I have a DC fan inside of a weather instrument shelter that runs 24/7 off a solar panel and rechargeable battery. Attached picture is a basic sketch of how it's wired (forgive my poor drawing skills). Basically, the solar panel is wired to a charge controller which is then wired to the leads of a 12v battery. The fan is also wired directly to the leads of this 12v battery.



For the most part, this solution works as long as there's sunny days. Battery partially discharges at night and fully recharges with only a few hours of direct sun on the panel the next day with plenty of extra power from the solar panel to run the fan while charging the battery. Problem comes when there's a cloudy day or two. There's not enough current from the panel to recharge the battery on a cloudy day and ultimately the battery gets drained.

I could always add more panels to increase the limited charging on cloudy days or get a higher amp hour battery that could last longer between recharges...But an alternative solution would be to slow the fan a bit to use less power. The fan is a variable-speed based on it's voltage with operating voltage range from 3v (very slow speed) to 14v (high speed). In this particular case, since the motor spins at a slower RPM with lower voltage, the amps and watts also decrease as the voltage lowers using overall less power...The way it's currently wired, the fan is running about 14 volts (full speed) on a sunny day which drops to about 12.2v (still a relatively fast fan speed) on the battery by the next morning. In reality, these voltages produce way more airflow than I need. I could easily get by with lowering the fan to 9 or 10 volts during the day and maybe down to 6 or 7 volts at night. This would get the job done with less drain on the battery.

I was told that since I just need to lower by "a few volts" without a specific value needed, that the cheapest and most simple solution would be to add a resistor between the battery and the fan with enough resistance to cause an average voltage drop of somewhere around 3 or 4 volts. Makes sense, but considering the way I'm wired, would that added resistance in the line also get applied to the solar charger and interfere with charging/charging rate of the battery? I feel sure than an in-line resistor would be a problem if the fan and solar charger were wired in series to the battery, but since they're both wired in parallel with the battery, then I'm not sure if it would be an issue or not?

I realize another option would be to use a buck converter, but my understanding is that would give the fan a specific voltage at all times. Whereas a resistor could allow for more voltage during the day when it's original input is more (~14v) and less voltage at night when it's original input value from the battery is less (~12v).

Thank you in advance for any advice!

 

Although a resistor will reduce the power taken, it will waste power in heat. With the resistor in the power lead between the fan and the battery, in parallel with the solar charger, it will not affect the charger.

A better solution would be a PWM motor controller which can be bought for a few dollars these days - it will reduce the power to the fan by switching, hence no heating losses. Adjusting the switching duty cycle controls the speed.

 

Well, if I go the resistor route, I'm not sure what ohm value to go for. I'm getting a bit stumped on Ohm's law, because the fan isn't a constant power device. As voltage rises, the current also rises as it spins faster; lower voltage uses less current as it spins slower. As a rough average, based on my multimeter testing, you can take the voltage, move the decimal place to the left 2 places and you'll get the amps it's drawing at that voltage....Example: at 14v it pulls close to 0.14amp; at 12v it pulls close to 0.12amp; 10volts = ~0.10amp; 3v = ~0.03amp......So, say it's daytime and the fan is running @ 0.14amp and I put in a 40 ohm resistor......40 ohms at 0.14amp would give me a 5.6 voltage drop...But once that voltage starts dropping, the fan goes slower and is no longer pulling that 0.14amp, which alters the ohms law result. So I don't know what actual amps I'll be using until I know my end voltage and I can't calculate my voltage drop without knowing how many amps I'll be using. So for this situation, I guess there's no way to know what my end numbers will be until I put in a resistor and take a measurement?

As far as the power wasted by the resistor, how would I know how much is being wasted? If it's only a little bit and significantly less than what the fan would be using at full power, then it still may be doable. But if it's wasting almost as much as the fan would be using at full power, then it would defeat the purpose.

The heat the resistor gives off would be another potential issue. Only place I have to put the resistor out of the weather would be in the junction box where the battery is. It already gets rather hot in there during our summer afternoons and certainly wouldn't want to add so much extra heat inside that box as to make it too hot for the battery.

That PWM controller sounds like it would be the better long-term solution for me, but something I would have to order. I already have resistors that I could experiment with right away, as long as it doesn't drain too much power or give off too much heat.

 

If the fan is for cooling, and it needs more in the day than at night, why not just use a thermostat?

 

If the fan is for cooling, and it needs more in the day than at night, why not just use a thermostat?

It's a weather thermometer and humidity sensor that's housed inside a plastic shield with vents/louvers all around it. The purpose of the fan is to keep air constantly pulling in one side of the shield and blowing out the other side so that the thermometer and humidity sensor inside is constantly getting a breeze of outside air blowing across it. Otherwise, air sits inside the shield too long and that air gets conductively heated (in the sun) or cooled (radiational cooling at night) to the temperature of the plastic case. Essentially, we're trying to measure the temperature of the air in the yard. The plastic case gets hotter than the air during the day and cooler than the air at night. That constant airflow from the fan across the thermometer assures that we're constantly measuring "new" air from outside rather than air whose temperature has been contaminated from being in contact with the plastic case too long.....But, as the physics work, the surface of the plastic case will get a lot hotter than the air during direct sun but only slightly cooler than the air at night, so the fan doesn't have to spin as fast at night because there's less temperature contamination. But this trend of the plastic being warmer during the day and cooler at night happens regardless of what the temperature of the outside air is; So even a bitter cold day, the thermometer will still read several degrees too warm during the day without the fan. I can't use a thermostat because the fan needs to run regardless of how cold or hot it is.

 

Okay, that makes aense.

 

12V DC fans come in a variety of speeds and powers. Replace yours with a lower power one.

 

Another option is to get a Stevenson screen and eliminate the fan altogether. I built a DIY screen with inexpensive white plastic saucers.

 

12V DC fans come in a variety of speeds and powers. Replace yours with a lower power one.

This was the lowest power I could find that would reach the airflow that I needed (~30 CFM). Although it's a bit more airflow than I needed, anything lower power I found wasn't nearly enough.

Another option is to get a Stevenson screen and eliminate the fan altogether. I built a DIY screen with inexpensive white plastic saucers.

That's basically what I've got and as long as there's at least a light wind, it works well. But when winds go calm, the air still gets stagnant and subject to conductive and radiation errors by up to a few degrees. The fan really makes a noticeable difference in those times. Since I'm trying to keep "formal" weather records, I'm trying to minimize artificial error as much as possible and the fan seems to be doing that. Just trying to conserve a little energy on the cloudy days as cheaply as possible without having to spring for another solar panel or another battery and junction box (though I may still be forced to go that route if slowing the fan doesn't save enough power)

 

The Stevenson screen is supposed to solve that problem.
Can you post a picture of your setup?

 

It's the Davis Vantage Pro2 with passive radiation shield (Davis' version of a Stevenson screen), like in this picture.



Inside of the white radiation shield plates looks like this (of course, these are the middle plates with solid plates for the top and bottom to keep rain out:


Accuracy rated +/- 4 degrees in calm wind (much better with at least a light wind). They also make a fan aspirated version that costs a couple hundred dollars more that's within 0.5 degree accuracy at all times. I've made a cheap replication of their fan version (like a lot of other's in the industry) by adding additional plates and using a 12v computer fan...It works just as well as the factory version for a small fraction of the cost. Most that have done this hack simply run a power cord from the house/building, but there's no AC power on-site at this location. Hence, I'm relying on a solar/battery solution.

 

Ah! Good to know. If I were in your position, I would experiment with extending the length of the screen.
I would search for white toy plastic saucers or bowls and make the screen longer, something like this.

 

I've got 6 plates on it right now. I can buy extra plates from the factory for a couple dollars a piece, though would need longer screws. Just wondering if it would make any difference because the plates themselves have some radiational affect, even though they're solid white. I checked with an IR thermometer the other day and it showed the outer surface of the plates were about 7 degrees warmer than the air temp. So that's naturally going to apply a little error to any air sitting stagnant inside the shield. I can see where a larger space inside the shield could potentially minimize it some (more air inside the shield to collect the heat of the plastic = less heat absorption per parcel of air). Yet, on the flip side, more plates = more outer surface area being heated.

 

Also worth noting, the temperature and humidity sensors are then housed inside a small cap with a fine mesh screen around it to filter dust, pollen and other pollutants; Another place for still air to become stagnant. So not only are we using the fan (or natural wind, when available) to constantly replace the air inside the shield, a breeze is also needed to force airflow through that fine mesh to reach the sensors. This is the sensors (inside that white filter cap)

 

Get one of these or a similar module and wire the output contacts as shown. R1 and R2 are to reduce the Fan voltage.
R1 is for during the day and R2 for at night. You will have to experiment to find the values needed for the suggested voltages.
Do you have any specs on the Fan?
https://protosupplies.com/product/light-sensitive-ldr-relay-module-12v/

 

Get one of these or a similar module and wire the output contacts as shown. R1 and R2 are to reduce the Fan voltage.
R1 is for during the day and R2 for at night. You will have to experiment to find the values needed for the suggested voltages.
Do you have any specs on the Fan?
https://protosupplies.com/product/light-sensitive-ldr-relay-module-12v/
View attachment 300630

Looks interesting, but I don't quite understand how it would work? Would I have to use my own resistors for R1 and R2?

 

Correct

Looks interesting, but I don't quite understand how it would work? Would I have to use my own resistors for R1 and R2?

Correct and the resistor values you will have to determine. It's not difficult. Do you a link to the fan or a model #?

 

Correct

Correct and the resistor values you will have to determine. It's not difficult. Do you a link to the fan or a model #?

Ok, that makes more sense! So the only benefit here would be that I could use a stronger resistor at night. Although I'm actually not sure that I would need a stronger resistor at night since my input voltage is already about 2 volts lower at night. So I should have some extra voltage drop at night even using the same value resistor.

This is the fan: https://www.amazon.com/dp/B0C96HN76V

But, as mentioned in a past post, it's not a constant-power device so the rated 0.12amp is only valid when running the fan at 12 volts. When running the fan at 14 volts, I'm measuring about 0.14amp. At 10 volts, about 0.10 amp...At 5 volts, about 0.05amp....That's what's throwing me off. Like I say, it's running at 14 volts; 0.14amp during the day right now. If I put in a 40 ohm resistor, 40 ohms at 0.14amp would give me a 5.6 voltage drop...But once that voltage starts dropping, the fan goes slower and is no longer pulling that 0.14amp, which alters the ohms law result (if that resistor lowers it to, say, 9 volts, then I'm only pulling about 0.09amp instead of 0.14amp). The fact that both my volts and amps lower together, I won't know what my final amps are until I know my final voltage and my final voltage is dependent on my final amps. I can't solve ohm's law with both my final voltage and final current being unknown variables.

 

The fan shows a linear impedance which measures 14/.14 or 100 ohms. Divide the final voltage by 100 to find current.

 

A 20 ohm 1 watt resistor in series with the positive lead of the fan motor would drop the voltage to appx 11.6 volts during the day and 10 volts at night according to your measurements. Saving 17% of battery demand.