I don't understand.

It seems like you are saying that when the C8 is fully charged nothing is returned to the battery.

This is clearly incorrect. The first scope photo shows that. You can't argue that just by saying it dosen't work that way. "Who you gonna believe, me or your lying eyes".

By large, I would agree to your description. On the point 6, I would rather say : the energy remaining in the couple C1/L(W1) is returned to the battery.
Nice simulation. The overall shape is close to what may be seen on a scope.
In fact there are such spikes on my pictures !

I confess that I do not know. If T1 would remain open, it would be an attenuated sinusoidal signal. So is it a half sinusoid ? is this important ?

There is something that does not correspond to reality with I(L1) (black) :
The current cannot decrease instantly to zero when T1 shuts off.

At this voltage level and with C1 charged to Vout / 6 it operates like the flyback Loren is talking about. The flux in the core forces I(L2) positive (I(L2) = 1/n * I(L1) where n = turns ratio) so the flux does not change instantaneously.

It's simple:

Energy In ~= Energy Consumed

Watt-second = joule

Power In = 1.0 amps X 10 volts = 10 watts

Osc. Freq = 3.3 KHZ

Energy In = 10 watts X 300us = 3.0mj

Energy stored in 1uf capacitor when T1 switches off = 80V X 80V X 1uf / 2 = 3.2mj

Therefore all the energy intially stored in the osc. system is consumed and basically
none is returned. The energy is consumed by the forward biased zener (during the
dumping of the L current - after C discharges) and T1.

[ No one is saying it doesn't consume energy.

C8 charges to 460 volts to provide energy for the spark
There are also switching losses, core losses, leakage yada yada yada...

What is being discussed is how when C8 is charged energy is returned to the battery.

The mechanism is exactly like how the flyback in a horizontal deflection coil in a CRT works. Ramp up the current, turn off the Horizontal Output Transistor which dumps the energy into a capacitor and the L-C reverses the current and rapidly returns the beam

If energy is not returned to the battery when C8 us fully charged how do you explain this?
Are you talking about the zener diode being in reverse breakdown or forward bias? For a normal input voltage range I don't think the zener ever breaks down. I don't have one to experiment with but I am pretty sure you could replace the zener with a 1n4003 and with a 12 volt input the output would still be ~460 volts

You do agree that these voltages on the primary during flyback would limit the output correct?

"What is being discussed is how when C8 is charged energy is returned to the battery."

It's not. The proof indicates such!

"Are you talking about the zener diode being in reverse breakdown or forward bias?"

It's forward biased. The LC oscillation ends because of the zener across the capacitor,
i.e. the zener acts as a diode (forward bias) and clamps the capacitor from charging
in the reverse direction when the inductive current is negative. The base emitter
diode and the collector base junction of T1 can act as a diode without the zener.

"I don't have one to experiment with but I am pretty sure you could replace the zener with a 1n4003 and with a 12 volt input the output would still be ~460 volts"

Correct!

"You do agree that these voltages on the primary during flyback would limit the output correct?"

No! As I said before, a .10uf still produces the same peak output. If the simulation
worked properly, it should indicate that too.

Proof ? Which proof ? The figures and scope pictures that I provided about the power consumption of a 6-pin CDI unit clearly show such a return.
I agree too. I presume that was the choice originally made by the designer to limit the Zener diode reverse conduction to the highest input voltage range (above 12 V for example).
A 0.1 µF C1 produces the same peak output as a 1 µF C1 ONLY because of the presence of the Zener diode.
If the Zener diode was replaced by a simple diode as in the Rick-I's simulation the output voltage would be much higher with a 0.1 µF C1, as I have shown in a previous post for some other values for C1.
With a 0.1 µF most of the energy in excess at low rpm is dissipated in the Zener diode rather than being returned to the battery as it is with a 1 µF C1.
This is clearly visible with the power consumption figures I have provided earlier.

"Proof ? Which proof ?"

Are you brain dead?

Energy In ~= Energy Consumed

Watt-second = joule

Power In (consumed) = 1.0 amps X 10 volts = 10 watts

Osc. Freq = 3.3 KHZ

Energy In = 10 watts X 300us = 3.0mj

Energy stored (and then consumed) in 1uf capacitor when T1 switches off = 80V X 80V X 1uf / 2 = 3.2mj

WHERE'S THE ENERGY RETURNED??????????

AND maybe you just can't properly understand what the scope traces actually MEAN!

What a waste of time.

If the current in the primary is negative isn't it going back into the battery?

The above text that you are repeating again and again has nothing to do with a "proof".
A real "proof" would have been to demonstrate that my pictures and figures were invented. I recognize that it is always difficult to contradict facts, but to treat me of "brain dead" does not seem to be a valid alternative.
Sorry. I had exactly the same feeling about you, if you are talking about my scope pictures.
For this very last point I agree with you.

"A real "proof" would have been to demonstrate that my pictures and figures were invented."

Not invented, just doesn't understand what's really happening.

"If the current in the primary is negative isn't it going back into the battery?"

Correct! But that doesn't mean the energy is saved. The coil's current just
flows thru the power supply (remember power supply is basically a voltage
source with zero ohms) as it charges the capacitor and as the capacitor
re-charges the coil (field builds up in the coil). The coil trys to re-charge
the capacitor but the coil current by-passes the capacitor because of the
forward bias diode which results in a loss of energy.

Furthermore:

Energy In = 10 watts X 300us = 3.0mj

Energy stored (and then consumed) in 1uf capacitor when T1 switches off = 80V X 80V X 1uf / 2 = 3.2mj

Energy stored in the coil = 4.6 amps X 4.6 amps X .30 mh X .5 ~= 3.2 mj

Iavg = 1.0 amp = Ipeak X .5 X Duty Cycle = 4.6 X .5 X D.C.
Duty Cycle ~ = 45%

Bottom line: All the calculations basically indicate the same; Little to no energy returned!
There's NO mis-interpretation or one's VIEW (scope traces) when it comes to equations.
The numbers don't LIE.

I have encountered a bunch of these 1980's vintage self resonant power supplies.

• In my broken (now fixed) Phillips oscilloscope
• modifying Compact Fluorescent Lights
• the electronic flash in disposable cameras

An Internet search doesn't yield much useful information about how these work.

jmchrist

I have learned alot from your scope photos and help with analysis

Bosch 6 Pin CDI
 

Where can I get a C8 1.4uF 600v capacitor for my 6 pin CDI? Is this just a standard capacitor??

Tim

I have used a this one successfully a couple of times in the past: http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?name=EF6155-ND. I have not been able to locate the original style capacitor. Let me know if you find something other that matches closer.

Cdi
 

Thankyou. I will if I find a better source.

Hi gents,
After testing I found that C8 was open circuit in my CDI. I replaced it (and also fitted a bigger SCR) and now it all works again. I wanted to leave it running on the bench a while before re-firtting it to the car. What input frequency should I use to simulate 8000RPM? I figure if it works OK like this then it will be fine when on the car (clearly I don't rev mine to 8k)

I reckon it is approx. 400Hz

Can anybody confirm this or tell me what it is?

Thanks for helping...

Steve

8000 rev/min * 3 sparks/rev = 24,000 sparks/ min divided by 60 = 400 sparks/second or 400 Hz

Thanks Rick :)

Regards

Steve

My C8 was bad also. I wonder if this is common.

What is the "SCR"? Sorry, I know zero about electronics.