How I use Crane Optical Trig with Bosch CDI (3-pin)

[mskala]

Hi,
Maybe everybody who would want to do this has already done it, or done it in a
different way. But, for archive, here is how I use just the Crane trigger, without
buying their ignition box, and send it to the 3-pin CDI.

Why? I always like fairly easy projects , but also there is one thing I'm not
impressed with about pertronix; I don't believe in the position accuracy of the
tiny magnets, I'm more comfortable with the plastic vanes.

The Crane optical trigger alone is ~$29. (Search 700-0020), and the kit
of vane wheels and the bracket is ~$34(Search 700-2231). The trigger comes
with the connector, but to get its mate and pins for it, I got Molex "1396PRT"
for ~$6.

To make it work, besides finding a switched 12V wire, you need to have a
resistor to limit current to the LED, and a transistor and resistor to make
the proper output signal. I put these in a small box filled with epoxy, with
the crane connector at one end and spade lugs going to +12, GND, and the
existing trigger in the harness. This is the picture of the box:


This is the schematic:


This is the distributor:


This is all that's in the box:


The pinout used is the same as theirs, so if there was ever a reason to go to the
'fireball' system, the distributor would just plug right in.

[RD911T]

Nice

[mysocal911]

The problem with an optical system is the heat sensitivity and the resulting reduced reliability.
Most all automotive sensor applications use Hall-Effect devices (magnetic) which are much less temp
sensitive.

[wwest]

Optical = oil misting

[jpnovak]

HMM I Have had optical triggers in all my cars including one for 14 years. NEVER had a problem.

They work great.

There is no oil misting inside a distributor unless you have major other problems. They use an IR LED and photoreceptor. The level of IR from heat (same thing) is minimal compared to the output of the LED. Don't see this as a problem.

Carry on. Great work on your circuit.

[mskala]

Originally Posted by mysocal911

The problem with an optical system is the heat sensitivity and the resulting reduced reliability.
Most all automotive sensor applications use Hall-Effect devices (magnetic) which are much less temp
sensitive.

No offense, but if you have any detailed test data or studies about problems
I would like to see it. Based on the very reduced max currents of a simple design
like this and the lack of serious heat inside the distributor, I don't see that it's
anywhere near the limits that would cause problems.

For a decade on this car, with a different distributor, I used just a random old
LED and photodiode from my parts bin, with no problems ever (~60Kmiles).

[304065]

Mark, great idea and execution!

A few questions.

I see you used an N-channel MOSFET for the output stage.

So when the gain-source voltage is HIGH, the drain gets pulled LOW, to ground.

This replicates the square wave of the points- in the old Kettering-ignition, when the points open is when the coil field breaks down, that is triggering on the falling (trailing) edge.

When the factory went to CDI they used a 33 ohm pull-up resistor in the trigger circuit in lieu of the coil. But same action-- when the points are closed the trigger gets +12V, limited by the resistor, then when they open it goes low, triggering the CDI to fire.

So that is how the CDI gets the square wave that drives it. You are replicating that here, the MOSFET just gives you a solid-state way to pull the "C" pin of the CDI down to ground, then open it to fire the CDI.

So backing up to the MOSFET's gate input. . .the Mallory is an optocoupler with an LED that presumably shines all the time (I haven't taken one apart) and when the window in the shutter lets light through, the detector emits a positive voltage.

The positive voltage hits the gate and this is what triggers the source/drain to go low.

Am I following your design correctly?

I have been wanting to understand this for a while, for me the big advantage of using an optical pickup is the ability to file the shutter to alter the individual cylinder timing events.

And eliminating the points bounce is a plus as well.

Thanks for posting this is terrific.

[mskala]

What you've said is right except for the optical receiver. I think it's
a photodiode, but not sure. It will conduct current when it sees light.
The 33K resistor is there to turn that current into the voltage that the
mosfet gate needs. With no light, the current is gone and the gate
sees high, output is like points being closed. When light makes it
through the slot, detector produces a small current, which causes a
low at the gate, and output is like points being open.

[dicklague]

great post.

I think I will build one of these and see if I like it better than the Pertronix.

Very ingenious set up. thanks a bunch

[304065]

Mark- I get it now.

The 33K ohm resistor is in series with the +12V power supply-- so it provides a slight positive bias to the gate of the MOSFET. This allows the MOSFET to conduct from Source to Drain- it connects TRIG_OUT to GND. As you say, this looks like points closed.

Then the shutter rotates and the window opens, allowing the phototransistor (or whatever) to become conductive. This then "diverts" the current AWAY from the gate of the MOSFET, which pulls the gate LOW, shutting off the Source to Drain-- which looks like the points opening. (Or are you saying that there is +12v coming from both sides of the gate, through the 33K resistor and from the phototransistor, so the potential difference goes to zero, driving the gate low?)

I think the light bulb went on for me when I read somewhere that the MOSFET gate doesn't like to float-- and it's susceptible to damage from static, etc. So it made sense that there would be some sort of bias all the time.

Is that right? I guess we don't know a huge amount about the phototransistor in the optical trigger but given that we know what the engine expects we don't have to.

I am going to also try this myself, I don't have the trigger but they are cheap.

Thanks again I learned something from this

[wwest]

Back when, 'WAY back when, Chrysler had CD ignition using a magnetic pickup to replace the points. I had a friend at Boeing that turned down the cam in my Ford distributor so the Chrysler "star" magnetic part slid right on the Ford distributor "cam". Home brew CDI (Popular Mechanics[?]).

Paid a wrecking yard to pick up the car after 250,000 miles.

[mysocal911]

"I think the light bulb went on for me when I read somewhere that the MOSFET gate doesn't like to float-- and it's susceptible to damage from static, etc. So it made sense that there would be some sort of bias all the time."

A MOSFET (Metal Oxide Semiconductor Field Effect Transistor) just like the name states
is a device whose conductance is affected be the electric field produced by the gate.
The MOSFET is a voltage switched device versus a bipolar transistor which is a current
switched device. The MOSFET will basically retain the voltage between the gate to
source causing it to continue to conduct unless the voltage removed, i.e. the voltage
between the gate and source is switched to zero volts. This is different than the bipolar
device where just disconnecting the base will switch off the device.

[mskala]

Originally Posted by 304065

Mark- I get it now.
...
Then the shutter rotates and the window opens, allowing the phototransistor (or whatever) to become conductive. This then "diverts" the current AWAY from the gate of the MOSFET, which pulls the gate LOW, shutting off the Source to Drain-- which looks like the points opening. (Or are you saying that there is +12v coming from both sides of the gate, through the 33K resistor and from the phototransistor, so the potential difference goes to zero, driving the gate low?)
...

Here's the way to look at it. The gate is essentially infinite impedance, so it really
doesn't affect the circuit that you connect to it.

There is a 33K resistor with one end at ~12V. At the other end there is a device that either allows current or not (to ground in this case).
deltaV = IR
Vgate = 12 - Vacross_resistor
Vacross_resistor = I*R

If photodiode does not conduct, then
Vacross_resistor = (0)*33K = 0 and Vgate = 12
If photodiode conducts, then
Vacross_resistor = I * 33K. Even with small current this voltage is large.
and Vgate ~ 0.

[gereed75]

Mark, I am about to put Crane optical points in car with a 3 pin CDI. Sorta understand what your circuit does (two semesters of EE along time ago), but do I really need your circuit??

It seems others use this set up without it. Does the XR700 do the same or similar to your circuit?? Thanks

[mskala]

Sorry I didn't see this when you posted. The Crane optical 'head' does not have anything
inside it except for an LED and a photodiode. The Bosch CDI does not have anything to
interface to it; it expects a signal that is basically shorted to ground or totally open.
So to use the Bosch you need to power it and produce the correct signal. I'm sure the
Crane ignition box just has it built in so you can simply connect the 3 wires to it.

[gereed75]

Put the Crane optical points in my 3 pin CDI car. Up until then, I had played around with the points multiple times and could never get it to run right. The tach was erratic and there was an intermittent miss that I could never completely get rid of.

With the Crane opticals and their "fireball XR700" unit, the thing runs great. Pulls cleanly to 7K without a hicup and the tach is rock steady.

I have a 72T distributor with the vacumm retard. I think the play in the retard plate was enough to up set the points but it does not seem to effect the optical points. Works like a champ.

[jimmyjimmy]

This electronics conversation is way up my knowledge.
As I have the crane unit ,do i really need the Bosh cdi (3 pin) ?
If yes , someone has this running and could take a look how it is wired ?

thanks

[mysocal911]

Originally Posted by jimmyjimmy

This electronics conversation is way up my knowledge.
As I have the crane unit ,do i really need the Bosh cdi (3 pin) ?
If yes , someone has this running and could take a look how it is wired ?

thanks

Yes, remember the Crane unit just provides a trigger signal like a set of points to an ignition unit, it does not drive an ignition coil.