CurtPalme.com Home Theater Forum

ecrabb

I need to mod my new Crown amps and I want to verify that I figured the calculation right. They have a 12vdc 80mm fan in them.

Here's the fan specs:
RDH8025S1, 12vdc rated, 6vdc startup, .23A, 3000RPM, 42CFM

When you power the amp on, it spins the fan up to full speed then after a second or two, backs the fan down to probably 1/2 speed. Even at that speed, it's still way too loud. Some guys just disable the fan altogether. Apparently, in home HT duty, there's enough air that the amps never overheat. That bugs me a little because I know what heat does long-term. Even if the amp isn't overheating, it could still be much hotter than necessary. So, I want to see if I can slow them down a little to make them mostly silent, but still have a little air moving to keep the air inside the amps from getting stagnant.

Now, if the amp is firing the fan with the full 12v blast at startup, but the fan is rated for a 6vdc startup, I think I'll shoot for a little above that 6v - say 7v.

P = .23a * 12v is about 2.76w...

Problem is, I'm having a brain fart and I can't remember how to figure the voltage drop. Let's say I want to drop 5 or 6 volts. I can't remember how to figure the resistor value for the voltage drop. Can somebody help?

Thanks,
SC

WTS · Joined: 08 Mar 2006 Posts: 1276 Location: Calgary

V=IxE, so voltage drop is the current flowing through the R times the valve in ohms.

Chuck27 · Posted: Tue Nov 20, 2007 2:04 pm Post subject:

The resistance of the fan is 12v / .23A = ~52 ohms (no omega key on my kb). You may want to verify the resistance using your own instruments.

If you put a 52 ohm resistor of appropriate wattage in series with the fan, you will drop the 12v to 6 by the time it gets to the fan. The key is appropriate wattage because the resistor will get hot while it's doing this. I'd go at least double and get a 5-watt or even a 10-watt wire-wound resistor.

Voltage drop is proportional to the value of the series resistor.

52 ohms from the resistor plus 52 ohms from the fan = 104 ohms.

12 v / 104 ohms, A = 0.116 Amps

52 ohms x 0.116 A = 6 v, which is the voltage drop across that section of the circuit.

Hope this helps...

Chuck

**Curt Palme** · Posted: Tue Nov 20, 2007 2:48 pm Post subject:

One thing to keep in mind: Some of these fans have regulators/electronics in them, so you can't go by the resistance measurements that you get. I'd get a 100 ohm 2 watt pot and see where you want to be noise and speed-wise. THen measure the pot.

WanMan · Joined: 19 Mar 2006 Posts: 10270

I do not think this board or BB software is supporting special characters, even though I've posted the HTML for Omega.

<TR><TD>937</TD></TR>

ecrabb · Posted: Tue Nov 20, 2007 3:45 pm Post subject:

Awesome. Thanks guys. I came up with 52 ohms, too but assumed I was doing something wrong because it matched the resistance of the fan. Now that I think about it, that makes sense that it would be the same. Duh.

Great idea to test and measure a pot, Curt. It's a REALLY basic fan. Looks very cheap and no-frills, so I'm betting/hoping on no special regulation. Hopefully, RatShack has a suitable pot so I don't have to spend an hour of my time and $3 in gas just to drive across town to go to the only electronic wholesale supplier in the area.

Here's my special character test:
52Ω

Thanks for the help, guys!

SC

EDIT - WanMan, I edited my post and the forum changed my special character to read:

Chuck27 · Posted: Tue Nov 20, 2007 3:49 pm Post subject:

cmjohnson · Joined: 03 Apr 2006 Posts: 5180 Location: Buried under G90s

Curt's suggestion is the best way to go. Tune for highest fan speed that you can tolerate, noise-wise, measure the pot, and get a suitable resistor of that value. Or just leave the pot in the circuit and put it on the back panel.

CJ