differential amplifier analysis

Discussion in 'Homework Help' started by TheSpArK505, Sep 29, 2016.

  1. TheSpArK505

    Thread Starter Member

    Sep 25, 2013
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    Hi guys, hope u fine.

    I have encountered a differential circuit of two transistors Q1 and Q2,
    upload_2016-9-29_21-19-52.png

    The first sentence of the answer says we assume Q1 off and Q2 on. Can I know what's this assumption based on and when am I allowed to make assume one transistor to be on and the other is off. Also how to validate my assumption?

    Thank you in advance.
     
  2. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    The problem writer assumed this was apparent from inspection.
    The normal Vbe of a BJT is about 0.7V.
    Since the base of Q1 is biased at +0.5V, the emitter voltage would have to be about 1.2V for Q1 to be on.
    But the emitter voltage can only be about 0.7V since the base of Q2 is grounded, turning on Q2.
    Thus the base-emitter junction of Q1 has only about 0.2V forward bias which is not sufficient to turn it on.
     
    Last edited: Sep 29, 2016
  3. TheSpArK505

    Thread Starter Member

    Sep 25, 2013
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    0
    So I'm confused, I can get two values of VE, each one can turn ON one transistor, so which value should I go with for VE, is it 1.2 V or 0.7 V and WHY??
     
  4. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    You are confused because that isn't what I said. It's not either or.
    I said you would need a VE of 1.2V to turn on Q1.
    But it never gets there because the base-emitter junction of Q2 keeps VE at about 0.7V with its base grounded.
     
  5. TheSpArK505

    Thread Starter Member

    Sep 25, 2013
    92
    0
    so why the grounded VB has the advantage on 0.5 VB when it comes to calculating VE
     
    Last edited: Sep 30, 2016
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,098
    Did you ever see a BJT with Vbe =1.2V ?? (what collector current will flow ?)

    But in real life all you need is to do a single KVL around the input.
    BJT11.png
    And notice that Vin = Vbe1 + Veb2 (from KVL loop). And this means that as Vbe1 increase Vbe2 must decrease by the same amount.
    For example if Vbe1 changes from |0.60V| to |0.61V| the Vbe2 will drop by |0.01V| from |0.6V| to |0.59V|
    And for your case if Vin = 0.5V we have Vbe2 = |0.7V| and Vbe1 = -0.2V. This mean that Q1 is cut-off. In fact any differential input voltage larger then 100mV will cut-off one of the transistors.
    http://www.te.kmutnb.ac.th/~msn/224510diffamp.pdf (page 4)
     
  7. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492

    The base emitter voltage is assumed to be a constant 0.7v when forward current flows through it.
    Thus, you can temporarily replace the base emitter of Q2 with a 0.7v battery, with positive at VE and negative at the Q2 base. That puts +0.7v at VE referenced to ground.

    Q1 needs the same base emitter voltage, 0.7v, but it can not get that because the base is connected to a 0.5v source. That makes the Vbe of Q1 only 0.2v which is too low to allow it to turn on. Transistors dont start to turn on until around 0.4v but once biased correctly they show around 0.7v.

    So basically you have to figure out the differential voltages across the base emitters of both transistors to see why one happens to be on and one off.
     
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