OK, what is the input voltage? I will give you a ruff diagram here. (volt in)
Vcc--sw's-on/off/IR-control board (microprocessor)---ccfl (cold cathode
fluorescent lamp) inverter (changes DC voltage tow switching DC (such as
on/off/on/off/on/off but at a much faster pace (20Khz)---toroidal coil (this
coil will step the switch DC voltage up to about 1000 to 1500 switch DC volt,
then this coil will change the orientation of the voltage to current
relationship by shifting the current by 90 degree's where it lag the
voltage---this process has some changes on circuit and how it works)----OK, from
the ccfl inverter------voltage multiplier (all this some HV capacitors and HV
diodes hook up an configuration where each stage will double the voltage that go
into... EXAMPLE: 4 stage voltage multiplier for each stage now it double the
voltage 1500V input to 1st stage--output to 2nd stage 3000V output to 3rd stage
6000V out put to 4th stage 12000V switching DC, they will also refer to this
voltage AC voltage because AC switch to from - negative voltage to + positive
voltage.
Like in you case, the two volts from the controller to ccfl. Out put volt
from ccfl with be around 2.4 to 2.6 due capacitance discharge, Now this voltage
will be divided in the ccfl by two input windings (I am not going to explain
this). Now the two input winding will induce voltage to the secondary winding
(step up). The output from secondary of toroidal coil will be multi switching to
the voltage multiplier. Probably in the neighborhood of 1500 switching volt. How
many stages the voltage multiplier has I don't know. The ignitier voltage to the cathode from the
multiplier would be around 12000 voltage to get image intensifier tube to fire.
stewbison
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