0
\$\begingroup\$

Say I have a motor that uses 9v @400ma, but it's powered by a 12v battery. I 'd use the ohm law and get a: 7.5 ohm resistor.

But, as soon as the battery is being drained, let's say now the battery is 10v the resistor should be 2.5 ohm?

What consideration should I have if the voltage is not fixed (because of the draining).

I have a 6x 1.5D = 9v portable fan and I want to make it rechargable with 18650 3s1p using its pcb module to charge it but I have had that doubt about resistor.

This is how I currently have it, but I've notice it draws current even when it's off, i guess too many "bridges" and connections.

enter image description here

\$\endgroup\$
2
  • 3
    \$\begingroup\$ Why are you not considering using a voltage regulator? \$\endgroup\$
    – Andy aka
    Commented Jun 6 at 9:30
  • 1
    \$\begingroup\$ This is one of the reasons why a resistor is not a suitable voltage regulator. \$\endgroup\$
    – StarCat
    Commented Jun 6 at 11:49

1 Answer 1

Reset to default
0
\$\begingroup\$

If you really care about the voltage, use a low drop out (LDO) voltage regulator, not a resistor. LDO because you want it to be able to take in 10V and give you 9V, which is only 1V difference.

If you don't care, calculate a resistor that drops about 2V. So the motor is running at about 10V with a new battery and 8V on a run-down one. Close enough for many purposes.

\$\endgroup\$
1
  • \$\begingroup\$ I guess resistors are ONLY for fixed voltages not variable on time voltages. \$\endgroup\$
    – Panamanian
    Commented Jun 6 at 21:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.