I'm thinking of doing the same thing. The only concern with this route is buck converters increase the current when stepping down the voltage. Some power is lost (buck converters aren't 100% efficient), but going past the maximum current rating on the fan is not recommended. According to Noctura for the NF-A6x25 5V that's a max input current of 0.26 A. I'm wondering what kind of current you buck converter is going to output when attached to the 7V fan wires on the ps2 and the hard drive is running, ramping up the power.

If i end up testing this ill add a post.

The fan will take its rated current when given the correct voltage. A buck converter does not increase the current taken by the load, that would be impossible.

The statement about dropping the voltage while increases the current means the buck converter outputs more current than it takes in. And that is only true on average.

 

I'm thinking of doing the same thing. The only concern with this route is buck converters increase the current when stepping down the voltage. Some power is lost (buck converters aren't 100% efficient), but going past the maximum current rating on the fan is not recommended. According to Noctura for the NF-A6x25 5V that's a max input current of 0.26 A. I'm wondering what kind of current you buck converter is going to output when attached to the 7V fan wires on the ps2 and the hard drive is running, ramping up the power.

If i end up testing this ill add a post.

If the fan draws 250 mA at 5 V, then if you use a switching regulator that is 100% efficient with an input voltage of 7 V, it would draw about 180 mA from the 7 V supply in order to deliver 250 mA to the fan. The fan sees a 5 V supply and will still draw 250 mA from that 5 V supply. It neither knows nor cares what kind of supply it is. If it were 85% efficient, it would draw about 210 mA from the input, but still deliver 250 mA to the fan.

A better way of thinking about it is not that a buck converter increases the current, but that it doesn't need to draw as much current from it's input to deliver the same current to its output. A boost converter is the opposite -- it will need to draw more current from it's input in order to deliver the same current at its output.

 

If the fan draws 250 mA at 5 V, then if you use a switching regulator that is 100% efficient with an input voltage of 7 V, it would draw about 180 mA from the 7 V supply in order to deliver 250 mA to the fan. The fan sees a 5 V supply and will still draw 250 mA from that 5 V supply. It neither knows nor cares what kind of supply it is. If it were 85% efficient, it would draw about 210 mA from the input, but still deliver 250 mA to the fan.

A better way of thinking about it is not that a buck converter increases the current, but that it doesn't need to draw as much current from it's input to deliver the same current to its output. A boost converter is the opposite -- it will need to draw more current from it's input in order to deliver the same current at its output.

I see. Thank you for the clarification.

 

Ohms law simply states that the voltage divided by the resistance equals the current. So if you have a 5 volt supply of any kind capable of 250mA (or 10 amps or more) connected to a load (fan motor) that appears as a 20Ω resistance then 5V ÷ 20Ω = 0.25A (250mA). As long as the source is capable of supplying the needed current (or more) it doesn't matter. The load will draw what it will and no more. Even if the source voltage is capable of 100 amps. It will still draw only 0.25A (250mA).

 

it would draw about 180 mA from the 7 V supply in order to deliver 250 mA to the fan.

Actually, it would draw 250mA 71% of the time and zero the other 29% of the time. With a capacitor, it could actually be more than 250mA for part of the cycle.