so now I try to explayn what is going on :
by the Fuse F 1.5 Amps the nominaly 12 Volts coming to the unit.
this onboard voltage can variate from 25 to 0
volts, up and down witouth a danger for the unit.
this incoming voltage needs to be at least 7.6 volts to get the unit started.
once the unit started, bike onboard voltage dropes lower
then 5.9 volts
, the Relay switches the unit OFF
to protect the Transistore T1 from overheating because the BASIS B would not get enough mAmps ( 350 mA are recommendet ) to drive the 4 Amps the coil is needing on Transistore line Collector C to Emmiter E.
and it gos like this : power comes to pin 4 and 9 of the relay by the same time. pin 4 is in contact to pin 3 so current runs trough the resistor R 1 to fill the capacitor C1. when C1 is filled,( friction of a second, current comes to pin 1 and the relay operates. to avoid flickering of the relay, C1 is giving wheilst switching the power to the relays coil.
when relay is in ON position, current is now on pin 5 and on pin 8 as well. pin 8 serves now the actualy Ignition unit,.
theweil pin 5 is giving the current trough diode D1 and the Zener Diode DZ2 to the relays coil.
the two Diodes are set to reduce
this onboard voltage by around a fix voltage of 2.7 volts.
as the relays coil is usualy releasing by 3.2 volts, we have to add the 2.7 what it operates back to release now, so it releases in reality by this 5.9 onboard voltage now.
so this is the LOW VOLTAGE SAVETY CUT OFF PART.
looking at the relay in operated position, from pin 8 the current enters the LDO Type LM 1084 iT.
this thing is to restrict a maximum voltage on the pin "out".
this maximum output voltage is set by the Resistore R3, so output on pin "out" is 10.0 Volts.
this LDO is NOT restricting voltage to the low side
, this means, it will give out any voltage down from 10 V.
and this voltage on "out" will be always 0.8 volt lower than the incoming voltage!
so, when onboard voltage drops to 5.9 volts, ( wich is the lowest possible ) "out" on the LDO willbe only 5.1 volts.
but still, by 5.1 volts to the B on the Transistore there is enough Amps ,370 mAmps, to drive the transistore corectly.
the diode D3 makes shure there is no higher voltage on pin "out" in compare to pin "in" of the LM1084, because this is something it may brake the LM.
Resistore R2 and the fallowing LED 1 ( green color ) are to indicate is there a fault in the unit like Fuse F brocken ore Relay not "ON" ore LM1084 brocken.
the 10 volts are sentd to the points, and hopefully coming back to the Resistores R6 and R 7.
they are both from a strong type 30 Watts each, so the 10 Watts they have to cop with are savely devidet for good heat transmition.
together in parallel they have 8.5 Ohms, this means as a current of 950 mAmps is going in to the Base of the Transistore BUT 34.
BUT 34 is a strong thing, 500 Volts on line C to E, up to 50 Amps is possible to switch and on line B to E it cops with 10 Volts and max 10 Amps as well.
the Resistore R4 is a pull down for the Base of T1 to make it switching fast and savely.
the Resistore R5 and the LED 2 ( red color ) are set to indicate is there voltage coming from the points, so it is helpfull to set Ignition timing.
when points open, ( moment of the Spark occuring ) the red LED will go "OFF".
if the green LED will be on and the red will never going on for a example, the line to the points must be interrupted. ( maybe by a mouse ore something like....)
puhh, it tooks me a weil to get it, but now comes the more handy part to bring it on to the bike, wich I do prefere to do of cause...