Fantastic Fan and battery draw

System

This discussion was created from comments split from: Norcold 30L cooler and Solar question.

LuckyJ

And do I remember reading somewhere that the way the fan resisitor works, that the draw is almost identical no matter the setting?

WilliamA

@Fourman110 said:
I'm struggling on a campouts right now
Had 13.1V upon arriving at the campsite. Ran the fan for 15 minutes and the Nordcolds been running and set for 36. Voltage dropped to 12.4. Had frozen items in the fridge to help minimize the cooler power and added ice about 4 hrs after arriving. My solar setup keeps raising the voltage but I drop so quickly I'm concerned. I've got two 6v golf cart batteries and still struggling to keep voltage up. The batteries were in my heated basement and hooked to a trickle charger for a few days once a month. I've turned the jvc lighting to as dim ad it will go and added the toggle switch to the TV and power is still dropping to 12.3V at times. I've been monitoring with a cigarette lighter battery monitor. Come to think of it I should drag our the two probe volt meter tomorrow to double check accuracy.

Interesting,
The first thing to do is determine if there actually IS a problem. As Sharon_is_Sam said, the meter does not show actual battery state while voltage is flowing. It (they) need to be in a resting state to accurately measure their charge. She is also exactly correct in that the best way to measure "running" voltage is by monitoring current draw. This won't give battery state either unless one does the math and calculates the current capacity and then subtracts the average current draw from that. If you want to eliminate walking around camp with a slide rule and pencil, you need a shunted smart monitor that does the math for you in real time.
The biggest problem with dual (or more) batteries is that it creates an even more complex problem in that the best monitor won't give you the battery voltage but rather will show an "average" of the two. There has never in the history of batteries been a set so perfectly matched that they don't leak voltage back and forth. One battery will almost always be more efficient than the other. Then what you see is current flow from battery to battery, even in a resting state, as the two work to level out with each other. Think of it as two buckets hooked together with a hose at the bottom. Turn on the spigot and one will almost surely drain out a little faster than the other. Turn the water flow off and you will still see water flow as the two buckets work to level out. Even though there isn't any water going "out" it's still flowing between them until they are the same level again.
So where to start? At the beginning! Disconnect the batteries from each other, wait 20 minutes or so, then check their individual voltage and compare them. You'll immediately be able to determine which battery is weaker as it's static voltage will be lower. Even a tiny bit makes a difference. If both test "good" , then hook it up and get a static voltage of both batteries in series. That should equal the sum voltage of each battery added together. It's possible that you have one gimpy battery (common). It's also possible (probable) that there's no problem and what you're reading is the normal flow-state of 2 busy batteries.
LuckyJ is also correct. Running an electric motor in series with a resister slows the motor speed, but still uses the same current draw regardless of the setting. The only way to reduce current draw in a motor or light is to shut it off. That's exactly what a PWM does.

WilliamA

Ianstag

WilliamA, great explanation of the 6 volt battery issue and the problem that might be happening. A little disappointing that you confirm my fear that the Fantastic Fan in the ceiling is drawing the same current regardless of the setting. The rest of the current is just dissipated as heat through through a resistor. We must find a way around this. Perhaps isolating the circuit and using a smaller rechargeable 6 volt battery mounted somewhere would help. The website for the fan list the amo draws for the different settings and they are lower for a lower speed. Is this just false advertising or are they getting the lower amp draw some other way?

WilliamA

There is some confusion about this and while the calculation is simple (Ohms law: E=i*R) the results are not "linear" per se and there does result some reduction in amperage draw. I can take a stab at a simple explanation:
Let's say you take a 12V motor and run it with 12V current. It will run at a fixed speed based upon its design and will always try to accelerate up to that speed, regardless of the load, in this case, the size of the fan blade. Putting a resistor in-line with the motor reduces the voltage. The motor will reduce its speed because, like the accelerator in your car, it is getting less voltage and hence produces less torque to spin the fan. The resistor, however, is using its resistance to lower the voltage and in the process, gets hot. One might think that by adding the 2 numbers: motor current (amperage) draw and resister amperage draw, resulting in the same total power use regardless of the size of the resistor. However, Ohms law dictates that since the motor is receiving less voltage, it actually needs less amperage. So as resistance is added, there will be some reduction of total (motor+resistor) amperage draw. The equation is actually quite simple to do and if you run it with resistance variables, you will be able to plot a sort of funky amperage curve.
The short answer is yes, the amperage does drop as speed goes down, but the current "wasted" by the resistor in the form of heat is not proportionate to the motor speed. Think of this analogy: if your car gets 25 MPG at 60 miles per hour, it would make sense that at 30 miles per hour, it would get 50 MPG. Unfortunately, both Newton's law and Ohm's law are weighted against us in that the factors determining efficiency are "not" so linear. Spinning the fan at 1/2 of its maximum speed will result in some amperage draw reduction, but 1/2 the fan speed does not equate directly to 1/2 the power needed to do so. It's also true that 1/2 the fan speed result in 1/2 the air moved. It's much more complicated than that. You are saving power, but at the expense of fan efficiency and wasted heat...

WilliamA

Sharon_is_SAM

For those of you with access to the TaB forum, check out an answer to your fan issues.

https://tab-rv.vanillacommunity.com/discussion/8556/fantastic-fan-upgrade-to-7350-dramatic-improvement-in-efficiency

DaveB

@WilliamA you seem to be quite up on your electronics. I have a question/theory. I agree totally with the resistor method just burning up the extra as heat, and for sure PWM is the way to go. My question is if you were to determine what the voltage drop is with the resistor method could you not then just use a voltage regulator which would give you its 'set' voltage and just bleed the leftover to ground? Or is heat still wasting energy in this scenario as well, since all these little regulators have a heat sink albeit small.
Not that it would be worth the work involved to modify the fan switch wiring, but just a thought.

WilliamA

@DaveB said:
@WilliamA you seem to be quite up on your electronics. I have a question/theory. I agree totally with the resistor method just burning up the extra as heat, and for sure PWM is the way to go. My question is if you were to determine what the voltage drop is with the resistor method could you not then just use a voltage regulator which would give you its 'set' voltage and just bleed the leftover to ground? Or is heat still wasting energy in this scenario as well, since all these little regulators have a heat sink albeit small.
Not that it would be worth the work involved to modify the fan switch wiring, but just a thought.

The short answer is no. It's easy to reduce D.C. Voltage, but not without the attendant waste heat through resistance. A regulator doesn't control voltage, it controls amperage output. The voltage output is regulated by the battery cell count and the windings of the generator. I think when you are referring to the resistor on a regulator you are talking about old-school generators with a separate firewall mounted regulator. That technology was a waste-heat system too. The resistor wasn't there to regulate output but rather to provide an idling generator with a "shock absorber" of sorts. (No pun intended). Generators work as an "either off or on" system. Voltage output was fixed and amperage was controlled by the regulator via a large solenoid-style switch. When the switch (relay) opened, the points would arc at the moment of loss of contact causing a massive (up to 10 times battery voltage) spike in voltage output. This is a fundamental law of capacitance in all electrical systems. That resistor served to "absorb" the voltage spike when breaking contact, but also robbed output from the charging system as waste heat. An alternator, by comparison, doesn't ever stop charging but varies amperage output by controlling the strength of the magnetic field in the rotor winding. The weaker the field in the rotor winding, the less amperage is induced in the stator. Generators have permanent magnets providing the field strength while alternators do not. That's why alternators need voltage in to make power while generators need only be spun with no voltage input needed to get power out. In addition (There's More!) a generator has almost 0 ohms resistance so If it stops turning and the regulator relay "sticks" closed, it would be a very large dead-short in the electrical system with the resulting cloud of billowing smoke. That resistor prevents it, should that occur. A regulator is as dumb a device as you'll ever find this side of a light bulb. The coil that opens and closes the regulator is hooked to the battery on one side and the generato output on the other. As long as the voltages are different, current flows and the relay closes ( charging). As soon as the voltages equalize, (battery voltage=generator output) current stops flowing and the regulator opens, stopping output.

You're probably sorry you asked...

PWM controllers do not change voltage but rather switch rapidly off and on. This cycling "frequency" is what controls fan speed. The longer the "off", the slower the fan. It's also interesting to note that in PWM systems, it's common to get attendant noise. Some manage it better than others. I can hear it in my Maxxfan at certain speeds. It would seem that there really is no free lunch.....

WilliamA

DaveB

@WilliamA I think you misunderstood my meaning of regulaltor, I was referring to the tiny solid state type. Think 7805 found in just about anything now days, it does have internal resistors but perhaps they are more efficient, or maybe even this is how they are already controlling speed on non PWM fans.
My electronics knowledge is okay, but not guru status by any stretch.

WilliamA

@DaveB said:
@WilliamA I think you misunderstood my meaning of regulaltor, I was referring to the tiny solid state type. Think 7805 found in just about anything now days, it does have internal resistors but perhaps they are more efficient, or maybe even this is how they are already controlling speed on non PWM fans.
My electronics knowledge is okay, but not guru status by any stretch.

Dave,
You are correct to state that the 7805 is indeed a "waste heat" regulator. It reduces voltage to a fixed output and "wastes" the difference in voltage as heat. As we have been discussing, whether mechanical as I described above or solid state as the 7805, It's pretty much the only way to reduce output of DC. PWM controllers don't reduce voltage as in, 13.2V input will still be 13.2V output but the "frequency" (on time vs off time) pulses so fan speed slows down. The only non-resistive way to lower (or raise) DC voltage is to first convert it to AC, then change the voltage, then RE-convert it back to DC. This is done all the time for increasing output but seldom for reducing it. An example of voltage increase in automotive applications is HID headlights where output voltage is 20,000 or more volts, but at tiny amperage. Sort of the same principle as flourescent lights. I can't think of any automotive-based system where voltage is reduced without using a waste-heat resistor of some kind. The 7805 style regulated output semiconductors are common in automotive applications (it's the primary component in 5V USB chargers) but are notoriously unreliable and prone to failure because they get too hot and have heat sinks that are too small for the application. They also are not used where output amperage is at or above about 1 amp, therefore wouldn't work for long on a fan system, or anything with a motor for that matter.

Hope this helps...

WilliamA

DaveB

Thanks for clearing that up @WilliamA and confirming my suspicions. I would never think of using 7805 for a motor, just an example because its common and easy to remember kinda like the 4007 diode or 555 timer. Again, thank you for taking the time and hopefully others learn from your explanations as well. A mentor once said to me, "If you don't learn something every day, you're not living life to its fullest"

WilliamA

My motto has been:
"If you didn't learn something new,
you probably won't need the First Aid Kit"

WilliamA

JamesDow

My testing shows that the Fan-Tastic Fan consumes about the same power of the Norcold refrigerator. 53/54 watts. I recently went desert camping (108 degrees) and utilized a OPOLAR 9 inch battery powered rechargable fan with 6,700mAh capacity (Amazon $34.99) along with a power brick and ran it close to my body all night, multiple nights and it did the job great. Since it does not draw from the trailer battery, I thought it was a great solution. Attached is an Excel spreadsheet of the various electrical systems I tested using a Poniie PN2000 Plug-in Kilowatt Electricity Usage Monitor Electrical Power Consumption Watt Meter Tester (Amazon $ 26.99) As you will note, the Fan-Tastic fan consumption drops almost 50% when run on low. I hope this information is usefull.

WilliamA

@JamesDow said:

"As you will note, the Fan-Tastic fan consumption drops almost 50% when run on low. I hope this information is usefull."

There has been much discussion and confusion about the voltage/amperage draw of the fantastic fan (and other devices) when turned down. As this is a common user modification to try and reduce energy usage, and as I've probably added to the confusion by not clearly explaining myself, I would like to try and clear the air a bit about the topic. Here's my attempt at that:

As James Dow said, reducing the fan speed does indeed reduce the energy needed to run it. That is easy to see using just about any kind of test meter. What's harder to get your head around is the percentage of drop as compared to the reduction. For example, if I take a motor and reduce its speed by 50% using a resistor, I should see a 50% reduction in amperage usage. James' very nice chart seems to bear this out. If your goal (as all of us are trying to do) is to reduce the amperage draw, then it's a success. That's not the whole story however. A very simple (sic) test of the theory is to add a resistor to the line, test its voltage drop, find it to be 50% less, then touch the resistor (CAREFULLY!!!!) If it's warm (HOT!!!! OUCH!!!!) then you are "wasting" voltage. The voltage used to heat the resistor is "waste" heat energy. It's a relatively small amount to be sure, but nevertheless, it is waste energy.
What is not discussed in this kind of conversation is work done vs. energy used. If the fans' airflow at both full power and at the reduced output is not measured, then the equation is incomplete. Using a resistor to reduce the fan speed does indeed reduce necessary amperage, but what you get is voltage used to turn the fan + voltage used to heat the resistor. The end result is yes, you are using less amperage to run the fan, but the fan speed is not as efficient as the total voltage (amperage) consumed. For example, the amperage draw might be half as much, but the fan efficiency might be as low as 10 or 20% of the "compared" amperage between its maximum speed and resistance-reduced speed. The fan "should" be running at 50% of its' capacity when in face, it's actually "working" at much less than that. Pur another way; If the amperage goes down by 50% but the airflow goes down by 80%, the result is that 30% of the reduction in energy is wasted. By comparison, a PWM controller would reduce the fan speed to 20% using (about) 20% of the energy as opposed to 50% with a resistance circuit. The amperage charts of various PWM fans at different speeds bear this out. What that means is with resistance reduction, the difference in "work" the fan is doing is much less than the amperage used to do it.
This principle is as common as dust in the electrical industry and resistors are commonly used to accomplish all manner of speed/illumination/etc reduction because they are cheap and easy to apply. To think, however, that they are "efficient" is wrong. Do they reduce the amperage draw of a device? Yes. But by comparing the amount of "work" to the amount of power used, you quickly find that they are frightfully inefficient. If you are a stickler for the details (as I am) you'll discover that there are much more efficient ways to do the same job.
Now that you are thoroughly confused, here's my best advice:
If your goal is to reduce amperage draw of the fan and therefore, reduce overall energy usage, then yes, a resistor-style speed controller (or light dimmer) will accomplish that.
If, on the other hand, you truly want to be "efficient" in usage, then some other method (PWM) is the only option.
Don't for a moment think that a resistor anywhere isn't wasting amperage. It's how they work.
There's no free lunch in the equation....

WilliamA

LauraC

JamesDow ... how much power did you use recharging the fan?

JamesDow

LauraC, - I assume you are asking about recharging my portable OPOLAR battery powered fan.
Bottom line do not know, since I have never needed to charge it on the road while camping.
I have yet to run down the two 3350mAh lithium batteries included with the fan.
At night run the fan using a USB power block with the lithium batteries disconnected.
OPOLAR claims the two lithium batteries that comes with the fan can last for 8 to 11 hours.
I have not run down a power block (10,000mAh) with the fan.
I carry multiple (3-5) power blocks and also use them to charge cameras and phones.
Occasionally, I do use the USB charger socket (replaced defective original sockets) during a sunny day while I am on solar.

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LauraC

Thanks, Jim - I asked this before I realized that the power block/bank is used to store power rather than transfer power. Thanks again!!