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Simple circuit with a transistor

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Here's a crude description of the problem until my attachment is approved: it's essentially a voltage divided connected to a common-emitter amplifier (I think). There's a +10 V line connected via an 8.4 kOhm resistor to the base of the NPN transistor, and a 1.6 kOhm resistor connecting the base to ground. Then there is a resistor R_load connecting the +10 V line to the collector of the transistor, and a 1.0 kOhm resistor connecting the emitter of the transistor to ground. I'm supposed to calculate the current through R_load, and say whether the result depends on the value of R_load. We're assuming the simplest model of a transistor: the base current is always 0, and when the B-E junction is forward-biased at least 0.6 V, then current flows from C to E, where V_E = V_C - 0.6.


    2. Relevant equations
    When V_BE > 0.6:
    I_C = I_E and V_E = V_C - 0.6


    3. The attempt at a solution
    I think V_B = V_in*(1.6/(1.6+8.4)) = 1.6 V, so the B-E junction is forward-biased, and current flows from the collector to the emitter. Then V_C = 10 - I_C*R_load, but I_C = I_E so V_C = 10 - I_E*R_load. Then V_E = V_C - 0.6, so V_E = 9.4 - I_E*R_load. Then I_E = (9.4 - I_E*R_load)/1000, so I_E = 9.4/(1000+R_load). But I don't think this is right, because I'm pretty sure the current through R_load doesn't depend on the value of R_load. What am I doing wrong here?
     

    Attached Files:

  2. jcsd
  3. Feb 17, 2009 #2
    Last edited by a moderator: May 4, 2017
  4. Feb 17, 2009 #3

    LowlyPion

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    Isn't your Rload constrained to be greater than 9kΩ ?

    If the current through the 1kΩ resistor is greater than 100 ma doesn't the C-E open because the base is at 1.6v?
     
  5. Feb 17, 2009 #4
    I'm not sure I follow.. why is R_load constrained to be greater than 9kΩ?
     
  6. Feb 18, 2009 #5

    LowlyPion

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    Maybe I'm not understanding, I just thought that you said the voltage from the base to the emitter needed to be greater than .6v. Since the base is held at 1.6v I thought that meant the Ve couldn't be greater than 1v. So long as the Rload is greater than 9kΩ then the Ve will be <1v.
     
  7. Feb 18, 2009 #6

    Delphi51

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    I shouldn't comment, having done nothing with transistors for many years, but it seems to me the absence of any specs on the transistor - such as its beta - means a crude answer is desired. I was taught that the emitter-base voltage on a working transistor is always 0.6 V. The voltage divider on the base puts the base at 1.6 V. So there must be 1 volt on the 1k emitter resistor. That determines the current regardless of the collector resistor!
     
  8. Feb 18, 2009 #7
    That's right - we're using a very simple model right now, where beta is ignored and we assume that no current flows from base to emitter. What you say makes sense though - so if there's 1 volt on the 1k emitter resistor, then I_E = I_C = 1/1000 = 1 mA, regardless of the value of R_load.
     
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