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

"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."

Wrong!

1. There's NO regulation. There's just a peak
voltage limitation controlled by the zener.

2. C1 is used to reduce secondary breakdown
in T1. (Get a power semiconductor data book
and read about secondary breakdown and the
damage it causes.)

The energy not tranferred to C8 is dissipated in the zener and T1.
That's why there's a current when NO spark is developed.

POWER IS BEING USED!!!!!!!!!!!!!!!!

I agree with that, and the Zener diode is mainly there for limiting the voltage on T1 and consequently on the output voltage for the highest values of the battery voltage.

This is only a marginal benefit.
As this converter may be called a "pseudo-resonant" flyback converter, C1 has a major role in shaping the pulse on the collector of T1.
The peak value of the pulse determines the maximum value of the voltage on the output capacitor C8 (the turn ratio W2/W1 is approximatively 6/1).

In steady state conditions (no sparks) everything else being equal, the smaller the value of the capacitance, the higher the peak voltage, the narrower the pulse and consequently the longer the time necessary to fully charge C8.

Just to give an example I have observed the variations in the shape of the pulse on the collector of T1 when changing the value of C1 :

C1 = 0.5 µF pulse height = 88 V pulse width = 40 µs
C1 = 1 µF pulse height = 67 V pulse width = 55 µs (original value)
C1 = 2 µF pulse height = 50 V pulse width = 80 µs

I have made these test with an input voltage low enough in order to prevent the Zener to enter in action with the lowest value of C1.

C8 needs to be charged fast enough to provide good sparks even at the highest rpm of the engine. This is not the case with C1 having a value lower than 1 µF.

Sorry, I disagree with that. If there is some energy dissipated in the Zener with the highest values of the input voltage, the energy not transferred to C8 is not dissipated in T1. With some unavoidable losses it is returned in the battery, as visible in the pictures I provided earlier, thus lowering the current consumption.

For example, I have measured the input current in the CDI unit I have in hands (under 12 V) , and I found 2.6 A at 6500 rpm and only 1.2 A with no sparks.
Of course ! But not always in the same quantity ;)

The energy stored on a 1 uF capacitor at 82 volts = 3.4 millijoules
If the voltage on the inductor is 12 volts the inductor should be about 0.36 millihenries.
The energy stored in that inductor at 5 amps = 4.5 millijoules.


When I = 5 Vc=0 and when Vc=82 I=0.
Do you think that the difference is the energy transfered to the C8?

I am having trouble visualizing the second picture.

I do not think that we can evaluate the transfer like that.
The energy transferred to C8 is highly variable between two successive sparks.
C8 is charged progressively by successive pulses. The load on the converter is maximum immediately after a spark and asymptotically decreases as C8 is being charged.
Please have a look to my answer to Lorenfb about the role of C1.

The details in second picture are questionable because I made it mainly to show the overall decrease in the negative part of the W1 current, but the load was a resistor in parallel with the SCR instead of a spark plug at 6000/7000 rpm. Is that your question ?

"As this converter may be called a "pseudo-resonant" flyback converter, C1 has a major role in shaping the pulse on the collector of T1."

Only from a safety standpoint of T1.

"The peak value of the pulse determines the maximum value of the voltage on the output capacitor C8 (the turn ratio W2/W1 is approximatively 6/1)."

Nope!

The peak voltage on the C8 is detemined by the energy STORED in the transformer primary.
That's what a flyback system DOES!

"C8 is charged progressively by successive pulses."

Wrong! See Ingo and Bob's scope images of the SCR anode.

"the narrower the pulse and consequently the longer the time necessary to fully charge C8."

Again nope!

The charge time is a simple LC (L of the transformer) time constant.

"Do you think that the difference is the energy transfered to the C8?"

The energy stored in C8 is over 100mj!

"With some unavoidable losses it is returned in the battery, as visible in the pictures I provided earlier, thus lowering the current consumption. "

Totally incorrect, i.e. if the current is about 1 amp @ 12 volts then power is consumed and
NOT returned and so 12 watts is being dissipated:

Where? now let's guess;

1. the C1 ?, NO
2. the transformer ?, NO
3. the zener ?, YES (above a certain input voltage)
4. the wires ?, Hardly
5. the power transistor ?, YES

TOO MUCH GUESSING!!!!!!!!

OK cut me some slack here, is this how it works?

Say C8 is charged to 460 volts

1. T1 turns on until either current limit or saturation turns off the transistor at what appears from the scope photo to be 5 amps in the forward direction.
2. For any current change to happen the voltage at the collector of T1 has to rise but it is restricted by the fairly large capacitance C1.
3. With C1 at zero volts and T1 off the current continues to rise with 12 volts across it. It is at 5 amps and even if the core is saturated the permeability will still be that of free space and it will increase. This current starts charging C1.
4. No current will flow through the diodes into C8 until the secondary exceeds 460 volts. For this to happen the voltage on the primary has to be 60 to 80 volts assuming the 6 to 1 turns ratio of the secondary to the primary.
5. For the primary to increase this much C1 must increase this much
6. Using the general rule that where there is flux there is current flow, C1 charges to 60-80 volts using the energy stored in the core.
7. When C8 is fully charged the energy required to increase C1 to the primary required to support the secondary current is almost all the energy stored in the inductor.
8. With T1 still off C1 has to discharge through the primary back into the battery
9. When the current reverses the transistor turns back on

When C8 is discharged C1 won’t have to increase much and a large percentage of the energy stored in the inductor will go into C8. The amount of energy transferred per cycle will go down as the voltage on C8 increases.

Sorry, I disagree. I have scope pictures to support what I said on the relation of value of C1 with the shape of the pulse. No guessing.

Of course, the energy stored in the primary is the source of the energy transferable to C8, but the value of C1 has an important role in the shape of the pulse on T1 collector and thus on the way the energy is transferred to C8. I have scope pictures to support the figures I gave on duration and height of the pulse in relation with changes in the value of C1. No guessing here.

Right. Look below (voltage on SCR anode, 1/100 probe)
http://forums.pelicanparts.com/uploa...1198063102.jpg

I did not say the contrary. If C is changed so does the time constant.

You are mixing here text from two persons and that does not help to understand your point. Considering what I wrote I totally maintain it.
The measurements figures and the pictures I gave in my previous posts cannot be ignored.
Too much unfunded criticisms !
My posts are based on observations and measurements. No guessing here.

"With T1 still off C1 has to discharge through the primary back into the battery"

Hopeless!!!!!!!

You do see the 5 amps coming out ot the primary and going back into the battery in this scope photo correct? Where does it come from?

I have a few comments along your points.

1. BTW I forgot to tell that my experiments were done with an input voltage of only 10 V, in order to prevent ZD1 to enter into action.
   This would slightly modify your estimation of the value of the inductance of W1 to about 0.3 mH.
   
   Quote:

   For any current change to happen the voltage at the collector of T1 has to rise but it is restricted by the fairly large capacitance C1.

   Yes and the rate of change depends largely on the value of C1.
   
   Quote:

   With C1 at zero volts and T1 off the current continues to rise with 12 volts across it. It is at 5 amps and even if the core is saturated the permeability will still be that of free space and it will increase. This current starts charging C1.

   I am not sure to understand your point. As soon as saturation occurs and T1 is consequently blocked, the current begins to decrease (see my picture), at a rate which essentially depends on the value of C1. This is the pseudo-resonant part of the cycle with the inductance L of W1.
   
   Quote:

   No current will flow through the diodes into C8 until the secondary exceeds 460 volts. For this to happen the voltage on the primary has to be 60 to 80 volts assuming the 6 to 1 turns ratio of the secondary to the primary. For the primary to increase this much C1 must increase this much Using the general rule that where there is flux there is current flow, C1 charges to 60-80 volts using the energy stored in the core. When C8 is fully charged the energy required to increase C1 to the primary required to support the secondary current is almost all the energy stored in the inductor.

   Sorry, I am a little lost in your reasoning, with a sentence like "...the energy required to increase C1...
   
   Quote:

   With T1 still off C1 has to discharge through the primary back into the battery

   I would not say like that. C1 and the inductance of W1 are not separable in this process. The unused energy of W1 is sent back to the battery and this occurs for a time longer than the blocking period of T1.
   
   Quote:

   When the current reverses the transistor turns back on

As seen on the picture, T1 turns back on before the current into W1 becomes positive. This negative part of the current through W1 does not come from C1 (since the potential at the collector of T1 is practically zero) but surprisingly from T1 and from ZD1 (in the forward direction).
See the pictures hereunder, all with T1 collector voltage as reference.
The first one shows the current across T1 (10 mV/A) :

http://forums.pelicanparts.com/uploa...1198080859.jpg

This one shows the current through ZD1, in the forward direction. The small peak in the zener direction is the beginning of the voltage limitation occurring at the top of the impulse on T1 collector.

http://forums.pelicanparts.com/uploa...1198081029.jpg

The current in W1, as shown in the picture I provided earlier, is the combination of the currents from T1 and ZD1 shown above.

The current in C1 shown hereunder is, as could be expected, the derivative of the pulse voltage on T1 collector.

http://forums.pelicanparts.com/uploa...1198081093.jpg

I would write more simply your first sentence "When C8 is discharged, a large percentage of the energy stored in the inductor will go into C8".
I fully agree with your second sentence.

I can't exactly duplicate the scope photo, but I can come close.

Note that is not a zenner diode, it is the spice default.

http://forums.pelicanparts.com/uploa...1198086028.jpg
http://forums.pelicanparts.com/uploa...1198086041.jpg

Interesting simulation.

May be you would come closer if taking into account my correction about the input voltage I used for the scope picture (10 V) and consequently the estimated value of W1 inductance (0.3 mH instead of 0.36).

"Just to give an example I have observed the variations in the shape of the pulse on the collector of T1 when changing the value of C1"

So? The charging dv/dt of C8 is UNAFFECTED by C1 whether it's .10uf or 1.0uf.
It's not a "shape" issue, i.e. basically no effect on energy transfer! This is NOT
a pulsed type of voltage step-up circuit dependent on the pulse shape on the
primary of a transformer. It's just an energy transfer circuit - a flyback system
JUST like what's found in the inductive discharge ignition system, i.e. energy
is stored when the points close and tranferred to the distributor when the points
open. It's SIMPLE!

"C8 needs to be charged fast enough to provide good sparks even at the highest rpm of the engine. This is not the case with C1 having a value lower than 1 µF."

Not really! The C8 charge time is JUST a function of: C8, Ipeak(primary), L(secondary),
L(ignition coil) and Vzener (determines the max voltage).

"The unused energy of W1 is sent back to the battery and this occurs for a time longer than the blocking period of T1."

Another incorrect statement! If that were the case, the average input current would
be zero, i.e. no power dissipation when there's no spark output, which is NOT the
case as the average is greater than 1 amp. Power (energy loss over time) occurs
from the forward bias of the zener (during the T1 off) and the losses of T1 as it's
switched off. Thus, energy is NOT conserved as implied in a number of posts when
no sparks occur.

Furthermore, C1 has little affect when charging C8 as C8 is reflected to the primary
as the square of the turns ratio times C8 to the primary. Without C1, the circuit energy
from the primary current ramp STILL transfers fully to C8. C1 can be reduced to a
.10uf which allows the zener to limit the peak voltage at a much lower input
voltage. Thus, the C8 charges to a voltage determined by the the turns ratio
times the V(zener) even with Vin = 10 volts. So, the claim that C1 is critical
for the so-called resonant charging has no validity. As defined previously, the
circuit is JUST a basic flyback circuit where C1 limits the inductive energy seen
by T1 when switching off.

The above should be obvious when using the proper simulation values and
elements. Better yet, just replace C1 with a .10uf and see the effect. It's
very simple! Actually, the circuit works very well using .10uf, i.e. notwithstanding
the potential for secondary breakdown in T1, and better at low input voltages
which occur at cold temp cranking.

Hey I learned something today
 

1. I didn't think SPICE would handle a feedback element like the mutual inductance L3
2. Since the ideal transformer in SPICE doesn't saturate and the simulation somewhat mimics the device, I think the core saturating is not what turns off the transistor.
3. The zener diode doesn't seem to do anything at normal voltage levels and if you think about it I don't think that diode would be a happy camper dissipating all the energy stored in the inductor meant for the spark with Key On Engine Off.
4. Looking at the first sim you can see more current is transferred to the secondary when Vcol (Voltage on C1) is low and tapers off to almost nothing when the peak voltage on C1 goes up.
   
   I think this relation between the secondary voltage reflected back in the primary matching the energy stored in the capacitor to balance the energy stored in the inductor at full voltage was chosen on purpose.
   
5. This isn't a typical fly back and a lot more thought went into the design than I first assumed.

I got the simulation below to run. I am guessing at the transformer parameters so some more tweaking is required but I can duplicate most of jmchrist's scope photos although the timing is not exact.

Note the timing in the simulation is determined by the time it takes the primary to reach the peak current determined by the base current and the transistors beta.

If anyone wants to play with LTSpice PM me your e-mail and I'll send you the .asc file (schematic).

http://forums.pelicanparts.com/uploa...1198126847.jpg

http://forums.pelicanparts.com/uploa...1198126872.jpg

http://forums.pelicanparts.com/uploa...1198126895.jpg

Yes it is (basic electricity law ;) ).
Do you think the figures I gave on the variations of duration and height of the pulse on T1 collector in relation with changes in the value of C1 are invented ?
I am beginning to wonder what is the "issue" in this discussion.
I am talking about the consequences of modifying the value of C1 in the converter of a 6-pin CDI, and accessorily to show that the value of 1 µF chosen by the Bosch engineers was likely the best one once the other compelling parameters are fixed.

You seem to talk about the general principle of flyback converters, in which I agree that they can be built with any value of capacitor at this place.
No. You are forgetting to add the duration of charge at each pulse.
This duration depends on the duration of the pulse.
You should not introduce Vzener in this discussion. Its action is not fundamental here and just prevents overvoltage to occur with the highest possible values of the battery input voltage.
I never implied that the consumption would be zero. Look at my picture. The average is obviously positive on one cycle.
There is of course various sources of losses in any design.
I have just shown that some energy is sent back toward the battery during a part of each converter cycle when in idle conditions, so that the consumption of a CDI unit is significantly lower in idle condition than at maximum rpm.
This is a FACT you cannot ignore.
Again you are mixing general flyback principle considerations with the constraints of optimisation in the design of the 6-pin CDI.
Just for fun I have put a 0.1 µF capacitor instead of the original 1 µF for C1.
I have found for the consumption (at 12 V) the following figures :
0 rpm : 2.4 A
600 rpm : 2.4 A
6000 rpm : 2.4 A

With the original value of 1 µF the consumption is :
0 rpm : 1.2 A
600 rpm : 1.5 A
6000 rpm : 2.4 A

I just wonder how much time will live ZD1 with a 0.1 µF C1 ;)

"If anyone wants to play with LTSpice PM me your e-mail and I'll send you the .asc file (schematic)."

Total waste of time! Just buy:

1. a Bosch CDI
2. a scope
3. a basic electronics book

As finite beta value and core saturation would have the same action, it is difficult to say that. May be you were just lucky in choosing a beta value ;) .
I fully share your opinion. The lesser it conduct, the better the efficiency is.
But its presence is necessitated to attenuate the consequences of an input voltage varying between 10 to 15 V by large.
Sorry, I do not read your simulation like that. The current injected in C8 is maximum when Vcol is maximum too, and this reflect the reality.

This is what I was tryng to say. I was thinking of it in terms of energy

1. It appears that the inductor charges until it reaches 5 amps.
2. This stores a finite fixed amount of energy in the inductor
3. The voltage on the capacitor (peak) when the transistor is off is 1/6 the output voltage.
4. nothing is transfered to the output until the capacitor is 1/6 th the output voltage
5. the energy to charge C1 up to 1/6th the output voltage comes from the energy stored in the core
6. the energy that is stored in the capacitor, 1/6th the output voltage is returned to the battery.
   • peak capacitor voltage low = most all of the energy stored in inductor is transfered to the output
   • peak capacitor voltage high = most all of the energy initially stored in inductor is returned to the source

You can kind of see this is the sim below when the capacitor voltage goes up current in the inductor goes more and more negative. Also in the previous V out plot the output takes bigger steps when the peak voltage on C1 is low

http://forums.pelicanparts.com/uploa...1198180411.jpg

I could be wrong in item #5 above but it seems logical.

It seems like if the core saturated you would see some other effects also, current, voltage spikes etc.

When the thing is fully charged and the energy is going from the battery to the inductor to the capacitor and back to the battery is the shape of the voltage on the collector = Vout * Sin(sqrt(1/(L* C1))) for a half cycle?

Here is a confusing example at partial charge

• I(L1) (BLACK) ramps up to 4.5 and transistor shuts off
• I(C1) (BLUE) charges C1 to 35 volts
  • I(L1) goes down as C1 charges - less energy
• Now that V(col) = V(out)/6 + 2 * 6*Vd (should be but isn't exact) the output diodes conduct
  • I(L2) goes positive
  • V(out) rises
• When v(out) = 6 * V(col) etc I(L2) = 0
• C1 starts to return energy to source
• T1 turns on and dumps remaining C1 charge

http://forums.pelicanparts.com/uploa...1198181901.jpg

"the energy that is stored in the capacitor, 1/6th the output voltage is returned to the battery.
peak capacitor voltage high = most all of the energy initially stored in inductor is returned to the source"

"T1 turns on and dumps remaining C1 charge"

"Hum??? What could this be, may be an energy loss?
But I just said: "most all of the energy initially stored in inductor is returned to the source".
Let's get the electronics book back out!"

Still doesn't understand the basics of the circuit. So when trying to do a simulation:

GARBAGE IN = GARBAGE OUT