tach input

[Krister Wikstrom]

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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?


Mystery
input
voltage       R1             R2

   0--------WWWW-----------WWWWW------ opto
                      |
                      | /
                     ---
                    / ^   Zener
                     / \
     
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