[madmickiii]

opssssssss, was thinking about something else then, the 78L12 is only a 100ma device,

Mike

[gar]

060111-1539 EST USA

GAWnCA:

You have a power supply that can provide 50 Watts output (25 * 2).

Your fan requires 12 * 0.43 = 5.16 Watts. Suppose this fan has about the same apparent resistance 12/0.43 = 27.9 Ohms independent of fan mechanical loading, then put a 27.9 Ohm resistor in series. This is not a good idea for at least two reasons --- you will use 10.32 Watts of your available 50 Watts and you will have this extra power to get rid of. This is exactly what will happen if you use a linear series pass regulator (the 7812). A resistor instead of the series pass in not good because the load probably is not a constant resistance.

The best idea is the suggested 24 V fan. The power load in this case is 5.16 Watts. Actually somewhat more because the source voltage is 25 V.

As an in between choice use a switching regulator from the 24 V to 12 V. Now your load power on the supply will probably be in the range of 7 Watts.

To be precise the series resitor to put in series with the motor is 13/0.43 = 30.2 ohms, and it power dissipation is 13 * 0.43 = 5.59 Watts. But it does not matter for this illustration because none of these values are actually that precise.

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