I was playing with a BJT based circuit and I learned about the breakdown voltage of BE junction.
I came out with a solution of connecting a diode in series, with higher breakdown voltage. But I am not sure if connecting in the bottom and top are both same
where the zener BE is the model of the base-emitter junction. If I connect the protection diode in the emitter (D2) I will certainly avoid the stress in BE junction. Does it act in same way if I connect the dioe to the base? (D3)?
The bjt has a breakdown of 6V. so if I apply like 12V (limited with a high resistor) I see a voltage of around 7V, which makes sense since 6V is just the minimum specified. The current through the gate is very low, since I work with 12V and limiting resistors of 270k. when I breakdown the junction and 47k as pullup. The BJT in the bench is a BC846ALT1G.
EDIT: since the complete circuit could depends on the collector state, here is a sketch of the circuit:
The emitter of Q2 is the one which I try to connect to 12V (VCC) with 270k, while the collector of Q2 is always connected to 12V. Basically I want to bypass a resistor with a BJT where the OUTPUT goes to another circuitry which uses a pull down of the same size. The load driven by the Q2 emitter is 270k/2 Ohm, since the other pulldown is 270k as well, there is nothing else.
I was trying to understand the situation more deeply in order to apply less changes as possible.
Q1 is kept conducting or open (applying a current limited voltage on base, like 5V or 0V) and works as it is supposed to work.
The reverse junction which I am referring to is on the Q2 BE junction.