"I failed to include earlier that diode D2 clips the positive peak"

Actually, the diode "blocks" (the postive portion) and doesn't clip.
A forward biased diode "clips" which is not the case here.

There's a major confusion here:

The Hall input will work on the 3 pin CDI (as long as it can sink 30-100ma),
which is the case as shown in the image. It WON'T work on the 6 pin CDI.
The 6 pin CDI requires about -1.00 volts to trigger (because the way the
input is initially biased at final test during assembly).

• D2 does not clip the input waveform
• D2 is always forward biased
• The base of T2 is what clips the input waveform when it becomes saturated
• The coil in the sensor has a DC current through it by design.

long tailed pair analysis from some .edu

I back annotated the spice schematic with the PCB values. If anyone wants to play PM me with your e-mail and I will send you the LTSpice .asc file.

http://forums.pelicanparts.com/uploa...1194627133.jpg

"* D2 does not clip the input waveform
* D2 is always forward biased
* The base of T2 is what clips the input waveform when it becomes saturated
* The coil in the sensor has a DC current through it by design."

Time to "drag" the circuit simulator to the "trash can" and enroll in a H.S. circuit
design class. The grade on the above is 25% or a F-:

1. D2 does not clip the input waveform - correct
2. D2 is always forwrded biased - wrong
It's only forward biased when the input goes negative (that's the purpose of the diode,
i.e. to only allow NEGATIVE pulses).
3. The base of T2 is what clips the input waveform when it becomes saturated - wrong
The input transistor is saturated because of the bias from the base resistors WITHOUT any
input signal. There's NO clipping of the input signal, i.e. the signal can go as negative as
it likes, or as positive (based on the breakdown voltage of D2).
4. The coil in the sensor has a DC current through it by design - wrong
There's NO DC current! There's just a pulse current when the signal goes negative:
I(signal) =(V(signal) - V(D2) - V(D5)) / 3.3K

OK

My observations are based on the output of an accepted circuit simulation technique. What are yours based on?


http://forums.pelicanparts.com/uploa...1194630308.jpg

"My observations are based on the output of an accepted circuit simulation technique."

Right, but like the saying goes; "Garbage in equals garbage out."

"What are yours based on?"

1. BSEE degree
2. 25+ years designing digital/analog circuits
3. Doing a proper circuit design analysis with CORRECT inputs and understanding
the functionality of ALL components.
4. Actual lab testing of both 3 & 6 pin CDIs using an oscilloscope.

Any chance you could point to the garbage?

What is in error in the above simulation?

And although I never took an analog/digital circuit design class in High School I believe the way you would find the current through D2 would be to apply Kirchhoff's Voltage Law around the loop to the left.

The algebraic sum of the voltage differences in any loop must equal zero

The analysis is very simple:

1. Remove the signal generator from the simulator.
2. The simulator should indicate that T2 is "on", because it has an external current
source which T3 doesn't (it's dependent on T2).
T2 base voltage is just = Vbase (T2) = 6.8 X 150 / (560 + 150) + Vbe(T2) =~ 2.0
3. To turn T2 "off" & then T3 "on", the input must sink the base
current of T2 which is determined by basically the 3.3K, i.e. the
input must go minus about .5 volts. Just grounding the input
just reduces the base drive of T2 but doesn't completely turn it "off".

When the above is verified with the simulator, then use a negative
ramp input simulator to determine where the actual switching point is
for T2 "off" & T3 "on". This will easily determine how much negative the
distributor signal must go, i.e. to eliminate any mis-triggering of the SCR
from noise.

With the sensor shorted (V_SENSOR = 0 volts), the condition the sensor coil of wire would be in if the distributor was not turning, the voltage at the base of T2 is determined by .

(6.8 -3*Vd)/(11k + 3.3K) * 3.3k + Vd ~= 1.7 volts [Vd is the diode forward voltage]

which agrees with the simulation. T2 is ON but not saturated and it sets the current through R8 at ~6.4 mA acting like an emitter follower. If it were saturated from your calculations above the current would be 9.5 mA through R8


If I sweep the input (V_SENSOR ) from 1 volt to -1 volt it appear T3 switches (fires the SCR) at about -450 mVolts.

Also notice the nice jump though the switching point (zero volts) of v(base2) - v(base1) caused by the hysteresis voltage added into the reference of the difference amplifier V(base2).

http://forums.pelicanparts.com/uploa...1194676681.jpg

Hey HKZ Bob

I gather from your other posts on the 3 pin CDI and transformers that you are not economically motivated to fix these but like myself are more interested in learning how they work.

Have you made any progress yet or found anything interesting?

Rick I don´t want to be disrepectful or non motivated
but I guess I am lost.

My simple question was why the my system works with an R10 of 10K & not 12K. I replaced the metioned parts and my assumption was that my 6.8 V Zener is not stable. The value I saw is between 6.5V to 6.8V.

What can be a potenial fault??
I think it is a good ID to simulate the schematic. Then we would have scope pictures on any point of the schematic. But could be another solution bring the same result ?
For example I we would have screenshots of BASE T2 & T3 as well anode of the thyristor and Base of 2N3055 to compare an OK unit with against my trouble unit. I really apresciate your interest. What I was figuring out is a simple example out of my training programm 20 years ago.

The other point is that i am not a native. So on some point it gets hard to follow.

Down a schematic of an usual Schmitt Trigger. As I understand we have two dynamic situations.

If the Hall Distributor (is high 5V ) is not pulling the Base1 to 0.6 V than T1 is closed & T2 is open. The 6.8 V is dropping over R7/R10/R9. There is a little current (1 to 10 mA) IB2 going in the base of T2.

If the Trigger is going against ground the 6.8V is dropping over (R5*R6)/(R5+R6) the D3&D4 (1.4V) (IB1=1mA) another 0.7V on D2 & R4 (3.3K) The base droppes below 0.7V and T1 opens and all currents runs through R8 (150R) It is paraelle to R10 & R9 which are togeher (12K+47K) 59K. This is a ratio of 1/400: So the current on the 59k path is so low that there is low voltage on Base 2.

Rick that´s what I understood.

When T3 is closed it pulles down Base T4 Base 0.7v more into the negative than the emitter voltage which is 6.8V.
On the Thyristor anode is now a 6.8V shaped pulse.

Can somebody confirm my thoughts.
Bob



http://forums.pelicanparts.com/uploa...1194689171.jpg
http://forums.pelicanparts.com/uploa...1194689952.jpg

Zener will limit the voltage from going above 6.8, not prevent it from dropping below. Either the input voltage is low, or the circuit is drawing excessive current through r3 so more than 7.2v is dropped across R3(14v-6.8v), or there is some other resistance in the circuit (cold solder joint, etc.)

"If it were saturated from your calculations above the current would be 9.5 mA through R8"

Read again and COMPLETELY, as there are two conditions
defined; input "open" & input shorted to ground. Remember, that when
the input goes positive, that an OPEN input condition exists.

Bottom line: Again, "Garbage in equals garbage out." especially when one
doesn't undestand the BASICS, i.e. no amount of simulation will help.

So:
1. Buy a 6 pin CDI from Pelican Parts.
2. Buy an oscilliscope.
3. Setup a bench test and PROPERLY analyze the CDI system without GUESSING.

HKZ Bob
That is exactly the circuit that is in the CDI. All the simulations are just for me to understand how the circuit works and see what the thresholds were.

I did not understand. Are you trying to modify the 6 pin CDI so it will work with a Hall effect sensor distributor? I thought you had a CDI that just didn't work. Is the Hall module in the distributor an open collector output? If so it would seem to be simple to get it to work by getting rid of D2 and R4.

Which transistor do you have for T3, BCY or 2N ?

A Hall sensor input is NOT the proper way to test the 6 pin CDI.
In actual use, the input signal must be negative (~.50) to switch
the input circuit (trigger SCR) and then it must go positive (~.50)
to reset the input circuit for the NEXT negative pulse. This assures
that false triggering is minimized, e.g. multiple negative pulses.
The best approach is to use a sine input.

HKZ Bob

Is this is what is in the distributor?

var reluctance sensor See page 2 of the PDF file.

This shows what happens to the trigger point as R10 is decreased form 12k to 6k in steps of 1k. The trigger goes from -0.5 at the input to about -0.1,

So if your are using the magnet/coil of wire to sense ignition and it takes reducing R10 to get it to work I would say your sensor is not puttin out enough voltage. Is the gap right? What is the resistance of the coil? What should it be/

http://forums.pelicanparts.com/uploa...1194726999.jpg

"So if your are using the magnet/coil of wire to sense ignition and it takes reducing R10 to get it to work I would say yo5{uttin out enough voltage."

You don't understand the problem.

He has the problem because he's using a Hall device as an input source.
As was said before, one must use a signal which goes negative & replicates the true signal,
i.e. basically a sine wave input. The problem
will NOT occur then.

Again, the circuit is very simple and doesn't have any critical components.

Rick, I do not have a 911 Hall dizzy thats why I took this GM Unit.
I thought they are using the same hall device?

bob

I use a BCY for T3.

bob