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1980 - 1986 Bullnose F100, F150 & Larger F-Series Trucks Discuss the Early Eighties Bullnose Ford Truck

Tachometer signal?

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Old Feb 2, 2013 | 10:03 AM
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Tachometer signal?

I've recently decided to get back into fiddling with electronics, and picked up an Arduino to play with. While trying to come up with some project ideas to work on, it hit me that I could probably build a digital tachometer for my truck.

Given that I've got a 300 engine with the Duraspark II ignition, what exactly is the signal coming off the tach wire attached to the coil? Am I correct in guessing that it is one pulse per cylinder firing? If so, at what voltage?

I realize I could probably figure some of that out on my own if I went out and put my meter on the truck. But, (a) my meter is analog, and thus not optimal for pulsed signals, and (b) it's cold outside.
 
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Old Feb 2, 2013 | 11:09 AM
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The green wire is coil primary ground.
The DSII box opens this wire in order to fire the coil.
 
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Old Feb 2, 2013 | 11:11 AM
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Originally Posted by ArdWrknTrk
The DSII box opens this wire in order to fire the coil.
Wouldn't opening (disconnecting) the wire un-fire the coil?
 
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Old Feb 2, 2013 | 11:13 AM
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Originally Posted by ArdWrknTrk
The green wire is coil primary ground.
The DSII box opens this wire in order to fire the coil.
In my truck (which does not have a tach in the instrument cluster), there is a green wire that comes off of the coil that ends in an unused plug. When I am using my tach/dwell meter, I attach one lead to this wire and the other to ground. Is this the same wire you are referring to?
 
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Old Feb 2, 2013 | 11:27 AM
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No Chris.
By "disconnecting" interrupting the flow of electrons in the primary side of the coil, the field collapses and induces a high voltage spike in the secondary windings.

Here is the schematic;



Dorsai, are you saying the green wire does not go to your DSII module?
 
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Old Feb 2, 2013 | 11:29 AM
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At any rate coil voltage should be (nominally)12V in start and 7-8V in run, due to the resistor wire in the harness near the ignition switch.
 
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Old Feb 2, 2013 | 11:43 AM
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Originally Posted by ctubutis
Wouldn't opening (disconnecting) the wire un-fire the coil?
A points style ignition system fires when the points open, right?

The transistorized ignition module operates the same way, it just uses a tiny electrical pulse from the reluctor pickup to switch the transistors gate and disconnect the coil from ground.
 
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Old Feb 2, 2013 | 11:47 AM
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Originally Posted by ArdWrknTrk
A points style ignition system fires when the points open, right?
Yes, that is correct.
 
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Old Feb 2, 2013 | 11:48 AM
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Originally Posted by ArdWrknTrk


Dorsai, are you saying the green wire does not go to your DSII module?
I just went out and looked more closely at how the ignition is wired; it is exactly like that (very helpful!) diagram you posted. What isn't shown on the diagram, though, is the green wire that I mentioned above; I am guessing from both the color as well as the fact that the green wire to the coil is labeled 'tach' in your diagram, that this wire is spliced into the green wire that runs from the ignition module to the coil.

So from this as well as your later comment, am I correct in assuming that (a) this wire pulses once per coil firing (well, technically, it would have voltage between firings, but for my purposes I don't think this distinction matters), and (b) that these pulses will be 12v at start, and 7-8v during run?
 
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Old Feb 2, 2013 | 12:46 PM
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Originally Posted by Dorsai
So from this as well as your later comment, am I correct in assuming that (a) this wire pulses once per coil firing
Talking about the tach test wire, then yes it does.


(b) that these pulses will be 12v at start, and 7-8v during run?
Yes and no...

The starter takes a lot of "power" to operate the engine. With the starter cranking the Voltage at the Coil evens out. What is more correct is that you have unresisted voltage at start, and resisted voltage in run.


Just to make something more clear...

Chris:

It's correct to say if the tach lead is "totally disconnected" that the coil won't fire. Same thing with it being "totally connected" the coil won't fire.

Ardwrkintrk:

It's also correct to say that the coil has to build up charge, and the act of the tach signal being disconnected is what sends the stored charge to the distributor.

It has to reconnect, and disconnect for every firing. Connect builds the charge, disconnect releases it. This is what increases the 12v to 20,000v aproximate.
 
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Old Feb 2, 2013 | 12:49 PM
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The connected time being called "Dwell"

Yes I understand the point about voltage, that's why I said 'nominal'.
 
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Old Feb 2, 2013 | 01:09 PM
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Originally Posted by 81-F-150-Explorer
The starter takes a lot of "power" to operate the engine. With the starter cranking the Voltage at the Coil evens out. What is more correct is that you have unresisted voltage at start, and resisted voltage in run.
OK, that makes sense. So in theory at start I could see voltage as high as whatever the battery puts out (~13.6v), dropping to that minus whatever the voltage drop caused by the starter is, and then 7-8v in run.

I'm thinking that if I build my circuit to account for a peak voltage around 14v, I should be good under normal conditions. Right?
 
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Old Feb 2, 2013 | 01:18 PM
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Originally Posted by 81-F-150-Explorer
Ardwrkintrk: Connect builds the charge, disconnect releases it. This is what increases the 12v to 20,000v aproximate.
This is not correct.
The coil is a simple transformer.
It is the ratio of primary to secondary windings that steps up the voltage.
The coil must have a certain(enough) time to saturate before the circuit is broken again. (dwell)
The coil creates a magnetic field.
Breaking the circuit collapses the primary field and induces a charge in the secondary winding.
The only thing that remains connected is a condenser. (capacitor)
This quells reverberations in the circuit.
 
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Old Feb 2, 2013 | 01:19 PM
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Your actual voltage will be far in excess of 14v. If you want I'll put my scope on this afternoon and see what it really is. But, when the module turns off, meaning let's the voltage go up to whatever it wants to go to, it'll go way above battery voltage due to the voltage induced while the secondary discharges.

Typically I put a zener diode in the input to the tach to protect it. Something like a 20 volt zener would do it, and it takes the induced voltage to ground. But, having said that, I've not worked on a DS-II so don't know for sure what the voltage looks like.
 
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Old Feb 2, 2013 | 01:20 PM
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Originally Posted by Dorsai
OK, that makes sense. So in theory at start I could see voltage as high as whatever the battery puts out (~13.6v), dropping to that minus whatever the voltage drop caused by the starter is, and then 7-8v in run.

I'm thinking that if I build my circuit to account for a peak voltage around 14v, I should be good under normal conditions. Right?
I would go 14.5V and would use a cap here to, in order to decouple it from spikes.
Most components in a nominal 12V vehicle are designed to operate at charging voltage.
 
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