Question on wiring stir plate

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massdestruction

massdestruction

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I'm going to attempt a DIY stir plate like many plans I've seen online. I've got an unregulated AC-DC 12V power supply (1A) that I can use. When measuring it with a multimeter (under no load) it gives about 19V actual voltage. The way I understand the unregulated part is that when producing maximum current (1A) it will actually be at 12V. My fan only draws 0.3A of current, so my guess is that the voltage will still be above nominal--about 16V? My rheostat is rated at 3W (25 ohm). My question is, do I have any worries about the rheostat not being able to handle the power/heat dissipation given that the voltage will be higher than nominal?
 
The power dissipated in a resistor is I*I*R so even if the fan managed to draw the maximum current when at the maximum resistance setting the power would be .09*25 = 2.25W and that is the maximum it could ever be.

From the POV of the power supply: its internal impedance drops 7 volts at 1 amp load and so has a value of 7 ohms. To drop the 7 volts at 0.3 amp (thus presenting 12 V to the fan motor) the total impedance would have to be 7/0.3 = 23 ohms with 16 from the pot and the other 7 from the power supply i.e. setting the pot to 16 ohms will put 12 V across the motor terminals causing it to draw 0.3 amp. Thus the dissipation in the pot will be 0.09*16 = 1.44 watt.
 
ajdelange said:
The power dissipated in a resistor is I*I*R so even if the fan managed to draw the maximum current when at the maximum resistance setting the power would be .09*25 = 2.25W and that is the maximum it could ever be.

From the POV of the power supply: its internal impedance drops 7 volts at 1 amp load and so has a value of 7 ohms. To drop the 7 volts at 0.3 amp (thus presenting 12 V to the fan motor) the total impedance would have to be 7/0.3 = 23 ohms with 16 from the pot and the other 7 from the power supply i.e. setting the pot to 16 ohms will put 12 V across the motor terminals causing it to draw 0.3 amp. Thus the dissipation in the pot will be 0.09*16 = 1.44 watt.
Click to expand...

1. Is there anything AJ doesn't know?
2. Will this work for the OP?
 
LOL. Not sure about #1, but #2 helped me out quite a bit. Once I got the stir plate set up, however, it turned out that 12V made the motor go way too fast. So I ended up using a different power supply from my box o' tangled and forgotten wires (5.2 V). It gives a nice smooth stir with a 1.5-2" funnel. Sweet. Thanks, guys.
 

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