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Topic: Charge capacitor using higher voltage (Read 12277 times) previous topic - next topic

Pavouk106

BillO: Some more information - I have tried to connect 56 ohm resistor to the solars and the voltage dropped from 6,5V (open) to something like "almost nothing" and current through resistor was at "almost nothing too". Cells provided 5,5V @ 25mA for Arduino at the same light intensity (when I disconnected resistor and connected Arduino back).

Can I measure them better? I have two digital multimeters and one analog (really cheap) and some parts like resistors and LEDs (rated 30mA).

Thanks for your calculations, have read them at work. I will have to go through them again at home. I will probably go for something between 150 and 330 ohms and increase cutoff voltage to get at least 5V on capacitor. I can calculate this myself thanks to you ;-) Thanks.

BillO

#31
Jul 10, 2013, 05:25 pm Last Edit: Jul 10, 2013, 05:58 pm by BillO Reason: 1
Put one (1) of the cells in full sunlight, connect a 100 ohm resistor across it, then measure the voltage.  Also measure the voltage without the resistor.  Let us know what you get.

Also, that capacitor is a complicated device.  I would suggest taking one of the scenarios and trying it out.  It may not indeed be impossible to tell exactly what will happen before you do some experiments as the specs in the datasheet are intended for an entirely different application.

Try the scenario with the 150 ohm resistor.  Just make sure you have a 500mw 5.6V zener across the capacitor and program the Arduino to cut off at the 5.2V and to begin charging a 4V.  This will be safe.  Make measurements of voltage across the capacitor and see what the final voltage is a few seconds after charging has stopped.  After that, use a 100 ohm resistor to discharge the cap as you keep measuring the voltage to make sure it switches back on a 4V.  See how long each part of the cycle takes (charge time, discharge time).

Pavouk106

Will do.

I won't propably measure times sooner than on weekend :-( Voltage on one solar with 100 ohm resistor should be here tommorow around 11AM (I'm 5:30PM now).

BillO

#33
Jul 10, 2013, 06:46 pm Last Edit: Jul 10, 2013, 06:51 pm by BillO Reason: 1
I have to apologize to everyone who is following this.  I made a algebraical error in my first attempt.  Here are the calculations again.

The error I made was in estimating Io = (Vi - Vo) / Rt, when in fact, Io = (Vi - Vo) / (Rt - Rc) = (Vi - Vo) / (Ri + Rs)

Given this, Rs = 50 is not longer valid as Vo will exceed 5.5V slightly, but for a lengthy period.  So, doing the calculations again, this time correctly the differences are not huge, but....

For the following calculations, we define:
Vc = Final voltage across capacitor (to be calculated).  
Vo = Voltage across capacitor at cutoff
Vi = Open circuit voltage of solar cells
Io = Current at cutoff
Rc = internal resistance of capacitor  = 30 ohms, from the capacitor datasheet (should be measured)
Rs = Added series resistance
Ri = Internal resistance of solar cells = 3 ohms (estimated)

Rt = Rc + Ri + Rs
Vi = 6.7V, measured
Vo = 5V, initial arbitrary value
Io = (Vi - Vo) / (Ri + Rs)
And finally Vc = Vo - (Rc * Io)


Case 1:  High charge rate -> Rs = 75 ohms.
Then:
Ri + Rs = 78 ohms
Io = (6.7 - 5) / 78 = 22ma
Vc = 5 - (30 * 0.02) = 4.35V

4.4V 4.35V is not enough for the final capacitor voltage. Using a cutoff voltage of 5.4 will yield:
Io = (6.7 - 5.4) / 78 = 16.7ma
Vc = 5.4 - (30 * 0.0154) = 4.9V

Now, for a quick charge rate use a 75 ohm series resistor, stop charging at 5.4V and use a 5.6V protection zener

Case 2:  Medium charge rate -> Rs = 180 ohms.
Then:
Ri + Rs = 183
Io = (6.7 - 5) / 183 = 9.3ma
Vc = 5 - (30 * 0.0093) = 4.7V

4.7V is not enough for the final capacitor voltage. Using a cutoff voltage of 5.2 will yield:
Io = (6.7 - 5.2) / 183 = 8.2ma
Vc = 5.2 - (30 * 0.0082) = 4.95V

So, for a medium charge rate use a 180 ohm series resistor, stop charging at 5.2V and use a 5.6V protection zener


Case 3:  Low charge rate -> Rs = 330 ohms.
Then:
Ri + Rs= 333
Io = (6.7 - 5) / 333 = 5.1ma
Vc = 5 - (30 * 0.0051) = 4.85V

if 4.85V is not enough for the final capacitor voltage. Using a cutoff voltage of 5.1 will yield:
Io = (6.7 - 5.1) / 333 = 4.8ma
Vc = 5.1 - (30 * 0.0048) = 4.95V

So, for a low charge rate use a 330 ohm series resistor, stop charging at 5.1V and use a 5.3V protection zener

Again, these are based on estimates for the internal resistance of the solar cells and the capacitor.  Both should be properly determined before actual values can be calculated.

Pavouk106

Thanks for recalculating.

Values promised are here:
1 solar cell (open circuit) - 6.95V
The same one with 2x 56 ohms resistors in series (112 ohms) - 5.8V

Is it possible to calculate internal resistance from this?

I will test other thing at weekend probably. Have to buy some parts and have some time for it through the day (when sun shines). Thanks so far.

dc42

#35
Jul 11, 2013, 11:22 am Last Edit: Jul 11, 2013, 11:23 am by dc42 Reason: 1
A solar cell does not really have a quantifiable internal resistance, even for a fixed level of sunlight. Or, to put it another way, the internal resistance of a solar cell varies greatly with load. See the V-I curve at https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcRtwzlv1Qu0dSJWYtH_gwRyMNL9Gm-PUHeBvWCo-MWhCjBYsfPK4g. It's probably more useful to know the short-circuit current in bright sunlight.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

Pavouk106

dc42: Well, I can try to connect various resistors and make more measurements and try to point out some kind of graph of my solar cell. That would be pretty good to know if I want to use them for anything in the future ;-)

I can measure the short circuit current. But I will do it at ~22 hours from now. The only time, I'm at home when sun is shining.

BillO


Values promised are here:
1 solar cell (open circuit) - 6.95V
The same one with 2x 56 ohms resistors in series (112 ohms) - 5.8V

Is it possible to calculate internal resistance from this?


As dc42 says, there is no single internal resistance to a solar cell.  They must certainly have resistance, but they also have a pretty hard current limit which is why I think he was asking about the sort circuit current.  Personally,  I do not think that will be a useful figure in this case.

All that said, solar cells do exhibit a fairly linear voltage/current line until they begin to approach their maximum power point and as long as you do not use them past that point I have found it useful to assign them a "characteristic" internal resistance, if you will.  Yes, it will be off at some points along the power curve, but it will be useful for us in this case.

So, your measurements tell us the cells will deliver about 52ma into 112 ohms.  That gives us an internal resistance of about (6.95-5.8) / .052 = 22 ohms at that point.  I was hoping to get a better agreement with the "5.5V, 100mA" spec.  However, that indicates an internal resistance of about 14.5 ohms (given your updated open circuit voltage).  The internal resistance should not appear to go up as the load increases, so I am now not inclined to trust those figures.  So, in order to get at least three points on the curve I am going to ask you to do do the same thing again at 56 ohms if you don't mind.  Same solar cell, full sunlight, measure the voltage across 56 ohms.  Also, please use the same meter.  I recall you said you had several.  Let's stick to just one throughout this.

Pavouk106

To be honest, I haven't believed in the written power rating right at the time I have ordered them (from that shop with extreme deals, you know ;-) ). But they were rather cheap and I will not have much use for them. Maybe some solar battery charger in future or something like that, just to leave it be and use it occasionally.

I will measure it again ASAP. I will use the same DMM (Pro'sKit MT-1280, don't know how good it is, but it's the best from those three) and will measure short circuit current, 56, 100, 150 and 330 ohms - I should have all these - then I will have some idea about solar cell's output power. I could also try to measure two 56 ohms in paralel to increase load.

Numbers should be here at around the time they were today.

polymorph

Very often, sellers give the open circuit voltage with the shorted current. Which is meaningless for telling you what you'll get out.
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - https://tinyurl.com/Technote8
Multitasking: forum.arduino.cc/index.php?topic=223286.0
gammon.com.au/blink - gammon.com.au/serial - gammon.com.au/interrupts

Pavouk106

#40
Jul 12, 2013, 01:42 pm Last Edit: Jul 12, 2013, 01:51 pm by Pavouk106 Reason: 1
polymorph: looks like this one is just as you said.

I have made many measurements with the same one solar cell and the same resistors and multimeter at all times on direct sunlight. At start (9AM, from angle), solar was giving me 6,9V open, so I made measurements. But at 11:30 (solar on the same place, but this time facing sun much better), I was reading open voltage at ~6,65V. I really don't understand that :smiley-eek: Well, I had to re-measure everything, because I was getting way off readings compared with the ones from 9AM.

Attached should be table with measurements and graph like the one that on dc42's link. These are measurements taken within 10 minutes around 11:30 AM with solar facing sun pretty well. Looks like I can get 0,4W out of it, that is not really bad, considering the questionable quality.

BillO

Okay, good data.

From this I would characterize the internal resistance at around 24 ohms for the area of interest for one cell and about 6 ohms for all four paralleled.

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