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Common Base

Discussion in 'Electronics Homework Help' started by Wade Yaden, Jun 25, 2016.

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  1. Wade Yaden

    Wade Yaden

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    Jun 25, 2016
    1. For the circuit shown in Figure 3 below, draw the DC load line and locate its quiescent or DC working point. Show all work.
    I have never done a common base problem before and our book does not go over this. Our book says to mentally "short" out the transistor to find Current Saturation Point. And "open" transistor to find the Voltage Cutoff. Once I short the transistor not sure how to calculate the current. Little lost on this..Just want to be able to understand what to do...

    upload_2016-6-24_19-13-51.png
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,482
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    Jan 21, 2010
    if you short out the transistor, what is the combined voltage across Vb and Vl?

    for the purposes of calculation you might also want to include Ib in your final calculation, but this may be both unknown and small
     
  3. Wade Yaden

    Wade Yaden

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    Jun 25, 2016
    Thats the question Im asking. If I short the transistor out. How do I treat each resistor (label). The same - different. This is our first week in transistors. Can I do KVL in a loop and treat them the same. All the problems I have to reference in the one chaper we started are all common emitter.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Take the transistor out and place a line between where the emitter and collector were connected.
     
  5. Wade Yaden

    Wade Yaden

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    Jun 25, 2016
    And?? I think we have established the part of shorting out the transistor. I stated this in the original post. Yes a line across the collector and emitter. How do I treat the problem( 2 voltages 2 resistors). If that was misunderstood. That was my original question.
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Two voltages mean one voltage difference does it not?
     
  7. Herschel Peeler

    Herschel Peeler

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    Feb 21, 2016
    If you take the transistor out you lose the E-B voltage. So short Emitter to Collector. W
    Is the circuit drawn correctly? I would expect the higher resistor value to be on the Base. NPN or PNP transistor?

    I would expect a circuit more like this.

    Design 779 common base.PNG
     
    Last edited: Jun 25, 2016
  8. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Drawing the load line does not involve 2 voltages & 2 resistors. The load line is all about the load resistor, and finding two points of corresponding load current and load voltage in order to draw a straight line between those two points which will then be the load line. So think in terms of the current through the load resistor (i.e. collector current) and the voltage drop across the load resistor, Vlr, except that you make the load line using the collector voltage which is Vcc-Vlr. So the two points usually employed to draw the load line are when the collector current is zero & collector voltage is maximum, and when the collector voltage is zero & collector current is maximum. Note that in this problem the collector voltage is zero when the collector is shorted to the base (which is grounded).

    But in reviewing the circuit, does it not seem a little strange that VEE is driving the collector and Vcc is driving the emitter?
     
    (*steve*) likes this.
  9. Wade Yaden

    Wade Yaden

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    Jun 25, 2016
    Picture was copied straight from practice problem..
     
  10. Wade Yaden

    Wade Yaden

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    0
    Jun 25, 2016
    Sheet says NPN
     
  11. Wade Yaden

    Wade Yaden

    7
    0
    Jun 25, 2016
    Even tried to play around with it in Multisim and I wasnt able to pick up any readings anywhere.
     
  12. Herschel Peeler

    Herschel Peeler

    401
    65
    Feb 21, 2016
    You have an NPN transistor with the emitter more positive than the base. It will never turn on. Bad picture.
     
  13. Wade Yaden

    Wade Yaden

    7
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    Jun 25, 2016
    I agree..it's a bad pic...Yes that was what I was getting in Multisim (0) ...when I flip the transistor (which is the way I think it should be drawn) I start getting data. Didn't catch that Vcc and Vee label...waiting from teacher on response...
     
  14. Herschel Peeler

    Herschel Peeler

    401
    65
    Feb 21, 2016
    Okay, having flipped it, go back to your questions. What will the voltage on the emitter be? Assuming the output is on the collector (it doesn't show that either). Shorting the transistor (transistor on and in saturation) what will the voltage on the collector be? What is the maximum collector current? With the transistor fully off (ground the emitter) what will the voltage on the collector be? What will the minimum collector current be?
    Given the values shown what would the emitter-base current be? What might the collector current be? What is the current gain of the 2N1711 at that collector current?

    This is just a badly made question if you ask me. The transistor is flipped. No output is shown. I would have made the emitter resistor about 430,000 ohms.

    Interesting questions ...
    What is the emitter-base current?
    what is the maximum collector current?
    What is the gain of the 2N1711 at that collector current?
    At that gain and that emitter-base current what might the collector current be?
    The task falls apart here.
     
    Last edited: Jun 27, 2016
  15. Herschel Peeler

    Herschel Peeler

    401
    65
    Feb 21, 2016
    Not that it is really relevant to your question,, just because it is a nice place to mention transistor characteristics ... The data sheet for a 2N1711 gives the following info on the gain of the transistor.

    hFE = 20 (@ IC = 10 µA)
    hFE = 35 (@ IC = 0.10 mA)
    hFE = 75 (@ IC = 10 mA)
    hFE = 100 to 300 (@ IC = 150 mA)
    hFE = 40 (@ IC = 500 mA)

    The entry at 150 mA shows how much variation there could be for any given transistor. At low collector currents the gain is small. It peaks at some point, then goes down again at saturation. Given any specific current we can only interpolate what it might be at any specific collector current. So, hFE about 50 (@ IC = 6.66 mA maybe)

    VCEsat = 500 mV (IC – 150 mA, IB = 15 mA) We are well below a collector current of 150 mA so we can at least realize VCE will not go all the way to zero volts. We inject a small error when we short the transistor Emitter to collector for setting the load line.

    VBEsat = 1.3 V (IC – 150 mA, IB = 15 mA) We are well below 15 mA on IB but we can estimate our VBE to be about 600 mV.

    If you actually have a 2N1711 you can confirm these statistics.
     
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