🚀 go-pugleaf

DaveC wrote:
> 
> On Sat, 17 Aug 2002 2:24:04 -0700, Terry Pinnell wrote
> (in message <so4slug6g5r6etb42fjckvf3622prkov2i@4ax.com>):
> 
> > DaveC <anon@example.net> wrote:
> 
> >> Using a small gp transistor, I think I can eliminate the relay altogether:

Neat idea!  It does sacrifice crisp turn on of the enable
circuit and the ability to drive a higher current alarm,
but neither of may be of concern.

> 
> How and why does the zener enter in to the calculation?
> Doesn't 4.7K and 680uF give ~3 seconds?

Yes - 3.196 is the RC, but not the delay time.

The zener sets the voltage to which the capacitor must charge.
Typically RC is around 63 percent of Vcc.  With a 9v zener,
the base of the transistor doesn't see any voltage until
75% of Vcc (using 12 volts as Vcc.  Actual voltage depends
on the state of charge of the battery and any Vd in the
circuit).  Even once it starts seeing voltage, it doesn't
switch on (i.e. saturate) until the voltage rises high enough
above the emitter voltage to drive it into saturation.  Lets
say the base has to be .6 volts above the emitter to drive
the transistor into saturation. That means the cap has to
charge to 9.6 volts, or 80 percent of Vcc.  You can see the
added time delay at:
http://micro.magnet.fsu.edu/electromag/java/timeconstant/

In practice, the buzzer will start sounding at some point
in time after RC but before the transistor is driven into
saturation.  And RC applies on the breadboard.  In the car
the precise Vcc voltage is unknown and can vary, so the 
exact RC time and the exact time the piezo starts sounding 
can't be determined.

A bigger C and smaller R was recommended, and is a good idea
with a single transistor. As was pointed out, the thing
won't work with a 1 meg resistor due to insufficient base
current.
Another idea, which will allow say a 10 uf cap and a higher R
(in the range between roughly 15K and 6 meg, depending
on what kind of delay you want and transistor gain) is to use 
2 transistors in a darlington pair.  The darlington pair greatly 
reduces the base current needed. To pair two gp NPN transistors
as a darlington, connect the emitter of one to the base of 
the other. Connect the collectors together, and to the load,
and the other side of the load to +.  Connect the unconnected
emitter lead to ground, and the unconnected base lead to
the zener.

There is still a problem that this design does not
overcome:  you get in your car, turn the ignition
on but do not start the car. You fidlle around with
adjusting the mirror, fastening the seat belt, whatever,
and the time delay expires. Your circuit is now enabled,
and if any of the connected idiot lites are energized
during cranking, the alarm sounds.

I wrote an alternative approach in another reply.
What you choose and why you choose it is an 
"engineering decision" where you weigh factors
such as cost, parts count, reliability etc.
against the function required, and decide which
approach to use.

Playing around with the circuit on a breadboard
is a good idea.  You can vary the Vcc and see
what effect it has on the delay. You can measure
base current required to turn a transistor on
in a given circuit, then add a transistor to
make it into a darlington pair and use a resistor
substitution box or potentiometer in series with 
the base to see how you can now turn the transistor
on with a much lower current.  If you use an LED
as the load, you can watch the transistor start to
conduct when the LED first starts glowing dimly
until it glows at full brightness. You can listen
to the piezo to see how it behaves as the transistor
goes from the start of Ice to full saturation.

Have fun with this project!




> > I'd breadboard it and try it first in some similarly hot environments,
> > like my oven!
> 
> Thanks!
> Any ideas on how to accomplish this with a different circuit design?
> --
> Note that my return address is corrupted in an attempt to reduce spam. If you
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> 
> Thanks,
> Dave
> ---
> Dave Carpenter
> Sound Logic
> voicebox@d8n8a8i.com
> 
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