Negative voltage on emitter follower

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In an emitter follower configuration, the npn transistor can only source current, limiting the output voltage swing to a maximum of +9.9V and a minimum of -5V. The transistor turns off at an input voltage of -4.4V, which is due to the reverse biasing of the emitter-base junction when the output voltage drops below -5V. A voltage divider formed by resistors defines the minimum output voltage, allowing the emitter to handle negative voltages. For voltages lower than -5V, the transistor cannot conduct, effectively preventing further negative output. Understanding these principles is crucial for proper circuit analysis and design.
srinivasbakki
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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 ?
 

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srinivasbakki said:
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 ?

Welcome to the PF.

The voltage divider formed with the 2nd 1k resistor is what defines how low the output voltage (Ve) can go. For voltages lower than 5V at Ve, the E-B transistor junction is reverse biased.
 
The emitter node can have negative voltage, notice the negative voltage source and the voltage drop across the leftmost resistor. When you calculate all the currents in the emitter node (the currents through the emitter and the two resistors) you can notice that the emitter 'sinks' current only if the voltage is above -5V.
 
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