Reduce voltage with a diode

Got a 5v fan (40mm) with the following specs.
I used these specs to calculate the required potentiometer needed to control the speed

Vf = minimum fan voltage (2.5)
Vs = (power supply), (5)
Rf = fan resistance in ohms (35.7)
Rr = potentiometer resistance in ohms
Vf1 = fan voltage, (5)
Va = fan amps, (0.14)
Calculate Pot size:
Rf = Vf1/Va (36)
Rr = ((Vs*Rf)/Vf) – Rf ((5*36)/2.5) - 36 = 36ohms IVE GOT A 50R

Ok thats done now the issue I have is the pot rating is .5 watts and the fan is .7 (amps x volts) or IxV (.14 x 5) =.7
Is this a problem? .2watts higher? if so can I lower the volts to the fan by 1.4 using a pair of diodes? I understand most silicone diodes use .7volts
here is my proposed schematic.
i have 1n4002, 1n4007, bat85, bat46

Hi. A potentiometer is meant to pass much smaller currents as to drive a power transistor base/gate to control speed properly. Why is the supply to the fan dropped from 5V to 3.6V with D1 + D2 ? Do you realize doing that again with more diodes in series can step down the voltage to 2.9V, 2.2V, 1.6V, 0.9V, 0,2V which with a multi position switch can provide stepped voltages to 0V instead of destroying potentiometers ?

Much better to drop the voltage through diodes, as the reduction is much less dependent on the stability of the current drawn.

ak

Externet said:

Do you realize doing that again with more diodes in series can step down the voltage to 2.9V, 2.2V, 1.6V, 0.9V, 0,2V which with a multi position switch can provide stepped voltages to 0V

Click to expand...

Thats a pretty cool suggestion.

AnalogKid said:

drop the voltage through diodes

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In this project the fan is used a few minutes at a time once per week. Is any silicone diode better than another? ie 2004,2007

The 1N4004 is rated at 400V and 1A.
The 1N4007 is rated at 1000V and 1A.

So either will work fine.

Mike.

ThomsCircuit said:

silicone diode

Click to expand...

Those are kind of rubbery.
Silicon diodes are better.