Electrical – Two grounds when using transistor switching

If I understand correctly, you are trying to show the below circuit and want to know what happens if the user raises the user control.

What happens will depend on which component gives up first. The devices are each trying to maintain certain states, but they conflict, so they cannot all win. In theory, we can write equations for each of them, but you asked real world, so I'll speak in generalities here.

1. The bottom transistor is trying to turn on hard. About the best it can do due to the way a BJT works is to pull the collector to about 0.3 V of the emitter. This is called saturation.

2. However, the top transistor wants to keep it's base to emitter to about 0.7 V. This is very similar to how a diode operates. As the base to emitter voltage is forced larger, the current will rise exponentially.

3. The voltage source for the power supply will attempt to maintain the voltage across it. But in real life, power supplies have limited current capabilities.

In a real circuit, the exponentially rising current across the upper base to emitter will either pull the power supply down if it cannot handle the current or the heat generated across the transistors will destroy one of them.

2

I will start with a shift of terminology here -- instead of "grounds", we have "returns" -- namely a power return and a signal return. The power return is the easy case -- it should be a star topology with the star point at the power supply output, tying to the signal returns of each board at a single point in turn, and also to the chassis ground at a single point.

Now, we have the RF and processing boards to deal with. Each board has a signal return, which is handled through a ground grid at a minimum, if not a full ground plane. This return is contiguous, even on the RF board, and also is interconnected between the two boards alongside the digital signals passing between them.

The digital board can be laid out freely, provided the loops are kept small. This is not true for the RF board, though, which needs to have a layout that is strictly partitioned between analog nets and digital nets -- there should be an "analog area" on the board that has only analog signals, and a "digital area" on the board that has only digital signals. The return currents will then proceed to stay with the signals they are paired with, thus keeping them separated without the need to do any ground plane splitting.

Finally, connector shields need to be returned to the chassis ground by the shortest path possible, and there also needs to be a "bridge" between the chassis (I/O) ground and the signal return located with the I/O signals going to the outside world. (ESD/EMI protection devices, as well, return to the chassis aka I/O ground.)

For more information on this topic, I recommend reading Henry Ott's fabulous book, Electromagnetic Compatibility Engineering, and in particular, Chapters 16 and 17 on circuit board layout.