Hi lovely All About Circuit Community,

This is my first post, so I'm new to this super community and also pretty new to electronics. So please don't be too harsh on me.

About my litte project:
I'm trying to build a solar powered audio amplifier. It should also include batteries, so i can keep listening even after the sun went down.
So far I have build (with the help of Instructables and Co.) an audio amplifier based on the LM386 (see the green part in the schematic below). It works fine during daytime, when I plug the solar panel straight to the amplifier. But of course, as soon as the sun is gone, the sound is gone as well. That's why I'm trying to include some batteries, which should be charged while the sun is up and when the sun went down the batteries should be used as power supply for the amplifier (see the red part in the schematic below).

So far I read quite a bit about charging batteries and pretty much everywhere is written that it's a really difficult topic and you should rather just buy a quickcharger, instead of making your own. That's probably true, but I would like to build my own anyway, because I want to understand what's going on. I'm also aware of the fact that "my" charger won't be as efficient or good for the batteries as one you buy, but at least I made it myself.

I made my plan for the charger based on the "Solar Charger Circuit" from http://electroschematics.com/4746/solar-charger-circuit/ . I pretty much just changed the values.

The plan is to use 5 x AA batteries, which should have a voltage of around 6V if connected in series. I read that you can charge a AA battery with 1,6V and around 0.2C. Assuming that my batteries gonna have a capacity of 1200mAh, I could charge them with 240mA and 8V (5x1.6V). To achieve the 240mA the resistor R5 needs a resistance of 33Ohm and the 8V can be achieved by adjusting the LM317. When the batteries are fully charged they'll have a voltage of >6.2V and the Zener-diode will conduct and provide base current to T1, which will then ground the ouput of the LM317.

Would this more or less work or am I totally wrong? Sorry, but I'm quite new to electronics.

Thank you so much for any response.

 

When it is dark, the battery discharges through the charger because the diode is not connected between the charger and the battery.

For the charger to charge the battery to 8V then the output of the LM317 must be 8.7V.
Then the minimum input to the LM317 must be 10.5V which is not produced by the 9V solar panel.

The volume control is connected backwards and is shorting the signal source. The slider should connect to pin 3 of the LM386.

With a 6V supply, the output of the LM386 into one 8 ohm speaker is only 200mW at clipping. Into two 8 ohm speakers in series like you have the total power is only 116 mW (58mW in each speaker) which is almost nothing.

 

A .1 μF cap from pin 4 to pin 6 of the LM386 is often required to prevent power noise from corrupting the audio. If you are going to use the amplifier while charging the batteries, this might be needed.

 

Assuming that my batteries gonna have a capacity of 1200mAh, I could charge them with 240mA and 8V (5x1.6V).

Most AAs you'll find at Walmart et al are more like double that capacity.

IMHO, you should experiment with using nothing more than a single blocking diode. At full charge, your panel voltage (less the diode drop) is barley more than the batteries (and you could experiment with adding another cell). The current flow under those conditions may be so small as to be well tolerated by your batteries. Adding another diode would be less wasteful of power than the voltage regulator, and by dropping another 0.6V would further cut the current if you need to.

If you can't get the current down enough at full charge, then I would consider adding in your current dump circuit, but I'd ignore the regulator. Your batteries will hold the voltage down.

 

First of all Thanks for all the fast replies.

For the charger to charge the battery to 8V then the output of the LM317 must be 8.7V.
Then the minimum input to the LM317 must be 10.5V which is not produced by the 9V solar panel.

Ok, but if I would have a 12V panel it would be fine?

The volume control is connected backwards and is shorting the signal source. The slider should connect to pin 3 of the LM386.

I think I actually soldered it right, but draw it wrong in the schematics. Or at least it is working.

With a 6V supply, the output of the LM386 into one 8 ohm speaker is only 200mW at clipping. Into two 8 ohm speakers in series like you have the total power is only 116 mW (58mW in each speaker) which is almost nothing.

These are actually 2 x 3,2Ohm speakers, but you're right that it gets very noisy.

A .1 μF cap from pin 4 to pin 6 of the LM386 is often required to prevent power noise from corrupting the audio. If you are going to use the amplifier while charging the batteries, this might be needed.

Thanks, I'll keep it in mind if I'll get any noise.

Most AAs you'll find at Walmart et al are more like double that capacity.

IMHO, you should experiment with using nothing more than a single blocking diode. At full charge, your panel voltage (less the diode drop) is barley more than the batteries (and you could experiment with adding another cell). The current flow under those conditions may be so small as to be well tolerated by your batteries. Adding another diode would be less wasteful of power than the voltage regulator, and by dropping another 0.6V would further cut the current if you need to.

If you can't get the current down enough at full charge, then I would consider adding in your current dump circuit, but I'd ignore the regulator. Your batteries will hold the voltage down.

Thanks, that is a very helpful reply. You mean something like the schematic below? I changed the value of the batteries to 2400mAh, so they can be charged with 480mA. Can I just put a 18Ohm resistor after the diode to limit the current to 460mA (8,3v/18Ohm)? I also added a switch, so I can run the amp with the full 9V from the solar panel without "wasting" the 0,7V in the diode. Is this ok?

 

Your batteries won't be at 0V so the voltage drop across the resistor is more like 8.3-5=3.3V (assuming deeply discharged batteries). I very much doubt your panel can output more than 480mA into 5.7 volts, and therefore the resistor is not needed. Of course you should measure to confirm. I'll bet it'll be less than 200mA and therefore on scale of a cheap ammeter.

A switch to bypass the diode is not worth it, IMHO, once the resistor is out of the picture.

 

Energizer and a Japanese Ni-MH battery manufacturer say that overcharging a Ni-MH cell should be avoided to stop high temperature damage and over-pressure damage.
They say that trickle-charging current is safe at or below the capacity of the battery divided by 40.
Battery charger ICs sense when the battery is fully charged then shut off.

 

They say that trickle-charging current is safe at or below the capacity of the battery divided by 40.

Since that's only 60mA for a 2400mAH battery, there's a fair chance the OP will need a circuit to avoid overcharge. He'll have a much better idea once he measures the current from the panel into his battery pack. He may just need a dump circuit like he showed earlier, if the excess current is only 100mA or so.