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Negative voltage on emitter follower

  1. Jul 21, 2011 #1
    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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  3. Jul 21, 2011 #2

    berkeman

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    Staff: Mentor

    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.
     
  4. Jul 21, 2011 #3
    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: Jul 21, 2011
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