View Single Post
Old 03-03-2018, 10:49 AM   #2
Curious.George
Badcaps Veteran
 
Join Date: Nov 2011
Posts: 1,161
Default Re: How many uf do i need to cope with the HDD inrush current

Quote:
Originally Posted by evilkitty View Post
In my computer i have 2 hot swap bays, i made my power cable and i used some "18 awg copper" wire, i think it is actually 20awg (got it on ebay...)

anyway i have some legit 18awg copper wire i was going to remake it and stick a couple caps at the end
my issue is when i am using one bay and i go to use the second the other bay looses power for a moment (if i use that drive to boot it as annoying)
i have some 470 low esr caps and i have some 10uf general purpose caps i could use as well as some 10uf low esr ones
my PSU is a Seasonic M12II 520W
You shouldn't be seeing a "dropout" like this -- unless your drives have some hellacious spinup currents and/or you have high IR drops in your cabling/connectors.

But, to answer your original question...

The voltage on a capacitor is a function of current and time (and capacitance). Specifically, CV=It.

"I" coming INTO the capacitor (remember, there's current flowing in while it is also being drawn out) will be limitted by the peak capabilities of your power supply as seen behind the impedance of your interconnect to the powersupply. "I" exitting the capacitor will be defined by the spinup requirements of your disk drive. So, you have a "net" I flowing out of the capacitor that reflects the difference in these two.

Similarly, "t" will depend on how long this "transient" lasts (presumably, your power supply can meet the steady-state operating demands of the disk drive).

"V" should be treated as "dV" -- a change of voltage over a particular change in time (dt -- time interval).

So, if your drive specs an input range of [Vmin,Vmax] and your power supply puts out Vout... and, the drive actually "sees" Vin (Vout behind the IR losses of your cables) then:
- Vin must be > Vmin (else you won't be able to run the drive)
- Vin must be < Vmax (d'uh)
- Vout must be < Vmax (i.e., when the load is minimized and IR ~= 0)
- Vin - Vmin is the dV that you have to work with; you can afford for the voltage at the drive to drop from Vin to Vmin and still be assured the drive will function.

[The capacitor starts out at Vmax before you plug in the drive -- cuz there are no IR losses due to the absence of a load!]

This is really just a crude approximation of the situation. In reality, your drive will probably work with a lower Vin on the 12V supply (motor) during spinup. OTOH, the 5V supply probably is a lot more fussy about Vmin as "it can lose its train of thought"

[One "simple" fix is to run a second power pigtail from the power supply so the load from "disk #1" isn't being handled by the cable that is ALSO powering disk #2 (see your photo)]

Last edited by Curious.George; 03-03-2018 at 10:51 AM..
Curious.George is offline   Reply With Quote