Solar Circuit Killing Battery Pack

Thread Starter

Mukkles

Joined Feb 14, 2017
6
Be gentle, I'm still new to circuit analysis...so I'll run my project by you keen folks and perhaps you can give some advice.

I've built the second schematic in question here...right now there is no lights attached because I wanted to testing the charging circuit side. Now, from what I've gathered, because I don't have the lights connected there will be a drain from the battery to the solar panel despite there being a diode on the Emitter side. So right now, if I charge the batteries all day but leave the circuit alone all night, the batteries get drained. What I'm gathering from this circuit is because it's a PNP transistor, the current is flowing backdown through it at night (or low charge) through the 470 & 1K resistors assisting in draining my battery pack (perhaps through the panel as well?). But if I had my LED's connected with the small 47R resistor, then in theory the solar panel & voltage divider would not be draining the batteries, only my lights. Does that seem like a fair assessment? I've also swapped out the BC557 with a 2N3906 since I don't have any BC557's.

Say I wanted to outright prevent any current back feeding through the PNP if there was no lights, would a Schottky be a good way to prevent that with minimal voltage drop across that end? What about using a Schottky instead of a 1N4001, it would eliminate a "soft" transition from on to off correct?

Thanks in advance :)
 

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wayneh

Joined Sep 9, 2010
16,400
So which circuit are we talking about?

You could fix these problems by using a MOSFET switch instead of a BJT. And yes, a schottky will give you less power loss from the ∆V across the blocking diode.
 

Thread Starter

Mukkles

Joined Feb 14, 2017
6
It's the second one that I have built.

Ok, I'll look into how to implement a MOSFET instead of the BJT (I didn't care for how the PNP worked in this circuit anyway). I just wondered if there was merit to having a sharper on/off of the Schottky vs the standard.
 

wayneh

Joined Sep 9, 2010
16,400
I haven't thought much about this but you might look into using a zener diode above the gate of the MOSFET, with a resistor - say 10K - to ground below it. The resistor would hold the MOSFET off unless/until the voltage at the panel exceeds the zener voltage plus the drop across that resistor, which needs to provide the gate voltage. Just thinking out loud. I'm sure this has been solved before. Somebody that knows may come by soon.
 

Bernard

Joined Aug 7, 2008
5,686
It seems that at night without LEDs that base current would be around 10 mA, not enough to drain charged batteries ? P-MOSFET should be logic level.
 

ronv

Joined Nov 12, 2008
3,770
It's the second one that I have built.

Ok, I'll look into how to implement a MOSFET instead of the BJT (I didn't care for how the PNP worked in this circuit anyway). I just wondered if there was merit to having a sharper on/off of the Schottky vs the standard.
It needs to be a logic level FET so it can turn on with the 4.5 volt battery minus the diode.
Depending on how many leds you are driving you can make the drain on the battery much lower by making the base resistors larger.
 

Thread Starter

Mukkles

Joined Feb 14, 2017
6
Actually your original circuit, minus the 1k resistor, ought to work fairly well. Did you try that?
I haven't yet. The odd thing with this current circuit is I'm getting 50% of the battery voltage at the solar panel leads (I guess because of the PNP). I'll try removing the 1K resistor and applying a load and see what happens.
 

Thread Starter

Mukkles

Joined Feb 14, 2017
6
Too late to edit, but would I want to put a diode inline with the 470R to prevent a voltage going back through the dark panel?
 

wayneh

Joined Sep 9, 2010
16,400
Too late to edit, but would I want to put a diode inline with the 470R to prevent a voltage going back through the dark panel?
No, a BJT requires a base current in order to pass the larger collector-emitter current through to the load. You can however experiment and probably cut that current in half or more. Keep raising the resistance until your LEDs dim.

And it's never too late to edit. Maybe for newer members.
 

Thread Starter

Mukkles

Joined Feb 14, 2017
6
No, a BJT requires a base current in order to pass the larger collector-emitter current through to the load. You can however experiment and probably cut that current in half or more. Keep raising the resistance until your LEDs dim.

And it's never too late to edit. Maybe for newer members.
Thanks for all the help. I really like how responsive and helpful this community is...I'll mess around a bit with this over the weekend and see what I can do with it.
 

ScottWang

Joined Aug 23, 2012
6,963
You better in series with a resistor for each led, otherwise the leds may damage sooner than usual, and you also need to recalculate the values of resistor.
 
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