What Size cap for this task?

[leham]

Hi everyone
Quick question, I have a dell optiplex SFF 990 here and it does this odd thing (I've tested on other units from the same line of machines, we use them at work)..
OK so when you press the power button, the system powers on for 2 seconds, then off like it's lost power (leds fade out) then back on as normal. As I've said this seems normal as this happens on other machines of this same line..

Now this shouldn't be a problem BUT I have this device attached called a "light pack" (it's an active backlight system for the display). What happens is when the system initially powers on, the light pack turns on but when the system turns off then back on it messes with the light pack (making it have static random colors which dont go away) the only way to get it going again is to unplug the usb cable and plug it back in.

This gets annoying really fast as you could imagine

So my guess is, if I put a beefy capacitor in series on the 5v line from the usb port, I can keep the 5v line stable for the few seconds the system is doing it's dell-Voodo.

But how do I work out the capacitance I need?


I'm going to be generous and say I'd need to supply 5v @500ma draw for 15 seconds


Thanks
 

[David_AVD]

The capacitor solution is not going to be practical.

A better solution is to add a relay in series with the light pack's power, and only energise it once the power has been on for 5 seconds.

[Galaxyrise]

First, I want to echo what David said.  Prevent the brief on time from getting to the light pack rather than trying to extend it.

Second, a capacitor in series would definitely not do what you want. I'm going to assume you meant in parallel.

But being able to compute that number is worth understanding.  It's fairly simple, thankfully, but you need to specify another parameter: what's the lowest voltage your light pack is ok with?  Capacitor voltage goes down as you take charge out, so what we're going to be solving for is how much capacitor it takes to maintain your minimum voltage after 15s of 500mA.  If the answer is 5V, then this simple approach is out, since you'd need an infinite capacitor not to drop at all.  Let's suppose it's 4.5V.

The main formula for this Q = CV.  Since current times time is charge, then we have a simple system of equations: Q(0) = C * 5 and Q(0)-0.5*15 = C * 4.5. 
Substituting for Q(0) yields C*5-7.5 = C*4.5.  Or C = 15F.  Getting 15F capacitance that can supply 500mA at 5V is almost certainly harder than what you had in mind.

But it gets worse.  You stipulated charging that capacitor in 2 seconds.  Q = CV again; Q = 15*5 = 75 coulombs. I = Q/t, so I = 75/2=37.5 Amps.  Your usb port isn't going to be able to do that.   (It's not trivial getting the capacitor to do that, for that matter.)

[Ian.M]

Even if you could get the charge into the capacitor without blowing out the USB port, it would back-feed the port when the supply dropped, and would be likely to blow the USB power management circuit. 

N.B. the USB standards effectively prohibit more than 10uF on Vbus.