Negative voltage on emitter follower

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SUMMARY

The discussion centers on the behavior of an emitter follower circuit as described in Horowitz and Hill's "The Art of Electronics." Specifically, it highlights that an NPN transistor in an emitter follower configuration can only source current and cannot output voltages lower than -5 volts. The transistor turns off at an input voltage of -4.4 volts, which is due to the reverse biasing of the emitter-base junction when the output voltage drops below this threshold. The voltage divider formed by the resistors in the circuit plays a crucial role in determining the minimum output voltage.

PREREQUISITES
  • Understanding of NPN transistor operation
  • Familiarity with emitter follower configurations
  • Knowledge of voltage dividers and biasing techniques
  • Basic circuit analysis skills
NEXT STEPS
  • Study the principles of transistor biasing and its effects on output voltage
  • Learn about voltage divider circuits and their applications in biasing
  • Explore the characteristics of NPN transistors in various configurations
  • Investigate the impact of load conditions on emitter follower performance
USEFUL FOR

Electronics students, circuit designers, and engineers interested in understanding transistor behavior in amplifier circuits, particularly those working with emitter follower configurations.

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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