Solar panels produce their rated open circuit voltage and short circuit current with the rated amount of power. Where the load falls on the V-I curve determines the power output.

 

So I played with this in some good light I was getting almost enough to run a R pie 90 milliamp at 17v but it was left open circuit for a while and dropped quickly back down to about 75. I had the Multi meter set to the 200 ma setting. This thing I am making is mad of pannels meant to trickle charge 12v batteries. I am thinking that at peak I should be able to buck it down and have enough. for now I am just going to hook up a battery and see where its at.

I got one of these and I am going to wire it to a 12v charger I grabbed for the auto parts store.

https://smile.amazon.com/dp/B095SF992Q?psc=1&ref=ppx_yo2_dt_b_product_details

 

Solar panels produce their rated open circuit voltage and short circuit current with the rated amount of power. Where the load falls on the V-I curve determines the power output.

If I want to increase current I probably need to put some resistors in parallel. Now the question is do I need a higher rated resistor or lower rated resistor.

 

Where will you connect resistors "to increase the current"? Series resistors reduce current and parallel resistors waste current.
You need bright continuous sunshine on the solar panels and have them face the sun to produce more current.

 

Where will you connect resistors "to increase the current"? Series resistors reduce current and parallel resistors waste current.
You need bright continuous sunshine on the solar panels and have them face the sun to produce more current.

Ok if you say so. I = v/r right So if I have extra V I should be able to increase I. I am pushing like 17v. So if r goes down I goes up parallel resistors do that right? The other way is what a transistor?

 

You want 19V at a fairly high current but instead you are using low current solar panels in the dark to charge a 12V battery that feeds a voltage booster. Then the current is much too low.
You do not want to throw away some of the low current to heat some resistors.

If you increase the voltage of the solar panels then their total overloaded voltage will drop and their current might increase only a small amount.
You need much more current.

 

For a DC-DC converter: Power In = Power Out - switching / resistive losses.

By reducing the voltage at the output via a buck converter will yield an increase in available amperage at the output (ignoring losses). Likewise, increasing voltage at the output via a boost converter will come at cost of more amperage at the input. So it's a matter of trading one for the other. Thus the only way to get more power at the output is to increase the power at the input. There exists more ways to trade voltage for amperage and vise versa but the above formula must always be obeyed.

For example, I have a boost converter which operates to less than 1V. As the voltage slowly drops at the input (draining battery), the device must slowly draw more current to maintain a steady voltage at the output.

 

This curve shows the typical response of a solar panel. Two important data points are the short circuit current and the open circuit voltage. The point of maximum power delivery happens at a very specific point between those points. Higher voltage and lower current or lower voltage and higher current will deliver less power.

 

I think he said that his solar panels are indoors behind a window and with no sunshine so the current is very low.

 

I think he said that his solar panels are indoors behind a window and with no sunshine so the current is very low.

No not really (I assume) you kinda scanned the post. When I went back and looked another time during better light I got a better result sadly though I am still no where near the current I think the Laptop will need. For now the plan is to continue with the laptop but my expectations are to eventually abandon that and settle for a R-PI.

THE REAL POINT of this whole thing is a learning exercise for me. Basically I am going to create a renewable BMS system throght trial and error. I want to learn allot more about powering things with solar and wind etc. Thanks for the responses they have all been helpful.

For a DC-DC converter: Power In = Power Out - switching / resistive losses.

By reducing the voltage at the output via a buck converter will yield an increase in available amperage at the output (ignoring losses).

From what your telling me I will get a increase in amperage. That is good. however is there any formula that would allow me to estimate this? Managing expectations I don't expect to get 5 amps fom these little battery tinder solar panels even with a buck converter. It would be nice to know how much though. what I can do is wire up a couple extra cells my self and drop those into the mix if I am way to short. None the less you answered my most important question and gave me direction thank you.

 

No not really (I assume) you kinda scanned the post. When I went back and looked another time during better light I got a better result sadly though I am still no where near the current I think the Laptop will need. For now the plan is to continue with the laptop but my expectations are to eventually abandon that and settle for a R-PI.

THE REAL POINT of this whole thing is a learning exercise for me. Basically I am going to create a renewable BMS system throght trial and error. I want to learn allot more about powering things with solar and wind etc. Thanks for the responses they have all been helpful.



From what your telling me I will get a increase in amperage. That is good. however is there any formula that would allow me to estimate this? Managing expectations I don't expect to get 5 amps fom these little battery tinder solar panels even with a buck converter. It would be nice to know how much though. what I can do is wire up a couple extra cells my self and drop those into the mix if I am way to short. None the less you answered my most important question and gave me direction thank you.

Formula for using a buck converter in general (voltage source input) or using one with a solar panel? If the former case then you use the formula I presented but if you are talking about the latter it becomes more complicated because the nature of solar panels is such that the open circuit voltage and closed circuit amperage that you measure while unloaded will change as the panel is loaded. Like motors (inductive loads), solar panels have complicated characteristics that place them in a class of their own which requires unique analysis which is beyond my scope. As you may have noticed a solar panel will show a near full voltage in any light condition but will produce little to no current unless well lit. Plus a DC-DC converter needs some current of its own just to turn on.

If you have a buck converter on hand I suggest hooking it up to some.voltage sources and different loads to first understand it's characteristics as well as studying the theory. And if you are asking how to apply the formula I presented I can walk you through that as well but my experience is limited compared to what other users here can advise you on so my solutions may not be practical.

 

You said you measured only 30mA from your two paralleled solar panels that are rated for a total of 13W at 12V which is 1.08A. That is 36 times lower than rated. You said "it is winter and the light was indirect".
The rated power from a solar panel is when it is with the sun directly above it (in the tropics where there is no winter), at noon and on a sunny day. Frequently a solar panel is turning and facing the sun all day but yours is on and behind a window and is not even pointing up.

In the morning, afternoon and night the battery powers the laptop then the solar panel power must be a lot more than double for the laptop to operate 24/7. Also, don't you have cloudy days with no sunshine?

If the laptop needs 19V then the 12V from the solar panels that drive the voltage stepup converter need their current increased 19V/12V= 1.6 times more plus even more current for efficiency losses.

 

You said you measured only 30mA from your two paralleled solar panels that are rated for a total of 13W at 12V which is 1.08A. That is 36 times lower than rated. You said "it is winter and the light was indirect".
The rated power from a solar panel is when it is with the sun directly above it (in the tropics where there is no winter), at noon and on a sunny day. Frequently a solar panel is turning and facing the sun all day but yours is on and behind a window and is not even pointing up.

In the morning, afternoon and night the battery powers the laptop then the solar panel power must be a lot more than double for the laptop to operate 24/7. Also, don't you have cloudy days with no sunshine?

If the laptop needs 19V then the 12V from the solar panels that drive the voltage stepup converter need their current increased 19V/12V= 1.6 times more plus even more current for efficiency losses.

Is there an approximation used to estimate the average power output of a panel over 24 hours / 365 days a year? Clearly it's dependent on latitude, cloud cover, angle of the sun from noon and snow coverage but I imagine someone has calculated the mean / median / mode of common consumer panels over lifetime. It seems the rated output as you described is a useless quantity most of the time as many users like myself have been woefully disappointed at the actual output over the coarse of the day

 

Everybody I know use city electricity and do not live "off-grid".

 

Search for the average insolation. This data shows the average solar irradiance and may include the effect of average clould cover.

To learn about solar cell characteristics, you might perform a small experiment. With the solar panel exposed to a constant bright light source, measure the open circuit voltage and short circuit current. Then using some resistors, rated for the appropriate power dissipation, measure the voltage and current at maybe 10 points across the range and create a curve like the one I posted above.

From there, calculate the power produced at each point. You'll find the maximum power isn't deliver at the maximum voltage or current, but at a condition somewhere near the middle of this curve. This point is the maximum power delivered for the light source you're using, and operating at greater or less voltage reduces the power output possible.

That point is the maximum power available to the DC-DC convertor × an efficieny of 75% – 85%. If you increase the voltage, the current available will be less. More current means less current. Power = volts × amps and there isn't any way to get more power out than you have to put in.

 

That point is the maximum power available to the DC-DC convertor × an efficieny of 75% – 85%. If you increase the voltage, the current available will be less. More current means less current. Power = volts × amps and there isn't any way to get more power out than you have to put in.

I know your lying man I know you have the seceret to quantum power just tell the common folks the truth MAN. JK LOL

I just wanted to know if I would increase current and by roughly about how much. Your comment was helpful thank you.

 

I just got the little charge manager working the other day by hooking the panel's up direct to the battery for a day. For what ever reason the battery was to low to even power the thing.

https://smile.amazon.com/Sunway-Solar-Controller-7A-Maintainer

The eventual plan is to draw from what ever source is good at the time with full expectation that the total power wont keep the laptop up and running. Then start adding new components till I get to at least a raspberry pie 24/7. Before that the circuit that I have to learn to build is the bridge between the battery and the Load be it laptop or R pi.

First I need to know everything is working proper. The Gel acid battery was from an old jumper that stopped working and while I am 90% that the little bugger is good but its been siting on shelf for a few years at this point. If its not I am going to replace it with a big Lithium ION.

 

Your results are obvious:
1) You have little solar panels in the dark producing almost no current.
2) You have an old lead-acid battery that was never upkept. A topping-up charge must be applied to the battery every 6 months but you did not do it so your battery is finished.

 

Your results are obvious:
1) You have little solar panels in the dark producing almost no current.
2) You have an old lead-acid battery that was never upkept. A topping-up charge must be applied to the battery every 6 months but you did not do it so your battery is finished.

I have said this a few times now. This is an educational exercise. Its not that results oriented.

I have said this twice now. I went back to the panels the other day in better light and got some better results higher voltage more milli amps. It wasn't terribly high but better, enough to play with. Certainly enough to try and build a buck circuit or power a buck converter. So I can stick the Multi meter on it and learn.

If I wanted to just get the result I would buy a bunch of solar panels hang them all over the dam place punch a hole in my wall and an I am sure i could get enough going to power a whole room. If the battery is dead Ill get a new one. If the solar panels are not enough I have extra to compensate.

I have a box of these begging to be wired up and hung in the window.

https://smile.amazon.com/gp/product/B0895XHPKT/ref=ppx_yo_dt_b_search_asin_image?ie=UTF8&psc=1

so no worries.

 

I'm not an expert but I do feel you are jumping ahead without a solid foundation in theory in each of the three categories of which will be combined in the end design:

1) Power produced by a panel is intermittent and unreliable

2) Energy storage is complicated

3) Line and load regulation is complicated

It's for these reasons I've placed my free energy projects on hold as there is significant theory to be understood. If you intend to use commercial charge controllers and DC-DC converters then things get easier but if you are like me and want to build from scratch then there is a lot to consider. Setting up 1, 2 and 3 each separately is relatively simple but once you consider how to interpret and implement feedback a proper analysis becomes critical. I've made topics like yours in the past and have gotten similar replies basically saying not to do it because they know from experience it is a long and hard road ahead especially if you are not willing to extensively educate yourself, observe, calculate and experiment.

That being said if I were you I would look into the theory before moving ahead and in the meantime wire up your panels and load them with simple loads such as a fan or LEDs and at the very least identify the dangers involved for whatever you decide to do.