Re: Electronics Theory: System 80 Pop Bumper Driver
- From: Bob Russo <pinballDontSpamBob@xxxxxxxxxxx>
- Date: Thu, 05 Jul 2007 23:08:14 -0400
Without looking up the specs on the transistor, I can offer this:
The inverter (7416) is an open collector device, it's essentially not there when the output is high, and ground when the output is low. It can sink 40ma, the 74121 can only sink 16ma. The circuit requires 23 ma of sink current.
The diode both raises the noise immunity when the Darlington is off (keeps noise from triggering the Darlington, lotsa noise in a pin), and has the nice effect of protecting the TTL stuff if and when the Darlington does self destruct.
TVisitor wrote:
A lot of my basics seem to have disappeared, while some of the more.
complex knowledge seems intact. Odd. That's what you get when you
don't handle circuits for years, but you do logic-based programming.
So, you've basically got a one-shot that pulses for a time when the
pop-bumper driver board is activated. That one-shot feeds the coil,
to insure that it is only on for that duration, even if the trigger is
held high. Gotcha!
I realize the output of the one-shot is taken from Q*, so it's
normally high when idle, but it's fed through an invertor, so it's
normally low (why not just use the Q output?).
The pop bumper solenoids require some voltage across them. Normally
it would seem that one end of the solenoid is tied to +something
voltage (+38VDC, for example) and the other end is under control of
the driver board. If the board grounds it, the pop bumper goes on.
If the board "floats" it, it is off.
The +5 from the inverter goes through a diode then to the base of the
Darlington on the pop bumper board.
When this hits +5, I'm assuming (and this is where I get a little
fuzzy) that it exceeds the breakdown voltage of the diode + the Base-
Emitter pairs of the Darlington, turning on the transistor and
effectively tying the emitter to the collector, which ends up being
ground - and viola, the pop bumper solenoid engages.
When it (the output of the inverter) goes back to 0V, the transistor
switches off and the solenoid is not grounded, and it opens up and
releases.
So, through that rambling:
1. Why not use the Q out of the one shot? [And I know it has one,
it's pin 6...] It just doesn't make a lot of sense to me why they'd
put on an extra part (the inverter) if they didn't need it.
2. Is my rambling about the operation accurate?
3. If so, why do you need the diode between the inverter & the base
of the transistor?
Thanks all.
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