I have the following setup with a voltage divider that steps-down a voltage from 7V to around 6V, so I can plug it into a DC motor.


simulate this circuit – Schematic created using CircuitLab

When I measure the voltage across the point A and B with my voltmeter, I get around 5,95V, but as soon as I plug my 6V DC motor across those same points, my previous voltage of 5,95V seemingly disappear and I can't understand why.

PS: I also tried to replace the LiPo and the XL6009E1 voltage regulator with my power supply at 7V but I get the same result.

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    \$\begingroup\$ measure the voltage across R1 when motor is connected ... don't forget that the motor is equivalent to a low value resistor \$\endgroup\$ – jsotola Nov 28 '20 at 18:53
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    \$\begingroup\$ Once you connect the motor it draws additional current. The current is also going through R1 which causes additional voltage drop. \$\endgroup\$ – Lars Hankeln Nov 28 '20 at 18:53
  • \$\begingroup\$ One of those problems many beginners have faced with motors :) \$\endgroup\$ – Mitu Raj Nov 29 '20 at 0:13

See that 1k resistor? At 6V, you can get about 6 milliamperes of current through it.

Your motor needs far more than 6 milliamperes of current to run. Your motor looks like a short circuit when compared to that 1k resistor. The voltage drop across the motor will be very small - almost all of the voltage drops across the resistor.

Remove the resistors, and don't sweat that 1 volt difference between the power supply output and the motor rating. The motor will run a little bit faster, but it shouldn't bother anything.

  • \$\begingroup\$ If I understand correctly, the increased size of current drawn by the motor cause the voltage drop across R1 to increase, letting little voltage left for the motor, right? \$\endgroup\$ – geauser Nov 28 '20 at 19:09
  • \$\begingroup\$ Correct. The resistor limits the current. V=IR. I is about 6 milliamperes. Since R is very small for the motor, IR is a very small voltage. \$\endgroup\$ – JRE Nov 28 '20 at 19:11
  • \$\begingroup\$ I tried simulating the circuit with smaller resistances, namely R1=10 and and R2=60 the motor R3 at a resistance of 35 ohms (measured at 6V unloaded), but I get only 4,8V to the motor... Is it correct to say that my voltage divider here can't really be used as I the resistance of the motor keeps changing? \$\endgroup\$ – geauser Nov 28 '20 at 19:37
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    \$\begingroup\$ Yes. The motor resistance isn't fixed. It is lower when the motor is stopped, and gets higher as the motor turns faster. If you block the motor from turning then the resistance is at its lowest. A series resistor is a bad way to regulate a DC motor. \$\endgroup\$ – JRE Nov 28 '20 at 19:40
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    \$\begingroup\$ It's more complicated than just "the motor resistance isn't fixed" -- this may trip you up later. But for now that's a good working explanation. Just remember that later someone is going to say "it's more complicated than that" and then you'll need to listen and learn. \$\endgroup\$ – TimWescott Nov 28 '20 at 20:19

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