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Forum Index : Electronics : 24 hour Weather Station Fan.
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| Phil23 Guru  Joined: 27/03/2016 Location: AustraliaPosts: 1664 |
Hi All, After thoughts on a better solution. Wanting to run a 2V 100mA fan 24 hours. At this point I have cobbled together the following. TP 4056 charger module; fed by a 4.5W 6V solar panel. It charges & protects a 3400mAh 18650, (A real one, Sanyo). The battery then connects to a board robbed from a "Juice Bank", 18650 in, 5V out. The 5V out goes to an LM2596 buck converter to be regulated to the 2V to run the fan. Very inefficient, with lots of losses I know..... BUT, what I discovered today is that one the batteries reach full charge, the solar input is disconnected, and the fan is then drawing on the batteries. Occurred at 3pm today, when theoretically the fan could have still have been on solar power. The panel seems capable of delivering 450mA of charge to the cells between 9:30am & 2:30pm, so recovery seems to be Ok, but it would be better if the fan wasn't draining battery power until solar was on longer available. It seems like the TP4056 does not reconnect until the battery voltage drops below 4.0V & Testing with the 3400mAh 18650 that seem to be about 3 hours. It did cross my mind to parallel the original 2V solar panel in the weather station across the output of the LM2596 module, but I don't know if it becomes a load on the output once the sun is gone. Thoughts? Thanks Phil. |
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| Phil23 Guru Joined: 27/03/2016 Location: AustraliaPosts: 1664 |
Further info, The battery was at 3.675V this morning & still losing 80-90mA. It had been drawing that current for almost 15 hours. At 7:00am the battery is receiving charge, just 50mA. By 7:30am the battery is receiving 150mA of charge. The USB meter show 450mAh supplied by the step up module. So about 1300mAh out of the battery @ average 3.8V. Power out of the battery 3.8V x 90mA x 15h = 5130mWh Power out of the 5V Boost 5V x 450mAh = 2250mWh The USB Meters mAh reading appears low.... Power consumed by the fan 2V x 100mA x 15Hrs = 3000mWh. So to use the 3000/5130, is only 58% efficient. The other issue is that once the battery returns to full charge, the fan will soley on battery power, not using available power from the panel. Surely the solution is a much simpler circuit with a bypass so once charged the battery is not used until a solar panel output is no longer available. Phil |
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yahoo2 Guru Joined: 05/04/2011 Location: AustraliaPosts: 1166 |
Are you trying to get better temp and humidity readings or is it for something else? I'm confused, no wait... maybe I'm not... |
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| Phil23 Guru Joined: 27/03/2016 Location: AustraliaPosts: 1664 |
Yeah basically, 24 hour aspirated sensor shields are supposed to report better that no fan or daytime only aspiration. I could try & reproduce the Genuine Davis 24 hour fan setup, but not sure of it's components; seems like nothing more than a couple of resistors & a diode. Solar panel; I don't know the voltage or wattage; guessing about 2 watts+. Batteries they use is a pair of C size Ni-Cads; odd. Surely it's not too hard to come up with a better solution using lithium cells. Phil. Edit; Living in town; ColourBond fences & other heat sources tend to distort readings if you measure stale air. |
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| yahoo2 Guru Joined: 05/04/2011 Location: AustraliaPosts: 1166 |
I fitted the davis daytime only fan and the longer radiation shield and noticed my peaks on very hot days was about 0.5-1.5 deg C cooler depending on the wind. I decided against fitting the 24 hour fan because at night there was always a temperature gradient between the weather station and the ground. So even with a fan it would not show an accurate dew point or freezing at ground level. I figured I would be better off fitting an extra temp sensor or moisture sensor close to the ground. which I never did, of course. In the end I was only interested in mid afternoon dewpoint and humidity as that was my prediction of the next mornings minimum temperature if the humidity was up and the chance of frost if the humidity was low (providing a front is not coming through). I was told to expect to replace the fan every 2-3 years, mine is 9 years and still spinning, I guess the dry weather is good for something. if it had broken in two years I probably wouldn't have fixed it, I am not that in love with it to tell the truth. I'm confused, no wait... maybe I'm not... |
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| yahoo2 Guru Joined: 05/04/2011 Location: AustraliaPosts: 1166 |
I do remember somebody saying, possibly Stan Swann ?? that solar garden lights could be purchased far cheaper that their individual component parts. maybe there is a possibility there to adapt something. laptop batteries, cordless tool cells, buy a couple of lights for a few bucks and double up the solar panels. I have seen some pretty cheap security lights with quite large panels. I thought one solar light I pulled to pieces had a low voltage cutout mechanism just by taking advantage of the quirks in the components. Perhaps I imagined that, it did have a rusty switch. I know some of them use light hitting the solar panel to switch the transistor for the LED, perhaps you could direct power the fan during the day with a similar technique. I'm confused, no wait... maybe I'm not... |
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Madness Guru Joined: 08/10/2011 Location: AustraliaPosts: 2498 |
I bought a couple solar security lights from Aldi recently for $15 each, they have 2W of LED lights and are supposed to be able to power the lights for 2 hours. Without the lights and infrared sensor, they might power the fan 24/7. You have to wait for them to come into stock again but would be many similar available. There are only 10 types of people in the world: those who understand binary, and those who don't. |
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| yahoo2 Guru Joined: 05/04/2011 Location: AustraliaPosts: 1166 |
 I had a quick look on google image at some circuits and they get quite complex quickly, last count I was up to 13 components and it still was not really suitable.I'm confused, no wait... maybe I'm not... |
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| Phil23 Guru Joined: 27/03/2016 Location: AustraliaPosts: 1664 |
The TP4056 modules I have are fine for looking after the 18650's. It has over discharge protection.  Part I need next is a decent step down to 2.0V. An LM2596 module is a bit useless on it's own as it starts to lose regulation below 4V with a far bit of drop in output by 3.5 & cutting right out at 3.3. Ordered a few LM1117T-Adj's for that part, so it only leaves a lack of by-pass to run it on solar once the batteries a fully charged. Phil. |
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| Boppa Guru Joined: 08/11/2016 Location: AustraliaPosts: 814 |
I dont see the problem Phil, if the panel is charging the battery and the fan is discharging the battery at the same time, whats the advantage of running the fan directly off the panel once the battery is full, rather than just letting the panel continue to top up the battery as the fan pulls current from it? |
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| Boppa Guru Joined: 08/11/2016 Location: AustraliaPosts: 814 |
here's a silly idea, we used to use a circuit in alarm panels that needed a standby battery and had no provision for one, but could handle a small variation in supply voltage We would put a 12v battery pack with a diode in series with it, and a psu that had a higher voltage than the battery voltage, also with a diode in series with it, then join the two diodes together at the alarm panel Basically we had a pair of 6v lantern bats in series as the '12v battery', and a 13.8v plugpack, all negatives hooked together at the alarm panel neg, both diodes anodes hooked together and in the alarm panel +, and the battery pack + connected to one diode cathode, and the plugpack + to the other diodes cathode When the plugpack voltage dropped below the battery voltage or went away entirely, the alarm would run off the lantern battery pack, otherwise it ran off the plugpack and the batterys never drained- we had some that never had blackouts that the batterys were still charged after several years You have excess voltage you need to get rid of anyway, so use the diode to drop the voltage down (maybe even a couple in series would be enough voltage drop) While ever the panel output is above the battery voltage, it will supply the fan, once the panel power levels drop below the battery voltage, it starts to provide current instead |
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| Phil23 Guru Joined: 27/03/2016 Location: AustraliaPosts: 1664 |
I don't know quite what to expect by the change, but I do notice a change in the graph stability when shade hits the ISS & the fan stops. Other thing I wonder about is the fact that condensation builds up on the shield on may nights, & that could falsify humidity readings. I'd assume if the air movement was maintained, I may not get the condensation on the shield plates. My VP2 has an SHT11 sensor..... Got an SHT31 sitting in the office.... It's been there quite a few weeks. Cheers. |
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| Phil23 Guru Joined: 27/03/2016 Location: AustraliaPosts: 1664 |
One the batteries reach full charge, 4.2V, the panel disconnects & does not connect again until the cells drop below 4V. I have now seen it run in that state since mid-day. It would only become an issue when we get a few days of bad weather & then the battery supply would be depleted a bit earlier. Guessing with the 2 3400mAh cells I'd be getting about 2 days runtime with no charge. The Sanyo cells only measure about 2500mAh when discharged to 3.0V, The Spec sheet rating of 3400 is based on 2.8V end of discharge. Phil. |
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| Boppa Guru Joined: 08/11/2016 Location: AustraliaPosts: 814 |
Seriously, I'd try that two diode trick panel (+) connects to battery charger (+), from bat (+)(diode K), diode A to your voltage drop/reg (+) panel (+) also goes to other diode K, other diode A goes to voltage drop/reg (+)  AFAIK that will keep the voltage reg/dropper to the fan fed from the panel whilever the panel voltage exceeds that of the battery, once the panel voltage drops under the battery voltage, it will start feeding from the battery instead worth a try at only two diodes.... |
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| Phil23 Guru Joined: 27/03/2016 Location: AustraliaPosts: 1664 |
Thanks @Boppa, Will give that a try. Want to knock up the output reg first, for the 2.0V output. Based on this, if I'm starting with 4.2V from the 18650's, & maybe around 6 from the panel with the bypass diodes, do I just stick with these suggestions, 121 for R1 & the 1k pot, or am I better off to up R1 a bit? Or drop the pot value? I'm not that up on design requirements / best practices.  Thanks Phil. |
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| yahoo2 Guru Joined: 05/04/2011 Location: AustraliaPosts: 1166 |
the equation is saying the ratio of r2/r1 sets the voltage out. Those two resistors should give a Vout of 11.6 volts according to my feeble braincells. 2.0 = 1.25( 1+ 1000/1666) or 2.0 = 1.25 (1+ 72/121) something like that, as long as the ratio is right I'm confused, no wait... maybe I'm not... |
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| Phil23 Guru Joined: 27/03/2016 Location: AustraliaPosts: 1664 |
Thank's Yahoo, I Understood the formula, & got the same 72 ohm, but was wondering if I could up the 121 ohm resistor, considering I probably can only get 1k pots locally. Don't know if the current that resistor supply's is important or not. Phil. |
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| yahoo2 Guru Joined: 05/04/2011 Location: AustraliaPosts: 1166 |
I think you could raise the value of R1 quite a lot I have seen plenty of circuits that use a 240 ohm resistor for R1 with a lm117. The lm317 is not as forgiving. Vadj is going to be a few microamps and the flow through both resistors will be a few milliamps. in theory the voltage regulation gets unstable with higher resistor values and higher load impedance on Vout. I would be tempted to temporarily fit two 10k linear pots and have a play. I'm confused, no wait... maybe I'm not... |
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| Phil23 Guru Joined: 27/03/2016 Location: AustraliaPosts: 1664 |
That's what I'm hoping would work. Higher would have it's advantages. 72/121 would have a 10.3mA load on the 2V output. So consume almost 250mAh in a 24 hour period of bad weather. We've had very little sun on the last 4 days, so 1000mAh of the batteries would have been wasted in the voltage divider. The batteries are 3250mAh, but the protection chip cuts them off at 3.0V I think, so that gives me a total of about 5200mAh from 2 cells. |
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