Understanding this Circuit


#1

I’m having some issues understanding how this circuit works and I was hoping that I could get some help with figuring it out.

This UGLY THING:

The main issue I have is Q2. Looking up the 2SC2383 datasheet, the pinout is:

  • Pin 1: Emitter
  • Pin 2: Collector
  • Pin 3: Base

But attempting to set it up that way in LTspice gives no output for VDD. Which I realize isn’t labeled but is where the TVS diode is.

Changing the pinout around to be (more common) I get the following output at the TVS:

  • Pin 1: Base
  • Pin 2: Collector
  • Pin 3: Emitter

Lastly, I have redrawn it already (with the corrected(?) pinout of Q2) here so maybe it helps understand it a little.

THOUGHTS? Cause uhh, yeah that pinout doesn’t make any sense to me.


#2

Could it be a Reverse Avalanche pulse generator? The “backwards” emitter strongly reminds me of that. This link has more info: http://www.kerrywong.com/2014/03/19/bjt-in-reverse-avalanche-mode/


#3

Ah! I should have said (derp, I was writing it quickly before leaving work) that it’s being used to power a PIC micro. I’m not 100% sure what’s it’s doing. I think it’s running a 30 minute timer. But it needs to have stable power.

Sorry about that! That’s why I was trying to mark Vdd, since that is for the PIC. R1 also goes off to one of the pins of the PIC as well.


#4

This seems like a linear regulator with a reverse polarity protection diode D4. Transistor is in a common collector configuration https://en.m.wikipedia.org/wiki/Common_collector with Vin and equal to the zener diode voltage. I would guess R1 is used in part to measure the supply voltage but not sure without the entire schematic.


#5

@chia.brian That’s basically what I’ve come to understand from it too, but the pinout seems wrong as shown in the original schematic and I wanted to make sure that I wasn’t going crazy or something. Based on the common collector pinout, Base is input and Emitter is output (and that works for me in the LTspice files).

I’ve taken the PCB and done some measurements. Based on the image below (so we all know what I did)

BJT

  • Pin 1 was 5V
  • Pin 2 was 20V (or input voltage)
  • Pin 3 was 5.6V

Basically, the issue I’m trying to understand is the difference in pinout in the datasheet and what… well it is based on measurements.

I think I just have to ignore it and move on though. I think the Schematic is just somehow wrong and the BJT should be left to right 3, 2, 1. Not 1, 2, 3 like it is now.

Based on that re-ordering, it would make sense. I’m just going to drop this. WEEEEEEEEE

haha, thanks for the help though!


#6

Well, assuming your drawing is top down through the part to the PCB (like an “x-ray”), and looking at your voltages and assuming an NPN - I would guess that 1 was the base, 3, was the emitter and that 2 would be collector by default.

That happens to match this which shows what looks like a view looking at the bottom of the part before installation (ie: non-x-ray)

Keith