From: Vladimir Pantelic <pan_at_nt.tu-darmstadt.de>
Date: Tue, 17 Sep 2002 12:03:33 +0200

Uwe Freese wrote:
> Hello Vladimir,
>
> Vladimir Pantelic wrote on Tuesday, September 17, 2002, 10:00:03 AM:
>
>
>>>>2 MB : 50mA + 300mA * 2/100 + 500 * 3/100 = 71mA average
>>>>8 MB : 50mA + 300mA * 4/450 + 500 * 3/450 = 56mA average
>
>
> The correct power consumptions, as I measured (Recorder 20), are:
>
> HDD off, LED off, idle 94 mA
> HDD off, LED off, play 97 mA
> HDD off, LED on, idle 129 mA
> HDD off, LED on, play 131 mA
> HDD on, LED on, play 230 mA
> HD on, reading, LED off ~ 600 mA
> HD spin up before read max 800 mA

So we have:

2MB: 100mA + 600mA * 2/100 + 800 * 3/100 = 136mA
8MB: 100mA + 600mA * 4/450 + 800 * 3/450 = 110mA -> 20% less

So this is the theoretical value, let's see how much it is in real life.

Regards,

Vladimir
Received on 2002-09-17