Ingo and Bob,

The #7 input normally normally triggers all the way down to 10 Hz to 20 Hz, i.e., 200 - 400 engine rpm, and the engine can also run in the 450 - 600 rpm range with mal-adjusted retard settings.

Of course. My apologies for asking this stupid question.
Very interesting simulation indeed, with a very good agreement with the scope observation, as you said before.
The picture I provided in a post above is apparently also in agreement when having a closer look to it. However the picture is not precise enough. Next time I have a 6-pin CDI to repair I'll make a closer area picture.

Now your simulation incite me to ask a new question (hopefully less stupid as my previous one) :
What could have been the motivation of the designers of this CDI for providing for a fine tuning of R5/R6 (at final stage of manufacturing ?) since the influence on the triggering input voltage level is so weak ?

Ingo isn´t the C3 value 0,0068µF=6,8nFf=6800pF

bob

Dear all &
Warren what is the signal amplitude and frequenz to do the R6 evalutation?
Question 1:
I read that this evalution has to be measured an pin 7. 100mV above zero crossing is the factory set up? I don´t understand why to measure at pin 7 not at base of the BCY58X.

Observation on my box.
I see a relation between the supply 12V vs. 14,5V and the amplitude of the trigger signal.
Low supply (10V til 12V) & 4V pp trigger signal brings less misfire. The problem get amplified when the box gets hot.

High supply 14V & 15V & with a 2Vpp trigger brings a lot misfre. I also battery supplied the schmitt trigger to seperate the DC/DC noise, but it did not worked out. I assume the noise was tranfered though the ground line.

If if supply with more that 15V with 33Hz and 4Vpp the SCR gets permanent conductive. Without a trigger the box does not goes into permanent conductive state at 15V and up.

I use 33Hz sine for triggering with an amplitude of 4V pp my box.

I would like to replace the diodes above R5/R6. Can I use 1N4148 as a replacement?

Regards Bob, have a nice christmas pre-season.

"since the influence on the triggering input voltage level is so weak"

IT'S NOT!!!!!!!!

The bias can set the level over a wide range.

Not so wide if we look at the simulation provided by Ischmitz above.
It can be seen a maximum variation of about 0.4V in the triggering input voltage for the various values of R6 he tested.
This is to compare with the minimum 4 V peak-to-peak input signal (as far as I know, do you confirm ?).
This is not to be ignored but does not seem to deserve a "fine tuning" since the variations in signals between distributors might be of the same order of magnitude.
This was my point (just for curiosity...).

Oops, my bad. You are right it is C3 is 6.8nF and C6 is 0.01uF. I corrected that and here are the new traces. C6 makes the SCR gate signal decay a lot quicker. C3 has a similar effect. I agree that shifting the trigger point away from 0 volts makes the box less susceptible to noise.

http://forums.pelicanparts.com/uploa...1196982745.jpg

The values for R6 are: 14k 22k 30k 38k 42k 54k

http://forums.pelicanparts.com/uploa...1196982756.jpg

http://forums.pelicanparts.com/uploa...1196983343.jpg

"Not so wide if we look at the simulation provided by Ischmitz above."

Please! It doesn't take a simulation to EASILY determine the effect of R6:

1. R6 = 0, the Schmitt NEVER triggers, as the input basically can't turn Q2 "off".
2. R6 = R5 = open, Q2 is biased off and the Schmitt never triggers.

It's really not that complex.

Great simultion, my failure mode is that I have a double spike. D,F occur at my gate paraelle. It could be caused by a problem of R5,R6, D3,D4.
can you show me your input value on the Q2 with and adjusted R6, so with the 100mV offset.

Bob

Ingo what is the impact of C3. What will happen if I C3 has leakage. Can you simulate different values for C3 and check the output on SCR Gate?

A little off topic of the thread but can someone tell me how the self resonating inverter works?

If the "A"s are the dots the primary and secondary (W1 and W2) make this look like a fly-back converter with the 460 volts being the turns ratio between that and the zenner and batt voltage.

But how does T1 switch? It looks like winding terminal 3A is pretty firmly held at about 2 volts so when T1 is on the induced voltage would tend to subtract from the 2 volts and turn T1 off. Does this make sense? What am I missing?

p.s. I'll admit being a moron up front to save Loren the trouble.

Rick I am not the expert her, but if you ask such question you have to consider that you get nailed by the real experts. LOL.

The resistor R1 & the inductance W3 are causing a different phase shift than only the inductance on W1= Base T1. T=R/L.

The winding ration on W1 & W3 will cause a high pulse. This pulse loads the 1mF capacitor which gets limited by the 82 Zenner.

Experts am i correct?

bob

"A little off topic of the thread but can someone tell me how the self resonating inverter works?"

The winding W3 provides positive feedback to saturate (turn fully on) T1 and
ramp the current in W1. Once the transformer core saturates, the feedback
ends and the W3 voltage reverses to fully turn off T1. Once the negative
feedback ends, T1 turns on again. The circuit is a flyback circuit as used in
the old CRT TV sets to generate a high voltage.

That is kind of what I figured.
Is the "A" and "E" labeled correctly for W3 on the schematic? Seems like when T1 is on the feedback tries to turn T1 off.

Is C1 just a snubber or part of the oscillator circuit?

"Is the "A" and "E" labeled correctly for W3 on the schematic? Seems like when T1 is on the feedback tries to turn T1 off."

They're wrong, but what's new.

"Is C1 just a snubber or part of the oscillator circuit?"

Yes (snubber), just to reduce secondary breakdown in T1
and partial primary peak voltage limitation with ZD1.

Why is A & E not labeled correct? It means Anfang & Ende.

BOB

This is the way I thought the 460 volt supply worked.

1. When the ignition switch is turned on, the transistor T1 is biased on by the resistors R1, R2 (2 volts, 8.5 ohms) and a small voltage is generated across the primary of the transformer (W1).
   
2. The transformer action induces a voltage in the feedback winding (W3) which turns the transistor hard-on.
   
3. When the transistor is hard-on or saturated, practically the whole supply voltage (Vbatt) is applied across the primary winding. A linearly increasing collector current flows through the transistor (v = L di/dt and v is constant so I is a ramp).
   
4. The controlling base current IB flows into the base determined by the voltage of the turns ratio in series with the bias circuit. The transistor remains saturated until the collector current exceeds a value beta * IB (beta is the static common emitter current gain of the transistor). The transistor begins to come out of saturation and the collector voltage rises. The voltage across the primary thus decreases which also encourages a further decrease in base current. This action continues until the voltages across the windings are reversed cutting off the transistor. The stored energy is now dumped into the output capacitor and secondary current decays to zero.
5. When this happens the cycle repeats.

If "A" represents the dot of the winding (W3) the reinforcing action in step 2 would not happen. The induced voltage would oppose turning T1 on.

"The transistor remains saturated until the collector current exceeds a value beta * IB (beta is the static common emitter current gain of the transistor)."

Nope! It was explained clearly in a previous post, i.e. The frequency is NOT a function of Beta.

I think you are right when referring to the schematics.
I had today the opportunity to look closer to a 6-pin CDI and I have found that the schematics is incorrect as to the "A" and "E" identification of W3 winding.

On the photography of the 6-pin CDI printed circuit the labeling "W3/A T1/B" is partially hidden by the capacitance C1. The letters "A" and "B" are not visible.
The other side of W3 is labeled "W3/4" (why ?) It should have been labeled "W3/E".

I think that explain why this error was not found earlier.

The same inversion of "A" and "E" labeling for W3 winding exists on the schematics of the 3-pin CDI.

Operation of the flyback dc/dc converter
 

I just made a few scope pictures about the operation of the the DC/DC converter. As this subject was discussed some days ago, this might be of some interest here.
My original curiosity was to observe how energy is sent back to the battery in relation with the load of the converter.

The first picture shows the current flowing through the W1 winding in no load condition (no sparks generated) together with the voltage on the collector of T1 as a time reference. This shows an important negative part that correspond to a reinjection of current in the battery ! Of course the average is positive, as the surface above the zero line is larger as the surface below.

The current measurement in done with a 10 mV/A DC current probe.
The voltage measurement is done with a 1/10 voltage probe.

http://forums.pelicanparts.com/uploa...1197913270.jpg

In the second picture the load is maximum (equivalent to 6000/7000 rpm) and the negative part has disappeared during the current ramp.

http://forums.pelicanparts.com/uploa...1197913739.jpg

I have made some other pictures about the individual contributions of T1, ZD1 and C1 in the current flow that I could provide if found of interest here.
Please let me know.

Merry Christmas to all (with special health wishes to Early_S_Man).

Wow

So the energy stored in the inductor is transfered to C1 until the two diodes in the secondary breakdown. C1 has a lot more to do with the regulation of the ouput than I thought.

That isn't the way they explain a flyback regulator works :)

edit: not breakdown, conduct