Negative voltage on emitter follower

AI Thread Summary
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
Messages
1
Reaction score
0
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 ?
 

Attachments

  • Emitter-follower.png
    Emitter-follower.png
    1.1 KB · Views: 817
Engineering news on Phys.org
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.
 
Last edited:
Thread 'Weird near-field phenomenon I get in my EM simulation'
I recently made a basic simulation of wire antennas and I am not sure if the near field in my simulation is modeled correctly. One of the things that worry me is the fact that sometimes I see in my simulation "movements" in the near field that seems to be faster than the speed of wave propagation I defined (the speed of light in the simulation). Specifically I see "nodes" of low amplitude in the E field that are quickly "emitted" from the antenna and then slow down as they approach the far...
Hello dear reader, a brief introduction: Some 4 years ago someone started developing health related issues, apparently due to exposure to RF & ELF related frequencies and/or fields (Magnetic). This is currently becoming known as EHS. (Electromagnetic hypersensitivity is a claimed sensitivity to electromagnetic fields, to which adverse symptoms are attributed.) She experiences a deep burning sensation throughout her entire body, leaving her in pain and exhausted after a pulse has occurred...
Back
Top