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For a project I'm working on I need a power supply of 3.3 V to supply power and current for 14 mini vibration motors that work on 3.3 V. Each motor is rated at 85 mA so I need to be able to deliver ~1.2 A.

A voltage regulator such as this one can deliver up to 800 mA current so I guess it is not a good choice. I was thinking of using voltage divider and a voltage follower circuit using an LM358 op-amp. What do you think about that solution? Here is my schematic for that:

enter image description here

With that circuit the output voltage at LM358 pin 1 will be fixed at ~3.3 V and the current will be as much as the phone charger can deliver(?), so if my phone charge is rated @5 V with 2 A, will I be able to deliver almost 2 A of current to the motors?

Full schematic in here:

enter image description here

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  • \$\begingroup\$ Is my updated schematic will work? the input voltage to the usb breakout module will be 5V @2A phone charger. \$\endgroup\$
    – Omer
    Nov 24, 2022 at 10:19
  • \$\begingroup\$ This is now a mess, what is the question really about after the edit? \$\endgroup\$
    – pipe
    Nov 24, 2022 at 10:50
  • \$\begingroup\$ you are right. My question is 1) how can I power 14 motors that consume 85mA each with a 5V 2A phone charger? I think the best is to use two of the 3v3 module voltage regulator so each 7 motors will consume around 600mA (within the 800mA max current) and within the 2A phone charger output current. 2) Do I need to add a bypass capacitor in parallel with the motor and the 1N4001 diode? if so what value? 100nF ceramic cap is ok? \$\endgroup\$
    – Omer
    Nov 24, 2022 at 10:53
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    \$\begingroup\$ you should ask yourself if you really need regulated voltage for the motors. I assume you don't and in that case you get no benefit from voltage regulators. They will waste 1.7 V as heat just as resistors would do while being 10-100 times more expensive. \$\endgroup\$
    – Sim Son
    Nov 24, 2022 at 13:20
  • \$\begingroup\$ Can I replace the 2N3904 npn transistors with bc337 npn transistors? \$\endgroup\$
    – Omer
    Nov 28, 2022 at 12:02

7 Answers 7

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One option could be to just use two of the 800ma regulators that you all ready referenced then run the 14 motors in 2 banks of 7 ? Each bank of 7 should draw 595ma.

Yet another option is to find a 3.3v step-down module with a higher output current, such as this or even this.

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Your circuit won't work. Those op-amps are used for signals or very weak loads. They can't provide enough current for the motors.

If you are not explicitly after a sophisticated solution I'd actually recommend to just put a resistance of 20 ohm in series with each motor and supply them with 5V. At 85 mA each resistor will dissipate around 150 mW. Resistors rated for 250 mW or even 500 mW are common and cheap.

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  • \$\begingroup\$ if I'm putting 20ohm resistors with 5V power supply the current is 250mA . shouldn't I use ~50ohm resistor? 5V/50R = 85mA (?) \$\endgroup\$
    – Omer
    Nov 24, 2022 at 7:11
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    \$\begingroup\$ If the motor actually draws 85ma at 3.3v then the 20 ohm resistor is the correct value to use. At 85ma a 20 ohm resistor will drop 20 x 0.085 = 1.7v, So 5v - 1.7v = 3.3v. However a motor will vary in voltage from start up till spinning at full speed, so do a bit of testing to find the actual resistor value that works best overall. \$\endgroup\$
    – Nedd
    Nov 24, 2022 at 8:27
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    \$\begingroup\$ Note that this is only usable if the current is constant; if it varies, or you don't know what it will be, you can't just use a resistor like this. Helping somewhat is the fact that motors are very unpicky loads; they won't break from a few milliseconds of too high voltage. \$\endgroup\$
    – Hearth
    Nov 24, 2022 at 19:50
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    \$\begingroup\$ Would using two diodes rated 2A in series with the 5Vsupply in order to drop the voltage to ~3.3v is better solution? \$\endgroup\$
    – Omer
    Nov 25, 2022 at 12:44
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It won't work.

The LM358 op-amp can only output few tens of milliamps, so it can't drive even one motor.

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  • \$\begingroup\$ and If I will use tl072 opamp instead? What are my option to deliver 3v3 voltage with at least 1.2A of current to those 14 motors? \$\endgroup\$
    – Omer
    Nov 24, 2022 at 6:47
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    \$\begingroup\$ No standard opamp can give you 1.2A output. Why can't you just use two or multiple regulators if each can provide 800mA? At any rate 1.2A drop for 1.7V will waste about 2 watts so a single component or even two might not handle it. \$\endgroup\$
    – Justme
    Nov 24, 2022 at 6:52
  • \$\begingroup\$ As such: imgur.com/a/JhSTsS0 ? \$\endgroup\$
    – Omer
    Nov 24, 2022 at 7:15
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    \$\begingroup\$ @Omer You MUST put that imgur link into your question and probably the diagram. Did you originate that specifically for this task? Do you mean that you want to be able to control the motors individually - that is not at all obvious from your question. The imgur diagram will do what you want AND allow individual control of each motor. \$\endgroup\$
    – Russell McMahon ♦
    Nov 24, 2022 at 8:47
  • \$\begingroup\$ I did so just now. \$\endgroup\$
    – Omer
    Nov 24, 2022 at 10:18
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If you must have an op-amp circuit, use a power device, such as OPA548, which supply sufficient current.

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Do you really, really, really need to power these motors from a regulated source?

I don't think so.

  1. Chances are that the motors have a bit higher current draw at start. If you power them at once you will get a lot more than 1.2A inrush current. You will need a big-big capacitor or a better power supply or some logic not to allow them to start together.
  2. Chances are the motors don't have very tight specs, so the vibration frequency and amplitude vary between items even when the voltage is fixed.

In short, use 20 ohm resistors in series with the motors and power them with 5.0V directly.

  • inrush current limited to 250ma (maybe less, depending on the unknown motor properties)
  • simple
  • you may use vary the resistance in order to get lower frequency or higher fault-tolerance.
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  • \$\begingroup\$ What about dropping the voltage using two diodes in series? less parts same results as 20R resistor for each motor? \$\endgroup\$
    – Omer
    Nov 26, 2022 at 10:49
  • \$\begingroup\$ @Omer Pretty much reasonable idea, as long as you take care to dissipate ~2.5W at two adjacent elements. You should also keep in mind that at low current, Si diodes are closer to 0.6-0.7V forward voltage so you will get 3.6-3.8V when you power a single motor. This is somewhat out of spec, but I don't expect any issues with these motors. They are specified up to 3.7, but most circuits power them off a single Li-ion cell that is 3.7V nominal but can go up to 4.4V when well charged. \$\endgroup\$
    – fraxinus
    Nov 26, 2022 at 18:44
  • \$\begingroup\$ Thanks, I might drop the voltage using three diodes. I wanted to ask: what do you mean by taking care regarding the ~2.5W? where this power is dissipated at? \$\endgroup\$
    – Omer
    Nov 27, 2022 at 7:16
  • \$\begingroup\$ Each of your diodes will get about 1.2W of heat at the maximum load (1.2A) \$\endgroup\$
    – fraxinus
    Nov 27, 2022 at 13:20
  • \$\begingroup\$ and is that a problem for a 1N4004 diodes? should I use different diode? \$\endgroup\$
    – Omer
    Nov 27, 2022 at 16:30
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Feedback the voltage from the output transistor to the op amp to regulate the voltage Yep you do need a 5 watt transistor(maybe more)to feed all of those motors. Or as Henros states a TI OPA548

But if one of the motors locks. You also need a trigger to shutdown the motor current.

Or you can use a cheap (2N550, BSR13?) transistor (much less than 5 watts) as a fuse to each motor. I don't know of any fast blow small current fuses.

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Opamp you use is not good choice for your application. LM358 can source 40-50 mA.

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