tach input

[tom cloud]

A few more morsels:  The ign. coil _IS_ inductive, and it stores _LOTS_
of energy, so you can't ignore its voltage excursion which can be
_30 TIMES_ as big as the 12 volt supply.  A zener will probably work
OK.  Only one prob with a zener: it's not very fast!  Two switching
diodes (as I described incorrectly / then correctly in previous posts)
will switch off in about 4 nS (use 1N914B or 1N4148 or such).  A note
about all PN junctions (that's diodes, transistor base-emitter junctions,
JFET gate-channel junctions, SCR and TRIAC gates, UJT emitter - channel)
-- all PN junctions turn on reasonably fast -- it's the turn-off that
varies.  Big, high current devices are slow.  Zeners, rectifiers are
slow to turn off (slow means hundreds of nanoseconds or a few millisecs).
This may seem insignificant, but a voltage spike a few nanoseconds
long can be enough to destroy a junction -- and certainly a MOS gate.
That's why an opto-isolator works fairly well: besides giving isolation
(maybe -- you're probably not running a separate ground -- you can't,
since the coil is grounded to the chassis), its input is an LED which
is reasonably immune to transients.  Since you've gotta protect the
opto anyway, I still say a ten cent 2N4401 or 2N3904 or 2N2222, etc
is a more reasonable choice (consider: the opto requires several
milliamps to make it work.  The xstor can be made to switch with
microamperes.  The power dissipated in the resistor can be quite high.
To get 5 mA or less with a 300 volt spike requires R=300/.005=60k.
100k limits to 3 mA.  P=E^2 / R so for 60k P = (300)2 / 60k = 1.5 watt.
For 100k, P~1 W.  This may be for short periods, but at high rpm
those periods get real close together and a 1/2 watt R can get real hot!
So, restricting the current to, maybe, 1 mA gives R=300k and P=.3 W.

Sorry, just a little engineering rambling .. You can also drive a
schmitt trigger input inverter (7414) instead of the transistor.
Actually, this would be better .... use a low power Schottky device
(74LS14).  It still has bipolar inputs, so is less immune to transient
voltage damage than MOS / CMOS, requires similarly low currents to the
transistor, and its input is protected the same way (two diodes: one
with cathode to Vcc, the other with anode to GND).  The supreme advantage
to the Schmitt device is a little more relief from multiple signals
from one ignition event due to coil ringing (you'll have to put a
monostable / one-shot in to get rid of the rest).
>
>Like I said, I'm no expert on this, but: If there were no inductance in the
coil, you'll get square wave of
>battery voltage. The lowest battery voltage sets the highest value for the
resistors to give enough
>current for the opto. Let's say 9V - this leads to around 4k to give 2mA.
Max. current for the opto then
>defines R2, 10mA gives around 1k, so R1 is around 3k.
>The inductance introduces large voltage spikes up to 300-400V, but this is
no static state. If it was,
>R1 should dissipate 40W, but I guess .5W should be enough, at least I used
.25W, and it don't warm up.
>
>Come to think of it, the voltage spike is negative?, and so the zener not
actually required...
>Anyway it is there for safety.
>
>Some filtering is also required to smooth it up.
>
>----------
>From: 	Grant Beattie
>Sent: 	Monday, October 21, 1996 19:34
>To: 	Krister Wikstrom
>Cc: 	'DIY_EFI'
>Subject: 	Re: tach input
>
>
>I guess I was a little vague.  The only peice of info I'm missing is 
>what's the original input voltage?  You have a 15v zener that limits the 
>voltage excursions and an input resistor as well.  So if the voltage on 
>the zener is 15v, what's the voltage on the input res?
>

Tom Cloud <cloud at peaches.ph.utexas.edu>