study of CSI1802X schematic, 12V rail question

[cellularmitosis]

Hey all,

I've been studying the CSI1802X schematic, and I decided to sketch it up in KiCAD so I could rearrange it to try to understand it.

One part which I don't understand yet is the way they float the 12V supply section.  The entire 12V section is floating via a transformer, and the only point which ties it to the rest of the circuit is where the relay transistor base pin connects to the main negative output pin.

With the way I have it sketched up, I don't see how that base pin will ever move anywhere (i.e. I don't see how the relay will ever switch).

Did I make a mistake in transliterating this schematic?

Here's the github page: https://github.com/pepaslabs/CSI1802X

The CSI1802X: http://www.circuitspecialists.com/bench-power-supply-csi1802x.html

The manufacturer's schematic: http://www.circuitspecialists.com/pdf/1802X_schematic.pdf

[pa2ees]

With the way I have it sketched up, I don't see how that base pin will ever move anywhere (i.e. I don't see how the relay will ever switch).

Did I make a mistake in transliterating this schematic?

I think your schematic is correct as far as the relay goes (I have not looked too much into the rest of the circuit).

I agree that the base will not move anywhere *with respect to the negative output pin*. 

However, because the negative output pin is not shorted to the 12V circuit section's ground rail, the base of the transistor and the 12V circuit section GND rail are not necessarily the same voltage.   In fact, if there is indeed +12V from IC5 Vout to IC5 GND, that means that there should be current flowing through R6, and into the base of the transistor Q3.  This should turn on the transistor and subsequently turn on the relay.

An interesting point to note is that the GND rail of the 12V circuit section will probably be lower potential than the negative output pin.

In summary, the voltage of the base of the transistor with respect to the emitter is the important bit, as it will determine whether there will be Base-Emitter current, or not.  The voltage of the base of the transistor with respect to the negative output pin is not important.

I hope that explains it clearly enough.

 - Erik

[cellularmitosis]

Thanks!  I'll put on my thinking cap and think about that for a while.