How to calculate motor parts and energy requirements?

Question

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 a 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 the engineer had a smaller magnet, and he 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?

For example, 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.

Images of my parts

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. Thank you!

Answer 1

The DC motor torque is proportional to current which results in acceleration unless opposed by a spring whose torque increases with deflection.

The result is the calibrated deflection = a calibrated current. When added with a scaled series resistor, then it measures voltage.

The analog meter is often measured in kΩ/V. Large meters were 50kΩ/V while small meters were 1kΩ/V.

BLDC motors use a Hall magnetic field detector to sense a position used to commutate the coil current with electronic toggled power switches called a "Half & Full bridge driver". There are many different types using FETs.

The rules for ammeters are similar to motors with variables like magnetic field strength, number of turns, diameter and gap between the coil and magnet, as discussed in white paper.

http://www.ni.com/white-paper/14921/en/

As far as using solar power of tiny photo diode (PD) to drive a tiny motor, I recall a small 5mm PD response of 0.5mA per mW of optical power, thus the power in the motor cannot exceed this.