AC vs. DC for solar powered fridge

[chand-o]

Being in a "circle" means voltage doesn't drop makes no sense at all. Connect the circuit and measure the voltage at the panel, then measure it 100ft away at the battery and it will be lower = voltage drop.

The current on the other hand will be the same at any point in the circuit, but voltage no.

 

[dwh]

Being in a "circle" means voltage doesn't drop makes no sense at all.

Of course it makes sense once you understand that the voltage can't drop while something on the circuit is forcing the voltage to rise.

You can drag the voltage down, or you can force it up, but not both at the same time.

Connect the circuit and measure the voltage at the panel, then measure it 100ft away at the battery and it will be lower = voltage drop.

As I said, people fool themselves with their meters.

The measurements will be different because you are measuring two different circuits.

1. Battery to meter and back to battery.
Shows you battery voltage.

2. Solar to meter and back to solar.
Shows you solar potential voltage.

Neither one measures the voltage of the actual charging circuit. And it doesn't matter if you measure 18v solar potential voltage, that's not the voltage of the circuit with the battery in it.

But because the entire circuit will be at battery voltage (ignoring MPPT for simplicity), measuring battery voltage also tells you the voltage of the entire charging circuit.

And as I said, a circle has no ends. The solar is not "at one end" and the battery is not "at the other end". They are both inline in a circuit (circle) that has no "ends".

The current on the other hand will be the same at any point in the circuit, but voltage no.

The voltage will also be the same at any point. But good luck trying to measure it by creating bypass circuits with your meter.

 

[Rando]

I generally agree with you on electrical issues - but you are flat out wrong in this case. You can easily measure the voltage drop across the set of wires in this circuit without 'creating a bypass' that 'isolates' the circuit. Take your volt meter and measure from one end of one conductor to the other - the observed potential difference is the voltage drop across that wire. If you want to get really, really pedantic you can consider the input impedance to your voltmeter - which or decent multimeter is ~1 Gigaohm. Given the resistance of the wire will be on the order of 0.1 ohms, the current flow through your 1,000,000,000 ohm volt meter will impact the measured voltage drop by a factor of 1/10,000,000,000. The wire is still in the circuit, 99.9999999999% of the current is still flowing on that wire, and the voltage drop is still proportional to the current * resistance of the wire.

Theoretically yes, of course it would - if you could measure that - but you can't do it without creating a bypass that isolates part the circuit and taking a measurement of an isolated part of the circuit.

In which case you are only measuring what is happening on the newly created bypass circuit, not what is happening over on the complete circuit without the bypass.


Which is why I say people often fool themselves with their meters. They take readings at various places, each time creating a bypass circuit with their meters and reading what is happening on the bypass circuit.

They see a higher voltage in one place, and a lower voltage somewhere else, which reinforces the false notion that there are end points in the middle of a circle, and that voltage drops between these end points.

Which also reinforces the dogma that wire resistance always drops the voltage regardless of the situation.



I blame the internet solar gurus. They sink their teeth into this idea of voltage drop due to wire resistance, and then apply that idea across the board to any situation that involves a wire. Even when the wire resistance results in a different net effect.

And then that gets parroted endlessly until everyone and his mother thinks it's gospel.

 

[dwh]

Pretty sure we said the same thing.

Here is what I said:

Theoretically yes, of course it would - if you could measure that - but you can't do it without creating a bypass that isolates part the circuit and taking a measurement of an isolated part of the circuit.

And then you:

Take your volt meter and measure from one end of one conductor to the other - the observed potential difference is the voltage drop across that wire.

What you just described is exactly what I described - taking a measurement of a part of the circuit, not measuring the entire circuit.


And yes, you can do that, but...


The next thing I said was:

In which case you are only measuring what is happening on the newly created bypass circuit, not what is happening over on the complete circuit without the bypass.

In other words, if you do what you described and get some reading, say 0.5 or whatever, that's only telling you what is happening on that section, it's not telling you anything else.

For instance it isn't telling you the operating voltage of the circuit.

So you end up with:

(?) - 0.5 = (?)



Perhaps it's the word "bypass" that's the problem? Should I have described it as "a section" or some such?

 

[ajmaudio]

I gotta say.. this thread has been interesting. I'm with all the people who understand that an amperage or wattage drop is the result of a voltage drop. You cant cheat the physics... V=IR .... P=IV etc .. they are all connected... there is not special case for solar charging a battery etc. This is actually a very simple case as a battery is essentially a purely resistive load. Also..if one wanted to prove it you could simply meter the wire in circuit and there will be a voltage there.. that is the measured amount of drop. IE... place probe at panel positive, and the other where the same wire enters the controller. You will me measuring the voltage drop across that length of the wire, which is half of the trip. All that said I hope the OP is getting what he needs from all of this?

 

[Rando]

You cannot "measure an entire circuit" as the voltage drop around any circuit is always zero. Of course a volt meter is only telling you the potential (voltage difference) between two points, that is what a voltmeter does. There is no 'operating voltage' of a circuit, the voltage at every node in the circuit will be different. Measure the voltage across each of the four elements in this circuit - the panel, one length of wire, the battery, the second length of wire, add them up and they will equal zero. As the lengths of wire get longer (or skinnier) their resistance increases, and the voltage drop across each wire increases. The battery voltage is approximately fixed as it is mainly a function of SOC, therefore as the voltage drop across the wires increases, the solar panel voltage increases (it is seeing a higher load resistance) moving to the right on the solar panel I-V curve, the current decreases and the battery charges (slightly) less quickly.

This is all Circuit Theory 101. Just because there is a non-linear device in the circuit (the solar panel), doesn't invalidate Kirchhoff's laws, or Ohm's law.

As for the equation, it is: Vpanel - Vwire1 - Vwire2 = Vbattery or to use example numbers 14.7 - 0.5 - 0.5 = 13.7 . Look at the I-V curve posted earlier, raising the solar panel voltage from 13.7 - to 14.7 V could drop the charge current by 0.5A or so. This get backs to my original point - yes there is a voltage drop, yes it impacts the charging rate, but no it is not a major impact.

Pretty sure we said the same thing.

Here is what I said:




And then you:



What you just described is exactly what I described - taking a measurement of a part of the circuit, not measuring the entire circuit.


And yes, you can do that, but...


The next thing I said was:



In other words, if you do what you described and get some reading, say 0.5 or whatever, that's only telling you what is happening on that section, it's not telling you anything else.

For instance it isn't telling you the operating voltage of the circuit.

So you end up with:

(?) - 0.5 = (?)



Perhaps it's the word "bypass" that's the problem? Should I have described it as "a section" or some such?

 

[dwh]

"PWM Charging:

Traditional solar regulators featuring PWM (Pulse Width Modulation) charging operate by making a connection directly from the solar array to the battery bank. During bulk charging when there is a continuous connection from the array to the battery bank, the array output voltage is ‘pulled down’ to the battery voltage. The battery voltage adjusts slightly up depending on the amount of current provided by the array and the size and characteristics of the battery."

And again,

"Using a nominal 12V system as an example, the battery voltage will normally be somewhere between 10 – 15 VDC. However, 12V nominal solar modules commonly have a Vmp(STC) of about 17V. When the array (having Vmp of 17V) is connected to the batteries for charging, the batteries pull down the output voltage of the array. Thus, the array is not operating at its most efficient voltage of 17V, but rather at somewhere between 10 and 15V."

http://www.morningstarcorp.com/wp-c...WM-vs-TrakStar-MPPT-Whitepaper-March-2015.pdf


Solar at battery voltage.

Not at battery voltage minus (voltage_drop_value).

Solar and battery both at battery voltage. Voltage drop = ?

 

[Bbasso]

It's a single cable, so he'd need two. From the image it looks like a low strand count, probably not very flexible and certainly not intended to be rolled/unrolled a bunch of times.

Omg... so he buys two. As for the strand count... find another cable.
I've undone my MC4 cables by hand a few times, not hard.

 

[Rando]

"PWM Charging:

Traditional solar regulators featuring PWM (Pulse Width Modulation) charging operate by making a connection directly from the solar array to the battery bank. During bulk charging when there is a continuous connection from the array to the battery bank, the array output voltage is ‘pulled down' to the battery voltage. The battery voltage adjusts slightly up depending on the amount of current provided by the array and the size and characteristics of the battery."

And again,

"Using a nominal 12V system as an example, the battery voltage will normally be somewhere between 10 – 15 VDC. However, 12V nominal solar modules commonly have a Vmp(STC) of about 17V. When the array (having Vmp of 17V) is connected to the batteries for charging, the batteries pull down the output voltage of the array. Thus, the array is not operating at its most efficient voltage of 17V, but rather at somewhere between 10 and 15V."

http://www.morningstarcorp.com/wp-c...WM-vs-TrakStar-MPPT-Whitepaper-March-2015.pdf


Solar at battery voltage.

Not at battery voltage minus (voltage_drop_value).

Solar and battery both at battery voltage. Voltage drop = ?

So some marketing material from Morningstar doesn't happen to mention voltage drop (which we all agree is a second order effect in most situations), therefore physics doesn't apply in this situation? This is really becoming an absurd discussion - the wires have resistance and there is current flowing through the wires therefore there is a voltage drop on the wires.

P.S. as was explained previously the solar panel is at battery voltage plus voltage drop.

 

[chand-o]

So some marketing material from Morningstar doesn't happen to mention voltage drop (which we all agree is a second order effect in most situations), therefore physics doesn't apply in this situation? This is really becoming an absurd discussion - the wires have resistance and there is current flowing through the wires therefore there is a voltage drop on the wires.

P.S. as was explained previously the solar panel is at battery voltage plus voltage drop.

Agreed

And the voltage drop will increase as the load increases, upto the max (current) the panel can supply.

 

[goatherder]

WOW. LOTS of "information" in this thread.

OP, I think you had the right idea. Park your rig in the shade, park your solar panels out in the sun, park your battery (and not a small one) as close as possible to your panels, connect your inverter close to the battery w/ 2ga wire, and run 12ga extension cords to your fridge. Should be OK...but 100w panel aint alot of input.

Don't expect that fridge to run off a small battery all night in a hot climate. My 12v dorm fridge draws 350w when running, and when I left it connected to the two starting batteries in my diesel shortbus...it emptied the batteries in a day. And it weren't even hot out.

And think of electricity as water thru a pipe. Voltage is pressure, and current is the actual liquid moving. Running an inductive load (electric fridge motor) at the end of a 100ft extension cord ain't ideal. But it may work ok..for a while, long as you use good extension cords.

 

[ajmaudio]

So some marketing material from Morningstar doesn't happen to mention voltage drop (which we all agree is a second order effect in most situations), therefore physics doesn't apply in this situation? This is really becoming an absurd discussion - the wires have resistance and there is current flowing through the wires therefore there is a voltage drop on the wires.

P.S. as was explained previously the solar panel is at battery voltage plus voltage drop.

Thank you...

 

[e60ral]

WOW. LOTS of "information" in this thread.

OP, I think you had the right idea. Park your rig in the shade, park your solar panels out in the sun, park your battery (and not a small one) as close as possible to your panels, connect your inverter close to the battery w/ 2ga wire, and run 12ga extension cords to your fridge. Should be OK...but 100w panel aint alot of input.

Don't expect that fridge to run off a small battery all night in a hot climate. My 12v dorm fridge draws 350w when running, and when I left it connected to the two starting batteries in my diesel shortbus...it emptied the batteries in a day. And it weren't even hot out.

And think of electricity as water thru a pipe. Voltage is pressure, and current is the actual liquid moving. Running an inductive load (electric fridge motor) at the end of a 100ft extension cord ain't ideal. But it may work ok..for a while, long as you use good extension cords.

Dorm fridges have awful efficiency ratings, you should upgrade. OP has an ARB fridge, it is using much less than half that when running; they didn't specify the size but it probably has a max of <100 watts with a typical usage of maybe 40 watts when running.

 

[RonaldPottol]

Many years ago Real Goods had AC systems for getting your hydro power back to your house, because DC was kinda crazy over the distance. But that was for longer runs, and more power than you are talking about. It might be worth it, but I doubt it.

Sent from my Nexus 6P using Tapatalk

 

[rkfoote]

So in the end, I bought a cheap 6ft extension cord, cut in two, and wired in my SAE-2 to one end and MC4 to the other. (truck side is SAE-2)
Hot+Ground = DC+
Neutral = DC-

While camping I used a 16-gauge 100ft extension cord with my adapters and my PWM controller was reading up to 5.2Amp @ 13.5V off my 100watt panel (when the fridge was running). This is the most I've ever pulled off the panel (generally mounted to the roof rack parallel to the ground, aiming towards the sun always helps )