Zur Info an ALLE,
schaut Euch mal die Zündschaltung an,
ca. 8 Grad nach Top Dead Center muss eingestellt werden,
damit der Notstromagegat-Motor-Generator gut läuft.
Wir brauchen aber eine einfachere Zündschaltung.
Die von Les Banki ist zu komplizeirt und aufwending !
Gruss, Stefan.
>Woopecker schrieb:
>Bis jetzt noch nicht, aber würde ich die Zündung verstellen auf nach OT oder
mehr, dann sofort. Der waste spark würde sich dann nämlich genau im ersten
Takt ereignen, wenn das Einlass Ventil offen ist.
Damit nun der Motor effizienter mit dem Gas läuft ,muss aber die Zündung
verstellt werden.
Gutes Beispiel dafür ist ja dieses Video von dem Ami den einen Clio mit nur
5 Litern Gas hat laufen lassen, vor der Verstellung hätte er 20 Liter
gebraucht!!
Ich habe natürlich den Generator auch mit Last betrieben, bei 300 Watt ist
er noch schön gelaufen aber bei 1200 Watt war ich dann hart an der Grenze,
er lief eine Zeit, aber wenn dann nur etwas weniger Gas kam stellte er
sofort ab. Ich schloss daraus, dass ich sicher noch ein paar Liter mehr
brauchen würde.
Aber anstatt jetzt weiter krampfhaft zu versuchen noch mehr Gas zu erzeugen,
will ich lieber erst den Motor für Gas optimieren.
Es gäbe ja Generatoren, die mit LPG Gas laufen, die nicht mal teuer sind (
469.- 2,5kw) Aber die haben dasselbe Problem, dieselbe Zündung, Zündtiming
geht auch nicht über den Todpunkt und LPG Gas zündet fast so spät wie
Benzin.
Flipflop ist mir neu, würde sicher die 60 Hz Frequenz die jetzt ist, auf
gewünschte 30 Hz, verringern, aber wie weiss der Flipflop dann welchen der
zwei Zündfunken er durchlassen soll und welchen nicht? wenn es nur einmal
der falsche ist, dann Bum! Man weiss ja nicht in welchem Takt sich der Motor
befindet, wenn man ihn anlässt, dh welches Signal zuerst kommt.
Bei zwei Zahnrädern kann man das Timing genau so abnehmen, wie es jetzt
auch geschieht, da bin ich mir sicher dass es funktioniert,
Auch die "Les Banki" Schaltung würde funktionieren, ist halt etwas
aufwendig, der Mann hat wirklich an alles gedacht . (Schaltplan anbei)
Ich hoffe immer noch auf etwas das noch einfacher ist, vielleicht würde es
ja gehen mit einem Flipflop, ich kenne mich da leider nicht so gut aus und
meine Augen lassen es auch nicht mehr zu, dass ich sowas noch selber
zusammenlöte, leider.
Aber das sind alles jetzt keine Schwierigkeiten mehr, das kann alles getan
werden. Wenn die Leute mehr zusammenarbeiten würden, würd´s auch viel
schneller gehen. Sobald ich diese Zündung optimiert habe, kann ich mein
nächstes Video dann unter Vollast machen.
Liebe Grüsse
Woodpecker
==================================================
Circuit description for Electronic Ignition/Injection design
Background
In my thesis, "Ignition system for small engines", I briefly outlined why a new ignition system is needed with Hydroxy as the ONLY fuel and what are the technical requirements for such an ignition system.
To start with, here are a couple of quotes from an EXCELLENT web site which briefly explains (with moving animation!) ignition technology:
"It is interesting to note that one complete engine cycle takes two revolutions but that individual valves and spark plugs only operate once in this time. Hence their timing needs to be taken from a half engine speed signal, which is the camshafts speed."
"If the timing disk is attached to the crankshaft, there is a need in some engine configurations to have a sensor on the camshaft so that the igniter knows which ½ of the four-stroke cycle the engine is in."
(
http://www.gill.co.uk/products/digital_ignition/Introduction/6_4stroke.asp)Most (if not all) existing small engines use a magnet mounted on the fly wheel which gives two pulses for every engine cycle. (thus generating "waste" sparks) Electronically dividing by two would not solve the problem since another signal would be needed to determine which one of the two pulses we want and which one we don't.
Only ONE sensor is needed IF it gives only ONE pulse for every ENGINE cycle.
It does not matter WHERE in the engine's cycle this pulse originates because it can be electronically 'moved' anywhere in the engine's 360º (100%) cycle.
The obvious choice of a sensor is a 'Hall effect' switch. Since modern engine blocks are non-ferrous (aluminium) alloys, the Hall switch can be placed on the outside of the engine block. For example, it can detect the position of a magnet which is fitted to the valve's 'rocker' arm.
As the 'rocker' arm/magnet moves in and out of certain positions, the Hall switch turns on/off.
My choice is to place the magnet on the EXHAUST valve's (rocker) arm for the following reason:
The EXHAUST stroke is the LAST of the engine's working cycle.
From this point, everything is in the correct order. First the INTAKE stroke, during which the gas will be injected at the correct time (and duration, determining the speed)
Now the pulses from the Hall switch can be correctly delayed by the required amounts for the Injection and Ignition functions.
Following the INTAKE stroke is COMPRESSION, at the end of which IGNITION takes place.
With the help of a Frequency to Voltage converter, a linear sawtooth VCO is created which is synchronized and phase locked to the incoming pulses.
As the pulses from this single Hall switch usually do not occur exactly where we want them, they need to be 'moved' (delayed) to deliver the desired injection/ignition functions at the correct times.
First, some engine basics:
Engine speed is expressed as RPM (Revolutions Per Minute).
In electronics, however, the unit of time is SECOND.
Since there is 60 seconds in a minute, the engine's RPM is divided by 60 to get the engine's 'frequency', in Hz.
Example; an engine running at 3600RPM (crankshaft speed), divided by two is 1800 engine cycles per minute. Divide that by 60 gives 30Hz.
It means that the spark plug is going to fire 30 times per second.
Principle of operation:
1. The first task is to convert the pulse train from the Hall switch to 'frequency'.
[(so that ONE 'period' is ONE engine cycle. (4 stroke)]
This can be done by either digital or analog means. (or a combination of the two)
Both have advantages and disadvantages.
While the modern "buzz" word is "microprocessor", there is no need for it here with its complex software programming.
The two main issues with any design are: simplicity and cost.
My choice for this design is analog. It is a relatively simple and low cost design.
2. The engine's frequency is transformed into a LINEAR saw tooth waveform.
This saw tooth is fed to a comparator. The output is a variable duty cycle square wave.
[This is the basic principle of the analog Pulse Width Modulator (PWM)]
3. The rising or falling edge of this variable duty cycle square wave is used to trigger the desired ignition/injection action.
Details of operation:
The pulse train from the Hall switch, IC1, is fed to a Frequency to Voltage converter, IC2A's rising edge input A (pin 4, 4538 dual monostable)
The pulse width is set to 18ms by R3 (180k) and C3 (0.1µF).
The output pulses from pin 6 are integrated by R4 (47k) and C4 (2.2µF) and buffered by IC3A (TL074), wired as a unity gain voltage follower.
The output of IC3A is fed to an inverting, DC summing amplifier (attenuator) IC3B.
The 'gain' is set by R9 (100k), R6 (100k) and P1 (100k).
P1 sets the 'gain' between about -0.5 and - 1.0.
P2 (10k) sets the level shift.
The output of this stage (pin 7 of IC3B) is fed to a special, inverting Sample & Hold stage, built around IC3C and IC4A. (quad electronic switch 4066).
It works like this:
When the switch IC4A is closed, the stage behaves as a normal inverting amplifier
with a 'gain' of -1.0.
The sample & hold capacitor C7 (10n) is charged by the output of IC3C (pin
.
Since the inverting (-) input (pin 9) is at a 'virtual ground' potential, the voltage across C7 (and pin 8, the output of IC3C) is at the correct signal level, referenced to ground.
When the switch IC4A is opened, IC3C no longer behaves as an amplifier and the voltage across the capacitor "freezes" at the sampling point and is held indefinitely since the output of the IC keeps the capacitor charged AND supplies the current demand of the next stage! Thus, the capacitor will not lose any charge.
Since the sampling pulses are short (100µs) compared to period of the frequencies involved (10Hz -40Hz, 100ms - 25ms), ripple is virtually eliminated.
This is essential as the DC voltage supplied by the Frequency to Voltage converter stage
has a large saw tooth ripple, despite some filtering.
(This is due to the low frequencies of approx. 10Hz - 40Hz)
While the average voltage is still linearly proportional to the input frequency, the ripple MUST be eliminated as it would modulate the control voltage to the VCO!
It could be 'smoothed' by further filtering but the response time becomes far too slow!
The 'Sample & Hold' technique described above has virtually eliminated ripple, WITHOUT affecting response time!
Note that since both the level shift & gain set stage (IC3B) and the 'Sample & Hold' stage (IC3C) are inverting, the result is once again a positive voltage.
This (now smooth) DC voltage - level shifted AND slightly amplified, [then attenuated by the voltage divider R15 (4k7) and R16 (470ohms) to match the control voltage range and level required by the saw tooth VCO (Voltage Controlled Oscillator)]
before it is applied to the voltage-to-current converter IC3D & Q1 (BC547).
This current LINEARLY discharges C8 (0.1uF) through timing resistor R18 (15k).
The VCO is somewhat unusual. It is NOT free running. If it was, sparks would be generated even when the engine was not running.
Extra circuitry would then be needed to prevent that.
In other words, it is deliberately made NOT to work in the absence of SYNC pulses.
Both the FREQUENCY and the PHASE of the VCO must be precisely the same as that of the Hall switch pulse train.
The two signals are Synchronized AND Phase Locked in the following manner:
A 'sync' pulse is created by triggering the monostable pulse generator IC2B (4538) with the rising edge of the pulses from the Hall switch. These short pulses (100µs) operate the electronic switch IC4B (4066) which fast charges the timing capacitor C8 of the saw tooth VCO. Zener diode ZD1 (3.3V) limits the charging voltage on C8 to about 9V.
This is necessary since the type of OPs used would go into saturation (non-linear range with large amount of distortion) if operated too close to the supply rail. (12V)
(To sum up the operation of this VCO: In a standard saw tooth oscillator, a comparator is used to fast charge the timing capacitor. A voltage controlled current sink then linearly discharges it. However, in this application, the comparator is in fact NOT wanted since its presence would mean a 'free running' oscillator, without 'sync pulses', which would cause the spark plug to fire when the engine is not running!!)
The sync (charge) pulse cannot be shorter than about 100µs since the ON resistance of one CMOS 4066 electronic switch is about 80 ohms.
[It does take a certain time to fully charge the rather large (0.1µF) timing capacitor C8.]
As this trigger pulse is initiated by the rising edge of the Hall switch pulse, it is also used to control the 'Sample & Hold' circuit. Thus, it samples the peak value of the voltage output of the F/V converter.
(Sampling could be done at any point and the voltage would still represent the correct frequency.)
The output of the VCO is buffered by IC5C (LM324), wired as a unity gain voltage follower. The now low impedance saw tooth is fed to comparator IC5D' (LM324) inverting (-) input, pin13.
Range of firing is adjustable by P6 (10k).
Minimum and maximum limits of the range are set by R30 (36k) and R31 (15k).
The adjustment range of the ignition timing MUST always relate to the piston's position, NOT time! This is achieved by the comparator stage described above. The output signal (pin 14) is a square wave which is fed to monostable IC7A (pin 4, rising edge trigger)
Pulse width is set by C14 (10n) and R33 (10k) to 100µs.
'Q' output (pin 6) is fed to the base of 'emitter follower' ignition trigger transistor Q9 (BC 547) through R34 (2k2) and R28 (1k).
(Q9 and R28 are on the CDI module)
Q9' collector resistor R29 (390 ohms) supplies the necessary trigger current for SCR1 (TYN816).
R30 (470 ohms) & C11 (0.1µF) help to reduce/eliminate spurious noise pulses from the SCR's gate.
Capacitor C12A,B,C (1µF 400V) is continuously charged to 330V by a DC-DC converter.
The stored energy is then dumped into the primary winding of the ignition coil.
Thus, it is a CDI (Capacitor Discharge Ignition) system.
Please note that the system described above has a limitation.
It is more of a theoretical nature, rather than a practical one.
It is due to the very low frequencies required at the bottom end of the engine RPM range.
The value of the 'charge pump' capacitor C4 (2.2uF) in the Frequency to Voltage converter is a compromise.
The difficulty arises ONLY if we wish to provide an RPM range of 600 to 6000RPM, which translates to 5-50Hz ignition frequency. That is 10:1 span.
It is difficult to achieve a LINEAR voltage output from the F-V converter at the low end.
However, in practice, engines don't usually run at those limits!
If we are satisfied with an RPM range of 1200 to 4800, (the minimum frequency is 10Hz) the span is now 4:1, a much more manageable figure!
In practice, ALL generators are run at a CONSTANT speed somewhere in the range of 2000 to 4000RPM (depending on design).
(Remember, this ignition system is designed specifically for ONE cylinder generators.)
Having said all that, this does NOT mean that the ignition system will not work at the outer limits of 5Hz-50Hz (600 or 6000RPM)
It just means that the ignition point setting accuracy may be somewhat degraded.
Injection
To gain a little extra time for injection, the injection solenoid is usually turned on just before the intake valve opens. (But it must be turned off before the valve closes.)
The volume of gas injected is determined by the solenoid's ON-time.
Just as with the ignition, injection timing is related to the engine's work cycle.
This stage uses a 'window' comparator/PWM to set the minimum and maximum limits of the solenoid's ON time.
Adjusting the injected volume of gas (speed of the engine) thus can only be within those pre-set limits.
Injection circuit details:
The buffered sawtooth from IC5C (pin
is fed (through R21, 10k) to a 'window' comparator. (IC5A & IC5B, pins 3 & 6)
The 'closing point' (limit) of the injection solenoid is determined by P4 (5k), R22 (12k) and R23 (6k2).
'Opening point' (limit) is determined by P3 (10k), R19 (22k) and R20 (33k).
(P3 and P4 are pre-set adjustments.)
Engine speed (injection solenoid opening range) is controlled by P5 (10k).
The voltage at pin 2, the inverting (-) input of the upper comparator IC5A, is held steady by IC6B (LM358), wired as a unity gain buffer which has a very low output impedance.
This voltage, set by P3, represents the opening point of the solenoid.
Actually, P5 (the speed control) and the small value series resistor (R24, 330 ohms)
pulls up the voltage on pin 5 of IC5B, (the lower comparator of the 'window') to almost the same level as that on pin 2 of IC5A.
The difference is the voltage drop across R24 (330 ohms) which determines minimum engine speed at the minimum setting of P5.
The outputs of the 'window' comparators IC5A & IC5B (LM324, pins 1 & 7) are OR-ed by D1 and D2 (4148) are fed through R26 (1k) to the base of inverter/driver transistor Q2
(BC337)
R25 (10k) is the 'load' resistor for the window comparator's output and C12 (0.22uF) is a filter for switching transients which might upset the operation of the solenoid.
The solenoid driver MOSFET transistor (Q3, MTP3055E) has an in-built protection diode.
(If transistors without in-built diodes are used, an external diode MUST be fitted!)
Q2' collector resistor R27 (220 ohms) must be rated at least 1W since Q2 is ON most of the time but turns OFF only for the short duration of the solenoid opening time.
So, Gate drive to Q3 is only available when Q2 is OFF.
When Q2 turns ON, Q3's Gate capacitance is discharged very fast.
This means fast turn-off and reduced heath dissipation in Q3.
Testing and adjustments:
To test & adjust this circuit, a DUAL TRACE oscilloscope is essential.
First, a variable frequency (~8Hz to 40Hz) pulse generator is connected to TP4.
(Hall switch input)
Oscilloscope probe is connected to TP7. (buffered output of the sawtooth VCO)
Set the pulse frequency to 10Hz (1200RPM)
Adjust P1 (gain) and P2 (level shift) repeatedly to obtain a clean (undistorted) sawtooth waveform with the bottom 'tip' of the waveform about 2V from the zero line.
(These adjustments set the VCO's DC control voltage range)
Now set the frequency of the pulse generator to 40Hz.
Repeat the adjustments described above.
Go to 10Hz again and repeat the settings as necessary.
You may need to go over these adjustments at 10Hz & 40Hz several times,
until both the amplitude and the bottom level of the sawtooth are the same.
Please note: these adjustments are CRITICAL and must be done accurately!
Adjusting the 'opening' and 'closing' points (limits within which the injection solenoid allowed to operate) requires a DUAL TRACE OSCILLOSCOPE.
Connect probe no.1 to TP7. This LINEAR sawtooth is the reference waveform.
Connect probe no.2 to TP8. The pulse seen here is the solenoid ON time & position.
[When using BOTH probes, I suggest you connect only ONE of the ground probes as the circuit is sensitive to 'ground loops'. (they upset not only the measurements but also the operation of the circuit itself)]
While the adjustment ranges of the solenoid's 'opening' and 'closing' points, as well as the ignition potentiometers are fairly wide, it is still likely that component values need to be altered for different types (and makes) of engines.
Those values have to be determined experimentally for each engine TYPE.
For solenoid 'opening point' range, alter the values of R19 & R20
For 'closing point' range, alter the values of R22 & R23
For ignition point range, alter the values of R30 & R31
(Fine tunings are done with the potentiometers P3, P4 and P6, respectively.)
This Ignition/Injection control circuit, together with the CDI module and the Hall switch, forms a COMPLETE ignition system.
Before you decide to go ahead building/using it, you should also read
"Important notes about the Ignition & Injection control circuit and Test Oscillator".
Les Banki
(Electronic Design Engineer)
Water Fuel & LBE Technologies
CDI circuit description
Introduction
This brief explanation may assist those who are not familiar with
Capacitor Discharge Ignition.
A capacitor (usually about 0.5 - 2uF) is charged to about 300 - 350V.
The formula for the stored energy in each charge is: E=1/2·C·V²
In words: the energy (E, in Joules) = capacitance (C, in Farad) multiplied by the voltage (V) across the capacitor, squared, multiplied by 0.5 (or divided by two, same thing)
For example: in the CDI design I am presenting here, the capacitor is 1uF.
It is charged to 330V before it is allowed to discharge.
The stored energy (E) in each discharge is: 1/2·1·330² = 0.05445 joules or 54.45 mjoules.
[Incidentally, the required minimum spark energy for Internal Combustion Engines (ICE) is said to be about 25 mjoules. (millijoule = 1/1000 joule)]
As you can see, a 1uF capacitor delivers more than twice that minimum.
Of coarse, lowering the voltage decreases the energy.
(AND also the average charging power required!)
In case the capacitor is charged to only 300V (which will still deliver at least 20,000V to the spark plug), the energy in each charge is 45mjoules.
(Still almost twice the claimed minimum.)
Again, note that the voltage (V) in the formula is squared.
This means that even a relatively small change in the voltage (increase or decrease)
results in a significant increase or decrease of stored energy.
However, keep in mind that Hydroxy requires considerably LESS spark energy.
(Even a very low energy electrostatic discharge spark is sufficient to ignite Hydroxy!)
Capacitor Discharge Ignition differs significantly from the well known (and old!) 'Kettering' system.
Instead of feeding the primary winding of the ignition coil with LOW voltage
(12 - 14V) and HIGH current (5 -10A), HIGH voltage (300 - 350V) and LOW current is dumped into the primary winding from a charged capacitor.
Thus, the POWER requirement of the CDI system is only a fraction of the Kettering system!
The design I am presenting here uses about 6W while the Kettering type ignition use
50 - 120W, depending on the design.
The above mentioned 6W power consumption is for a system requiring a maximum of 50 discharges per second which corresponds to 6000RPM for a ONE cylinder engine.
Naturally, for multi cylinder engines the power consumption will increase somewhat as the number of charge/discharge cycles increase.
The high voltage (300 - 350V) needed for a CDI system is obtained by using a DC - DC converter. There are several types of DC - DC converters but all of them use an inductor or transformer of some sort.
Such a transformer (or inductor) usually has to be custom designed.
All designers face this problem.
Most (if not all) manufacturers are not willing to design/make just a few pieces.
Unless one is prepared to order large quantities, they are NOT interested.
Thus, the cost of any new design is very much an issue!
There is a very good reason for telling you all this.
Everyone who intends to duplicate this circuit needs the following information:
While investigating several options, I discovered that several types of commercially made
DC - DC converters are available to power CFL's (Compact Fluorescent Light).
One of these little beauties are sold here by Oatley Electronics for a grand sum of $4.00!!
www.oatleyelectronics.com(one could hardly get a transformer for that price!)
But there is a catch. Its output is over 500V! (unloaded)
That is WAY too high for a CDI unit!
Loading alone does NOT bring the voltage down to the desired value AND its output changes with load changes!
So, its output needs to be REGULATED.
There are basically two ways to do this. One is to regulate the bias to the two driver transistors, the other is to regulate the input voltage to the unit.
I have tried both.
When regulating the bias, both transistors need to be heath sinked since the they are no longer turned on fully and so they run hot.
I found regulating the input voltage to be a lot better option.
Since this design is based on this particular CFL inverter (or rather, its transformer), everyone who intends to duplicate this design will face the same practical 'problem'.
My circuit and pcb layout for this CDI system is built around this transformer.
I have actually bought a large number of these inverters.
(there is no point designing pcb's for just a few units).
I strip these units, discard the original (round) circuit board and transfer the components to my pcb.
I found this to be by FAR the easiest and cheapest way to obtain the desired DC - DC converter!
In any case, even if I choose a custom designed transformer, duplicators would still have no choice but obtaining THAT particular transformer.
In case this is not acceptable to some of you, you are on your own and you have to
"roll your own" design!
Needless to say, I will sell completed units and perhaps kits, too.
Regulating the output was relatively easy.
However, during extensive testing I discovered that in case of certain possible fault conditions (more on this later) the DC - DC converter draws excessive currents which over heaths the inverter transformer and destroys the driver transistors.
Therefore, I have added a fairly complex protection circuit which I developed/designed. It gives full protection!
Detailed circuit description
Let's start with the CFL inverter described above.
While the manufacturer/supplier does not offer any kind of description (they hardly ever do!), it is easy enough to figure out how it works. (it is not important)
All I want to say is that it is a clever, simple design which seems to be very efficient and works well. It runs at about 100kHz.
Looking at my circuit diagram, the components used from this inverter are:
TF1, L1, Q7, Q8, C9, R26 and R27.
The output is full wave rectified by HV ultra-fast diodes (UF4007) D3, D4, D5 and D6.
C10 (10n, 630V) is filtering the HV output.
This 330V (or 300V) output is connected to one side of capacitor C12 (1uF, 400V).
The other side of C12 is connected to the "hot" side of the ignition coil primary.
The other side of the coil is grounded. (as usual)
In other words, the other side of capacitor C12 is grounded through the ignition coil.
The capacitor's stored energy is discharged into the coil as follows:
SCR1 (TYN816) is connected between the high voltage output and ground.
Its Gate is triggered by transistor Q9 (BC547), wired as an emitter follower.
When ignition pulses (from the ignition module) are fed to its base (through R28, 1k),
it turns on and its emitter supplies the trigger current from the 12V supply rail, through collector resistor R29 (390 ohms).
When Q9 is turned on, some current also flows through R30 (470 ohms) in addition to the SCR' gate trigger current. The low value of R30 and C11 (0.1uF) shunt spurious transients which could cause false triggering.
When SCR1 is triggered, it becomes (for all practical purposes) a short circuit.
Through this 'short circuit' the capacitors energy is discharged to ground.
The discharge current also flows through the ignition coil's primary which is transformed (1:100) and creates a secondary voltage well in excess of 20.000V!
(depending on the type of coil used).
Regulating the inverter's output voltage
As I have stated above, I choose to regulate the inverter's input voltage.
It is a standard 'series pass', OP amp based regulator. (IC1B, LM324)
It drives Q5 (BC547) and Q6 (TIP31B) in a Darlington configuration.
The HV (300 - 330V) output is attenuated by R23 (680k) and R24 (15k) and connected to pin 6, IC1B's inverting (-) input.
It is also connected to the emitter of Q6, which is the output of the regulator.
The non-inverting input is connected to the slider of P1 (10k), which, with R25 (3k3) forms a voltage divider to restrict the adjustment range of P1.
This, in turn, limits the high voltage at the output of the inverter.
Since OP amp IC1B is a "virtual earth" amplifier, its inverting (-) and non-inverting (+) inputs are practically at the same voltage.
Therefore, the voltage appearing at pin 6 (regulator's output) will be the same as the voltage on pin 5, the slider of P1.
The voltage divider R21 (990k) and R22 (10k)/C6 (10n) provide a convenient low voltage, low impedance test point TP2 for adjustment/test purposes of the HV output.
Protection circuits
First, I will try to explain why the somewhat complex protection circuit is necessary.
Please look at the circuit diagram.
You will see that C12 (the 1uF capacitor which supplies the spark energy) is connected between the HV output and, through the ignition coil's primary winding, to ground.
Now consider what happens if C12 goes short circuit.
(In other words, there is a short placed on the DC - DC converter's output!)
The poor thing will try to supply power into a short circuit!
(with plenty of current but almost NO voltage!)
As a result, current draw from the power supply will increase dramatically. This causes the driver transistors AND the transformer to over heath, until something gives!
Consider now an open circuited C12.
There is NO stored energy to discharge. Then there is NO charge time to consider. Remember that SCR1 (TYN816) is also directly across the HV output.
Normally, when SCR1 fires to discharge the energy in C12, the current flowing through SCR1 is eventually reduced below its 'holding current' so it 'drops out'.
(stops conducting)
When there is NO capacitor, (same as an open circuit capacitor) there is NO periodic discharge, the DC - DC converter is continuously supplying current so SCR1 will NOT drop out.
This means an INDEFINITE 'short circuit' (in form of a continuously conducting SCR) across the HV output.
Further, the EXACT same condition will also occur if the wire to the ignition coil's primary is broken or disconnected. (or if the coil goes open circuit)
IC1A is used to detect the presence/absence of the HV.
R1 (12k) and R2 (680k) form a voltage divider between the HV output and ground.
The voltage developed across R1 is a fraction of the HV and it is fed to the non-inverting (+) input pin 3 of IC1A, used here as a comparator.
A fixed voltage (approx. 2.9V) is applied to the inverting (-) input (pin 2) from the voltage divider R4 (68k) and R5 (22k) which is filtered by C2 (10uF).
Under normal operating conditions the output of this comparator is HIGH.
Should the voltage on the non-inverting input (pin 3), which represents the HV output,
decrease significantly (below the voltage on pin 2, the inverting input) or disappear completely due to a fault condition, the output of the comparator IC1A (pin 1) will go
LOW.
This output is connected to the Gates of DMOS transistors Q2 and Q4 (2N7000), through R6 and R15, respectively (both 100 ohms).
(Note: for this application bipolar transistors are un-satisfactory.
Their off-state collector-emitter leakage is too high.)
IC1C and IC1D are wired as square wave oscillators. Since the normally conducting Q2 and Q4 are connected between the inverting (-) inputs and ground, both oscillators are DISABLED. (C3 - 3.3uF and C4 - 1uF are the timing capacitors)
In the (sampling) oscillator IC1C, the charge/discharge times are separated.
This gives (with the component values shown) approx. 2 seconds HIGH and about 25 seconds LOW signal at IC1C's output (pin
.
Through D2 (4148) and R13 (10k) this signal is fed to the base of Q3 (BC547) which is used as an inverter.
Q1 (2N7000) is connected between ground and the non-inverting (+) input (pin 5) of voltage regulator IC1B.
Its Gate is connected to the collector of Q3. When Q3 is conducting, Q1 is NOT.
(NO Gate voltage - it is shorted by Q3)
When Q3 is NOT conducting, Q1 gets its Gate drive from Q3's collector through R14
(10k).
Q1 is now conducting, bringing the voltage on the non-inverting input (pin 5) of the regulator (IC1B) to 0V.
As a result, the regulator's output is also zero.
NO INPUT VOLTAGE to the inverter means NO current draw.
In other words, this is NOT a current limiter.
The inverter is completely OFF, drawing NO current.
As long as the fault condition exists, oscillator IC1C continues its 2/25 seconds ON/OFF
cycle.
Its output is inverted by Q3 which then turns Q1 OFF/ON.
So, when Q1 is OFF, the regulator (and the inverter) is working normally.
When Q1 is ON (conducting), the regulator (and thus the inverter) is cut off.
In this condition, there is NO current draw.
In layman's terms, this is what happens:
Due to a fault condition, (capacitor C12 open or short circuit, ignition coil primary open circuit, wire to the coil broken or disconnected…) oscillator IC1C is ENABLED and is producing a 2 seconds ON and 25 seconds OFF signal.
This signal ENABLES/DISABLES the regulator supplying the inverter.
The 2 seconds ENABLE signal is for SAMPLING.
Is the fault still there? Yes. OK, cut power OFF for the next 25 seconds.
Then, SAMPLE again (for 2 seconds) to check if the fault has been cleared or not.
If not, this oscillator will continue its 2/25 second routine INDEFINITELY.
Since power is applied for only 2 seconds (SAMPLING) and there is NO power for 25 seconds, no harm is done!
If the fault has been cleared, the oscillator is disabled and the regulator/inverter once again works normally.
Since indicator LED1 for the sampling oscillator is only turned ON for 2 seconds (and OFF for 25 seconds) there is a need for continuous indication of a fault condition.
That is the role of oscillator IC1D. Under normal working conditions it is disabled by Q4 (2N7000) which is shorting its timing capacitor C4 (1uF).
It is wired as a square wave oscillator which, under fault conditions, flashes LED2 ON/OFF about 3 times per second (~3Hz).
Under normal operating conditions, the inverter's regulated supply voltage output is around 6.4V. Current draw is about 0.5A.
With a short circuit placed on the output, the current rises to around 1 - 1.2A.
Should the regulator transistor Q6 go open circuit, the inverter simply stops operating.
However, should it decide to go short circuit, (unlikely, due to the moderate current draw of only 0.5A) the full rail voltage of 12V would be applied to the inverter and its output would rise to over 500V!
This, in itself, should not be a problem, except for two things:
1. Capacitor C12 (rated at 400V) might go short circuit (which would activate the protection circuit described above).
2. The ignition coil would produce excessive secondary voltage which could cause internal insulation break down.
Testing and adjustment
A number of TP's (test points) are provided for testing and adjustment(s) purposes.
There is only ONE adjustment to be made on this pcb, to set the inverter's output voltage to the desired value (usually somewhere between 300 - 330V depending on the ignition coil used).
Connect a voltmeter (set to 600V or 1000V range, depending on the meter) between TP3 (ground) and TP1 (HV) and disable discharge triggering TP4 by shorting it to TP3.
Now adjust P1 to the desired voltage (300 - 330V)
You could also use TP2 (and TP3) to adjust to 3 - 3.3V (100:1 attenuator, provided mainly for oscilloscope connection to eliminate the risk of damaging the input)
The regulator's output voltage (supplying the inverter) can be measured at TP5.
For a 330V output it should be around 6.4V.
The operating temperature of the inverter transformer and its driver transistors are a very comfortable 47°C and 45°C, respectively, measured in ambient temperature of 30°C!
Les Banki
(Electronic Design Engineer)
Water Fuel & LBE Technologies
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