Re: Electronics Theory: System 80 Pop Bumper Driver
- From: TVisitor <tvisitor@xxxxxxxxx>
- Date: Fri, 06 Jul 2007 07:04:12 -0700
Sorry, I was the one who was talking about replacing the Darlington
with a TIP102. The only reason I was commenting on that was becuase I
had the TIP102 specs handy.
So really, in an ideal world, you could remove the Diode and the
circuit would operate. I guess my comments about the diode were not
really correct (i.e. you'd have +5 at the inverter output (not
correct), so you'd want to further reduce the voltage as +5 is the max
for the BE junction), but it does make perfect sense for the noise
immunity & protection when the Darlington goes to the winds.
I hadn't looked up the inverter specs, but the open collector makes
sense. I'll have to do that later and see. The Q / Q*, yeah, silly
me - the world is not an ideal place, reality keeps rearing it's ugly
head. These days I wonder if you could just rework the pop bumper
boards so that you could eliminate the inverter and just have the one
shot drive the diode to the darlington directly. Not that I am going
to do it... Just sayin'... :)
This is going to sound like a silly question, but just bear with me
and hopefully someone will answer it. sometimes things are unclear in
my head and I need an example to get things straight.
Let's say you had a circuit where you had the diode & the transistor,
as in this circuit. Pulling specs out of nowhere (just for
illustration sake), let's say you need 0.7V across the diode for it to
conduct, and you need 0.7 across the BE junction of the darlington
transistor for it to turn on. This means you'd need 1.4V from the
anode to ground to get the transistor to turn on, correct? As a
second example: you had 2 conventional transistors in a darlington
configuration, each needing 0.7V BE to turn on. This implies you'd
need 2.1V from Anode of the diode to ground to get the transistor to
turn on, correct?
Part of what puzzles me is this, and this may be due to my thinking of
"ideal" circuits.
Let's say you had a simple voltage source that is driving a resistor
and 2 diodes in series. the diode needs 0.7V forward voltage to turn
on.
(+ of supply) -> resistor -> anode #1, cathode1 -> anode #2, cathode
#2 to (- of supply).
You'd need 1.4V at the supply to get the diodes to turn on and
complete the circuit. I'm kind of confused as to how this works - you
can crank up the voltage on the supply, but since neither diode is on,
what is Cathode #1 at? It's not grounded, so how can you clearly get
the forward voltage across it to turn it on - what voltage is it at?
Argh.
.
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