1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook