Re: 6569 luminances

From: Gerrit Heitsch <*gerrit_at_laosinh.s.bawue.de*> Date: Tue, 15 May 2012 14:42:04 +0200 Message-ID: <4FB24F1C.8060808@laosinh.s.bawue.de> · gerrit_at_laosinh.s.bawue.de

On 05/15/2012 01:20 PM, Segher Boessenkool wrote:
>> You sure about those values? Because 20 Ohm (10 + 10) between 5V and
>> GND would mean 250mA and that's more than the whole SID uses.
>
> 20 Ohm? So you get 4.5V as well? :-P

Uhm, no... I just took a look at the first stage and did my calculation. 
I did see the second part but didn't evaluate it before clearing up the 
mystery.

All stages taken into account the upper half would be 3.3R (10R and 5R 
in parallel) which makes a divider of 1:3 with the other 10R, leaving 
the bottom of the '4R' at 3.75V. That means you have 1.25V to look at 
and that 1:4 means you lose 1V at the 4R so 'out' has to be at 4.75V.

All assuming that the +5V are still +5V in that part of the chip. :)

  Gerrit

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