transistor switch application circuit

Discussion in 'General Electronics Chat' started by donut, May 26, 2012.

  1. donut

    Thread Starter Member

    May 23, 2012
    51
    0
    Basically I am reading a transistor switch example fron Horowitz Hill book - The art of electronics.

    He starts by explaining that there are 4 golden rules for transistors:

    1) Collector must be more positive than the emitter
    2)base-emitter and base-collector act like diodes
    3)transistors have max values for Ic, IB and VCE that cannot be exceede
    4) when rules 1-3 are applied Ic is approximately IB that is (Ic=βIb)

    Going deeper into the example he explains when the switch is closed that.

    1) base rises to .6V
    2) voltage drop across base resistor is 9.4V
    3) base current is 9.4mA
    4) application of rule 4 leads to Ic = 940ma (for a typical β=100) which is wrong, wrong, wrong! Because rule 4 only holds when rule 1 is obeyed

    So here is where things get confusing for me.

    If Ic is 940ma how are you breaking the rule that collector voltage is not greater than emitter voltage?

    He says to have a current of 940mA can only occur if you would pull the collector below ground. I totally dont understand this statement!

    What exactly does this mean?

    How would you pull a collector below ground?

    Please help!
     
  2. aman92ullah

    New Member

    Jan 19, 2011
    12
    0
    See, first we calculate the resistance of the lamp as R=V/I=10/0.1=100ohms.
    Then, if according to 4, Ic=940mA, then voltage drop across that resistor(lamp) will be .94*100=94V!
    So, technically, collector is 10-94=-84V. But this will not occur, since the transistor will saturate when collector voltage is about 0.2V above emitter (which is ground). In saturation, current attains a maximum value and stays there.
     
  3. donut

    Thread Starter Member

    May 23, 2012
    51
    0
    doesn't saturation mode has to do also with the voltage potential between Collector/Base and Base/Emitter junctions?

    I noticed that the "diode representation" of a npn transistor is different between a transistor in saturation vs. a transistor in active mode.

    see attached file (saturation mode example on top, active mode on bottom)

    saturation mode (diode representation of transistor):
    collector base junction is forward biased
    base emitter junction is forward biased

    active mode (diode representation of transistor):
    collector base junction is forward biased
    base emitter junction is reversed biased

    Can you explain to me how the bias in the diode transistor representation is manipulated to change the mode of the transistor from active to saturation?
     
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