Transistor NPN BJT Question

Discussion in 'Homework Help' started by luv2draft, May 27, 2008.

  1. luv2draft

    Thread Starter New Member

    May 12, 2008
    2
    0
    I am working on 1 D from the homework attached. I have found Vce to be -0.5V.

    I found this knowing Vout=Vcc-(Bf)(Rc)((Vi-VBEon)/Rb)
    =5-(100)(2k)((1.25-.7)/20k)
    =5-5.5
    =-.5

    I am guessing the transistor is operating in the Reverse Active Region.

    My instructor said to "Switch the C & E Terminals and Bf->Br"

    Can someone explain what my instructor is trying to say by this??

    Thanks
    Daniel
    Cal Poly Pomona
     
  2. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The transistor does not have a negative supply so its collector cannot be negative. It is about +0.1V when the transistor is turned on hard.

    The base current is (1.25V -0.7V)/20k= 2.75uA.
    Its collector current will try to be 100 times more which is 2.75ma.
    The 2.75mA in the 2k collector resistor requires a supply voltage higher than 2.75mA x 2k= 5.5V but since the suppply voltage is only 5V then the transistor is turned on as hard as it can and its collector voltage is about +0.1V.
     
  3. silvrstring

    Active Member

    Mar 27, 2008
    159
    0
    I too would like to better understand this fixed bias circuit.

    luv2draft, I calculated V(rb)--the 20k--to be .55V, Ib to be 28uA, Ic to be 2.75mA, and V(rc) to be 5.5V. Leaving Vc at -.5V

    However, when I checked it out on Multisim, Vbe measures about .8V which I'm guessing correlates with the .1V collector voltage Audioguru mentioned.

    Audioguru, is this typical of all transistors when (Ic x Rc) > Vcc?

    It's important, because the .1V difference changes Rb to .45V which in turn changes Ic, and therefore Rc (4.5V now). Vc then becomes about .5V
     
  4. luv2draft

    Thread Starter New Member

    May 12, 2008
    2
    0
    Tell me if I am understanding this correctly,

    because the collector current 'wants' to be 100 times greater than the base current, this causes the collector to 'max out' or 'saturate' the power supply until the Collector voltage reaches it's saturation point?

    I am trying to understand the lesson behind Collector Saturation voltage. Is this trying to show us that we should always have a large Base resistance to keep the transistor operating in 'Active Mode' instead of saturation?

    Audioguru- You use the word 'Hard' when you talk about the transistor turning on... Is this a negative way to describe the operation of the transistor? In design, should I avoid this mode of operation?

    Thanks
    Daniel
     
  5. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Yes. You have the correct interpretation, barring any other external factors such as insufficient current from the power source and provided the transistor is not being required to deliver more current to the collector load than its maximum rated current.

    hgmjr
     
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