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I have a very simple question:

I am trying to power five 28BYJ-48 stepper motors attached to some ULN2003 driving boards but I am confused about the amperage I need. Each of the stepper motors needs 5V and 250mA to run and my driving boards take up to 500mA.

I'm wondering if I need to buy a power supply that's 5V/250mA or 5V/1.25A? If its the latter I'm worried that my driving boards may fry under the high amperage, what should I do about that?

Many thanks to anyone that can help.

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  • \$\begingroup\$ Are you aware that ULN4xxx rises about 1V when turned on, so current is only 100mA per phase x 2 = 200mA per motor thus max no load Revs/s increases from 1s to about 1.2s if full or (double?) step so 1A total but why not use a Molex output from a spare PC PSU for now. If you wanted more speed and torque use 10V with the ULN400x or use better switches with low Vce(sat) \$\endgroup\$ – Tony Stewart EE75 May 3 at 19:01
  • \$\begingroup\$ @TonyStewartEE75, Tony I have not clue what this means man. I do not know any of the technical words you just said \$\endgroup\$ – KidWithComputer May 3 at 22:54
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If each stepper requires 250 mA then your power supply current rating should be ≥ 5 × 0.25 A = 1.25 A in order to run all the motors simultaneously. The combination of motor and driving board will draw the required current and no more.

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  • \$\begingroup\$ So even if I supply 1.25A to a single board/motor they will be okay? \$\endgroup\$ – KidWithComputer May 3 at 22:50
  • \$\begingroup\$ Thought experiment: A 12 V car has a battery capable of supplying 100 A. What happens if you connect a 12 V, 6 W (0.5 A) bulb to the battery? How much current will the battery supply? \$\endgroup\$ – Transistor May 4 at 7:54
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Adafruit/Digikey specifies at DC R=42 Ohms per winding, but since it is a unipolar with a common tap, I believe that is per half winding and only 1 leg of each winding can be powered on so if using 5V across both coils it would be 5V/42ohms x2= 238mA per motor x5 =1.2A

But the ULN2003 Darlington has a Vce(sat)= 0.9 V typ. 1.1 V max. @ Ii =250μA and Ic =100mA so the motor winding only gets 4.1V typ or about 100mA/winding or 1A total.

But you should always use a supply rated for much more current than you need like 25% min. but if you have a spare PC PSU you can provide up to 10A from the Molex plug for HDD’s and use a decoupling cap near the load drivers.

Ref:

https://www.ti.com/lit/ds/symlink/uln2003a.pdf

https://www.digikey.ca/htmldatasheets/production/1825279/0/0/1/858.html Motor specs

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  • \$\begingroup\$ Tony, I really appreciate this good info man but I can not understand anything your are saying, could you dumb it down for me? I barely know ohms law \$\endgroup\$ – KidWithComputer May 3 at 22:52
  • \$\begingroup\$ Go learn Ohm’s Law , get familiar with switch voltage vs current from the specs. Do not believe Digikey tech or anyone else who says these motors draw 240mA .from a Darlington driving 42 Ohms from 5V. you only get 4V. go read some links in my profile. You must learn how to learn or find a good tutorial on steppers and power supplies. Do what I said about a PC surplus power supply and search to enable it without a MOBO with a simple jumper. \$\endgroup\$ – Tony Stewart EE75 May 3 at 23:30
  • \$\begingroup\$ Tony I appreciate it but I really I don't use electronics enough to make it worth reading about, I've tried to before and the stuff is so just so complex and honestly I find much more useful to talk to the friendly people here who are much better at explaining. Also what is a Darlington/MOBO enabling \$\endgroup\$ – KidWithComputer May 3 at 23:49
  • \$\begingroup\$ Can you search how to use the 5V from a surplus computer power supply to power it. The Darlington is the name for the dual NPN drivers in the ULN4002 in my 1st link . When it switches low the output is not 0V but rather rises to 0.9V because of the dual transistors cannot switch to ground all the way. Hence only 100mA per winding controlled in a sequence to step the motor angle in one direction or the other. It seems you don’t want all the detailed explanation of how it works , rather a good kit that shows what you have to do to make it work. \$\endgroup\$ – Tony Stewart EE75 May 4 at 2:22

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