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Wind Detecting Circuit

TKowalskii

Apr 13, 2019
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I’m a sophomore in high school in my first year electronics class. For a project I am trying to design a wind detecting circuit using a motor and a transistor. The motor is connected to the base of the transistor, and there is an LED connected to the collector. The circuit was made so that when the motor was spun, current would be sent through the base, activating the transistor and turning on the LED. The motor produces around 20-40mA but the issue is only around .2 V comes out, not enough to bypass the VBE. The goal is to get it working on a 9V battery. Any suggestions? I would add an image but I’m currently in a car and my data isn’t the greatest.
 

duke37

Jan 9, 2011
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The type of motor will determine the output. A DC motor with brushes will turn out a voltage proportional to speed but will have considerable friction. A fan motor may be AC and not suitable for this purpose.
You could use a thermistor to measure wind flow, the resistance will vary with temperature.
I have used a commercial hot wire sensor which was nicknamed the bee catcher.

You may be able to 'raise' the voltage of the motor output with a forward biased diode or you could use an op-amp which is capable of working with an input down to zero volts.
 

AnalogKid

Jun 10, 2015
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0.2 V is very low. How fast is the shaft turning for that voltage? Can you post images of the motor?

Detecting very low wind speeds is difficult. Professional units that have vanes or cups driving a small generator have very low friction bearings.

ak
 

hevans1944

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Jun 21, 2012
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You may want to reconsider your choice of sensor. Although a decent permanent magnet (PM) DC motor will produce an output voltage, which in turn can produce a current in a resistive load, that is more or less a linear function of the motor shaft speed, it may not be the "best" choice based on your limited experience.

Given very good, low-friction, bearings and a strong PM field, a three-cup anemometer vane attached to the motor shaft should produce a measurable output signal, even at very small wind velocities. How small depends on how "easy" it is to turn the motor shaft when wind blows on the anemometer cups. At some low wind velocity insufficient torque will be developed by the anemometer vanes to turn the motor shaft.

Of course there is also the possibility that what little signal does exist, prior to the motor shaft ceasing its rotation, is buried so far into the noise floor that it is not measurable. All sensors exhibit this problem... there is always a minimal detectable signal. If that minimum isn't small enough (insufficient sensor sensitivity), then either a different sensor is required, or more signal amplification may be necessary. Unfortunately, for signals below the noise floor, amplification amplifies the noise too, so there is no net gain in signal-to-noise ratio. It is beyond the scope of this comment to explain how to improve this situation by adding intelligence (modulation) to the signal.

There are many devices that respond to changes in temperature by changing an electrical characteristic. Tungsten wire, for example, increases in resistance as it becomes hotter. This attribute can be used to sense wind velocity because a heated wire loses heat at a rate that is a function of the velocity of air moving across the wire. Circuits that implement wind velocity measurements based on the rate of energy loss in a heated wire are called hot wire anemometers. I suggest that you research this method of measuring wind velocity. Here is a link to get you started.
 

Bluejets

Oct 5, 2014
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Hook the 3 cup unit to a shaft and an aluminium/timber(whatever...just not steel) disc with a dozen or more small magnets and a hall effect switch module on the side.
Surely that would detect even a light puff of wind.
 

AnalogKid

Jun 10, 2015
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Actually, steel will work just fine, and makes mounting and securing the magnets way easier. With all of the north or south poles facing outward, there still would be enough change in flux between them to be detected by a Hall sensor. Doesn't matter which pole is outward, as long as they're all the same.

ak
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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Surely that would detect even a light puff of wind.
I am pretty sure the lightness of the puff of wind would depend on the bearings used to support the shaft. But why use magnets and a Hall-effect sensor when an opaque slotted disk and an optical interrupter would perform the same function less expensively? You would still need some good bearings though... maybe a steel needle shaft supported between synthetic ruby "jewel" bearings?

The goal is to get it working on a 9V battery. Any suggestions? I would add an image but I’m currently in a car and my data isn’t the greatest.
Haven't heard from the OP in awhile. Hopefully texting while driving hasn't caused an accident. This could be a primo example of the drive-by poster looking for a "quick fix" to their problem. Which, BTW, I have in the form of a PDF file complete with pictures, schematic, annotated bill-of-materials (BOM) with vendor sources, estimated costs, and construction details. However, the file is too large to upload, so perhaps my data isn't the greatest either. Anyhoo, since this is homework, you can't have my solution... it's against forum policy. Happy, along with others here, to offer suggestions however... Remember, though, free advice is often worth less than what you pay for it.
 

Bluejets

Oct 5, 2014
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Must be a while since you bought a hall effect switch...5 here for $2.17 delivered. Used a couple now and work fine.

https://www.ebay.com.au/itm/5PCS-KY...889401?hash=item3d8422f879:g:xhsAAOSwmXpb-MVQ

Surely the friction of the bearings (using ball bearings) and mechanism would be negligible when accounting for the force applied by the 3 cups for arguments sake say on 300mm stems.

Price on some bearings here also just in case.(zz6000...10x26x8) $1.69 ea.

https://www.ebay.com.au/itm/Deep-Gr...hash=item36264dbf8c:m:mytAlBjabMo5hGYGXi11Z9g

If there is concern about the weight of the disc or the number of magnets that may be required, simply put one magnet on the back of the hall effect and use a thin "toothed" metal dics on the rotor.
 

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hevans1944

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Jun 21, 2012
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Must be a while since you bought a hall effect switch...
Sometime late in the previous century or the early part of this century (I forget exactly when), I purchased a commercial pair with built-in magnets for sensing the passage of steel spokes in a Welch foreline vacuum pump. The outputs were used to re-trigger one-shot multivibrators as each spoke passed by a sensor. If a certain period of time elapsed without a re-trigger event, an interlock relay tripped to shut down the vacuum system, thereby (hopefully) preventing atmospheric pressure from forcing hot diffusion pump fluid out into the vacuum system, where it became a real PITA to remove... weeks of work involved. They called me back about a year later, after I had "retired," because someone had tugged too hard on the Hall-effect connections and one of them had come loose. Any competent electronics technician would have discovered the problem in about ten seconds or less (that's how long it took me to figure out what was wrong), but perhaps the only help they had (an ex-Navy guy) wanted me to come back, if only for a little while. Didn't matter. I got paid a few hundred bux as a consultant, and a few years later they got rid of the machine and turned the laboratory floor space it occupied into office cubicles.
 

TKowalskii

Apr 13, 2019
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Hello once again. I apologize for not being active on this page. And to the one concerned, I was just a passenger in the car. With it being spring break in my school I have been more at the beach than working on this project. However over break my class was assigned to learn op amps so that has been on my mind. Thanks to a person on this thread recommending an op amp. Because of this I am currently in the process of using a non-inverting op amp circuit to take the approx .2 volts and amplify it so that it is large enough to pass the .7 V across VBE. To those recommending the three cup anemometer, I would have to talk to my teacher but the class suffers from zero funding, and all parts come out of the teachers pocket. The motor we are using is just a cheap dc motor that was in some toy car. I tried to upload my current schematic, but it seems to not wish to work on my phone. Thanks everyone so far, this is just an “extra-credit” (it’s more just for fun cause I don’t get anything out of it) project so the physics class kids get something to test.
Edit: I believe the image should work now
JhXFvMnZKfkaeRlWB6e1XvwGzjRuSaKNji8rb3YYLNctxaqDjgP-XOXHQu_rMAbyUPFHXJY0eeUO6eEsIIb6cfakXAbvkfaf2jAPcfmYtH6yLQHPI9M-wQLi8OceOX_cQdAqPyqU
 
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TKowalskii

Apr 13, 2019
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I also suppose now that I actually was able to post the schematic, I can give a better explanation of the reasoning behind all of it. This is more or less a proof of concept for another class to work on. The LED will instead be replaced with a release mechanism using a relay. The class that will receive this (they are called the engineering class, my class is just the electronics class) are in need of something that releases something else (sorry this is so vague the engineering class hasn't started their part of the project yet) when enough wind is detected. That class designs the fan blades and release mechanism. This project isn't to detect the amount of wind, just to detect the presence of a significant amount of wind. This project is also to increase interest in the class to keep it running, and to hopefully get funding for parts for later years. Thank you once again for all those showing interest. And please feel free to criticize ant part of the schematic above, from my extremely sloppy resistors or just problems in the schematics itself
 

hevans1944

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Jun 21, 2012
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Schematic? I don't see whatever you thought you posted.

Well, as long as you are having fun and preparing future engineers for work in the real world, you might as well learn something too. A non-inverting op-amp amplifier has a very high input impedance, which is good because you don't want to "load" your toy dc motor by drawing any appreciable current from it. To do so would increase the torque required to turn the motor shaft, since the motor is essentially operating as a dc generator. As you might surmise, any torque required to turn the motor shaft represents an unwanted load that sets a lower limit on the wind velocity you can measure.

The op-amp output represents a voltage source, and if applied to the base-emitter junction of a transistor, the op-amp output will have to be current-limited with a resistor to avoid damaging the transistor base-emitter junction. Since you are now playing around with op-amps, you might as well study how they can be wired as low-performance voltage comparators with adjustable thresholds and hysteresis. All comparator circuits need hysteresis to perform in a stable and repeatable manner. The nice thing about comparators is they are bang-bang type devices: either their output is ON or their output is OFF. You pick the state you want to light up your LED, sound an alarm, or release something with a relay to trigger the start of global thermonuclear war. You imagination will be your only limit! Well, that and money of course.

You can build your own three-cup or four-cup anemometer vane using ordinary ping-pong balls. Use a hot-wire cutter or a very sharp razor knife to slice the balls in half. Glue the halves to a central support made from light wooden sticks or plastic straws. Hot-glue the central support to the toy motor shaft. Now go do some electronics and report your results back here, so others may be helped by your experience, strength, and hope.:D
 

hevans1944

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Jun 21, 2012
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I also suppose now that I actually was able to post the schematic ...
It finally appeared. The schematic looks to be essentially correct, but what is the purpose of the voltage divider consisting of R3 and R4? Why are there no values associated with any of the parts? What is the transistor part number? What is the op-amp part number? The resistor, R6, in the emitter of the transistor serves no purpose and should be replaced with a wire connecting the emitter to "ground." The value of R7 should be small enough to drive the transistor into saturation, while large enough to limit the transistor base current to a safe value when the op-amp output is at its maximum level. The current through the LED will then be limited by the value of R5.

Your hand-drawn schematic is just fine. For comparison, Google Bob Pease and his columns named Pease Porridge. Bob was an MIT graduate and a wizard at op-amp circuit design. It's not how pretty you make the schematic that counts; what counts is how well the schematic conveys the intent of the circuit design. Bob was famous for drawing ten square feet of schematic on the back of a napkin and keeping what appeared to be a somewhat messy office. See here.
 

TKowalskii

Apr 13, 2019
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It finally appeared. The schematic looks to be essentially correct, but what is the purpose of the voltage divider consisting of R3 and R4? Why are there no values associated with any of the parts? What is the transistor part number? What is the op-amp part number? The resistor, R6, in the emitter of the transistor serves no purpose and should be replaced with a wire connecting the emitter to "ground." The value of R7 should be small enough to drive the transistor into saturation, while large enough to limit the transistor base current to a safe value when the op-amp output is at its maximum level. The current through the LED will then be limited by the value of R5
I apologize for the lack of labels, the transistor will most likely just be a 2n2222 due to their abundance and I will feel a bit less guilty frying one of them. The Op amp is a UA741. The resistor values are just tomorrow’s project and may have to be changed later because I don’t know what values are at my school. The voltage divider at R3 and R4 is to get 2 volts through the supply so if a hurricane force wind comes through, the output just peaks out at 2 volts. As for R6, no clue, thank you for telling me. Maybe it was just because of muscle memory of the 4 straight weeks of transistor math I did in class. And as for R7, I’ve looked through the internet and looked at the data sheet for the transistor and I can’t find a good answer, what is around a good value for current to put through the base of a transistor, in class we did math in the uA range, but no specific telling of when it would be too much. Thanks again
 

duke37

Jan 9, 2011
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Op-amp.
I am still in the 741 era but this amp will not respond to signal close to the negative supply.
Horowitz and Hill. "The Art of Electronics" gives details of various op-amps.
The 358 or 3130 series should do. Look up the data.
The op-amp output may be sufficiently powerful to drive the led directly.

If you use a relay, then a flywheel diode is imperative to protect the transistor.

If you are working in conjunction with engineers, then get them to make the cup driver. Co-operation is the name of the game.
Vertical axis turbines use cylindrical blades. Tin cans can cut to make these.
 

FuZZ1L0G1C

Mar 25, 2014
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Vertical axis turbine, as @duke37 said in #17, then instead of a motor, (which creates drag), drill 8 or more small holes vertically through both faces of the turbine, then use an LED / photo-transistor to detect the modulated beam, frequency dependant on wind velocity.
However, this may be steering away from your syllabus, if a DC motor-as-dynamo is specified.
Turbine's sensitivity can be amplified by means of conical 'ducts' that look like wave-guides.
The duct hack was seen on a DSTV reality series.
 
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