• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

How to calculate motor parts and energy requirements?

Thread starter #1
Hello Everybody, I’ve been wanting to ask this question for a very long time.


My main thing is knowing if a low power, low current motor is possible using a BPW34 photodiode. I hope you guys can help. Thank you!



I purchased a 3 in 1 soil tester meter https://www.ebay.com/p/3-in-1-Soil-Tester-Meter-for-Garden-Lawn-Plant-Pot-Moisture-Light-Ph-Sensor-Tool/1712433238?iid=311911276636


that comes with an analog panel meter


https://www.jameco.com/z/AVM7030-Velleman-Analog-Voltage-Panel-Meter-30-Volt-DC2-8-X-2-4_316654.html


In the center it has a photodiode BPW34


https://www.digikey.com/catalog/en/partgroup/bpw34/12351



The needle of this meter turns at very low voltages and current.

I measured everything and will post the exact info below.


My question is


How do engineers calculate how much voltage, current, coil resistance, and magnet strength a motor will need to be able to make a full turn?


Let’s suppose this magnet was smaller, and the engineer wanted to make a smaller meter. How would he know how many coils turns (resistance), voltage, and current he would need to make this thing turn?


Like in this science project


https://www.education.com/science-fair/article/no-frills-motor/


How did this person know how many turns (resistance) he needed, voltage, current, and magnet strength to make that motor turn?


In my meter, when you apply .5 mV and .5uA it will turn until stopped by a piece of metal at the bottom that holds the magnet.


The coil measures 794 ohms. The armature (coil) and needle weighs .415 grams.

The photodiode causes the coil to turn to full 180 degrees when it reaches .5 volts and 500uA (micro amps).

The bipolar circular magnet weighs 1.81 grams and has about 12 grams of lift force.


https://www.flickr.com/photos/153503439@N02/39315572332/in/dateposted-public/

https://www.flickr.com/photos/153503439@N02/39315594942/in/dateposted-public/

https://www.flickr.com/photos/153503439@N02/39315601022/in/dateposted-public/

https://www.flickr.com/photos/153503439@N02/38637464474/in/dateposted-public/




Another question that I have is


Since the magnet will stop at 180 degrees when the poles of the coil attract the opposite poles of the magnet, is there a small ic I could use to invert the current so that it would continue to spin? Don’t worry about the stopper.
 

kubeek

Well-Known Member
#2
The force that the coil makes is counteracted by a small spring, so that is where you have to look for the balance. Force of spring versus the force on the coil produced by coil current ad the field produced by the permanent magnet.
 

Latest threads

EE World Online Articles

Loading

 
Top