One good thing about music... (My 2.1 system!)

[cbraver]

thanks for the info. i don't think crown will give you the go signal to mod the amp, but the "workout" we home users do to these things, i don't think the cfm would matter much.

I agree, but, I wanted to see what they said. Crown has great support. The tech told me as long as I go with the same amperage or lower, I'll be fine (except for voiding warranty, of course). He says the specs are listed on the fans themselves. I'm going to take a look tonight.

 

[westcott]

I agree, but, I wanted to see what they said. Crown has great support. The tech told me as long as I go with the same amperage or lower, I'll be fine (except for voiding warranty, of course). He says the specs are listed on the fans themselves. I'm going to take a look tonight.

Im not familiar with your particular equipment but if space allows, go with a larger diameter fan turning at a lower rpm. It will move the same or more air with a lot less noise.

 

[cbraver]

Thanks for the motivation guys, I just finished installing the new fans. It's now SILENT!!!

Im not familiar with your particular equipment but if space allows, go with a larger diameter fan turning at a lower rpm. It will move the same or more air with a lot less noise.

It's only a 2U enclosure, 80mm is the maximum that I can fit. I ended up getting a Vantec 80mm double ball bearing "stealth" fan from my local electronics supply shop... here it is online:
http://www.directron.com/sf8025l.html

It's 0.10A instead of 0.17A, so I put a 160-ohm resistor in parallel of each fan to bring it back down to the speed Crown's original fan had.

 

[TLS Guy]

Thanks for the motivation guys, I just finished installing the new fans. It's now SILENT!!!



It's only a 2U enclosure, 80mm is the maximum that I can fit. I ended up getting a Vantec 80mm double ball bearing "stealth" fan from my local electronics supply shop... here it is online:
http://www.directron.com/sf8025l.html

It's 0.10A instead of 0.17A, so I put a 160-ohm resistor in parallel of each fan to bring it back down to the speed Crown's original fan had.

I sure hope that resistor is in series with the fan and not parallel!

 

[cbraver]

No, parallel. Why? My friend that's an EE did the calculation to make sure I wouldn't blow the resistor. I couldn't remember how.

 

[mike c]

so the fan was 12v ... so maybe I can use a pc fan?

 

[Mudcon]

nice amp in your set up

 

[TLS Guy]

No, parallel. Why?

Well if the fan is shunted by the resistor (parallel), then the resistor is loading the driving circuit and dropping the voltage by asking the driving circuit to provide more current than it is able. This will damage the driving circuit. You never drop voltage this way. Now if you put the resistor in series, the voltage will drop and the current will be lower.

Lets take a look.

Your previous fan took 0.1 amp at 12 volts. So the fan resistance from Ohms law is: - 12 volts/0.1 amps = 120 ohms.

Now your new fan takes 0.17 amps at 12 Volts. So the resistance of that fan is: - 12 volts/0.17 amps = 70.6 ohms.

Now you want the new fan to draw 0.1 amps. The voltage required to have it draw 0.1 amps is: - 12 volts X 120 ohms/170 ohms = 8.5 volts. The voltage drop required is 3.5 volts We therefore need a resistor to drop the current to 0.1 amps when a voltage of 3.5 volts is applied across it. Therefore the value of your SERIES dropping resistor is 3.5 volts X 0.1 amps = 35 ohms. So the power consumed by the resistor is 3.5 volts x 0.1 amps = 0.35 watts. So you need a 35 ohm 0.5 watt dropping resistor. That will draw the same current from the fan powering circuit as your old fan, and the fan will run slower as it will have 8.5 volts across it.

I hope this is clear.

 

[cbraver]

The new fan is 0.1A though, the old one is 0.17A. This should make it *look* like a 0.17A fan. Sorry, I wasn't clear about that.

 

[cbraver]

so the fan was 12v ... so maybe I can use a pc fan?

Yeah, that's all it is. Do you have a XLS you're going to mod?

nice amp in your set up

Thanks.

 

[mike c]

not yet, still deciding between a twice expensive crown or the cheaper behringer EP2500 (for subwoofer use)

but both of them need a fan mod.

 

[TLS Guy]

The new fan is 0.1A though, the old one is 0.17A. This should make it *look* like a 0.17A fan. Sorry, I wasn't clear about that.

The way you did it still is not correct. I would still make about a 23 to 30% voltage reduction. 40 ohms in series will reduce voltage by 3 volts and 60 ohms by 4 volts, so the 35 ohm resistor should a start point. You might have to play with the value until you get the fan speed you want. With the parallel resistor you have no idea how much you are affecting voltage because you don't know the source impedance of the driving circuit. A series dropper is the correct method.

 

[cbraver]

not yet, still deciding between a twice expensive crown or the cheaper behringer EP2500 (for subwoofer use)

but both of them need a fan mod.

I have a EP2500 I was going to use for a subwoofer build also, but, I have it all boxed up and I'm shipping it back. The one gain knob is sticking, and the XLR connector on the back is loose. That was right out of the box. I call Behringer and they weren't helpful, so, I'm just sending it back. They are inexpensive and there are some on this forum that absolutely love them, but, I guess it wasn't me. I like the crossover though (although one of the knobs is kinda sticky on that too, but other than that it is great). That said, a lot of people have had great experiences with them, so, I donno.

If you can afford an XTi you probably won't need to mod the fan. The fans don't come on until the amp gets to 50C, and you can set the fan triggers with the newer firmware. More expensive though. My dream amps. If you get an XLS, then you can just do the same mod I did.

 

[cbraver]

The way you did it still is not correct. I would still make about a 23 to 30% voltage reduction. 40 ohms in series will reduce voltage by 3 volts and 60 ohms by 4 volts, so the 35 ohm resistor should a start point. You might have to play with the value until you get the fan speed you want. With the parallel resistor you have no idea how much you are affecting voltage because you don't know the source impedance of the driving circuit. A series dropper is the correct method.

Thanks. I'm going to leave the amp off tonight and ask what his thought process was tomorrow. Worst case, I'll desoldier and do it in series, thanks.

 

[TLS Guy]

Thanks. I'm going to leave the amp off tonight and ask what his thought process was tomorrow. Worst case, I'll desoldier and do it in series, thanks.

One other thing I should have pointed out is that your 160 ohm shunt resistor is drawing 0.075 amps. So the total draw by motor and resistor is 0.175 amps. So the motor driving voltage is still 12 volts. So since you have not dropped the voltage, and you say the fan is quiet, you don't need the resistor at all. I would remove it.

 

[cbraver]

One other thing I should have pointed out is that your 160 ohm shunt resistor is drawing 0.075 amps. So the total draw by motor and resistor is 0.175 amps. So the motor driving voltage is still 12 volts. So since you have not dropped the voltage, and you say the fan is quiet, you don't need the resistor at all. I would remove it.

Wha?! 0.175amps total, which is basically 0.17A, and 12V. If I remove the resistor the fans speed up. There is a voltage drop across the resistor in parallel. Model the motor as a resistor and this will make more sense to you.

 

[TLS Guy]

Wha?! 0.175amps total, which is basically 0.17A, and 12V. If I remove the resistor the fans speed up. There is a voltage drop across the resistor in parallel. Model the motor as a resistor and this will make more sense to you.

I did. The motor according to you draws 0.1 amps. So the resistance as I showed before is 120 ohms. Now shunt it with a 160 ohm resistor, and the current through the resistor is 12volts/160 ohms = 0.075 amps. Since the 160 ohm resistor is in parallel with the motor (120 ohms) then the total current draw is 0.175 amps. This is almost identical to the current draw from the previous motor, 0.17 amps.

So if the previous motor was seeing 12 volts then this one must as well. The only thing that might confound it is the output impedance of the driver supply. If this impedance is unusually high then your new motor might be seeing a voltage higher than 12 volts. However the fact remains that the correct way to slow the motor is with a dropper series resistor, not a shunt.

 

[cbraver]

I did. The motor according to you draws 0.1 amps. So the resistance as I showed before is 120 ohms. Now shunt it with a 160 ohm resistor, and the current through the resistor is 12volts/160 ohms = 0.075 amps. Since the 160 ohm resistor is in parallel with the motor (120 ohms) then the total current draw is 0.175 amps. This is almost identical to the current draw from the previous motor, 0.17 amps.

So if the previous motor was seeing 12 volts then this one must as well. The only thing that might confound it is the output impedance of the driver supply. If this impedance is unusually high then your new motor might be seeing a voltage higher than 12 volts. However the fact remains that the correct way to slow the motor is with a dropper series resistor, not a shunt.

Ah, I think I know where you're confused now! It's a current source, voltage is constant from it. That's why the fan speeds up when there isn't a resistor in parallel. It adjusts the current based upon the speed it needs the fan to move. If you put the resistor in series, it will slow the fan down, but you're starving the fan of voltage. The minimum fan speed (current set) for this amp would probably be enough to start the fan if a resistor is in series, but if the minimum speed was low enough it could possibly even prevent fans from spolling up (epecially if the fan was closer to 0.17A). Parallel circuits are current dividers, 0.07 is going to the resistor, and 0.10 to the fan. Current adds in parallel, so you simplify to 0.17A (well, 0.175A exactly) and the amp has no idea the fan is any different.

Physical proof of this can be found from my experiment of removing the resistor in parallel, which in turn speeds up the fan.

The modification is correctly setup. I have seen some other threads about modding fans where they put the resistor in series. This is either a voltage source, or a current source in which case they are incorrectly modifying the circuit. I've now checked with multiple people on this, both electrical engineers, and they confirmed that my setup is correct. They even measured the current draw on the board for me.

Thanks for you're concern though!

 

[TLS Guy]

Ah, I think I know where you're confused now! It's a current source, voltage is constant from it. That's why the fan speeds up when there isn't a resistor in parallel. It adjusts the current based upon the speed it needs the fan to move. If you put the resistor in series, it will slow the fan down, but you're starving the fan of voltage. The minimum fan speed (current set) for this amp would probably be enough to start the fan if a resistor is in series, but if the minimum speed was low enough it could possibly even prevent fans from spolling up (epecially if the fan was closer to 0.17A). Parallel circuits are current dividers, 0.07 is going to the resistor, and 0.10 to the fan. Current adds in parallel, so you simplify to 0.17A (well, 0.175A exactly) and the amp has no idea the fan is any different.

Physical proof of this can be found from my experiment of removing the resistor in parallel, which in turn speeds up the fan.

The modification is correctly setup. I have seen some other threads about modding fans where they put the resistor in series. This is either a voltage source, or a current source in which case they are incorrectly modifying the circuit. I've now checked with multiple people on this, both electrical engineers, and they confirmed that my setup is correct. They even measured the current draw on the board for me.

Thanks for you're concern though!

That explains it. The driver has to see the same load it did before with a drive like that.