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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.
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.
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).
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