just "lost" .9 amps. Where did it go? To heat loss.
No, current is not lost, as heat or otherwise. There is just less current flowing, because the available voltage is not capable of pushing as much current through the larger resistance. Certainly some of the power being put out by the panel is being wasted as heat due to that resistance:

12 volts into 1 ohm load:
  • 12 volts across 1 ohm means 12 amps of current flows
  • 12 amps at 12 volts is 144 watts delivered
12 volts into 1 ohm load through 0.1 ohm wire resistance:
  • 12 volts across 1.1 ohms of resistance means 10.9 amps of current flows
  • 10.9 amps at 12 volts is 131 watts total, consisting of
    • 10.9 amps through 1 ohm is 119 watts into the battery, plus
    • the same 10.9 amps through 0.1 ohm is 12 watts wasted as heat from the wire

In practice, the panel itself is not an ideal zero-impedance voltage source. If the load has more resistance, the panel output voltage will be higher, and vice versa. In the example, the voltage would rise a bit when the resistance was added, so the delivered power would not drop so much.

The example is still a good illustration of why it is undesirable to have significant conductor resistance.

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Originally Posted by jwcolby54 View Post
My guess is that this was intended to be http://et.nmsu.edu/~etti/fall97/electronics/solder.html
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Old 07-13-2013, 10:47 PM   #63
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Well of course your math is absolutely correct.

And of course, even if you raise the voltage, you increase the current pushed through the resistance, which delivers more current to the battery but also wastes proportionately more watts in the wire. In any event, the less resistance in the wire, the more current delivered to the battery.

Someone mentioned using a 3 wire extension cord to deliver a two wire solution. If we simply short (the same) two of the three conductors at each end, then the resistance in one leg is halved - assuming equivalent gauge in all three wires. Thus the total resistance is 1.5 R instead of 2 R.

All of which shows why we try to use the right wire for the job, but as the OP pointed out, simply getting the panel in the sun over a 50' lossy extension cord will almost certainly be more efficient (deliver more current anyway) than the same panel in the shade over lossless wire.

Also I am looking at higher current fold-able panels which I certainly wouldn't want to hookup over a 16 gauge wire.
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