I have constructed 4 mini solar panels in series and on bright direct sunlight produces around 0.5 to 17 volts (very low milliamps). I would like to make an LDR circuit which would kick in the "charging" (the battery pack) as soon as the voltage hits 12 volts. I know that I could adjust the voltage through the resistors in the circuit pictured below.

My question is......where do I connect the positive and negative side of the battery in the circuit shown????? The picture depicts an LED coupled with a resistor (with a blue X) which will not be required. In addition, a low dropout diode is in series with the positive line from the solar panel so it will not drain the battery (although I believe that even then a .03v drop would occur).

Any suggestions or recommendations would be appreciated.

Thanks

 

Positive on the red line and negative to the collector of the transistor.

 

Thanks, that was what I was thinking and will try it out. If someone else has a better suggestion to improve the circuit I would most welcome.

Cheers.

"Nothing ventured, nothing gained"

 

So the voltage goes from .5 to 17 on the top and bottom rail of your circuit? Do you really need an LDR then? Couldn't you just sense the rail voltage itself? Perhaps you could do something like this (circuit attached). The downside is that R1 will draw (waste) a little current while the transistor is ON because the zener requires a little current to operate.

 

If you open up a simple garden solar light you will find the charging circuit consists entirely as a schottky diode.

That's in the older ones, the new ones have an application specific IC in them now (so you can't see that diode).

 

Or here's another idea. More parts, but you could adjust R2 to have some control over the amount of charging current.

 

As I understand the OP's circuit, its purpose is to provide less voltage drop when "on" than a simple blocking diode, to squeeze out a bit more efficiency. I doubt that it accomplishes that goal, and is likely worse.

The base-emitter current would need to be up to 10% of the charging current for the transistor to be fully on. That's a pretty big waste since that current is not going into the load. But any less base current leaves the transistor likely to be dropping more voltage that a diode would.

I'd be happy to be proven wrong. Let's see some data, such as the ∆V across the transistor during charging.

 

I am aware that there is a voltage drain using diodes and resistors. The purpose of the LDR is to shut off the circuit if it goes below a certain voltage threshold e.g. 12 volts and turn on above 12 volts. Understand also that am using small solar panels in series with small milliamp output although the peak voltage is 17 volts hence adding more resistors and/or IC's will definitely lower the voltage output and would be impratical. I will try some of your suggestions just to see what I come out with.

I was thinking though to bump up the current, I would do a simple ferrite (joule thief) with the solar panel as the power source, then to the charging battery.

Cheers