From: Jonathan Kirwan
Subject: Re: Roulette Wheel Project
References: <firstname.lastname@example.org> <email@example.com>
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NNTP-Posting-Date: Sat, 28 Dec 2002 18:39:53 GMT
Organization: AT&T Broadband
Date: Sat, 28 Dec 2002 18:39:53 GMT
On 28 Dec 2002 03:25:57 -0800, Winfield Hill
>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.
I had been looking at Daniel's circuit example posted in
a.b.s.e. In it, he included 7 current-limiting resistors.
You are right about only needing one, not two, of course. But
in the topology he illustrated, and if one assumes that the
output drivers are very strong (and not the somewhat more wimpy
ones usually found on real micros), then there is no avoiding
the resistor pair -- all possible pairs of output drivers need
to be permitted and that means you have to put resistors at each
I just took a look at your example, just now, to see the FETs
and so on. You've got 8+5=13 outputs involved. With a spin
switch, speaker, and red/green LED indicator pair, you've got 17
I/O pins. That pushes you to a 28-pin package, pretty much. It
has it's advantages, of course. But Daniel's can be done on a
16- or 18-pin package and (if one realizes that those current
limiting resistors aren't needed) with far fewer external parts.
>> 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
In Daniel's topology, the model I showed is about right. In the
model you show here for your arrangement, is that 68 ohm
resistor represented in your circuit shown elsewhere? Or is it
to model the LED?
You see, if you expand my model more fully:
| MSP430 | <--- LED ----> | MSP430 |
+3.6V <---/\/\/-----|>|----/\/\/-----/\/\/---> 0V
60 1.55V 21 60
That now includes the internal resistance to model the high
efficiency red. Current is (Vcc - 1.55)/(60+21+60).
Your model shown above doesn't illustrate that 21 ohm resistor,
from what I can see. So that's why I'm asking.
> 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.
For simplicity's sake and without some explicit need for the
external driving parts, I'd tend to prefer Daniel's circuit
modified to eliminate the current-limiting resistors. I think a
PIC16F628, for example, could do it using the internal RC
oscillator in a PDIP18 package. I just like fewer parts.