Hello,
I would like to use 2x12VDC 0,2A fans for cooling electronics, but I have several problems.
1.) PSU generates 40V 2x2A -> 40V 1x4A
2.) After rectifier are 2x10K uF caps (enought charge to blow up something)
I saw a few things that sould work, but best thing for (low amperage thing) it seem using a resistors. I would like to know how to calculate resistance and wattage for that, or if there is another better solution for that problem. (I mean something cheap mabye ?)
Thanks.
I would like to use 2x12VDC 0,2A fans for cooling electronics, but I have several problems.
1.) PSU generates 40V 2x2A -> 40V 1x4A
2.) After rectifier are 2x10K uF caps (enought charge to blow up something)
I saw a few things that sould work, but best thing for (low amperage thing) it seem using a resistors. I would like to know how to calculate resistance and wattage for that, or if there is another better solution for that problem. (I mean something cheap mabye ?)
Thanks.
Hello,
I would like to use 2x12VDC 0,2A fans for cooling electronics, but I have several problems.
1.) PSU generates 40V 2x2A -> 40V 1x4A
2.) After rectifier are 2x10K uF caps (enought charge to blow up something)
I saw a few things that sould work, but best thing for (low amperage thing) it seem using a resistors. I would like to know how to calculate resistance and wattage for that, or if there is another better solution for that problem. (I mean something cheap mabye ?)
Thanks.
Insert a resistor in one of the leads of the fan. The resistor needs to loose 28 V to reduce 40 V to 12 V. With a current of 0.2 A, that's R = 28 V / 0.2 A = 140 Ohm.
Note that the motor might mess up the otherwise clean DC you're using for the opamp.
First you would need to know the voltage at the capacitors, is it actually 40VDC? A 40VAC transformer going through a bridge rectifier and filter caps will give you more than 40VDC, something like 55V (40Vrms *1.414 = 56.56, subtract a little for diode losses).
You would need to measure the actual output voltage to be sure.
You say you have two fans at 12V/0.2A each. Again you would need to measure to confirm that the current draw the manufacturer states is correct.
Once you know the voltage and current it's just a matter of using ohms law. Suppose your supply has 55V across the capacitors, to get to 12V you'd need to lose 43V. Your fans are specified as 0.2A, so ohms law says R=E/I, 43V/0.2A = 215 ohms. So you'd need that amount of resistance in series with a 12V fan.
Now, how much power will that resistor have to dissipate? The formula for power is P=E*I, so 43V*0.2A = 8.6W. That's a lot of watts for a resistor, a 10W unit would be cutting it close, 15 or 20W would be better, and you'd need one for each fan.
Putting two fans in parallel fed by one resistor would halve the resistance value to 107.5, but double the watts to 17.2W.
Putting the fans in series reduces the voltage drop you need to 55V-24V = 31V. For that you'd need a resistor value of 31V/0.2A = 155ohms, and the wattage would be 31V*0.2A = 6.2W. This would be the best option for using a resistor with two fans. Still a pretty hefty resistor.
A linear voltage regulator would need to dissipate the same amount of power, 17.2W for fans in parallel, 6.2W for fans in series.
A switching regulator would dissipate less power, maybe a watt or two.
You would need to measure the actual output voltage to be sure.
You say you have two fans at 12V/0.2A each. Again you would need to measure to confirm that the current draw the manufacturer states is correct.
Once you know the voltage and current it's just a matter of using ohms law. Suppose your supply has 55V across the capacitors, to get to 12V you'd need to lose 43V. Your fans are specified as 0.2A, so ohms law says R=E/I, 43V/0.2A = 215 ohms. So you'd need that amount of resistance in series with a 12V fan.
Now, how much power will that resistor have to dissipate? The formula for power is P=E*I, so 43V*0.2A = 8.6W. That's a lot of watts for a resistor, a 10W unit would be cutting it close, 15 or 20W would be better, and you'd need one for each fan.
Putting two fans in parallel fed by one resistor would halve the resistance value to 107.5, but double the watts to 17.2W.
Putting the fans in series reduces the voltage drop you need to 55V-24V = 31V. For that you'd need a resistor value of 31V/0.2A = 155ohms, and the wattage would be 31V*0.2A = 6.2W. This would be the best option for using a resistor with two fans. Still a pretty hefty resistor.
A linear voltage regulator would need to dissipate the same amount of power, 17.2W for fans in parallel, 6.2W for fans in series.
A switching regulator would dissipate less power, maybe a watt or two.
With _two_ same-type 12V 0.2A fans from a 40V supply, put them in series!!
Now you have a 24V load, 'only' need to drop 16V. 16V 0.2A is 80 Ohms 3.2 Watts. I would use 100 Ohms 10 Watts, because the fans don't mind a little less, and 100r 10W is a VERY standard value.
Yes, fans in series will cause some odd results. If you stick your finger in one, the other one will run faster. However I have run series fans for years in adverse situations and they blow just fine.
Now you have a 24V load, 'only' need to drop 16V. 16V 0.2A is 80 Ohms 3.2 Watts. I would use 100 Ohms 10 Watts, because the fans don't mind a little less, and 100r 10W is a VERY standard value.
Yes, fans in series will cause some odd results. If you stick your finger in one, the other one will run faster. However I have run series fans for years in adverse situations and they blow just fine.
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