"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

Loren does the original porsche dizzy uses a special hall device which pulls the base into negative? Do you have a scope picture of the signal?

Bob

Loren there a two type of hall sensor.
Low over tooth & high over tooth. What technology is Porsche using.

Bob

"does the original porsche dizzy uses a special hall device which pulls the base into negative?"

No. There's no Hall device in the SC distributor. It's a magnetic pickup which basically produces a distorted sine wave. Just use a sine wave generator for testing and your headache will be solved.

Man so much trouble about nothing.

Thanks Loren,
so the magnetic pic up on the dizzy produces sinus peak to peak of 1 volt which goes to 133Hz which is 8000 RPM.

Or is the value higher than 1v peak to peak.

Bob

This is what the output of the variable reluctance sensor looks like (see Jimmcc). The signal has no DC component so 0 volts goes right through the middle. A sine wave would probably work (the 2 Vp-p in my simulation) but I remember reading from someone that fixes these (in France not the local vociferous TV repair entrepreneur) that the negative slope of the signal wasn't always quick enough to trigger the SCR

DC from the level shift input to the trigger does go through the sensor coil but does not show up as a voltage on the signal since dI/dt = 0

The signal is generated using Faraday's law v = N dflux/dt where the flux changes because the iron star wheel completes the magnetic circuit through the coil in the sensor.

The signal amplitude varies with the RPM, i.e. it increases as the RPM increases.
So use a sine wave with a 1.0 volt peak to peak. Since a six cylinder at 6000
RPMs produces a spark every 3.3ms, the corresponding freq is 300 HZ at 6000
and 400 HZ at 8000.

"the negative slope of the signal wasn't always quick enough to trigger the SCR"

Nope! Remember, the input circuit is a Schmitt Trigger which is INDEPENDENT
of the input slope, i.e. That's the reason for using it.

I get smarter now.

A page out of the "Bosch Bible".

http://forums.pelicanparts.com/uploa...1194902084.jpg

Bob

See I knew it was a problem with your induktionsgeber. :rolleyes:

I am going to Germany next week and don't speak a word of German. This is what I have to look forward to.

Is it eine beer bitta?

Loren, Waaren and all who can help & I changed the transistor T1 2N3055 but it was not the problem. I still have the failure.

See my scope pictures measured at the Zener at 12V without trigger.
What happens is at 12V the DC DC Inverter work. when I increase the voltage to 14V. Suddenly the Zenner Voltage cuts back. (20V/div). I get one spark.
It is repeatable without trigger. I have about 85V

http://forums.pelicanparts.com/uploa...1195758831.jpg

Here the picture with 14V and the current goes up to 3A.

http://forums.pelicanparts.com/uploa...1195758942.jpg

Here the pictures with trigger 12V & 14V.



http://forums.pelicanparts.com/uploa...1195759069.jpg

http://forums.pelicanparts.com/uploa...1195759102.jpg

Is this a Zener failure ?

Thanks for the help and have a nice Thanks Giving.

Bob

Bob, you probably have a dying SCR. The SCR becomes conductive (avalanche mode) when you increase the output of the DC/DC converter. This is why you see one spark. Next the frequency of the DC/DC goes much higher because its outout is shortened to ground by the SCR. Replace it and it should work fine again. Read here for more details.

http://forums.pelicanparts.com/porsche-911-technical-forum/378518-interesting-bosch-cdi-box-failure.html

Ingo```

Thanks Ingo.

So you mean my Thyristor is bad not the Zener?

Bob

I changed the Thyristor against an new OEM part. The failure is still there.
Without trigger I do increase the voltage to 14V. Suddenly the Zenner Voltage cuts back. (20V/div). I get one spark.

I will change now the Zener.

Bob

Ingo and all,
the Thyristor & Zener was not the root cause for the problem. Heat and 14.5 V causing the problem as explained before. The current goes up and the frequence changes.

Here all the parts circled red that I have changed.

I will change now the C2 capacitor. Any ID now?

bobhttp://forums.pelicanparts.com/uploa...1195849342.jpg

Bob, I wonder what SCR you put in there for replacement - if its tq is larger than 35usec it won't work! When you say you get one spark when rising the supply voltage can you measure (with a simple DVM) what voltage you see at the SRC right before it becomes conductive. Is it much greater than the 460 volts? If not it is the SCR going into premature avalanche breakdown. Also check with a scope if the voltage is mostly DC. Otherwise maybe the diodes D6 D7 are suspect.

Cheers,
Ingo

Ingo I used a T12N 800COB form AEG Brand.
Your Information with the 35 usec is not correct. Here original datasheets.


See my picture from the board. What do you think ?
New.http://forums.pelicanparts.com/uploa...1195887566.jpg
http://forums.pelicanparts.com/uploa...1195887771.jpg


Greetings Bob

Out of curiosity where did you get ahold of the AEG brand SCR? I was under the impression that AEG stopped to make semiconductors a long long time ago.

Back to your issue. The data sheet says tq = 60uS@125 degrees. I think this is borderline at best. I sometimes see 65 - 68 uS on the scope for the time the anode is negative. This is why I prefer a replacement with tq =35 uS. However, this does not really explain why your SCR turns on whith no trigger signal present when you raise the supply voltage. It should not happen even if you put 16volts or more into the box.

One reason for premature turnon is if dV/dt (the rise rate) of a voltage spike across the SCR is too high. This is called the Miller effect. Think of the voltage spike having to charge the capacitor formed by the junction in the SCR. The resulting current is proportional to dV/dt and if high enough you have static dV/dt breakdown. Assuming your SCR is in good working condition here is what you can do:

1. Check if the snubber network C7/R16 is working properly. It is to limit dV/dt across the SCR.

2. Check the voltage at the anode of the SCR with a DVM and a scope right before breakdown. If you are concerned about noise on the gate simply disconnect it for the test.

Let us know your findings.

Cheers,
Ingo

Dear all,
some information in advance. Measured at room temp.
My voltage at C8 is 433V without no noises on the DC voltage.
When I ramp though the 16.6 Volt level with my power supply. I get one spark.

When the box is getting warmed up this failure occures at 14V. that why I changed R1. The resistor was not stable at temp. I tested it.


I investigated that bosch was using at their 3 pin a 22V Zener and an R1 120R.
So I put and 120R as a replacement to 56R.

Result the single spark problem occured at 20.3 volt without a trigger hocked up.
Any ID what this causes. of cause I reduce the DC Base voltage on T1. But do I change the phase shift between W1 & W2 of the windings?

Bob

This is what I measure between T4 Emitter & R14.

Before the thyristor complete opens I am having an egde on my scope before the pulse comes.

Sorry for the bad pictures. Christmas comming soon.
http://forums.pelicanparts.com/uploa...1195906944.jpg
http://forums.pelicanparts.com/uploa...1195907001.jpg

This happens with triggering the box.
With trigger on it pulls down the output and locks up.

When the box is cold I have a nice spike on the Thyristor,
when its gets wamed up it is cutting out. See scope pictures.

Then it cuts out complete.

bob

[QUOTE=ischmitz;3604605]Out of curiosity where did you get ahold of the AEG brand SCR? I was under the impression that AEG stopped to make semiconductors a long long time ago.

I got one unit from ebay brand new old stock

Bob,

The 22v zener diode is a protection against over-voltage spikes on the supply. It has nothing to do with the DC/DC converter. Z1 (82 volt) is what limits the DC/DC converter output voltage. However, I am confused. Let's get some answers:

1. Are you saying the one-spark without trigger behavior changed because you changed R1 to a good working one or because you changed the value of R1 t0 120Ohm?

2. Do you see T4 producing a spike that fires the SCR? If yes simly disconnect the gate on the SCR and repeat the test. I am trying to isolate whether your signal stage fires the SCR or whether it breaks down.

Also is this single-spark event happening at 20 volts or at 14 volts and is it depending on box temperature?

Next, for the locking up with trigger signall can you measure the dV/dt rate for the SCR? See the scope picture as an example? In this case the anode was negative for 110uS. I have seen that time to be as short as 68uS in working boxes. So under temperature this is what challenges the SCR.

http://forums.pelicanparts.com/uploa...1195928974.jpg

Also, did you check the snubber?

By the way, even new old stock (NOS) is not an insurance for a good working part. This could well have been parts that did not meet one or several performance specs such as tq. If I were you and I wouldn't want to diagnose the problem but shotgun it I would drop in a suitable replacement SCR and check if this fixes the box.

Ingo