Timeline for Problems when powering a circuit directly from Solar Panel
Current License: CC BY-SA 3.0
32 events
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| Oct 15, 2019 at 7:30 | comment | added | Simon B | Can the solar panel actually produce enough current to start the motor you're using? Most motors require considerably more current to start them than they need to keep running. | |
| Oct 15, 2019 at 0:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
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| Jan 28, 2019 at 19:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 16, 2018 at 22:04 | answer | added | Bogdan | timeline score: 1 | |
| Aug 29, 2017 at 2:53 | comment | added | Frank Donald | @next-hack Had a thought of that. There will be a problem when the Voltage from solar panel falls below 5.1v breakdown voltage of zener diode isn't it? And another question the Schmitt trigger Vcc is powered directly from Solar panel and Vref is regulated to a point of 5.1v . Does it poses a issue since Vref and Vcc should be equal for setting up the thresholds. If it is, can i use a simple battery to for both Vref and VCC since current consumption will be relatively low? Kindly advice. Sorry if I am going round and round with this, thanks for your patience. | |
| Aug 28, 2017 at 16:44 | comment | added | next-hack | @FrankDonald, the circuit as it is now, has its thresholds that change as the Vcc changes (i.e. the voltage of the solar panel). Be aware of this, when you calculate the threshold! Please note also that the positive saturation voltage of that OP AMP is VCC-2V, which, on some MOSFETs, might be enough to make them slightly conductive. You could improve this figure either with an external pull-up, or using a resistor divider (toward VCC). Otherwise you should use the non-inverting schmitt trigger implementation, and use a NPN BJT to produce the correct voltage. | |
| Aug 28, 2017 at 5:30 | comment | added | Frank Donald | @next-hack That link really helps. Luckily i have got a 6v motor from a local store which will suit my purpose correctly. So instead of drawing 8v, 6v is my new requirement. I have drawn the circuit as per your suggestion imgur.com/a/1RTu6 . Have used a 5.1v zener diode for creating the reference voltage. Can you please kindly check the circuit and advice? P.S: I have made a rough calculation of the resistor values, just want to see the overall design is correct. | |
| Aug 27, 2017 at 21:45 | comment | added | Criticizing Israel not allowed | If the open circuit voltage is only 10.7V then you certainly won't get 12V when you connect the circuit. It will drop to something less than 10.7V (only way to know is to measure it). | |
| Aug 27, 2017 at 7:00 | comment | added | next-hack | @FrankDonald that circuit implies that you can create Vr. (If Vr=0, it won't work for you on a single supply, because one threshold will be 0). I suggest you this page (there you'll find an online calculator too): hyperphysics.phy-astr.gsu.edu/hbase/Electronic/schmitt.html By the way, using a 7808 regulator requires that the voltage at its input should be at least 10V, which is damn close... Instead of having too much trouble with electronics, consider the idea of winny, i.e. an integrated buck regulator with undervoltage lockout. | |
| Aug 27, 2017 at 3:21 | comment | added | Frank Donald | @immibis The open circuit voltage under good condition is 10.7v . Haven't tried powering the circuit directly from Solar panel to make that measurement. Just waiting for rainy climate to change before getting back to my testing. Will post the results once done. | |
| Aug 27, 2017 at 3:15 | comment | added | Frank Donald | @next-hack Haven't used a schmitt trigger before, kindly bear with me. If am gonna use this schmitt trigger goo.gl/zhaojh , the voltage from the solar panel under good conditions is around 10.7v. So I should be powering my circuit from this 10.7V right? I think its good to set the upper threshold voltage around 10.4 and lower as 10v for this trigger circuit. But am not really sure how to achieve this, can you please help me out by putting these things together in a single circuit? thanks in advance. | |
| Aug 25, 2017 at 7:07 | comment | added | Criticizing Israel not allowed | Did you measure the voltage and current coming from your solar panel when it's connected to the circuit? | |
| Aug 25, 2017 at 7:06 | comment | added | Criticizing Israel not allowed | AFAIK it's quite rare to permanently damage things from undervoltage. Often they might behave irregularly, but if you turn them off and on again with proper voltage, they should work fine. | |
| Aug 25, 2017 at 6:27 | comment | added | winny | @next-hack 's Solution is good for a fixed point solar maximum power point converter. There are comercial ICs too which do it for you. A straight buck converter with no input regulation would be a good choise too. You will however need an undervoltage lockout or there might be sporadious behaviour which in turn may or may not be acceptable in the application. | |
| Aug 25, 2017 at 6:08 | comment | added | next-hack | @FrankDonald, however, looking at the schematics, I would make these modifications: remove the ULN2804A (it's wasteful, unless you're using 8 channels), and use instead a discrete component (maybe mosfet or darlington. Be aware that a darlington has a high VceSat, about 1V). Then make sure that its base/gate goes high only if the the voltage at the 5V regulator is >4.5V. | |
| Aug 25, 2017 at 5:58 | comment | added | next-hack | @FrankDonald, no need for a battery. Just as an example (not optimized) use a schmitt trigger comparator powered by the unregulated 12 V. Then use any zener as a reference voltage. Compare a properly divided 12V with that zener. The ouput of that comparator drives a p-Channel mosfet, that powers on the circuit only if the voltage is large enough. Warning: if not correctly calculated, this might "oscillate" (it sees an open circuit voltage high enough, it turns on the system, and the voltage drops under threshold. Then the power is removed, and the VOC rises again. And the cycle continues). | |
| Aug 25, 2017 at 4:47 | comment | added | Frank Donald | @SimonB Yes i had that thought. But my application is not very critical it's ok if it doesn't work under cloudy conditions. But i Just want to make sure it will work under optimum conditions and won't damage the circuit when underpowered. | |
| Aug 25, 2017 at 4:44 | comment | added | Frank Donald | @winny Apologies added the circuit. | |
| Aug 25, 2017 at 4:44 | comment | added | Frank Donald | @next-hack So I should be using another piece of circuit powered by a battery to generate this power good signal isn't it? That might be a feasible option i believe | |
| Aug 25, 2017 at 4:34 | history | edited | Frank Donald | CC BY-SA 3.0 |
Added a circuit diagram
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| Aug 24, 2017 at 11:16 | comment | added | Simon B | 5W at 12V is only 416mA, so you haven't allowed a lot of headroom should a cloud drift across the sky. Don't assume a 12V solar panel will actually produce 12V. It's only a nominal voltage, and will depend on the amount of sunlight and the load current being drawn. | |
| Aug 24, 2017 at 8:59 | comment | added | winny | Post schematic or no help. | |
| Aug 24, 2017 at 8:58 | comment | added | next-hack | Some DC/DC converters have a signal that tells the rest of the circuit if the output is regulated. Typically this is called power good. (see for instance "special features" row in table here: ti.com/lsds/ti/power-management/step-down-buck-products.page#). Other have instead a under voltage lockout. They simply do not work if the voltage is under a certain threshold value. (You can also with a little external circuitry, generate those signal yourself). | |
| Aug 24, 2017 at 8:42 | comment | added | Frank Donald | Made the edit to show the main components of my circuit. "power good" signal? can you brief a bit about it please? | |
| Aug 24, 2017 at 8:39 | history | edited | Frank Donald | CC BY-SA 3.0 |
Added extra piece of information
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| Aug 24, 2017 at 7:38 | comment | added | next-hack | Are you, of course, using a 12V to 5V converter? And besides digital ICs, what is the rest of the system? In that case, I think that any problem could arise from the rest of the system (e.g. MOSFET or BJT used as switch, not being able to be fully ON, therefore dissipating power, when they shouldn't). Beside, to avoid this, you could add a "power-good" signal, that kills power in case of undervoltage. | |
| Aug 24, 2017 at 7:18 | history | asked | Frank Donald | CC BY-SA 3.0 |