Problem with solar panel charger

I have built a circuit that is supplied by 12Volt, 560mA, solar panel, I have modified the solar charger circuit you see on the attachments (solar-charger-550x274), and I conclude to the circuit (modified_2). I did this from suggestions I got, and because the "solar-charger-550x274" didn't work. In position D
I have connected a Sealed Lead Acid battery that supplies a working circuit for my project.

I have connected the INA219 module (on place D) of the foto.



The problem is that when I plug:

a) The 12 volts power supply (A,B) I get: 7,07Volt, 0 mA, 0 mW
b) The 12 Volts power (A,B) & 6Volt Sealed Lead Acid battery (D),I get 6,81Volt, -1407,8mA, 9782,00 mW
c) The 6Volt Sealed Lead Acid battery (D), I get: 5,98Volt, 480,00 mA, 2842 mW

The Sealed Lead Acid battery is connected in parallel with the power supply.

The LM317 goes really hot, although I have mounted a heat sink.
The "modified_2" is the circuit diagram.


What should I do?

Alternatively, can I use a simple step down DC/DC converter with 12 Volt input from the solar panel, 6 Volt output connected to the battery and charge it?

Can the 12 Volt, 560mA solar panel charge the 6V, 4Ah Sealed Lead Acid battery and supply the rest of the circuit?

Thank you...

The photo and schematic do not match:

There is no transistor in the photo,
the required potential divider across the LM317 isn't correct,
D1 ('E' I think) is across the 12V instead of in series with it although it serves no useful purpose anyway.
R2 is missing,
R3 is missing,
ZD is missing.

The design should work if built correctly but do bear in mind that the while idea of the LM317 is that is converts excess input voltage to heat so you should fit a heat sink to it. As shown in the photograph it certainly will never work.

Brian.

You should use a buck type regulator instead of the inefficient LM317. But how much current your solar panel can deliver?

c_mitra said:

You should use a buck type regulator instead of the inefficient LM317. But how much current your solar panel can deliver?

Click to expand...

It can give up to 560 mA

- - - Updated - - -

betwixt said:

The photo and schematic do not match:

There is no transistor in the photo,
the required potential divider across the LM317 isn't correct,
D1 ('E' I think) is across the 12V instead of in series with it although it serves no useful purpose anyway.
R2 is missing,
R3 is missing,
ZD is missing.

The design should work if built correctly but do bear in mind that the while idea of the LM317 is that is converts excess input voltage to heat so you should fit a heat sink to it. As shown in the photograph it certainly will never work.

Brian.

Click to expand...

The initial circuit ("solar-charger-550x274.jpg") is correct according to your opinion?

Thank you

If you fix the output at 6.6V (after the diode) then it should work fine even without the transistor and the zener.

The regulator shall dissipate around 3W and you can use a heat sink.

c_mitra said:

If you fix the output at 6.6V (after the diode) then it should work fine even without the transistor and the zener.

The regulator shall dissipate around 3W and you can use a heat sink.

Click to expand...

What do you mean? What should I do to fix the output?

What do you mean? What should I do to fix the output?

Click to expand...

I thought the VR in the circuit is used to adj the output voltage.

In fact, you can just adj the 317 output using fixed resistor and forget about everything else.

Your load (I guess a lead acid battery) may draw too much current (that will appear as a short to the regulator) and that may shut down the regulator but it will slowly get up (unless the battery is really down)

The initial circuit ("solar-charger-550x274.jpg") is correct according to your opinion?

Click to expand...

Well, it will work but it is hardly optimal.
As pointed out, for lead acid batteries, the normal charging method is to use a constant voltage a little higher than the rated battery voltage. The LM317 is an adjustable voltage regulator so you can use it to provide that voltage without the transistor, Zener and associated components. Look at the data sheet for the LM317 and it will tell you the formula to calculate the values. Also note the requirement for capacitors at the input and output pins of the LM317.

For example, if the battery should be charged at say 6.4V (check the battery specification for exact value) you could use a 240 Ohm resistor between 'Vout' and 'Adj' and a 1.1K resistor from 'Adj' to ground. That would produce about 7V from the regulator, remember you lose about 0.6V in the series blocking diode.

There is some merit in using a transistor like that to drop the voltage under high current, such as if the battery is completely discharged, so it doesn't overload the regulator but your circuit doesn't do that.


Brian.

Because I have spent too too much time to make this circuit work, I think of buying a ready solution.

If I use a 12 Volt, 40 Watt solar panel and I want to charge a 6V,4Ah Sealed Lead Acid battery and feed the rest of the circuit that the Sealed Lead Acid battery supplies. Can I use this product for this job?

**broken link removed**

Thank you...

No, it only works with 12V or 24 batteries and probably needs more solar power than 40W to work properly.

Just use the LM317 circuit in the data sheet or get yourself an adjustable voltage switch mode supply that can be set to battery charging voltage. For a 6V battery this is usually between 6.5V and 7.2V depending on whether you are trickle charging it or fast charging it:

(extracted from SGS-Thomson data sheet)
The second circuit is better but both will work. If you use the first one, change R2 to 1.1K so it produces 6V instead of 12V.

Brian.

Thanks a lot. If on the Figure 11, I use a 40 W, 12 Volt solar panel (instead of 7 Watt, 12 Volt I use know), it would charge the battery faster right?

Probably not, the rate the battery charges is mostly decided by the battery itself. Provided the current source (solar panel) can supply the current it needs, there is no advantage to using a bigger one. The power rating of a PV panel is a measure of how well it can hold its output voltage under a larger load, a higher power panel will still produce the same voltage under a heaver load. It follows that if the battery tries to draw too much current, a smaller panel may not produce enough voltage but a larger one might still work.

However, the LM317 will not allow more than about 1A to flow through it which is enough for most 6V lead acid cells and the 7W panel should be able to produce that much in direct sunlight. The second schematic (fig.11) has current limiting which also helps to prevent overloading the panel. Instead of connecting a fixed voltage across the battery, it raises it slowly from about 1.5V up to a maximum of 6V as the battery charges, this is kinder on the battery and gives some protection if a damaged battery is connected or the output is accidentally shorted out

Brian.

betwixt said:

Probably not, the rate the battery charges is mostly decided by the battery itself. Provided the current source (solar panel) can supply the current it needs, there is no advantage to using a bigger one. The power rating of a PV panel is a measure of how well it can hold its output voltage under a larger load, a higher power panel will still produce the same voltage under a heaver load. It follows that if the battery tries to draw too much current, a smaller panel may not produce enough voltage but a larger one might still work.

However, the LM317 will not allow more than about 1A to flow through it which is enough for most 6V lead acid cells and the 7W panel should be able to produce that much in direct sunlight. The second schematic (fig.11) has current limiting which also helps to prevent overloading the panel. Instead of connecting a fixed voltage across the battery, it raises it slowly from about 1.5V up to a maximum of 6V as the battery charges, this is kinder on the battery and gives some protection if a damaged battery is connected or the output is accidentally shorted out

Brian.

Click to expand...

Thank you. Because, the circuit I supply with the Sealed LEad Acid battery consumes ~450 mA, and lets say around ~50mA dropout on the Solar charging circuit (figure 11) I get ~500 mA. My solar panel at peak gives me 560 mA, that means 560 mA - 500 mA = 60 mA current for charging the battery, which means (4000 mAh) / (60 mA) = 80 h = 3,5 days. That is too long, isn't it?? If I go to a better solar panel that gives me 1A (so that can pass through the LM317, as you say), then I go to 1000 mA - 500 mA = 500 mA, and I go to (4000 mAh)/(500 mA) = 8 hours, right? If I want to use a solar panel that gives more current (2A) is there a chance to change the LM317 with other LM??? so that can let more current to pass through it (in order to give the battery "space" if it can get charged faster), or I need to change completely the figure 11 and use other circuit?

Thank you for your time...

You have changed your specification. You started with a battery charger and that is what you showed in the schematic but now you say you need an additional 450mA to power something else as well.

If you want to charge the battery AND power the equipment at the same time, I would suggest you use two different power supplies and a steering diode between their outputs. A discharged lead acid cell may try to take several Amps and will easily pull the supply feeding it down to almost zero volts, using a single supply will prevent your other equipment being powered as well.

Use a current limited circuit like Fig. 11 to charge the battery, it will limit the charging current to about 0.6A (about 6W from the PV) and use a standard LM317 circuit fed directly from the PV to provide a stable supply for your equipment. If you connect a power Schottky diode between their outputs, cathode end to the equipment output, it will let the battery supply power to the equipment when the PV output drops below about 8V. There is a slight drawback that the diode will drop the battery voltage by about 0.4V and when the battery is providing power, R3 (in fig. 11) will be in series with it. There isn't much you can do about that in a simple circuit I'm afraid. You will need around 15W of PV power to supply your equipment and a fully discharged battery at the same time, anything bigger will work but not better.

Incidentally, the current drawn by the fig 11 is only about 2mA, far less than the 50mA you quoted.

Brian.

If instead of the Sealed Lead Acid battery that needs several Amperes to get charged, as you said, can I use another kind of batteries, Li-Ion batteries and this charger here:

https://www.ebay.com/itm/3S-25A-18650-Li-ion-Lithium-Battery-BMS-Protection-PCB-Board-12-6V-With-Balance/332212297624?epid=933998449&hash=item4d59655f98:g:-7gAAOSwzppaMl4L

with the 40 Watt, 12 Volt solar panel to both charge the Li-Ion batteries and supply the circuit?

Thank you, I appreciate your support. I am trying to make this project work for too many weeks, without success.

What are the power demands of the "other" circuit?

It needs max 450-500mA. I use three step-down in parallel to the supply ( I mean that each Step down DC/DC will be supplied with 12 Volt input from the output of the Li-Ion charger I posted above) the "other" circuit. But the total current the "other" circuit needs including the 3 Step-down DC/DC consumption is maximum 500 mA.

(The 500 mA is a measurement I got using the Sealed Lead Acid battery, when I supplied the "other" circuit)

https://www.ebay.com/itm/Pololu-5V-Step-Up-Step-Down-Spannungsregler-Voltage-Regulator-S9V11F5/272622211607?epid=735528740&hash=item3f798cd217:g:HCcAAOSwL8xcm5EX

Thank you...

Hi,

adwnis123 said:

Instead of the Sealed Lead Acid battery that needs several Amperes to get charged, can I use Li-Ion batteries and charger with the 40 Watt, 12 Volt solar panel to both charge the Li-Ion batteries and supply the circuit?

Click to expand...

I think - and the words are well-meant - 30 minutes, or at worst a few hours, "wasted" on IC manufacturer's parametric search tables can save weeks of "hobbyist" (like me) non-ideal or inappropriate or just plain bad designs. e.g. It's worth browsing solar li-ion charger ic on Analog, TI, ST, Maxim and so on - there are many ICs that take 90% of the headaches and frustrations out of the project.

Good luck!

Is this a solar charger or not? I am really confused...

https://www.ebay.com/itm/3S-25A-18650-Li-ion-Lithium-Battery-BMS-Protection-PCB-Board-12-6V-With-Balance/332212297624?epid=933998449&hash=item4d59655f98:g:-7gAAOSwzppaMl4L

Hi,

May I ask why you think it has something to do with "solar", or why do you think this is a "charger" circuit?

It is what the headline tells: "3S 25A 18650 Li-ion Lithium Battery BMS Protection PCB Board 12.6V With Balance"

Klaus