6 pin CDI problem under temp.

[rick-l]

Ok what am I doing wrong?

• The blue trace is the simulated input (1 cycle sine wave)
• The black trace is the tach output
• The red is the current that goes through the PNP to trigger it

jimmcc - that is what I suspected the input looked like but it is too hard to simulate so I just used a sine wave.

HKZ Bob - if I put 4.7k in it does not seem to trigger. Did I get the wrong value for that resistor?

edit: adding the .0068 capacitor dosen't change much

[rick-l]

Originally Posted by Lorenfb

Time to get another circuit simulator (maybe ICE).

many of my cohorts use this instead of the one with a $7000 license fee
switchercad

[HKZ Bob]

Jim your scope picture is measured between R8 & C3.


What is the timebase & div on your scope.

Regards Bob

[HKZ Bob]

Rick 4.7K is the correct value. Not 47K



bob

[rick-l]

Originally Posted by HKZ Bob

Rick 4.7K is the correct value. Not 47K
bob

It sure looks like 47k to me here (Warren Halls thread with schematic)
but there is a conflict between the parts list and schematic. I wonder which one is right?

Oh yeah and what a nightmare this thing would be to produce. I would guess R8 must be selected to get about 1 volt on the emitter resistor and T3 to match the beta of T2.

[rick-l]

This what you get with 4.7K

wonder which one is correct they both give the desired result.

[Lorenfb]

"C3 feed back."

C3 is really not a feedback element, but just an input noise filter.
A feedback element would be connected from the base of Q1 to
the collector of Q2 and then you've have a one-shot.

Open a CDI unit and you'll see the R12 is 4.7K.
Don't rely on poorly drawn (incorrect) schematics posted on this website.

"I would guess R8 must be selected to get about 1 volt on the emitter resistor and T3 to match the beta of T2."

Not really!

This whole thread over complicates a VERY simple circuit design.

Bottom line:

If one doesn't understand (and never will) an electronics system (or any other system),
then a DIY effort will be a total waste of time!

Hey, pickup your phone and call John/Scott at Pelican Parts for a rebuilt and end this misery!

[rick-l]

Originally Posted by HKZ Bob

Rick 4.7K is the correct value. Not 47K



bob

This circuit makes a whole lot more sense if you use 4.7k

[rick-l]

Originally Posted by Lorenfb

Open a CDI unit and you'll see the R12 is 4.7K.
Don't rely on poorly drawn (incorrect) schematics posted on this website.

Like I said before I don't have one I was just curious how it worked the interst kindled by a buddies MSD unit I worked on.

Originally Posted by Lorenfb

"I would guess R8 must be selected to get about 1 volt on the emitter resistor and T3 to match the beta of T2."

Not really!

So when you refurb these units what criteria did you use to replace the Select In Test Parts like T3 and R8?

[Lorenfb]

"So when you refurb these units what criteria did you use to replace the Select In Test Parts like T3 and R8?"

Again, they're NOT selected. R8 is 180 ohms +/-5%, & Q1/2 are 2N2222As,
i.e. just generic. R8 just determines the hysteresis - Vhys = 6.8 X 180/(560 + 180).

[jimmcc]

Resistor R9 was inadvertantly changed by who ever finished up this schematic. I have an earlier version of this same schematic that has the component values written in by hand and R9 is listed as 4.7k.

Also have another schematic in PDF form that shows essentially the same schematic but with a few minor component value differences. Obviously a another version but no indication of revision date. I have been unable to convert it to a jpg as it is drawn in color on a black background.

It shows R9 as 4.7k.

[HKZ Bob]

Hallo,
what could be the potential drivers that with an(R10) 12K restistor on Base T3
the trigger won´t work. A bad capacitor whichs keeps the Voltage not constant.

It was working with 10K @R10.
See the schematic. All parts circled in read are changed.


Bob.

If T2 is open I have 6.8V /17,26K= 0.4mA. I have a voltage drop of 4,94V over R7&R10. I don´t know what IB is. But I have not enough voltage to switch the base through.

With an R10 10K I have 0,44mA current which gives me an drop of 4,68V.
which leaves 2.12V at T3 base.

[jimmcc]

Bob,

What are you using to drive (trigger) the CDI on your test bench?

Jim

[HKZ Bob]

Loren R8 is 150R. That`s what the Bosch document from 1976 says.
How comes you are changing values!

Bob

[HKZ Bob]

I use a new Bosch Hall distributor.

It is the std. hall device Bosch used for Hall distributors. I drive it with my Makita

[jimmcc]

Can you give me the part number for the Bosch Hall distributor? If you have a URL for an reference that would be ok also. I want to determine what the output signal looks like.

The Bosch CDI unit is designed for a reluctor sensor that produces a sinusoidal (alternating current) wave form (with both + and - components) whose amplitude varies with rpm The CDI triggers when the signal goes negative. The hall effect sensor you are using may be only outputting a positive pulse that never goes negative enough to trigger the unit.

As a test, place your scope or volt meter on the base of T2 which should show a positive voltage. Then ground pin 7 of the unit and the voltage shoud drop to zero. Do the same for the output of T3 which should be high until pin 7 goes below some low threshold (ground). Then you wil know if the circuit is triggering at least at low speed (like snail crawl). The base of T2 is biased high by the combination of R5-R5-D3-D4. Also the D2-R4 slightly forward biases the reluctor sensor and that could be interferring with the output of the hall sensor.

Don't expect to see a pulse on the TD pin when the unit triggers. D8 pulls the input of the tach to ground then the output of T3 goes low. Makes for an interesting scope image showing the diodes characteristics.

C3 is a positive feedback capacitor used to speed up the response time of the schmitt trigger for higher RPMs.

Let me know how it goes.

[rick-l]

Originally Posted by HKZ Bob

But I have not enough voltage to switch the base through.

With an R10 10K I have 0,44mA current which gives me an drop of 4,68V.
which leaves 2.12V at T3 base.

You don't think this circuit is a differential amplifier wired up as a comparator?

[jimmcc]

Originally Posted by rick-l

You don't think this circuit is a differential amplifier wired up as a comparator?

No, it's simply a pulse shaper/schmitt trigger/driver. A very common circuit.

I failed to include earlier that diode D2 clips the positive peak of the reluctor sensor output and allows the negative half of the waveform to pull the base of T2 low. It also has the effect of slightly forward biasing the reluctor which is no great problem.

I believe that forward bias is preventing the hall sensor output Bob is using from going low and triggering the circuit. Grounding pin 7 and watching the voltage level on the collector of T3 should prove if the cricuit is operating correctly.

[HKZ Bob]

Jim that what I use as my trigger.

It puls nicely the base voltage down to about 0.8V.

Bob

[jimmcc]

[QUOTE=HKZ Bob;3576773]
It puls nicely the base voltage down to about 0.8V.
QUOTE]

You have to drop below .6 to .65 volts for T2 to properly trigger. That occurs on the reluctor sensor when the waveform goes negative.

Did you try gounding pin 7 while observing the T3 collector output to determine if the circuit is triggering?

I haven't found any information on the sensor you are using but if it a hall sensor it will never drop low enough to trigger this circuit.

Jim