Why base and emitter are connected?

[EDA_hg81]

I saw a design like the attached picture.

Why base and emitter are connected?


Thanks.

 

[dick_freebird]

The 140ohm resistor provides base discharge, take
the case that your "Vout" has limited sink drive (a
unidirectional current source). A roughly 5mA sink
is provided (700mA/140ohm).

If the source can't be relied upon to sink current,
say under startup or fault conditions, the resistor is
going to make the output go to a known, 0 voltage.

 

[transbrother]

It is not shorted. Current first flows thru the 140 ohms and when the current is ~5mA, it creates a Voltage drop of 0.7V at which time the BJT will turn on.

The circuit ensures that the buffered output doesnt go much below 0.7V of Vout.

 

[EDA_hg81]

So many thinks for your sincere explanations.

I am wondering if the above and this attached design have the same functions.

Which one is better?

 

[dick_freebird]

Similar. But the first one has a lower effective base
impedance (so probably better transient behavior
especially low-going) and the input current varies
little w/ output voltage (always Vbe/140) , while the
second will have a significant variation (R1 sees 0 to
(VOH+Vbe), depending).

 

[Freddie777]

i'm a newbie but i'd like to ask...

is it something to do with negative feedback? i.e. feeding back a portion of the output back into the input signal.. like in an op amp. ?

 

[flatulent]

The input impedance is increased. The 140 ohm resistor is boosted by beta times.

The 100 0hm resistor in the second diagram is for oscillation prevention. With a capacitive load the input impedance can have a negative real part.

When I design RF power amplifiers, I always put a few ohm resistor in series with the input. So do many other designers.

 

[EDA_hg81]

Thank you so muck for your sincere answer.

Best Regards.