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I have loved my QSC K10's since I bought them 4 months ago but sometimes- and I swear it's really ROOM temperature that causes them to come on more than anything else (long story that's largely out of my control)- the fans are simply too loud. I've tried a new "quieter" fan, but it just doesn't get me where I want to be.
Now I've jammed 105+db, bass heavy, music through these for 30 minutes and the fans haven't moved, and the amp plate has remained cool, so I'm really not concerned with ~ halving the rotational speed of these things. I have read many have disabled them completely and gone years with no issues. I'm not willing to go that far, so...
The fans are 12V and marked 2.10W, so I figure that's .175 amps, seeing 2.10/12 = .175.
SO then if the target voltage is 7 volts (to ensure they start)...
QSC fan voltage: 12V
QSC fan amperage: 0.175A
V=IR (Ohm's Law), so 12 = 0.175R => R = 12 / 0.175 = 68.571
QSC fan resistance: 68.571 Ohms
Target voltage: 7V
V=IR (Ohm's Law), so 7 = 68.571I => I = 7 / 68.571 = 0.102
Target amperage: 0.102A
Voltage drop across resistor: 12V - 7V = 5V
V=IR (Ohm's Law), so 5 = 0.102R => R = 5 / 0.102 = 48.98
Target resistance: 48.98 Ohms
Power dissipated by resistor: 5V * 0.102A = 0.51W
So then I'm assuming a 50OHM 1 watt resistor inserted into the pathway should sort me? My question is, if I can only find 2 watt resistors at 50 OHMS ( that I can have in a couple days,
) I'm still OK, correct? That's (the wattage rating) just concerning heat dissipation?
Thanks for any help here, this is really not my strong suit!
James
Now I've jammed 105+db, bass heavy, music through these for 30 minutes and the fans haven't moved, and the amp plate has remained cool, so I'm really not concerned with ~ halving the rotational speed of these things. I have read many have disabled them completely and gone years with no issues. I'm not willing to go that far, so...
The fans are 12V and marked 2.10W, so I figure that's .175 amps, seeing 2.10/12 = .175.
SO then if the target voltage is 7 volts (to ensure they start)...
QSC fan voltage: 12V
QSC fan amperage: 0.175A
V=IR (Ohm's Law), so 12 = 0.175R => R = 12 / 0.175 = 68.571
QSC fan resistance: 68.571 Ohms
Target voltage: 7V
V=IR (Ohm's Law), so 7 = 68.571I => I = 7 / 68.571 = 0.102
Target amperage: 0.102A
Voltage drop across resistor: 12V - 7V = 5V
V=IR (Ohm's Law), so 5 = 0.102R => R = 5 / 0.102 = 48.98
Target resistance: 48.98 Ohms
Power dissipated by resistor: 5V * 0.102A = 0.51W
So then I'm assuming a 50OHM 1 watt resistor inserted into the pathway should sort me? My question is, if I can only find 2 watt resistors at 50 OHMS ( that I can have in a couple days,
Thanks for any help here, this is really not my strong suit!
James