I'm currently building a stir plate but I think I made a mistake when purchasing the parts....I bought a 7.5v dc power supply and a 24v fan. I thought the fan was 12v. Is it possible to power the 24v fan with my power supply ?
help stir plate !
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What matters is the power. Your fan will try to draw the power it needs to turn and it will compensate the voltage it is missing with current. If you use a power supply rated for 7.5 V, you run the risk of drawing too much current through the power supply transformer and/or through the wires from the supply to the fan.
P = I*V
So if your fan is rated at 1.44 W, you have
1.44 = I*V
The fan is rated for 24 volts, therefore:
1.44 = I*24
And is prepared to handle a current of I = 1.44/24 = 0.06 Amps
However, if the fan only receives 7.5 V, the current it will try to draw will be:
I = 1.44/7.5 = 0.192 Amps
This is 32 times more current than the fan is rated to handle adequately.
Also, the fan won't actually be able to draw the power it needs to spin at the speed at which it is designed for as it needs the higher voltage to overcome the internal impedances. This will result is a slower spin, and perhaps even no spin, while simultaneously risking burning up your wires and/or power transformer. If your fan is capable of handling the higher current, it is worth a shot and the fan may work just fine. Personally I wouldn't even bother as I don't like to waste equipment but it's entirelly your call. Just be sure to test it on an outlet that has a ground-fault interrupter.
P = I*V
So if your fan is rated at 1.44 W, you have
1.44 = I*V
The fan is rated for 24 volts, therefore:
1.44 = I*24
And is prepared to handle a current of I = 1.44/24 = 0.06 Amps
However, if the fan only receives 7.5 V, the current it will try to draw will be:
I = 1.44/7.5 = 0.192 Amps
This is 32 times more current than the fan is rated to handle adequately.
Also, the fan won't actually be able to draw the power it needs to spin at the speed at which it is designed for as it needs the higher voltage to overcome the internal impedances. This will result is a slower spin, and perhaps even no spin, while simultaneously risking burning up your wires and/or power transformer. If your fan is capable of handling the higher current, it is worth a shot and the fan may work just fine. Personally I wouldn't even bother as I don't like to waste equipment but it's entirelly your call. Just be sure to test it on an outlet that has a ground-fault interrupter.
Snood
Well-Known Member
Buy a new fan, or a new power supply, whichever is cheapest. I have no idea why you didn't get a 12v fan and a 12v supply. both very common and cheap.
Yes most door bell transformers are 24volt.
Well I don't know how to wire a transformer to my setup ! Are you telling me I won't be able to find a 24w wall adapter just like the 7.5v I used ?
Sure just cheaper you have 2 wires in black/ hot and white/neutral-common and 2 lines out to fan. Cut the end off an extension cord to connect to the line side of transformer, load side goes to the fan.
soccerpall86
Member
I am going to go ahead and challenge this analysis.
The fan is essentially a load (resistance R) powered by a the power supply. It is rated for use at 24 volts, which just tells you that the fan won't burn up with the current drawn/ power dissipated at that voltage. Using a 7.5V supply you are under powering the fan and V=IR is the most relevant equation. The 7.5V supply will only output enough current to get 7.5 V across the load, which is less current than the fan is rated for. The fan will be underpowered(slow/ maybe not moving). and the power supply will be perfectly happy.
Feel free to correct me, but I think one should be more concerned with using a power supply with a voltage higher than the fan is rated for. In that case, more current could flow through the power supply and fry a component.
Hang Glider
Beer Drinker
I'd hunt down a computer fan - 12V, cheap, easy to find (e-bay, computer stores in your area). And an old cellphone charger will accommodate the power. Don't have to make it any harder than that.
This is entirely correct assuming the fan's R is static. And Power doesn't necessarily have to be static either, I was operating under the assumption that the impedance would vary based on the fan's speed and voltage supply, and the fan would continue to try to draw a particular rated power. Considering fans are reactive impedances and not static resistances, this wasn't an unreasonable assumption. When the voltage supply is lowered, generally inductive load impedances also fall. Without knowing that relationship, I thought using the power equation would be more appropriate as a worst-case scenario, conservative thinking sort of approach.
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