Getting solar power into my truck

dwh

Tail-End Charlie
Thus...there is in reality no actual voltage drop. The MPPT is adjusting and compensating for that.

The increased resistance of smaller wire will instead manifest as a reduction in amperage..."amperage drop".
 

Photobug

Active member
Thus...there is in reality no actual voltage drop. The MPPT is adjusting and compensating for that.

The increased resistance of smaller wire will instead manifest as a reduction in amperage..."amperage drop".
Will this amperage drop create a real loss in charging efficiency?
 

dwh

Tail-End Charlie
Will this amperage drop create a real loss in charging efficiency?
Meh...not much. Figure if the loss was 3%, then if a full charge took 8 hours, which is 480 minutes, then 3% of that would be 14.4 minutes. So instead of 8 hours, it'd take 8 and a 1/4 hours.
 

rayra

Expedition Leader
you'll get more drop in potential efficiency from bad panel angle and clouds, than you will the wire.

low-volt landscape wiring is a great idea. I was going to suggest zip cord, but it's expensive compared to landscape wire.




10ga $1.04/ft
12ga $0.62/ft but nearly a buck/ft in shorter lengths.



12ga 100' $56. roughly 2/3 the price of an equal length of zip wire.
 

Gs WK2

New member
I don’t think @dwh is correct with the statement of same voltage everywhere in circuit.

The current is certainly the same everywhere (you are not loosing amps to the atmosphere) and if there is no current flowing then there is no voltage drop, but once you have current flowing through a resistance you will have voltage drop as line losses are creating heat which is the loss of voltage and efficiency.

Putting a volt meter along various points in a circuit reads the potential at that location. The volt meter is not confused. In a DC circuit if you measure amps and volts at the same time anywhere along that circuit you will find V= IR to be true.


Sent from my iPad using Tapatalk
 

dwh

Tail-End Charlie
I don’t think @dwh is correct with the statement of same voltage everywhere in circuit.
I am. A circuit is a circle.

Just as a thought exercise, imagine a loop of wire, two ends connected together. Now insert a diode between the two ends. Say it has a 1v drop.

So...where on that loop is the voltage 1v lower? The output of the diode right? But it's a circle, so the output of the diode loops around and is also the input of the diode.

The voltage is 1v lower - all the way around.


The current is certainly the same everywhere (you are not loosing amps to the atmosphere) and if there is no current flowing then there is no voltage drop, but once you have current flowing through a resistance you will have voltage drop as line losses are creating heat which is the loss of voltage and efficiency.
Yes you will. All the way around.


Putting a volt meter along various points in a circuit reads the potential at that location.
True. But that doesn't mean you are reading what you think you are reading. When you use a meter to read voltage, you are creating a new circuit (meter and probe leads) and reading what is happening on THAT circuit. If, by creating that new circuit, you bypassed some other circuit, you're looking at what is happening on the new circuit you just created, not the circuit you just bypassed.


The volt meter is not confused.
The meter is not confused. The user of the meter is confused, thinking he's reading a circuit, when actually he's reading the bypass he just created.


In a DC circuit if you measure amps and volts at the same time anywhere along that circuit you will find V= IR to be true.
Good luck reading volts. The only way to do that is to create a bypass and measure the volts on the bypass.

Measuring amps is easy; that is done inline. But measuring volts requires a bypass.
 
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dwh

Tail-End Charlie
Take a look at this whitepaper from Morningstar:



Note under the description of PWM where they twice state that the array operates at battery voltage. This is true. So if the battery is at say, 12v, the charging circuit, which includes the solar array is at 12v.

All the way around.

But if you put your meter leads across the solar terminals, you will see the max potential of the array. So it looks like a drop from 17v on one end to 12v on the other.

But that's wrong. There are now two circuits - one at 12v all the way around (battery and solar) and one at 17v all the way around (meter and solar) - but you're only measuring one of those circuits...and it ain't the one charging the battery.
 
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dwh

Tail-End Charlie
So let's say a solar panel at 17v and a battery at 12v. Keep it simple, direct connected, no charge controller.

How do you measure the voltage of the charging circuit?

You can't.

You can measure at the terminals of the solar and see 17v. You can measure at the terminals of the battery and see 12v.

But both measurements are creating a bypass and reading that instead of the charging circuit itself...

Well crap... So, how can you measure the voltage drop?

You can't.


Not that it really matters...because in that situation...there is no voltage drop anyway.

(This is the point where someone will start loudly quoting Ohm's Law while banging their fist on the table and spilling everyone's beer. :) )

Allow me to explain...

The solar has a potential of 17v. Given time, it will eventually push the voltage of the circuit up that high. But it can't because the battery is limiting the voltage of the circuit to battery voltage. Right now, that's 12v.

So you have one force trying to push the voltage up, and another holding it down.

You can calculate blah blah Xv Ya blah blah wire size length blah blah voltage drop blah blah half a volt.

So okay, you calculated half a volt drop. Fine.

Where is it?

It ain't in the charging circuit which is operating at battery voltage because the battery is limiting (you could say regulating) the circuit voltage.

So where is it? Gotta be there somewhere, Ohm's Law is a friggin' LAW.

Yeah, nah. Not in this situation. Doesn't apply.

The resistance of the wire still has an effect of course...it reduces the amperage flowing through the circuit.
 
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dwh

Tail-End Charlie
Which is a different situation than say an inverter running from a battery.

In that situation you do not have one force trying to push the circuit voltage up and another holding the circuit voltage down.

Instead, you have one force (the inverter) trying to drag the circuit voltage down, and another (the battery) trying to keep the voltage from going down.

But the battery can't keep the circuit voltage from dropping. It starts from a finite point, say 12.8v, and the only way voltage can go from there is down.

Wire resistance will make the problem worse, so in that situation, voltage drop due to wire resistance is absolutely gonna happen.
 

dwh

Tail-End Charlie
Picture it another way...

Two closed loop systems.

One has a device that extracts energy from the system, causing the voltage to drop.

The other has a device that injects energy into the system, causing the voltage to rise.

The voltage cannot drop while it is simultaneously being forced to rise. Resistance only slows the -rate- at which it rises.
 

Gs WK2

New member
@dwh, that’s a lot of reply which is probably directed at me.

Why do you say a volt meter is passing current when set to read voltage. By design the circuit in the meter would ideally have infinite resistance (open circuit) and in reality perhaps some fraction of a mA. This mA circuit in the meter is not enough to cause a false reading of voltage potential of the point of the circuit being monitored.

My second question is with the “it’s a circle so it’s all the same” where is the positive and negative of the battery in this circle?


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DaveInDenver

Expedition Leader
@Gs WK2, Kirchhoff's and Ohm's laws here require you to introduce dependent sources and superposition to get Thevenin equivalents. IOW a solar panel Voc and Isc yield an equivalent source voltage and resistance. The same would be true of a battery, dependent source and load simultaneously. Forget when putting a controller in there and how that further relates source(s) to load(s). So what you measure in reality and what you sketch up and analyze aren't easily compared without the maths predicting what your meter will show at some moment in time.
 
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dwh

Tail-End Charlie
@dwh, that’s a lot of reply which is probably directed at me.
No so much directed at you. More that your comment provided an opportunity for me to carry on with a lesson that I had begun with Photobug (and the "anonymous cowards"* who are silently watching from the peanut gallery :) ). He had obviously done his homework and understood some about resistance and voltage drop and the need for larger wire, and when I explained a bit more, asked pertinent questions.

So, what the hell, might as well take it to the next level; he seemed ready for it. Plus every time I explain it, it helps me hone it down so it's easier for the newbs to grasp.


Why do you say a volt meter is passing current when set to read voltage.
I don't recall saying that, unless you are referring to my comment that measuring amps is done inline, whereas measuring volts is done with a bypass.

Could you please quote what it was that I said that you are referring to?


My second question is with the “it’s a circle so it’s all the same” where is the positive and negative of the battery in this circle?
Depends on whether the battery is acting as a load/regulator, or a supply.

Though I'm not sure exactly what you're asking, so that might not be the answer you were looking for.




*Shout out to El Reg.
 

Gs WK2

New member
Yes you are. The voltage lost is dissipated as heat to the atmosphere, not easily detectable at these levels but there.
I agree that heat is lost to the atmosphere and measured in watts. But the electrons are all still in the wire flowing along from one end to the other. Power = current x voltage. The voltage drop caused by the skinny line means less watts available if the same current is assumed to be flowing.

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