Hi there,

I am planning on creating a solar emergancy charger for my phone. The idea is that I lay the project in the sun, this will charge the internal batteries in the gadget. Then over night I plug my phone in to charge the phone. Also in the day I might want to charge my phone, while the solar pannel is charging the batteries of the gadget.

I already made a schematic for this. I have a few questions about this and I would appriciate some feedback.

Some additional info:
As batteries I will be using 4x rechargable NiMH 2000Mah AA batteries(those might be switched out for 2700Mah ones. (I have a bunch of those lying around)
The USB resistor in the diagram is actually a DC-DC Converter where I can plug my phone in. ( link )
The solar pannel has nominal values of 5.5V and 320mA. After the solar pannel there will be a 1N4001 Diode(forgot to draw this one). ( link )


My questions:
1: Will this work? Won't I blow up my batteries?
2: I used 2 dtdp switches in the diagram. Is this neccesairy? If I line the batteries up in 2 pairs of 2 batteries(2.4V, 4000Mah) and I plug my phone in WHILE the solar pannel is charging the batteries, what will happen?
3: Is it neccecairy to add an indicator for when the batteries are full so I won't overcharge them?


Thanks in advance.

Ray

 

1: Will this work? Won't I blow up my batteries?

Not well. The panel voltage may be 5.5V but that is likely the open circuit, or maximum-ever-seen voltage when perfectly aligned in full sun. The current spec of 320mA is likely the short circuit current, measured when the panel voltage is near zero. At the 5V that a USB device needs, this panel might produce less than 100mA. That's barely enough for many devices, and charging a phone usually needs much more. Once you add in the voltage drop across your blocking diode, you're well below 5V. The DC-DC converter might help you overcome this problem though. What is it?

2: I used 2 dtdp switches in the diagram. Is this neccesairy? If I line the batteries up in 2 pairs of 2 batteries(2.4V, 4000Mah) and I plug my phone in WHILE the solar pannel is charging the batteries, what will happen?

The switches are not needed if the circuit is properly designed otherwise. For instance the blocking diode will prevent any current from the batteries into the solar panel.

3: Is it neccecairy to add an indicator for when the batteries are full so I won't overcharge them?

An automated charge controller would be better. I don't think you'd enjoy watching this setup for hours on end to notice when it's done! It's possible your batteries are large enough to tolerate a constant trickle from that small panel. You'd have to do a bit of research and make a calculation. Especially if you get a larger panel, you'll need to protect your batteries from overcharge.

 

Not well. The panel voltage may be 5.5V but that is likely the open circuit, or maximum-ever-seen voltage when perfectly aligned in full sun. The current spec of 320mA is likely the short circuit current, measured when the panel voltage is near zero. At the 5V that a USB device needs, this panel might produce less than 100mA. That's barely enough for many devices, and charging a phone usually needs much more. Once you add in the voltage drop across your blocking diode, you're well below 5V. The DC-DC converter might help you overcome this problem though. What is it?

It's more like I want the solar pannel to charge the batteries in the gadget(the project). And make the batteries in the gadget charge my phone. Following this reasoning the switches would always be in the opposite direction of each other. I do now know what would happen if you directly put the solar panel on the DC-DC converter.

That's why I thought I put the solar panel charge 2 batteries and let the other 2 batteries decharge while powering my phone. When they run out, flip the switch so the first 2 batteries power the phone and the 2nd 2 batteries get charged by the panel.


In short:
1) If I leave the USB part out, will this charge my batteries?

2) If I power the DC-DC converter with the solar cell and the batteries what happens in those 4 cases?
2a) the solar cell provides (more than) enough power for the DC-DC converter and the batteries are fully charged
2b) the solar cell provides (more than) enough power for the DC-DC converter and the batteries are not fully charged
2c) the solar cell provides not enough power for the DC-DC converter and the batteries have juice left in them
2d) the solar cell provides not enough power for the DC-DC converter and the batteries have no juice left in them

What I think that happens(please correct me if wrong)
2a: nothing? Battery gets a bit hot?
2b: batteries get charged
2c: batteries jump in to fill the gap
2d: nothing happens? too less power
But if my answers are correct I still have no idea for example how much power there will be drawn from the batteries in case 2c, as I don't have any knowledge about 2 different currents running over a wire.

I hope you can follow me,
Thanks in advance,
Ray

 

It's more like I want the solar pannel to charge the batteries in the gadget(the project). And make the batteries in the gadget charge my phone.

That's fine but you cannot get around the fact that your panel is small compared to the energy you need to power your phone. Many hours of sunlight will be needed.

Following this reasoning the switches would always be in the opposite direction of each other.

I don't think there's much value to dividing the batteries into two groups unless they need to be physically separated. Using two groups does not allow you to capture or store more energy.

I do now know what would happen if you directly put the solar panel on the DC-DC converter.

No one knows without seeing the specs.

1) If I leave the USB part out, will this charge my batteries?

Yes, but it might over-charge them also.

It's hard to say more without a schematic. It's not a problem to "OR" together multiple sources of power as long as you prevent reverse currents. It's like filling your swimming pool with your hose and your neighbor's hose at the same time. As long as your pressure does not force water backwards into your neighbor's, both hoses contribute to filling the pool.

 

So if I create a circuit without switches, where the positive of the solar pannel is connected to the positive of the batteries and to the positive of the DC-DC converter, and the negatives are connected to eachother. Then put a diode near the solar panel. I should be clear right

Image to illustrate:

The solar panel will charge the batteries using the (what they told me was the) 'tickle method'. When I plug my phone in both the solar panel and the batteries will power it. Because we know the solar pannel is to weak to power it on its own. Perhaps I can drop in a on-off switch to make sure the DC-DC converter doesn't draw any power when there's no USB cable plugged in. From what we know this will work right?

PS. Ebay links of the solar panel and DC-DC converter are in the first post. Perhaps you can find some specs there, but as we know those sellers don't add much details to their items...

 

So if I create a circuit without switches, where the positive of the solar pannel is connected to the positive of the batteries and to the positive of the DC-DC converter, and the negatives are connected to eachother. Then put a diode near the solar panel. I should be clear right

Yes, although you still have no over-charge protection. Don't forget that blocking diode. See here for an example charger. Another. If you ask for recommendations in this forum, I'm sure you'll receive many other alternatives.

The solar panel will charge the batteries using the (what they told me was the) 'tickle method'.

Haha, perhaps you mean "trickle".

When I plug my phone in both the solar panel and the batteries will power it. Because we know the solar pannel is to weak to power it on its own. Perhaps I can drop in a on-off switch to make sure the DC-DC converter doesn't draw any power when there's no USB cable plugged in. From what we know this will work right?

Sorry I missed that link earlier. Cute and cheap little device, assuming it works. I think many projects like your's we see here could benefit from that device. You may not need a switch if it draws very little juice when not loaded. Maybe wait and see.

 

First of all, thanks for everything so far!

Yes, although you still have no over-charge protection. Don't forget that blocking diode. See here for an example charger. Another. If you ask for recommendations in this forum, I'm sure you'll receive many other alternatives.

Those over-charge protection circuits are fairly complicated right? My knowledge kinda ends where the 'bridge thingys'(compartiments with legs, such as the overcharge circuit) starts(I really have no clue what the official name is). In addition I want to fit it all in an altoids tin. Adding a NTC to measure the temperature of the batteries just feels like a bit too much. And I probably have no clue on what I would be doing. Any chance you could tell me how I use those controllers?

Kinda got my idea from these guides: linkeh

 

Why not just buy one. Cheaper than building it.

 

Building is more fun, I learn some stuff from it and I have some strange needs on my charger.

I want to be able to attach the solar panel to my backpack. Be able to just charge the batteries when I'm at home. This far this will suit my needs.

but 2 times a year I'm going camping for a week or so, and there is no acces to a power grid there. As I live in holland, I can't really rely on the sun.
Right now I have a bunch of rechargeable AA batteries lying around here, so I charge a few pair of AA batteries and use one of those AA filled emergancy chargers where I just swap the batteries when they run low.( one like this one ).

It doesn't really make sense to walk around carrying 2 different devices when I can build one that suits all my needs. In addition if something breaks I just buy a new part, where in the bought ones I probably would have to buy a entire new unit.

 

That's a cute project, and I love building things into Altoid tins! The circuit relies on natural forces to avoid overcharging the batteries, and that's probably fine if you don't foresee using a larger solar panel. I think you'll be happy with the results if you go ahead and build it. I'd worry about protecting the solar panel though, a backpack is a pretty rugged environment and these things are fragile.

 

Ok, so I bought 2x0.75W(5V-150mA) and I want to hook them up in series.

This would mean in the optimal situation it would supply 5V and 300mA. Now I need to put a diode there, will a 1N914 suffice? The max current is 300mA but thats only in the optimal situation. Or could I better use another diode with a higher max current.

Thanks in advance.

 

You can add the voltage in series or the current in parallel but not both.

 

Oh sorry, I meant parallel. Still the question stands:

This would mean in the optimal situation it would supply 5V and 300mA. Now I need to put a diode there, will a 1N914 suffice? The max current is 300mA but thats only in the optimal situation. Or could I better use another diode with a higher max current.

 

I'd use a bigger diode, even though that one "should" be OK. Running anything at its rating makes me nervous and increases the odds of failure. Some diodes - not sure about this one - fail to a short. That'd be a bad thing and it's just not worth taking any risk when the proper diode is just pennies. I'd use a schottky diode rated to 1A or more to minimize power loss in the blocking diode.

 

Just to be sure a 1N4001 will suffice right??

 

Yup, although it's silicon and not a Schottky. It will drop ~0.7V versus ~0.4V.

 

1N5822 then ?

 

That'd be fine, but it's up to you. Either diode would work.

The percent of power lost in the blocking diode is proportional to the voltage drop. At low voltages like this, even a few tenths of a volt is significant. Half a volt lost in the diode is 10% of all the power from a 5V panel. So I do think it's worth using the Schottky. It should give you several percentage points better performance for little cost.

 

I'm soldering it as we speak, I only got the wrong dpdt switch deliverd(got one that only transmits when I hold it to the left but automaticly gets back to 0). I have 1 that does what I want, just not the 2 as in the schematic.

Now can I leave the bottom switch out? What will happen when I have both the usb device connected and the solar pannel out, since you have 2 power supplies, 1 of 5.5V 320mA(solar pannel), and the batteries(2.4V ....A). Is it efficient or will it break something?

I have a diode soldered to my solar pannel.

Thanks in advance

 

You can only safely trickle charge NiMH batteries at 0.05C. That would be 10mA for your 2000mAH batteries. If you put 300mA into them without detecting termination, they will likely get hot and vent.

Bob