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**jflongo** · 03-01-2013, 03:27 PM

Ok, I feel like I'm brain dead right now and stupid

I am using a 8V 500ma output adapter, and a 80mm computer fan, it runs it as is, but too fast.

I figured I could just hook in a resister to slow it down for now.

I have a 330 ohm resistor, so i figure V = (0.0005)(330) = 0.165 voltage drop

I must be doing this wrong, when I hook the resistor up, the fan won't spin at all, I'm a little confused. I know there are better solutions, I'm just wondering what I'm missing here. I have tried it on both wires, same result.

**jflongo** · 03-01-2013, 04:21 PM

I still would like to know what's wrong, but I think i'll scrap the resistor idea and buy this instead.

http://www.amazon.com/Zalman-Fanmate-2-Speed-Controller-Retail/dp/B000292DO0/ref=sr_1_4?ie=UTF8&qid=1362158287&sr=8-4&keywords=computer+fan+controller

**afr0byte** · 03-01-2013, 04:28 PM

Generally, afaik, voltage dividers use two resistors, like the diagram here: http://en.wikipedia.org/wiki/Voltage_divider

**jflongo** · 03-01-2013, 04:59 PM

Quote:

Originally Posted by afr0byte

Generally, afaik, voltage dividers use two resistors, like the diagram here: http://en.wikipedia.org/wiki/Voltage_divider

I'm just trying to drop the voltage by x amount, so not sure how to use that diagram with no ground?

**afr0byte** · 03-01-2013, 05:03 PM

Perhaps use figure 3-64 as a reference: http://www.tpub.com/neets/book1/chapter3/1-35.htm

**jflongo** · 03-01-2013, 05:18 PM

So is this saying by adding the resistor in, i'm unbalancing the circuit?

**afr0byte** · 03-01-2013, 05:21 PM

Quote:

Originally Posted by jflongo

So is this saying by adding the resistor in, i'm unbalancing the circuit?

My knowledge of electronics is limited, but I think you'd be decreasing the current draw by simply adding additional resistive load.

**IslandLizard** · 03-01-2013, 05:29 PM

Quote:

Originally Posted by jflongo

I have a 330 ohm resistor, so i figure V = (0.0005)(330) = 0.165 voltage drop

you're 3 decimals off:
V = .5 (A) x 330 (Ohm) = 165V, if you had that much to drop.

500mA is .5 A !

A divider is easiest, although you could build it with a single resistor. Start out with 0.33 Ohms perhaps?

And calculate the needed wattage of such resistor: V*V/R or V * I (0.165 * 0.5)

molybdenum26 · 03-01-2013, 05:32 PM

Quote:

Originally Posted by afr0byte

My knowledge of electronics is limited, but I think you'd be decreasing the current draw by simply adding additional resistive load.

That's correct. By only having one resistor in series with the fan, you're dropping pretty much the entire voltage across that resistor. A voltage divider would have to be used where the ration between the two resistors gives you the desired voltage drop across ONE of the resistors, which your fan would be hooked in parallel with.

The ground shown in the above diagram can be hooked up to the negative side of your power supply. It's a general diagram.

**IslandLizard** · 03-01-2013, 05:32 PM

...and that is IF the fan consumes 500mA. Your power brick's current rating has nothing to do with that.