AC vs. DC for solar powered fridge

[rkfoote]

Ok, here's the setup. I've got an ARB fridge and a 100 watt panel. I'm using the main battery on the truck. Solar panel is rack mounted and removable.
The truck will be parked in the shade so I can put the solar panel out in the sun. Let's say 50 - 100 ft away.

The DC voltage drop is too​ much unless I use some fairly heavy wire. If I put a small battery and inverter with the panel, I can bump up to 120v AC for the 100ft run to the truck (less than 1% drop with 16awg vs 35% drop @ 12v).

So, does the parasitic draw and the 12v - 120v - 12v conversion out way massive cabling? I can't be the first to try this, and I'm having trouble finding no-load ratings on 12v inverters. (And all this before taking about pure sine vs modified sine).

Discuss

 

[Bbasso]

With my limited knowledge I'd lean towards 10ga pure copper multi-strand MC4 cables in what ever length you need before trying to use an inverter setup.

 

[pugslyyy]

Ok, here's the setup. I've got an ARB fridge and a 100 watt panel. I'm using the main battery on the truck. Solar panel is rack mounted and removable.
The truck will be parked in the shade so I can put the solar panel out in the sun. Let's say 50 - 100 ft away.

The DC voltage drop is too​ much unless I use some fairly heavy wire. If I put a small battery and inverter with the panel, I can bump up to 120v AC for the 100ft run to the truck (less than 1% drop with 16awg vs 35% drop @ 12v).

So, does the parasitic draw and the 12v - 120v - 12v conversion out way massive cabling? I can't be the first to try this, and I'm having trouble finding no-load ratings on 12v inverters. (And all this before taking about pure sine vs modified sine).

Discuss

So if I am understanding you correctly, you want to have a solar system feeding an inverter which is powering your fridge - all completely separate from your batteries? That seems... suboptimal. Either the solar is going to be generating more power than you need (in which case you are just throwing those extra electrons away), or less (in which case the fridge won't run). Why not use the solar panel to charge the battery, and the battery to run the fridge?

 

[dwh]

The DC voltage drop is too​ much unless I use some fairly heavy wire.

There is no voltage drop.

With a PWM charge controller, the solar is trying to push the charging loop voltage up to its Vmp (probably around 18v), but it can't because the battery is limiting the voltage on the charging circuit to whatever the battery voltage is. So the solar panel is limited to operating at battery voltage instead of Vmp. The voltage of the entire charging loop will be at battery voltage - no voltage drop.

With MPPT the charge controller breaks the charging loop in two and allows the solar to operate at Vmp instead of being limited to battery voltage, which is more efficient and harvests more watts. But there is still no voltage drop.

Starting from a fixed maximum voltage, say the 12.8v of a fully charged battery, there is nowhere for the voltage to go but down. So there is voltage drop feeding a load, like a fridge, from a battery.

Charging is a different situation because you have a supply (charger or charge controller) trying to force the voltage higher and negating the drop.



Feed the fridge from the battery. Use the solar to replenish some (or maybe all) of what the fridge draws out.

For a100w panel, you could use 100' of 12 gauge Malibu low-voltage wire and be fine.

 

[ajmaudio]

There is no voltage drop.

With a PWM charge controller, the solar is trying to push the charging loop voltage up to its Vmp (probably around 18v), but it can't because the battery is limiting the voltage on the charging circuit to whatever the battery voltage is. So the solar panel is limited to operating at battery voltage instead of Vmp. The voltage of the entire charging loop will be at battery voltage - no voltage drop.

With MPPT the charge controller breaks the charging loop in two and allows the solar to operate at Vmp instead of being limited to battery voltage, which is more efficient and harvests more watts. But there is still no voltage drop.

Starting from a fixed maximum voltage, say the 12.8v of a fully charged battery, there is nowhere for the voltage to go but down. So there is voltage drop feeding a load, like a fridge, from a battery.

Charging is a different situation because you have a supply (charger or charge controller) trying to force the voltage higher and negating the drop.



Feed the fridge from the battery. Use the solar to replenish some (or maybe all) of what the fridge draws out.

For a100w panel, you could use 100' of 12 gauge Malibu low-voltage wire and be fine.

There is a voltage drop...

The voltage of the panel will indeed be brought down to battery voltage, but to bring it up you need current. More current into battery the faster its voltage comes up, which is the goal. The battery is essentially a resistive load to the panel, thus the drop that occurs in the lead from the panel to to the charger will affect how fast you can get that work done. You dont see it the normal way because the panel acts as a current source but the loss is there and should be kept to a reasonable minimum. A solar panel cannot "negate the drop" as it is not an active device. Any losses incurred in the wiring are just that... losses. This is mainly an issue in the bulk phase of charging a lead acid battery. Once the voltage of the battery has reached the set point for the absorption phase the current into the battery will be limited to keep the voltage constant. Current will be dropping at this point and the losses in the wiring will be decreasing in proportion to that drop. btw..100' of 12 gauge on a 100 watt panel toward the end of bulk phase would be about a 10 percent loss at full panel output.

 

[dwh]

There is a voltage drop...

The voltage of the panel will indeed be brought down to battery voltage,

If the panel is operating at battery voltage, then that's a voltage limit, not a voltage drop.

thus the drop that occurs in the lead from the panel to to the charger will affect how fast you can get that work done.

The drop that occurs is a drop in current, not a drop in voltage. And yes, it can certainly make charging take longer.

You dont see it the normal way because the panel acts as a current source but the loss is there and should be kept to a reasonable minimum.

Define reasonable. He says a 6a PV module. You figure a 10% drop. That's .6a. Times 6 hours of good sun, he's lost 3.6 amp hours into the battery.

Is that unreasonable in his situation?

If he had flat mounted the panel instead of ground deployed, he'd lose more than that.

So even if he did lose 10% from 100' of wire, he'd still get more hours of good sun so six of one, half dozen of the other.

A solar panel cannot "negate the drop" as it is not an active device.

It has a higher voltage potential, so as long as it's operating, it is *actively* trying to raise the voltage.

Any losses incurred in the wiring are just that... losses.

Sure, but not losses due to *voltage drop*.

This is mainly an issue in the bulk phase of charging a lead acid battery. Once the voltage of the battery has reached the set point for the absorption phase the current into the battery will be limited to keep the voltage constant. Current will be dropping at this point and the losses in the wiring will be decreasing in proportion to that drop.

Correct.

btw..100' of 12 gauge on a 100 watt panel toward the end of bulk phase would be about a 10 percent loss at full panel output.

10% loss of current, not voltage.

 

[ajmaudio]

current and voltage are directly tied and codependent .... ohms law here. However you look at it there is loss as you say.. that was my point... And if OP doesnt want 10 percent loss during good output then 100' of 12 gauge = bad idea. If thats an acceptable loss in his system and affords other benefits then it will certainly work without ill affect. Also... going through an inverter to power the fridge will generate more loss in the system and is not worth it imo just to get to thinner wire which I believe is what the OP was asking about? There are conversion losses from ac to dc and vice versa, some inverters better than others.. but there will be losses. A good solar setup charging a deep cycle battery which is then powering the cooler directly should be the most efficient path here.

 

[dwh]

Efficiency is a Holy Grail that all the gurus on the solar forums chase endlessly, down to the 10th of a percentage point. And if speccing out a system for a fixed location, I do it too.

But for vehicles, it's almost impossible anyway, so it's not worth getting headaches over. Most are flat mounted, which is inherently inefficient. Those that are ground deployed may or may not get moved during the day to track the sun, and are usually so small to begin with (~100w) that all they really achieve is to supplement the charging that the alternator​ does while driving.

Meh.


Trying to run an inverter directly off the PV/charge controller without a battery in the circuit to buffer the input to the inverter is just a bad idea to begin with - regardless of the AC-DC conversion inefficiency.


So yea, fridge to battery, solar to battery is the way to go. Which I think I already said once in this thread.

 

[e60ral]

The drop that occurs is a drop in current, not a drop in voltage. And yes, it can certainly make charging take longer.

Wires have voltage drop, they have a resistance/ft rating

 

[dwh]

Wires have voltage drop, they have a resistance/ft rating

And that resistance sometimes causes a drop in voltage, and somtimes a drop in amperage.

But a lot of people just assume that the resistance of the wire always drops the voltage.

Then they get their panties in a twist when I say, "Nope, no voltage drop in this particular situation".


Situation A:
There is a battery trying to hold a particular voltage, say 12.8v. The voltage of any circuit fed by the battery can never be higher than that, that is the maximum potential. Create a circuit​ with a load on it, and the voltage on that circuit will always be less than 12.8v. In this situation the resistance of the wire results in a reduction in the voltage of the circuit.

Situation B:
There is a supply source trying to force the voltage of a circuit to rise, but it cannot because it is limited by a battery in the circuit. The voltage of the circuit would be higher if it weren't for the limit, because the supply is trying to increase the voltage. In this situation, the resistance of the wire results in a reduction in the amperage flowing through the circuit.


In one situation, you start at 12.8v (or whatever) and pull the voltage down against a supply that tries to hold it up. Voltage drop.

In the other situation, you start at 12.8v (or whatever) and push the voltage up against a resistance that tries to hold it down. Amperage drop.

Pulling the voltage down causes voltage drop. Pushing the voltage up doesn't.

 

[e60ral]

No, that is not how it works, your voltage will always drop across the wire because the wire is a resistor and P=VI and V=IR and P=I[SUP]2[/SUP]R

In your second scenario the voltage at the start of the wire for your supply is still higher than the voltage at the and of your supply wire. You are never pushing voltage up across a wire.

Wanting to limit voltage drop by not having a long lead is reasonable although the inverter solution is not a good one, the real solution is using the appropriate gauge wire for your acceptable voltage drop.

 

[Joe917]

Both situations cause wattage drop.

And that resistance sometimes causes a drop in voltage, and somtimes a drop in amperage.

But a lot of people just assume that the resistance of the wire always drops the voltage.

Then they get their panties in a twist when I say, "Nope, no voltage drop in this particular situation".


Situation A:
There is a battery trying to hold a particular voltage, say 12.8v. The voltage of any circuit fed by the battery can never be higher than that, that is the maximum potential. Create a circuit​ with a load on it, and the voltage on that circuit will always be less than 12.8v. In this situation the resistance of the wire results in a reduction in the voltage of the circuit.

Situation B:
There is a supply source trying to force the voltage of a circuit to rise, but it cannot because it is limited by a battery in the circuit. The voltage of the circuit would be higher if it weren't for the limit, because the supply is trying to increase the voltage. In this situation, the resistance of the wire results in a reduction in the amperage flowing through the circuit.


In one situation, you start at 12.8v (or whatever) and pull the voltage down against a supply that tries to hold it up. Voltage drop.

In the other situation, you start at 12.8v (or whatever) and push the voltage up against a resistance that tries to hold it down. Amperage drop.

Pulling the voltage down causes voltage drop. Pushing the voltage up doesn't.

 

[dwh]

the voltage at the start of the wire for your supply is still higher than the voltage at the and of your supply wire

Sorry, but that's a common misconception.

A circuit is a loop, a circle. A circle has no ends. Voltage doesn't drop from one end of the wire to the other end of the wire - it drops around the entire loop.

People often fool themselves by the way they take measurements with their meters. They measure at one place and see a lower voltage than measuring at another place, and that makes it appear that voltage dropped "from one end to the other end".

But that's a (common) mistake in interpreting the measurements. A circle has no ends.

 

[dwh]

Both situations cause wattage drop.

Yup. But one manifests as a drop in voltage, the other as a drop in amperage.

 

[rkfoote]

Thanks to everyone that replied.

It sounds like the consensus is mostly KISS (keep it simple stupid) with a Solar -> Controller -> Battery -> Fridge where the "long" lead is between the solar and controller at a decent gauge (assume you want the controller/battery/fridge as close as possible).

The background on all this is a camping trip I do on a regular basis, truck is parked in the trees (campground) and the sun is available 100ft away or so. I had considered using extension cords to extend the solar. Basically build a small dongle adapter. Something like this: http://gridridder.com/buy-hardware/custom-parts/mc4-adapter-plug/ (but custom build since $45 seems overly pricey)

I can then build the male/female ends and get a 100' 10 gauge extension cord (and I can use it for power tools when not camping).
And yes, I understand the safety concern when someone tries to "plugin" my solar panel to 110v :Wow1: (I'd wire the + side of the panel to ground and hot, and the - side to neutral if someone accidentally plugged into an outlet it would blow the breaker as it's shorted to ground).