Negative voltage on emitter follower

[bakkis]

Hi,
Iam reading horowitz and hill and came through this paragraph in section 2.03 Emitter follower :
Notice (Section 2.01, rule 4) that in an emitter follower the npn transistor can only "source" current. For instance, in the loaded circuit shown in Figure 2.8 the output can swing to within a transistor saturation voltage drop of (about +9.9v) but it cannot go more negative than -5 volts. That is because on the extreme negative swing, the transistor can do no more than turn off, which it does at -4.4 volts input (-5v output).

How will the transistor turn off at -4.4, it should have turned off much before right ? Am i missing basics of it ?

Kindly help me understand this, i'll be greatful.

 

[ericgibbs]

hi,
Perhaps this simulation will help you understand.

Ask if you have a problem.

 

[alec_t]

When the transistor is off, the emitter and load resistors act as a simple voltage divider connected between 0V (ground) and -10V. As the two resistors are of the same value (1k) their junction point will be at -5V. Hence the emitter (which is connected to that point) wiil be at -5V. Hence the base must be at -5+0.6 = -4.4V (or less).

 

[bakkis]

Wow, That was as simple !!! Thank you.