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My question:

This is a simple design that I have created using everything that I know about electronics so far. Is there anything missing in my design or are there any concepts which I have misunderstood ?

My design:

Preliminary stage:

An AC to DC adaptor that was originally used to power a phone was repurposed as a power-supply for an ESC which controls a BLDC motor. This power supply was used as a Lipo battery was not available. This adaptor was able to power this motor to a satisfactory level. The setup was adapted from a youtube video (link is at the bottom). enter image description here

Proposed design

Since the adaptor worked with one motor, it will now be used with four. Although this would drastically reduce the current used by each motor, this should not matter that much as no load is being applied to the motor.

This adaptor will be connected in parallel to 4 electronic speed controllers which in turn power four 1000KV BLDC motors. The adaptor takes in a 220 - 240V 100mA current at 50/60Hz from the power socket in the wall and outputs a 5.5V DC current at 500mA.

Since the loads are connected in parallel they will all receive the supply voltage and equally share the current provided therefore, each motor will receive 125mA current at 5.5V. Since rpm = KV * volts therefore, the maximum rpm of each motor will be 5500rpm. No load will be applied to the motors.

I checked the data-sheet for the motor(link is here: https://www.rhydolabz.com/documents/26/BLDC_A2212_13T.pdf) but I couldn't find any information on the minimum current needed for the motor to spin at 5500rpm. The data-sheet did say that the No load current was 0.5A but apparently this was only true if the voltage was 10V.

Why am I using this design or asking these questions?

I made this design so that I could test whether I understood the principles behind choosing a power supply for my motors. If the logic that I used above is correct then I can figure out how heavy my drone will be -> how much lift force each motor needs to generate -> how much current each motor will draw -> how many mAh my lipo battery will need to be. So this design is essential for any work that I plan to do in the future.

If the logic is incorrect then I was hoping that someone could point out where I was going wrong or provide resources which could help me improve my understanding.

Links to equipment:

Motors: https://www.amazon.com.au/QWinOut-Brushless-Outrunner-Multi-Copter-Quadcopter/dp/B07CVDHQKS

ESC: https://qwinout.com/products/qwinout-2-4s-30a-rc-brushless-esc-simonk-firmware-electric-speed-controller-with-5v-3a-bec-for-2-to-4s-lipo-battery-diy-multicopter-quadcopter

Layout: https://www.youtube.com/watch?v=uOQk8SJso6Q Timestamp: 9:24

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  • $\begingroup$ You successfully powered that ESC on 5.5V? I think a higher voltage power supply is the first step here. Why do you want to use a 5.5v supply? $\endgroup$ Aug 15, 2022 at 12:37

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The no-load current doesn't change much with voltage.

If you try to run four motors at 5v, you'll need a 2amp supply.

As your supply is only rated for 0.5a, it could do one of two things. It could supply 2a and overheat, or you might find it's voltage drops.

Also, if you apply any load to the motor, it will draw a lot more power. Even during the second or so it takes to spin up, it will be drawing more than it's no-load power. So while you might be able to get one motor to run, you won't be able to get any useful work out of it.

I don't think it's a good idea to try to run a 150W motor from a 2.5W power supply - and trying to run 4 motors will definitely overload the power supply.

You would be better to use the power supply for a laptop, or from an old desktop computer case.

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  • $\begingroup$ >at 5v it will be 0.25a. This is a vast oversimplification. To run at half the speed does not necessarily take one fourth of the power. In fact, the current may actually go up since the voltage goes down, but wattage stays similar (no load current is more about overcoming friction and is relatively independent of RPM) $\endgroup$ Aug 15, 2022 at 12:36
  • $\begingroup$ @BrydonGibson - good point, I'll update my answer. $\endgroup$ Aug 18, 2022 at 8:12

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