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From: email@example.com (Tom Torfs)
Subject: Re: Limiting M5451 LED display driver power consumption
Date: 23 Dec 2002 04:36:01 -0800
NNTP-Posting-Date: 23 Dec 2002 12:36:01 GMT
> > This formula makes perfect sense, except I don't know how to get the
> > VOmin value? I'm probably overlooking something quite simple here, but
> > as I see it now the VO at the current sink terminals of the M5451
> > would depend on the resistor value I am trying to calculate, as in VO
> > = 6V - R*I*N - 2V. This equation with R in both sides is solvable but
> > I have the feeling something is wrong with my approach. Maybe one of
> > you has used this chip before and/or can tell me how to go about this?
> > Thanks!
> The VOmin value is from the spec sheet, and should be figured using figure 5
> and 6. It's the minimum voltage you want to operate the output at. For
> instance, if you're intending to run all 34 segments at 12mA, it would be
> about 1.2V.
Actually with those values I don't get acceptable results with a
single resistor. However, I have a bunch of 3.3V/1A linear regulators
lying around, so I am going to regulate the 6V battery voltage down to
3.3V and connect that to the common anode of the LEDs. This gives me a
constant Vo of 3.3V-2V = 1.3V, leading to a max. power dissipation of
28 * 0.015A * 1.3V = 0.546W (within spec). I am going to have airflow
with a ventilator anyway so this should be ok. Do you think it is a
good idea to attach some sort of heatsink? (not that convenient with
the dual inline package)
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