can i use a solar panel without using battery

[Sunking]

So, you're all telling me that if I want to run my 300 watt fridge during the "daylight" and I have 1000watts of solar panels, I can't use that to run my fridge with no battery in between? Off-grid.

Max you just do not understand the simple math problem I told you earlier. For a stand alone system, meaning batteries only, the energy from the panels has to have a place to go. Where that energy goes depends upon demand by the batteries state of charge, and the connected equipment load.

It can be 12 noon with bright sun, if the batteries are fully charged and no load demand by the equipment there is no power from the panels going anywhere. It is just not being used and all that potential energy is completely wasted..

If at 12 noon the batteries are discharged, and no equipment load, then current will flow to the bateries based upon the state of charge of the batteries.

But to answer your question you cannot do what you want to do. With a stand alone system you have to have batteries period. Thios batteries have to be sized at least 8 to 10 times larger the the charge current. If not you will burn them up. So if you are ignorant enough to run 5000 panel watts into 12 volt batteries will produce a maximum charge current of 5000 watts / 12 volts = 416 amps. That means you will need 10 x 416 amps = 4160 Amp Hour batteries. Put another way a battery that weighs 2700 pounds.

Change it to 24 or 48 volts same answer it takes a 2700 pound battery. Otherwise it goes BOOM! That is all you need to know for now.

[themaxx69]

So, you're all telling me that if I want to run my 300 watt fridge during the "daylight" and I have 1000watts of solar panels, I can't use that to run my fridge with no battery in between? Off-grid.

[russ]

Think this would be enough to not need batteries!

Nope - they have the grid to serve as a battery

[themaxx69]

Think this would be enough to not need batteries!

my house.JPG

[Mike90250]

...But simply looking at that graph, why would I need batteries if my power usage fell within those parameters?

That graph is in minute increments. See all the spikes on the left side, That's the power you can use, if you have a 4.5KW array (which is what the graph came off) The starting surge of everything, has to stay below the red line at all times.

Or just try it, and see what happens. Things may work ok for a while.

[themaxx69]

For what I want, that graph looks pretty good to me. Even though with my initial findings, I still think panels on a tracker provide more energy earlier and later. That I will need more time and more measurements.

But simply looking at that graph, why would I need batteries if my power usage fell within those parameters?

[Mike90250]

That answers some of my questions. And gives me more to think about.

May be this should have been my first question, can I have an off-grid PV system with no batteries. 100% of electrical needs will be covered by the panels.

Nope. The sun is too variable, and you would have to "overbuild" the system to run the 200w fridge, by nearly 150%, to supply starting surge, and capacity to keep it running while a cloud, haze, airplane shadow, or anything that affects the brightness of the sun. Otherwise, your gear will only run for about an hour per day (at solar noon)
5-13Chart.jpg

[themaxx69]

I didn't mention this before because to me it seemed like common sense, but weather I get a 3, 4, or 5kw system, it will be broken down to 2-3 smaller system. Just seems to be more logical to have say two 2kw arrays instead of one 4kw, or three 1kw arrays instead of one 3kw. Each designed for specific equipment, for the most part.

[themaxx69]

That answers some of my questions. And gives me more to think about.

May be this should have been my first question, can I have an off-grid PV system with no batteries. 100% of electrical needs will be covered by the panels.

[Mike90250]

I got an idea since no one is answering the question I asked 3 times.

I want a 5KW system. I know I won't use all that and only a fraction of that most of the time. And 99% of my electrical needs will be during sunlight hours and will fall way below 5KW and it's sunny every day and no clouds. Can I use say 2 Trojan L16RE-B for safe measure? That's only about $600 for batteries. I can live with that.

Simple question.

I'll have to make assumptions, since you have not given (in one place I can find it) all the info we need for the BEST answer, so here's a half assed answer:

L-16's a are about 400AH batteries, 2 in series would give you a 12V 400ah bank.

Proper charging would be 30 - 40 amps, to provide enough power to "stir" the electrolyte to pervent it from stratifing in the tall batteries.

Your inverter, will be powering unknown loads. While the sun shines ?

First problem:
5Kw of PV yields 333amps @ 15V
That will need 6, 60A charge controllers to manage the power Yilkes !

2nd problem:
Huge wires to carry the 333 amps You won't even be able to lift the wire.

3rd problem:
Much more esoteric, and problematic. Your large loads this 5Kw array will be powring , will need a HUGE inverter @ 12V. While you might find something on fleabay that lists 5Kw, it can't really exist, as the weight of the copper traces to carry 333amps, is probhitive. If we ignore that issue, we still have the 120hz ripple current, that the inverter imposes on the puny 400ah bank. It will charge & discharge the batteries at 120 hz, as the inverter draws power. Overall, the PV array will keep the batteries charged, but using the batteries as a capacitor, with 100 amps flowing in & out 120 times a second, they will heat up, shed plate material, and die, in a couple days. If your loads are less than 5KW, your 120hz ripple currents will be less.

4th problem:
If you run your batteries down 30% some evening (70% charge remaining), the high current the array can provide in the AM, when the sun comes up, will cook the batteries, max charge rate for the batteries should be in the 50-60A ballpark. Your 333 amps is much larger.

Scaled differently, a 5KW array in to:

24 v system = 180 amps, and just 3, 60 amp controllers (4ea, L-16 batteries, or you could easily use smaller, 200a golf cart batteries, and still have the same KWh storage as the 2 L-16s )

48 v system = 90a and only 2, 60a controllers, or one 80A controller may do it. (8, 6v batteries needed)

Higher voltage systems, give you lower system amps, lower losses, less parts count, except for batteries.

[russ]

I got an idea since no one is answering the question I asked 3 times.

I want a 5KW system. I know I won't use all that and only a fraction of that most of the time. And 99% of my electrical needs will be during sunlight hours and will fall way below 5KW and it's sunny every day and no clouds. Can I use say 2 Trojan L16RE-B for safe measure? That's only about $600 for batteries. I can live with that.

Simple question.

Simple answer? No

How have you defined and measured your loads? A strange situation and usually the bulk of power is consumed in the evening hours

[Mike90250]

Isn't this what a 'shunt' does?
Scenario: Your battery is full. The sun is shining brightly on the panels. The power being produced at that point by the panels is being 'shunted?' directly to any load you may be using. Completely avoiding any battery charge or discharge....

No, a shunt is a calibrated resistance used to measure amps

You are almost describing a "diversion load" or "opportunity load".

[themaxx69]

I got an idea since no one is answering the question I asked 3 times.

I want a 5KW system. I know I won't use all that and only a fraction of that most of the time. And 99% of my electrical needs will be during sunlight hours and will fall way below 5KW and it's sunny every day and no clouds. Can I use say 2 Trojan L16RE-B for safe measure? That's only about $600 for batteries. I can live with that.

Simple question.

[Sunking]

I'm totally new to this although might be researching a similar situation which may answer the ops question.

Isn't this what a 'shunt' does?

In terms that you are using it no a shunt is just a precision high wattage resistor used to measure current.

[TnAndy]

Hi guys,

I'm totally new to this although might be researching a similar situation which may answer the ops question.

Isn't this what a 'shunt' does?

Scenario: Your battery is full. The sun is shining brightly on the panels. The power being produced at that point by the panels is being 'shunted?' directly to any load you may be using. Completely avoiding any battery charge or discharge.

Disclaimer: The load would have to be less that equal to what the panel was generating at that time.

Disclaimer 2: I'm a total newb and this is my 2nd post after my introduction. Be gentle

Disclaimer: I'm not an electrical engineer, nor do I play one on TV.

A shunt can have quite a few purposes, but in a solar system, the most often use is to measure a flow of current you can't easily measure directly due it being a really HIGH current. In this case, the shunt is actually a resistor of a known value, and the meter reading in Amps is actually measuring a small voltage across the shunt, the reading value proportional to the current flowing at the time, and shows on the meter as "Amps".

The flow of power in a PV system is to the load...no load, no flow.

In a battery PV system, power flows from the panels to the charge controller, then to the batteries IF THEY NEED IT, then on to whatever other loads are connected to the system. The batteries are first in line of the list of needs.....the system is designed to take care of them as it's "primary mission".....as best I understand things.....which, self admitted, may be limited.