Did I fry my fans by wiring them backwards?

[Blakhash]

It happens, at least fans are cheap.

yup yup! looking on new egg for some now, apparently the artic 11 fans are cheap as ish because they do no provide a reverse protection diode in case of situations like these. and yea i am bummed but its whatevs, life goes on. and the show must go too. thank you guys for your help, hope my failure helps anyone else out in the future.

 

[alesh]

160mA is what that fan is supposed to take, and you tried putting 1250mA through them? No current limiting resistor?

This is why people need a basic benchtop power supply, and a multimeter.
Watching people "tinker" without them is like watching someone commit seppuku with a rusty letter-opener

BS. These are no diodes. You don't need no current limiting. Simply apply anything near the rated voltage and the fan will take what's it rated at. Reversed polarity (I actually think that he applied voltage on the PWM pin) probably burned PWM IC.

 

[Blakhash]

BS. These are no diodes. You don't need no current limiting. Simply apply anything near the rated voltage and the fan will take what's it rated at. Reversed polarity (I actually think that he applied voltage on the PWM pin) probably burned PWM IC.

No no no, I did not apply voltage to the pwm pin, I immediately cut off the other two wires for safety precaution. I reversed the current to the red and black wires chasing the small board to be instantly fried. There is no safety diode to prevent it being fried (El cheapo punks)

 

[alesh]

No no no, I did not apply voltage to the pwm pin, I immediately cut off the other two wires for safety precaution. I reversed the current to the red and black wires chasing the small board to be instantly fried. There is no safety diode to prevent it being fried (El cheapo punks)

There's no need for any safety diode since you can't plug it into the motherboard in the opposite polarity. And you didn't plug it into a motherboard el stupid punk, did you?

 

[Blakhash]

There's no need for any safety diode since you can't plug it into the motherboard in the opposite polarity. And you didn't plug it into a motherboard el stupid punk, did you?

Dip $hit, I never said anything about a motherboard... there is a small board under the fan. THAT board was fried when the polarity was reversed. Go ahead... grab a computer fan and look at the underside... I'll wait.

 

[heckler73]

In lieu of evidence to the contrary, I presume a basic wall-wart was used.

BS. These are no diodes. You don't need no current limiting. Simply apply anything near the rated voltage and the fan will take what's it rated at. Reversed polarity (I actually think that he applied voltage on the PWM pin) probably burned PWM IC.

http://www.digchip.com/datasheets/parts/datasheet/052/AH292Z-PL-pdf.php



Hypothesis: If those Hall sensors are fried, the coils won't switch; ergo, no movement. That's assuming the FETs aren't relieved of their smoke content. It is a likely candidate for failure in this case.

BTW Diode = Semiconductor = "PWM IC" (whatever that is)

 

[alesh]

Dip $hit, I never said anything about a motherboard... there is a small board under the fan. THAT board was fried when the polarity was reversed. Go ahead... grab a computer fan and look at the underside... I'll wait.

Point was that these fan+HS combos are intended to be used as a CPU cooler. That means that the connector is supposed to be plugged into the motherboard. And the connector is designed in a way that doesn't allow reversed polarity - no need for protection. I'm sorry for that I was a little harsh yesterday - I was drunk a bit

 

[alesh]

In lieu of evidence to the contrary, I presume a basic wall-wart was used.

http://www.digchip.com/datasheets/parts/datasheet/052/AH292Z-PL-pdf.php

View attachment 3420907
Hypothesis: If those Hall sensors are fried, the coils won't switch; ergo, no movement. That's assuming the FETs aren't relieved of their smoke content. It is a likely candidate for failure in this case.

BTW Diode = Semiconductor = "PWM IC" (whatever that is)

Thanks I didn't know exactly how these controls worked. But I still don't see any need for current limiting power supply for the fan.
BTW "PWM IC" =/= diode (but yes it's a semiconductor and the whole fan likely contains at least one diode)

 

[heckler73]

But I still don't see any need for current limiting power supply for the fan.

It's just a simple way to ensure the fan gets the current it needs. It's not a reverse polarity protection cct.
What's the equivalent resistance of the fan at operating values?
12 V / 0.16 A = 75 Ohms
But more importantly, what's the power rating?
1.92W or ~2W

If your supply is ready to crank out 15W, do you think there might be a problem?
The wart is just waiting for a circuit to hook-up to. You can be a fan and say, "Please sir, I only want 160mA", and the wart will say, "meh...fuck it, have about a pound".

Now that I think about it more, the resistor should be in parallel (instead of series), to act as a bleeder. That way, if 300mA is coming in and 140mA needs to be bled off, it makes the calculation trivial. I think my brain was stuck on voltages when I mentioned it earlier (due to some side experiment).

An 85 Ohm resistor (~2W) across each fan would allow them to run smoothly or a 21 Ohm across the lot (but it's pushing 10W+ ! ).

Although, a more proper "hack" in this case would be a JFET or Op-Amp as current regulator (still needs a resistor).
Or in line with the classics, use an LM317 as a current regulator by tying output and "ADJ" with a set resistor (which allows control to 1.5A). Everyone should have an LM317 in their "tinker box" for these sorts of occasions.

Regardless, I stand corrected. Thank you for bringing that to my attention.

 

[alesh]

It's just a simple way to ensure the fan gets the current it needs. It's not a reverse polarity protection cct.
What's the equivalent resistance of the fan at operating values?
12 V / 0.16 A = 75 Ohms
But more importantly, what's the power rating?
1.92W or ~2W

If your supply is ready to crank out 15W, do you think there might be a problem?
The wart is just waiting for a circuit to hook-up to. You can be a fan and say, "Please sir, I only want 160mA", and the wart will say, "meh...fuck it, have about a pound".

Now that I think about it more, the resistor should be in parallel (instead of series), to act as a bleeder. That way, if 300mA is coming in and 140mA needs to be bled off, it makes the calculation trivial. I think my brain was stuck on voltages when I mentioned it earlier (due to some side experiment).
An 85 Ohm resistor (~2W) across each fan would allow them to run smoothly or a 21 Ohm across the lot (but it's pushing 10W+ ! ).

Although, a more proper "hack" in this case would be a JFET or Op-Amp as current regulator (still needs a resistor).
Or in line with the classics, use an LM317 as a current regulator by tying output and "ADJ" with a set resistor (which allows control to 1.5A). Everyone should have an LM317 in their "tinker box" for these sorts of occasions.

Regardless, I stand corrected. Thank you for bringing that to my attention.

Why would you waste ~1.7W of power on resistor? The fan itself acts like a resistor (75ohm in this case). You apply 12V, the current is 160mA. You apply 13.2V (max value within 10% tolerance), the current is 176mA. No big deal.

To support my claim from the other perspective, have you ever seen someone using any resistors this way on a CPU cooler? In such application, the reliability is critical (as the CPU is pricey) and yet, I've never seen any resistors connected to the fan (with the exception of some pots for manual RPM regulation).

 

[churchhaze]

If your supply is ready to crank out 15W, do you think there might be a problem?
The wart is just waiting for a circuit to hook-up to. You can be a fan and say, "Please sir, I only want 160mA", and the wart will say, "meh...fuck it, have about a pound".

A voltage source rated 15W does not mean it "puts out 15W". If you have a 12V source that is rated for 15W, it means that it is considered to be overloaded when you draw over 1.25A.

The amount of current drawn by the circuit is dependent on the load and the source voltage, not the rated power. For a linear resistive load, like a resistor, the current drawn is based on ohms law. For an exponential resistive load like a diode, the current drawn is based on the characteristic curve of the diode.

When a voltage source is overloaded, the source voltage will be pulled down below its rated levels, there's a risk of fire.

 

[heckler73]

A voltage source rated 15W does not mean it "puts out 15W". If you have a 12V source that is rated for 15W, it means that it is considered to be overloaded when you draw over 1.25A.

While your logic has merit, I don't believe it is applicably clear.
Sure it depends on the load, but what is the real load here?
And to make sure we're on the same page, I have assumed from the start
his adapter in question was a standard, cheap, unregulated wart.

My particular confusion was coming from two facts about fans:
1) their draw of current is not constant (despite what it says on the label),
2) they have a (potentially) large inrush current at start-up

http://www.nmbtc.com/fans/white-papers/dc-brushless-cooling-behavior/


It would seem my (wayward) intuition got the better of me there. While grappling with the qualitative nature of the circuit, I mistakenly assumed a high-probability for a near-short to develop based on the dynamic resistance (or IV characteristics) of typical pn-junctions, which are littered throughout. The circuit board--upon closer examination--appears to have some SMD resistors in the path (going to the dual op-amp), so it could be buffered against such an event, during what would otherwise be low-impedance situations. However, I may have missed some paths or gotten lost on the visual trace attempt.

I suppose one could attach a fan to an unregulated wall-wart without limiting, but I still feel sketchy about doing that myself because issues such as the following are still lurking in the background:

Thus, if a wall wart is rated at 12 VDC and 500 ma., the voltage will be significantly above 12 VDC when the supply is powering a device that only uses 200 or 300 milliamps of current. Similarly, the voltage will drop well below the specified 12 volts if more than 500 milliamps of current is needed by the device to be powered.
http://www.dxing.info/equipment/wall_warts_bryant.dx

12 V 1250mA unregulated supply vs 12V 160mA dynamic load? Not a problem you say? I don't know. I'm a little hesitant over the load matching and power transfer in this situation. If the voltage goes up as a result of the current differences, will it overload some other sensitive components?

...fuck it...let's try burning something

I just tested a fan on a BOSS 9.6V 200mA adapter.
Knowing the fan is 0.15A 12V, what will the effect be?

It's a trick question. The BOSS adapters are regulated It pulled 9.6V at 130mA...as expected for a regulated supply.

How about my Pignose amp supply? Rated as 12V at 1500mA !!! With an open cct. voltage of 17.21V, I'd say it's a candidate for unregulated supply, eh?

So, care to hazard a guess what happens?

Spoiler: Answer here, surprised?

15.5V at 227mA the summum bonum from all this chit chat

you're welcome

​

 

[alesh]

[...]
his adapter in question was a standard, cheap, unregulated wart.
[...]

Is it just my impression or are these pretty obsolete nowadays? I'd say that most "power bricks" today are small switching supplies. At least I can't think of any in my house which isn't a switching PSU.
But if @Blakhash 's 'small wall charger' was indeed unregulated wart it's possible that voltage (and therefore current) was higher than expected. I'm pretty sure, though, that the fan could handle 15-18V easily for such a brief period of time.

 

[churchhaze]

Wow. This is really overthinking things. lol. We're talking about powering a 12V fan with a wallwart here.

You don't have to worry about load matching and reactive components. It won't be an issue.

The quote posted was a bit silly:

"Thus, if a wall wart is rated at 12 VDC and 500 ma., the voltage will be significantly above 12 VDC when the supply is powering a device that only uses 200 or 300 milliamps of current. Similarly, the voltage will drop well below the specified 12 volts if more than 500 milliamps of current is needed by the device to be powered."

Why on earth would you worry about the test case where the power supply is considered overloading? (over 500mA?) That's just absurd to even suggest as being a flaw. There's absolutely no problem with an unregulated supply being a few volts higher than its rating. The problem is when the voltage is pulled down too low (overloading). You're not even supposed to run a 12V wallwart rated at 500mA at 500mA. That's borderline overloading. Loading it with 200-300mA is actually a more realistic test case of an unregulated supply with max current of 500mA.

 

[unwine99]

While your logic has merit, I don't believe it is applicably clear.
Sure it depends on the load, but what is the real load here?
And to make sure we're on the same page, I have assumed from the start
his adapter in question was a standard, cheap, unregulated wart.
My particular confusion was coming from two facts about fans:
1) their draw of current is not constant (despite what it says on the label),
2) they have a (potentially) large inrush current at start-up

http://www.nmbtc.com/fans/white-papers/dc-brushless-cooling-behavior/
View attachment 3423100

It would seem my (wayward) intuition got the better of me there. While grappling with the qualitative nature of the circuit, I mistakenly assumed a high-probability for a near-short to develop based on the dynamic resistance (or IV characteristics) of typical pn-junctions, which are littered throughout. The circuit board--upon closer examination--appears to have some SMD resistors in the path (going to the dual op-amp), so it could be buffered against such an event, during what would otherwise be low-impedance situations. However, I may have missed some paths or gotten lost on the visual trace attempt.

I suppose one could attach a fan to an unregulated wall-wart without limiting, but I still feel sketchy about doing that myself because issues such as the following are still lurking in the background:

Thus, if a wall wart is rated at 12 VDC and 500 ma., the voltage will be significantly above 12 VDC when the supply is powering a device that only uses 200 or 300 milliamps of current. Similarly, the voltage will drop well below the specified 12 volts if more than 500 milliamps of current is needed by the device to be powered.
http://www.dxing.info/equipment/wall_warts_bryant.dx

12 V 1250mA unregulated supply vs 12V 160mA dynamic load? Not a problem you say? I don't know. I'm a little hesitant over the load matching and power transfer in this situation. If the voltage goes up as a result of the current differences, will it overload some other sensitive components?

...fuck it...let's try burning something

I just tested a fan on a BOSS 9.6V 200mA adapter.
Knowing the fan is 0.15A 12V, what will the effect be?

It's a trick question. The BOSS adapters are regulated It pulled 9.6V at 130mA...as expected for a regulated supply.

How about my Pignose amp supply? Rated as 12V at 1500mA !!! With an open cct. voltage of 17.21V, I'd say it's a candidate for unregulated supply, eh?

So, care to hazard a guess what happens?

Spoiler: Answer here, surprised?

15.5V at 227mA the summum bonum from all this chit chat

you're welcome

​

Wow. This is really overthinking things. lol. We're talking about powering a 12V fan with a wallwart here.

You don't have to worry about load matching and reactive components. It won't be an issue.

The quote posted was a bit silly:

"Thus, if a wall wart is rated at 12 VDC and 500 ma., the voltage will be significantly above 12 VDC when the supply is powering a device that only uses 200 or 300 milliamps of current. Similarly, the voltage will drop well below the specified 12 volts if more than 500 milliamps of current is needed by the device to be powered."

Why on earth would you worry about the test case where the power supply is considered overloading? (over 500mA?) That's just absurd to even suggest as being a flaw. There's absolutely no problem with an unregulated supply being a few volts higher than its rating. The problem is when the voltage is pulled down too low (overloading). You're not even supposed to run a 12V wallwart rated at 500mA at 500mA. That's borderline overloading. Loading it with 200-300mA is actually a more realistic test case of an unregulated supply with max current of 500mA.

 

[churchhaze]

Is it just my impression or are these pretty obsolete nowadays? I'd say that most "power bricks" today are small switching supplies. At least I can't think of any in my house which isn't a switching PSU.
But if @Blakhash 's 'small wall charger' was indeed unregulated wart it's possible that voltage (and therefore current) was higher than expected. I'm pretty sure, though, that the fan could handle 15-18V easily for such a brief period of time.

And then, with switch mode regulation of any kind, stability is dependent on a minimum load current. The switching coil will cause ringing in the circuit on power on if current is not sufficiently high to make the system overdamped. This means switch mode based regulator also suffers from the problem of not being able to run low currents. In that case the problem is actually a real issue, because system instability (ringing,oscillating) can mean large transient spikes greater than the rated voltage. We never notice this issue with our meanwells because we use constant current anyway.

 

[SPLFreak808]

Simply fried the inline/ic diode. Cheap fans usually have a low current diode for reversed polarity protection. The problem is a 2a pwmic board will not stop a 1a diode from frying at reversed polarity. See it all the time on cheap unregulated amplifiers

 

[Randomblame]

Wow, now there is much too complicated for me.
Sorry and shame on me than I've grilled some fans and have noticed the good brands like Noctua or Noiseblocker are protected against reverse polarity. The Arctic fans on the Alpine64 against burn through in seconds, Lol.

An easy way for me to power the fans without the risk to grill them is using a simple fan controller. No need to cut anything from the fans, just stick it together and take a small 12v driver to power them!
The black Gelid/Zalman Fan Mate costs 5-6$ and is good for 3 fans and the LianLi(12$) can drive 4 fans. Works great and gives me the opportunity to compensate higher ambient temperatures continuously.

 

[unwine99]

I wouldn't last long at all in these lighting threads -- y'all are way too smart for me. I would either burn my house down attempting some of this stuff or I would be reduced to a drooling vegetative state trying to comprehend what the hell you guys are talking about -- one or the other or both...... lol.......ok carry on.

 

[alesh]

And then, with switch mode regulation of any kind, stability is dependent on a minimum load current. The switching coil will cause ringing in the circuit on power on if current is not sufficiently high to make the system overdamped. This means switch mode based regulator also suffers from the problem of not being able to run low currents. In that case the problem is actually a real issue, because system instability (ringing,oscillating) can mean large transient spikes greater than the rated voltage. We never notice this issue with our meanwells because we use constant current anyway.

If I am not mistaken, the current draw needed for proper function is much lower, isn't it? Meaning that a single fan would draw current high enough for the PSU to properly work.