# Solar cell devices

Discussion in 'General Electronics Chat' started by Pyrdon, Jun 7, 2011.

1. ### Pyrdon Thread Starter Member

Mar 17, 2010
13
0
Hey,

I am interested in starting up some various hobby projects, and have gotten the interest for small solar cell driven devices. I have a masters degree in Computer Science and are currently working with FPGA and digital ASICS, but I want to increase my analog electronics knowledge.

Currently I am planning to make a cap with solar cells, capable of charging a cell phone or similar through USB.

I have been looking around for a electrical model of a solar cell and found http://www.pvresources.com/en/solarcells.php
This indicates that the maximum power output is achieved just at a little bit lower current than the short circuit current. Is this something that I should be thinking about when I design my circuit?

I think for charging a smartphone it would be enough with around 1-2 W. In my case, through the USB port at 5V. But there are various solar cells out there giving almost any voltage/current combination. I am assuming that the voltage needs to be regulated.

Which of the following methods are preferred to get the most charging power (less wastee?

1) 6V/250mA and using a linear regulator to get 5V. This would waste (6-5)*0.25 = 0.25 W.

2) 12V/125mA and using a switched regulator (buck converter?). Can I convert that to 5V and can that give me more than the 125mA I am putting in? Then I can normally get ~90% efficiency, right? That wastes 12*0.125*0.1 = 0.15 W.

Number 1) is alot easier, involving only a two capacitors and a regulator. Number 2) needs more components

In my application, does it matter if I would connect solar cells in series to get voltage or parallel to get current?

I have a number of other questions aswell, but I figured I could see what you say regarding this first

Thanx!

2. ### Dyslexicbloke Well-Known Member

Sep 4, 2010
546
34
What you are asking RE maximum output is essemtally called MPPT (Maximum Power Point Tracking)
In order to do this you will have to go with a switch mode power supply, two actually, one to manage thre array of cells, aquireing the maximum power by demanding 'some' current at 'some' voltage and the second to convert that to a usable voltage at a spaciffic level.

1. Small cell arrays will have limmited output so wht do you do if there isnt enough power to satisfy the phone chargers demands.

2. The best buck/Boost converters will only manage 90% efficiancy so two in series will only be 80% eficiant and that takes no account of the power requirements of the control circuitry necessarry to manage the MPPT algorthm.

I suggest you charge a super capacitor or even a battery and control over voltage with a shunt regulator, so no voltage drop when its off.
Enable the output, to the drvice being charged, when your cap voltage is slightly below the shunt thresghold and turn it off again when you are just above the minimum acceptable voltage for the device.

If you do that you will essentally store power untill you have enough for a burst of device charging and any exess in direct sun will be wasted by the shunt load.

Of course you could still use MPPT to charge thre cap/bat but for a small simple charger you will probably find that it is overkill.

Just a couple of thoughyts, hope they help
Al

3. ### Pyrdon Thread Starter Member

Mar 17, 2010
13
0
Hey and thank you for the answer!

Keeping that in mind, I guess that kind of optimization for power is nothing I want to put my hands on.
I would however try to get the maximum power acchievable, due to the fact that I want to keep it on my cap/hat or whatever you call it.

17% efficiency solar cell => 170W/m^2 =~> 11*11cm for 2W. Figuring I won't get the full 17% and losses in the conversion to 5V makes me think that the required cell would be around 16*16cm instead, which is perhaps a little bit too big.

Reading up a little bit on shunt regulation, I found that they could easily be created by a zener diode and a emitter follower. Is this the kind of regulation that you are referring too? This is actually according to the same theory as using a linear regulator as the extra power is just wasted? Which one is really preferred? I would prefer not to destroy my new phone
For this reason, would it also be better to use a solar cell of 6V/250mA than 12V/125mA?

What about my question if I use a switched mode supply? Will two equal solar cells in terms of power give me the same power output or should I still look for (or connect them) parallel cells giving high current/low voltage?

4. ### Dyslexicbloke Well-Known Member

Sep 4, 2010
546
34
The diference between a shunt and a series reg is what happens when you only have a limmited voltage or a small amount of power ...
If your reg is in seriese, switch mode or linear, then it will always drop some voltage and take some power. A 78xx or LM317 for example, both linear, need about 2 volts more than you are trying to regulate to before they will work. Look up dropout voltage for the above.
A shunt reg essentally dose nothing untill it sees an input above its setpoint, just like a zenner, and then shunts exess power to a dump load reducing the system voltage.

In your case if the cells were producing 5.5v at the current you need to charge the phone then exess current would be shunted to the dump to drop the voltage to your 5v setpoint.
In the same situation a 7805 reg wouldnt work at all because you would need 7V input to regulate to 5V. It my just stop or it may start to oscillate but whatever it did it wouldnt be good for the phone or the reg.

RE seriese and paralell cells ....
Rrmember the cells have their owne impedance which you need to considder.
Whatever arrangement you use it needs to be capable of supplying the current you need. a string of small cells may well be capable of generating suficiant voltage but they need to do it at your target current.
You can work out the impedance of a cell from its short circuit current which will be roughly proportional to its area.

If you had 5V worth of cells that delivered 300mA when shorted then they would have an impedance of about 16.7 Ohms.
So if you are drawing 150mA you will only have 2.49V accros your load because the other 2.51V will be dropped acros the cell its self.

Essentally you dont buy a 5v cell to run a 5v device because it simply isnt going to work unless the cell is huge and the required current is tiny.

If you have a look around at say unregulated 12v solar trickle chargers you will see that the open circuit voltage is usuly well in exess of 15v that translates into a volt or so to lift the battery terminal voltage and the rest dropped accros the cell at a few hundred mA.
Working with solar cells is not like working with abtteries, cell impedance is a huge concern and has to be designed for, although in low porer systems it can oftern mitigate the requirement for a regulator as in the example above.

Al

Dec 26, 2010
2,147
302
It does not work quite like that. The output impedance of a solar battery is very far from being a linear resistance - it is a diode characteristic. A current drain of half the short-circuit current would typically give an output voltage considerably more than half the open-circuit value.

Increasing the load current from zero has little effect on the output voltage initially, but as the optimum (maximum power point) loading is approached the voltage falls more and more quickly. Beyond the optimum load current the output voltage falls quite precipitously.

6. ### Dyslexicbloke Well-Known Member

Sep 4, 2010
546
34
I stand corrected .... I also apriciate the correction.

As my signiture states I have gained most of my knolidge from experimentation so its is flawed in some cases.

The comments RE small solar cels were based on my owne dabblings, trying to get useful power from tiny arrays, and I clearly missed a fundamenttal point.

Interesting article, which I am sure Pyrdon will find useful as well ... Thanks

Al