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Sir Hobby Junkie . . . . . . .
Session Deux . . . . .
I was able to make out enough on the first viewing of your reverse engineered drawing and mark in changes and add some info right on your schematic sheet .
First of all . . .might this just be an internal dedicated ancillary circuit to one of those " wun hunnerts ninety nines dollahs " tactical flashlights, since its being built on a round PCB.
Usually all of this ciruitry is built on a mini sized monolithic chip as the Laser driver circuitry.
Definitely a laser driver circuit with two positions, High and Low controlled by the 33kohm and 13 kohm resistors.
.Questions that I need further info . . .I am going to Bullet . . . . to be easily seen.
I am "reading" the board and seeing the battery wiring coming in to A and B copper lands that have NOTHING else connected to them, except for the foil path from A over to E.
Then there is the additional foil path from B to F.
These are the central pins on the DPDT switch, with the left C and D lands being unused / vacant.
On power on the E to G connection is being apparent for the + power insertion.
I think I answered these with the posting of the trace PCB and the switch. If not, let me know and I will try to do better.
You can see I placed the 1 ufd over where it belongs.
As for the F to H connection that will require a bit of pulling the battery and ohmming out by you.
Not sure I understand what you mean. If I read this correctly and forgetting about Ben Franklin's mistake about current flow against electrons, I THINK the photo part of the diode provides a current feedback that goes both to the base and the emitter of the PNP transistor, to the emitter through the resistors. Whether 1 or both ar used, it would control the amount of current to the collector which then feeds the base of the NPN transistor. This is just a guess and I am still reading up on NPN and PNP along with FET interactions in circuit design.
Look at the PD photo diodes output coming down to this circuitry . . . I am wanting to be seeing those two resistors being used in a voltage divider function with the final
output ending up at the base of 5AW.
So are they used as a voltage divider (drops voltage IIRC but current remains stable) or as current limiters?
As it stands switch F-> H is merely doing an illogical switching of a short across the 33K resistor. . . . . .ha . . . ha . . . like it needs to be in circuit while the unit is OFF but
needs to be shorted out when the unit is turned ON.
Knowing that the bottom row of pins is only used as switching for the resistors, it does not affect the neg side of the power structure as I see it. But I am probably wrong on this.
What I am now seeing is enough to give you:
CONCEPT OF DESIGN :
Flip the power switch and + power is applied to the Laser diode and it is wanting to find a ground path to activate.
Simultaneously with that power on, a 1.2 K resistor is giving base drive to 6CW so that ground path is then found via Collector -Emitter conduction of 6CW.
That base drive supplied is a bit more than needed, as the detected Laser output level is being monitored by the
photocell PD and will be supplying power correction via a feed back loop.
Note that the base of the 6CW transistor takes a path to the left and initially encounters a RC time constant which will give in the order an
~ 80 MS "hang" time on corrections. That path ends at the collector of 5AW . . . . . as the output from the PD increases it progressively
shifts base drive of 5AW and causes it to conduct more heavily and that progressively pulls down the voltage on the initial base drive of 6CW
to a stable correcting value..
This is what I thought: Using the ABC and DEF pin on the switch, with pin B always being "hot +", as the switch is thrown to the left, B connects to A activating the circuit through the 1.2kohm circuit initializing the base of 6CW. The emitter of 6CW feeds the BE output to the collector which drives the laser diode (controlled by the loop feedback of 5AW). So with the laser diode output as visible light, the photocell diode initiates a small current (or is it voltage?) that feeds the base of the 5CW as well as the emitter of 5CW but controlled through the combination of the two resistors as pins D&E do nothing or go nowhere forcing the series of 46Kohm to the emitter of 5AW. This in turn feeds the collector of 5AW which feeds the base of 6CW thus controlling the laser output in the LOW position. Now if the switch is thrown to the right, pin B activates pin C to activate the circuit on the + side through the 1.2kohm resistor and everything else is the same EXCEPT that the 33kohm resistor is bypassed thus upping the feed to the 5AW emitter increasing the collector output driving 6CW to a higher output, thus the HIGH position of the laser. I am very unsure about this......it is my best guess at this time.
- Pull the battery and ohm out and inspect to find out the real wiring path on the basic area within the RED dotted lines.
As seen in the picture of the PCB traces.
After that use a DVM in its diode test mode to ascertain and pass us back all of the Vf of all of the junctions of the two transistors, as well as the
slight possibility getting readings from of the Laser diode and PD.
The real work horses are being the 6CW and the Laser diode.
Can not do. The circuit burned before I stated this project. I think the 6CW was overpowered and fried thus frying the LD/PD so I can not test anything or the components. I will try to do the Vf readings on the old parts but I am not anywhere certain that these components are functioning. I did get that acrid silicon smell when the laser fried.
- Fulfilling that info . . . . .round up anywhere from 1/2 thru 2 watt . . . . . . 100 and 220 ohm resistors and I will tell you howtodoittoit next.
I believe that the low circuit provided a laser output of < 1mW while the high side provided a laser output of around 3mW. That is all I know about the output. I can not even say if it is current or voltage controlled, but I am assuming current because of the junction transistors. The new laser diode needs a threashold or around 35mA to 40mA to lase but I am waiting on the data sheet from the supplier.
Thasssit . . . . .
Addenda:
I see that my assignment of the switch Alfa designations differ from yours, because I used up my AB on the battery lands, and thereby created the resultant error on the CDE and FGH designations.
Redraw of your Techno Referencing:
73's de Edd
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Thanks. Nice drawing. It seems to make more sense than mine ever did. Let me know what I can do next.
