Resistor required to drop voltage?

Question

Hi,

I am currently doing an electronics project, and I've got mixed up in something quite simple... My project is composed of a motor (running on a max of 3V) and several circuits (eight, all of them exactly the same).

I have succeeded in operating a sample circuit using a 9V battery. Now, I am looking at how to connect everything (8 circuits and motor) together in parallel, so that they work at the same time using a 9V DC supply. However, I'm not sure what resistance I should put in series with the motor, to drop its voltage to 3V (so that I do not burn it!).

Help! Thanks!

Answer 1

As I understand it, you have successfully tested the configuration
9v battery --> circuit --> motor --> back to battery

This implies that the circuit is using up at least 6V of the potential difference because the motor did not burn out. ;-)

The desired configuration is

9V battery --> 8 copies of same circuit in parallel --> resistor --> motor --> back to battery

and the goal is to keep the total resistance the same as in the first case so that the current through and voltage across the motor are limited to the successfully tested values.

Unfortunately you haven't given enough information to answer this question. Fundamentally, the circuit resistance has dropped to 1/8 of its value, so the resistor has to equal 7/8 of the resistance of a single copy of the circuit. We know that this is twice as much as the resistance of the motor, but we don't have actual values for either. If you measure the current flowing through the motor in the single-circuit configuration, you can determine the overall resistance and hence the value of the resistor required.

Answer 2

Because of the variable current draw of a motor, you will need to use a voltage regulator. As the motor starts the current is high and then it will drop as the motor reaches full speed. Under load more current will be required etc. Use a voltage regulator for the maximum current required by 8 motors to give 3 volts from a 9V battery.

Answer 3

Since the motor current varies with the load applied on its shaft, use a voltage regulator IC like LM317. You can set the required voltage by adding a few components. This will give a constant voltage to the motor, irrespective of the load and you need not worry about the motor being burnt.

Answer 4

If you're talking powering everything from a 9 vdc battery, then I'd suggest you not try it. The current drain would probably empty the 9 v batt rather quickly.

Why not use a separate pair of d-cells (in series) for the motor?

Answer 5

Try a 3.5V zener diode across your supply for the motor