Old Paramount c4s/shunt regulator boards

[vetmed]

I made the change to the 5670 driver some months ago. This leaves 2 boards already stuffed that worked perfectly before they were removed. Looking at the old Paramount schematic there are 2 C4S sections with the regulator chip. If I remove the cap, TL431 regulator and 220 ohm CC resistor I should then have 2 C4S, one set at 10.5ma and one set at 3.5ma. I can then remove the K wire and the + wire on the  A side. Next to the + pad on the A side disconnect the red wire going to the O pad of the B side. This then becomes the output wire for the B side. Disconnect the red jumper at the O pad of the B side that goes to the I pad of the A side, this becomes your input for the A side. Finally disconnect the black jumper wire near the 300K resistor on the B side, this becomes the ground  for the A side.
   Understand that I have not done this yet but if I am correct then I will have  4 C4S circuits for experimentation, I am working on a small signal DHT circuit that these should be very useful for. If there are any problems with what I am proposing I'd love to hear about it. Thanks


Robert Lees

[Paul Birkeland]

Hello Robert,

Though the documentation on the various implementations of this PCB are not extensive, here are the general steps to experiment with these PCB's:

1.  Remove all the jumpers except the jumper connecting the G's down the center.
2.  Set R1 and R2 according to bias voltage and desired current (post here for more info).
3.  There's an "I" for input, "O" for output, and "G" for ground.

-PB

[vetmed]

Thanks Paul
   From the original Paramount schematic one half of the board is set for 3.5ma and the other half for 10.5ma, if I am understanding this correctly. The 3.5ma value is an excellent starting point for what I have in mind, but I would certainly be interested in knowing how to figure out how to set for different values of current.




Robert Lees

[Paul Joppa]

To a pretty good approximation, R1 = 0.855/current. Current is in amps, so for 3.5mA = 0.0035A you get 244 ohms (we use 237 ohms). 0.855 is the voltage difference between the LED voltage (about 1.57v) and the transistor base-emitter voltage (about 0.7v). All of these voltages are slightly temperature dependent so there's no point in great accuracy.

[vetmed]

Well, that's much easier than I thought it would be, thanks.




Robert Lees