Fan Header Question: Volts, Amps & Watts Multiplication & Multiple fans...

[JunkRoom]

Not sure where this should be posted so I've posted it to general discussion.

I want to use a fan-splitter to replace a single fan with two new ones.

The fan I've removed is 12V and 0.21A.

Assuming the header can support a fan of 12V and 0.21A:

Can I replace that fan with two 12V fans totaling 0.21A and not risk burning out the header, or will the wattage be too high?

E.G.

One fan of 12V and 0.11A + one fan of 12V and 0.10A equaling a total of 0.21A...

...but does that mean it's double the wattage?

e.g.

12V + 12V = 24V

24V x 0.21 = 5.04 Watts

Compered to the original single 12V, 0.21A fan:

12V x 0.21A = 2.5 Watts

I'm not sure if that's correct but it appears that putting two fans on the single header might double the number of Watts passing through the header even though their combined current matches that of the single 12V 0.21A fan.

Am I completely off bat with this? Is the above a correct assumption and can I assume it could be overloading and potentially damaging the fan header?

Thanks.

 

[duke37]

What is a fan header?
If the power supply has to provide 12V at 0.21A, then as you calculated the power is 2.5W. It doesn't matter what the physical load is. In your case with two fans taking 0.11A each, then each will use 1.32W, total 2.64W

 

[JunkRoom]

What is a fan header?
If the power supply has to provide 12V at 0.21A, then as you calculated the power is 2.5W. It doesn't matter what the physical load is. In your case with two fans taking 0.11A each, then each will use 1.32W, total 2.64W

Okay, I'm not sure I really understand it but, to simplify the question: what would happen if I plugged two 12V fans that require a total of 0.21 amps into it? Could they both reach full speed or would they be limited to 12V between them?

P.S. "Fan Header" is the popular term (among computer builders) for the 3 pins on a circuit board that the "Three-pin Molex connector KK family" (https://en.wikipedia.org/wiki/Computer_fan) on a computer fan plugs into (https://en.wikipedia.org/wiki/Computer_fan#/media/File:Three-pin_connector_on_a_computer_fan.jpg)

 

[garublador]

I'm guessing the "fan splitter" is just a little cable that plugs into the header and gives you two more headers to plug two fans into. If that's correct then the fans will be in parallel and the math duke37 gave will be correct. It will be the same power consumption as one fan running with higher current.

 

[JunkRoom]

I'm guessing the "fan splitter" is just a little cable that plugs into the header and gives you two more headers to plug two fans into. If that's correct then the fans will be in parallel and the math duke37 gave will be correct. It will be the same power consumption as one fan running with higher current.

Yes, that's correct, a "fan splitter" is just what you describe: a cable that plugs into the header and which splits into two, to provide two headers where before there was one.

I apologise, I assumed that the majority of people here would have built their computers and would know the terminology.

So if I plug two 12 volt fans drawing a 0.2A current between them into a header rated to support 0.2A they won't overload and burn out the header (i.e. the paths to the header)?

My other assumption is that the two won't be able to run at full speed because they'll be sharing the voltage? Is that correct?

 

[garublador]

Yes, that's correct, a "fan splitter" is just what you describe: a cable that plugs into the header and which splits into two, to provide two headers where before there was one.

I apologise, I assumed that the majority of people here would have built their computers and would know the terminology.

I think most would have a pretty good idea of what you're talking about, but most have also found that making assumptions like that can lead to very bad advice. We find it best to avoid making assumptions because each one made is another opportunity for an unnecessary mistake.

So if I plug two 12 volt fans drawing a 0.2A current between them into a header rated to support 0.2A they won't overload and burn out the header (i.e. the paths to the header)?

Correct. Just as a side note, I'd worry more about what is providing the current than the traces on the board. There's a good chance that will either just stop or get damaged before the traces will burn up. Designers usually want the parts to break (or stop working rather than break themselves) before the PCB gets damaged.

My other assumption is that the two won't be able to run at full speed because they'll be sharing the voltage? Is that correct?

No. They'll run just like normal but if they draw half the current each then I'd suspect that neither one will move as much air as the one fan that takes more current.

If you're interested in this kind of stuff it might be worth checking out some books or resources on basic circuit analysis.[/quote][/QUOTE]

 

[duke37]

The speed depends on the voltage. If both are 12V fans and fed on 12V they will run at design speed. They are in parallel so the voltage is the same and the total current is the sum of the two fan currents.

 

[JunkRoom]

I think most would have a pretty good idea of what you're talking about, but most have also found that making assumptions like that can lead to very bad advice. We find it best to avoid making assumptions because each one made is another opportunity for an unnecessary mistake.

I agree with you, the definitions and terminology are a mine-field, it's a huge headache. For instance, when you say I should worry about what's providing the current do you mean the fans? or the source of power running into the board? Technically the fans aren't providing current, they're dead until added to the circuit. Adding them provides a path along which current can flow but they're not themselves supplying power right?

I'm assuming you mean the person who designed the PCB would rather make it robust enough to carry enough power to destroy either the power source that's feeding it or the thing (in this case a fan) that's being plugged into the board?

I am interested in it and I am planning on learning more of it but I didn't think this was too advanced a question to post to a forum?

Ultimately I think I probably didn't clarify my question well enough.

I want to replace one 12V 0.21A fan with two new fans. If both of those new fans are both 12V and draw 0.1A amps each (for a total of 0.2A) will that overload the circuit? E.G. 12V + 12V x 0.2A = 4.8W (where the Wattage of the original single fan was 2.52 (i.e. 12V x 0.21A = 2.52W).

If both the new fans run at full speed then presumably that's 4.8 Watts of power, which is almost twice the 2.52 Watts of power that the original fan required. Will the increased wattage overload and/or damage the circuit?

The speed depends on the voltage. If both are 12V fans and fed on 12V they will run at design speed. They are in parallel so the voltage is the same and the total current is the sum of the two fan currents.

But doesn't that mean a higher wattage? (Volts x Amps) and if so, does that matter?

 

[Minder]

The common PC fan header should be able to support that current, modern PC fans are now mostly BLDC as opposed to the old brushed DC, these usually exhibit a fixed RPM range across a variation in voltage, and the current is fairly constant, as opposed to the old DC type, these have a BLDC IC inside similar to the Picmicro TC653 to control them.
M.

 

[JunkRoom]

The common PC fan header should be able to support that current, modern PC fans are now mostly BLDC as opposed to the old brushed DC, these usually exhibit a fixed RPM range across a variation in voltage, and the current is fairly constant, as opposed to the old DC type, these have a BLDC IC inside similar to the Picmicro TC653 to control them.
M.

Thanks for the input but I'm trying to modify the cooling system on an old Microsoft Xbox (the original xbox?). I've pulled the extractor and it's a 12V fan rated at 0.21 amps so I figured I'd just add two fans that didn't draw more than 0.21 amps powered by the header that powered the extractor but then it occurred to me that that added an additional multiplier...

...if you told me they wouldn't run at full speed I could understand it but for them both to run at full speed would mean that there's almost double the wattage. Wattage is a measure of Rate? I don't understand it because it seems like there's energy being produced as if from nowhere. I'm sure it's just a matter of clarifying the concepts and I was hoping someone here could help me to do that.

 

[Minder]

Wattage is a result of Voltage and Current and as already been posted, two fans in parallel double the wattage.
Across 12v they will each see no difference in performance.
M.

 

[JunkRoom]

Wattage is a result of Voltage and Current and as already been posted, two fans in parallel double the wattage.
Across 12v they will each see no difference in performance.
M.

So as long as I don't exceed 0.21 amps (which I'm assuming is roughly the limit the circuit can support without deteriorating) I can plug in as many 12V fans as I like and they'll run at full speed without damaging the circuit?

 

[garublador]

I agree with you, the definitions and terminology are a mine-field, it's a huge headache. For instance, when you say I should worry about what's providing the current do you mean the fans? or the source of power running into the board? Technically the fans aren't providing current, they're dead until added to the circuit. Adding them provides a path along which current can flow but they're not themselves supplying power right?

Right, the fans can't provide current. I'd have to make an assumption about what might be powering the fans (I don't have that great of a knowledge of how that's normally done) so I had to be vague.

I'm assuming you mean the person who designed the PCB would rather make it robust enough to carry enough power to destroy either the power source that's feeding it or the thing (in this case a fan) that's being plugged into the board?

Correct. Of course I can't guarantee that but that's what's normally done.

I am interested in it and I am planning on learning more of it but I didn't think this was too advanced a question to post to a forum?

It's not, and I don't think anyone has an issue with you posting it, I just thought I'd make that suggestion since you seem interested in why your assumptions weren't correct.

I want to replace one 12V 0.21A fan with two new fans. If both of those new fans are both 12V and draw 0.1A amps each (for a total of 0.2A) will that overload the circuit? E.G. 12V + 12V x 0.2A = 4.8W (where the Wattage of the original single fan was 2.52 (i.e. 12V x 0.21A = 2.52W).

This is why I made that suggestion. You are right that P = I * V. However, why are you adding the voltages? One fan is taking 12V * 0.1A and the other is taking 12V * 0.1A, so adding those together you get 2.4W. Why add the voltages together before multiplying?

 

[Minder]

Basically Yes, You can use as many that does not exceed the current rating of the circuit.
M.

 

[JunkRoom]

Basically Yes, You can use as many that does not exceed the current rating of the circuit.
M.

Thanks.

Why add the voltages together before multiplying?

I don't know. I've no idea. It just makes no sense to me how wattage can increase without the load on the circuit increasing. I just don't understand the logic underlying it, I was hoping to get a response along the lines of "No, what you've written is incorrect and here's why..." but it seems there's no one on the forum at the moment who's able to provide that kind of answer. :shrug:

 

[davenn]

I don't know. I've no idea. It just makes no sense to me how wattage can increase without the load on the circuit increasing. I just don't understand the logic underlying it, but it seems there's no one on the forum at the moment who's able to provide that kind of answer. :shrug:

They have told you but you haven't been reading / understanding what people have been typing

the load on the Power Supply IS INCREASING it has DOUBLED because you have added another identical fan

the power supply is supplying 12V

the existing fan is 12V @ 0.21A = 2.52W from the power supply

when you add another identical fan ( 12V @ 0.21A )
it is also using 2.52 W from the power supply

so there is now a total of 5.04W being drawn from the PSU


Dave

 

[JunkRoom]

the load on the Power Supply IS INCREASING it has DOUBLED because you have added another identical fan

Dave

But that's my point exactly.

 

[davenn]

But that's my point exactly.

what point ??

what about what I and several others have posted that you don't understand ? its pretty straight forward

 

[davenn]

e.g.

12V + 12V = 24V

24V x 0.21 = 5.04 Watts

Compered to the original single 12V, 0.21A fan:

12V x 0.21A = 2.5 Watts

I'm not sure if that's correct but it appears that putting two fans on the single header might double the number of Watts passing through the header even though their combined current matches that of the single 12V 0.21A fan.

Am I completely off bat with this? Is the above a correct assumption and can I assume it could be overloading and potentially damaging the fan header?

Thanks.

from your first post ....
no that is all wrong ... you are NOT doubling the voltage the power supply is still only 12V
it didn't magically turn into a 24V supply whilst you turned your back to grab the second fan

 

[JunkRoom]

from your first post ....
no that is all wrong ... you are NOT doubling the voltage the power supply is still only 12V
it didn't magically turn into a 24V supply whilst you turned your back to grab the second fan

That's what I'm asking...where I'm going wrong with those assumptions? It seems to me that the wattage is doubling even though the two fans are drawing the same current as the one fan referred to in my first post.