From: Winfield Hill
Subject: Re: Roulette Wheel Project
Date: 28 Dec 2002 03:25:57 -0800
Organization: Rowland Institute
References: <email@example.com> <firstname.lastname@example.org>
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Jonathan Kirwan wrote...
> Gary Lecomte wrote:
>> Actually, I don't even understand your simplified schematic
>> very well.
> I took a look at it, just now. What he means to imply is a
> 'simplex' arrangement. [snip]
> The external resistors, from each micro output pin to each
> simplex vertex, are there to 'allow' the voltage at the two
> active vertices to freely adjust per the LED requirements.
Since only one LED path is conducting at a time, only one
current-limiting resistor was required. In this case I put
it in the pullup FET switch drain bus.
> V(forward) = 1.55V + I(forward) * 21 ohms
> So we have this equivalent, when an LED is selected and ON:
> +3.6V <---/\/\/----|>|----/\/\/---> 0V
> 60 60
I'm imagining a 5V system, like this, say with a pulsed 25mA
in the high-efficiency LEDs.
. R 2n7000 LED PIC
. ___ ___
. +5 --/\/\--' '-----|>|-----' '--- gnd
. 68 10 1.55V 60
It's true that 25mA isn't an impressive pulsed current for LEDs,
especially since each pulsed bus has only a 20% duty cycle. It's
equivalent to only 5mA continuous current, but fortunately these
days one can get LEDs that're very bright running from only 5mA.
One issue, in this simplified scheme the MOSFET may have more
than 10 ohms of resistance, given that it only has a miserly
2.0V of gate drive. If the microprocessor had a lower Ron, say
20 ohms, then R would be 110 ohms to set 25mA, and the 2n7000
FET would have a more reasonable 3.0V of gate drive. Probably
a set of five NPN transistors with base resistors would be better
suited as pullups, to solve the FET's Vgs(on) issue.