Common emitter amp input impedance

Discussion in 'Homework Help' started by nik2009, Oct 4, 2011.

  1. nik2009

    Thread Starter New Member

    Jan 17, 2010
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    Hello guys,

    I'm learning how this simple circuit works just for a hobby.
    The circuit is a common emitter amp with fixed bias and grounded emitter.

    [​IMG]

    I'm trying to calculate its input impedance. As I understand the input signal current flows through Q1 base-emitter (re) to the ground and through R2 to the ground.

    So:
    Zin = R2 || βre

    β = 100
    re = 25/IE Ohm ≈ 25mV/50mA = 0.5 Ohm

    Then:
    Zin = 50KOhm || 50Ohm = 49.95Ohm

    But when I run a simulation the current through the input voltage source V1 is
    56μA PP.

    [​IMG]

    As V1 voltage is 5mV this gives
    Zin = 5mV / 56μA ≈ 89.29Ohm

    which is far away from 49.95Ohm I get from my calculations.

    Could anyone help to figure out what am I doing wrong?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    That's a hefty emitter current. Reduce the current by a factor of 10 and review the results. Set Rb=500K and Rc=2.5k. And put some resistance (say 1k) in series with the source to improve the linearity.
     
    Last edited: Oct 4, 2011
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Simply the current gain "beta" of your BJT is not equal 100.
    You must remember that beta is not a constant.
    The BJT current gain is very heavily collector current dependent.

    So the Zin = Xc + RB||(β+1)*re

    In you case we can plot the input impedance with the help of a ACsweep analyze
     
    nik2009 likes this.
  4. nik2009

    Thread Starter New Member

    Jan 17, 2010
    13
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    Hello t_n_k,

    I changed the circuit in the way you suggested.
    [​IMG]

    For this circuit I calculate the input impedance like this
    Zin = Rs + (Rb || βre)

    β = 100
    re = 25mV/IE ≈ 25mV / 5mA = 5 Ohm

    Zin = 1K + (500K + 100 * 5) = 1499.5 Ohm

    But when I run a simulation it show 6.4μA flowing through the input source V1. The V1 voltage amplitude is still 5mV so this gives

    Zin = 5mV / 6.4μA = 781.25 Ohm.

    [​IMG]

    Almost two times less than the result of my calculations. I think there's a mistake in my calculations by I can't see where exactly.
     
  5. nik2009

    Thread Starter New Member

    Jan 17, 2010
    13
    0
    Thank you, Jony130.

    I'm going to plot β with the help of ACSweep.
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For Vin = 5mV and F = 100Hz the capacitor reactance is equal to
    Xc = 1/( 2 * pi * F * C1) = 159.154943Ω ≈ 160Ω
    And the collector current is equal Ic = 4.837mA and Ib = 48.37μA
    For Rb = 500K; Rc= 2.5K and Vcc = 25V

    β = 4.837mA/48.37μA = 100

    Zin = Rs + Xc + (Rb|| (β+1)*re = Xc + 1K + 500k||101*5.16Ω =
    = Xc + 1KΩ + 500KΩ||522Ω = Xc + 1.521KΩ

    so the final

    Zin = √(1.52K^2 + 160^2) = 1.52KΩ

    The LTspice give me 1.54KΩ
    And form the transient analyze
    The input current is equal to 3.2μA
    Zin = 5mV/3.2μA = 1.56KΩ
     
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  7. MrChips

    Moderator

    Oct 2, 2009
    12,420
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    The circuit as drawn has poor temperature stability. The top end of Rb should be connected to the collector of Q1.
     
  8. nik2009

    Thread Starter New Member

    Jan 17, 2010
    13
    0
    Done it.

    For this circuit I estimated
    Zin = √(Xc^2 + (Rs + (Rb || (β + 1)re))^2)
    β = 100, Zin = 1527

    But the results from a simulation

    [​IMG]

    give 5mv / 3.9μA = 1282 Ohm


    I've read that negative feedback decreases the input impedance of a circuit. And there's a negative feedback in this circuit from the collector to the base through Rb.

    So I guess the actual Zin is less than estimated because of this feedback. Is it correct?
    (If it's correct that Zin is decreased because of the feedback, where can I find an explanation of why this happens? I know chapter 4 of the Art of electronics talks about feedback, should I search there?)
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    Ignoring the capacitive reactance and the source resistance for the moment, a reasonable estimate for the resistance "looking into" the base would be

    R_base = (1+β)re||Rb/(|Av|)

    Where |Av| is the small signal voltage gain magnitude going from from base to collector.

    In this circuit configuration |Av|≈Rc/re if Rc<<Rb {More accurately |Av|=(Rc[1/(1+Rc/Rb)])/re if β>>1}

    So yes, the input resistance looking into the base decreases due to the negative feedback provided by Rb, with the effective AC equivalent resistance of Rb being reduced by a factor of 1/|Av|.
     
    Last edited: Oct 5, 2011
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    1,097
    Yes , you right.
    Zin is smaller because you use Rb as a feedback resistor.
    Output voltage sampling - shunt (parallel) connections to the input--> shunt-shunt topology.

    And this imagine will help you understand why this type of feedback reduce the input impedance

    [​IMG]

    I1 = (1V - (-10V)/10Ω = 1.1A
    So the Zin = 1V/1.1A = 0.9Ω

    Or if we use |Ao|

    Zin = Rb/(Ao+1)

    Where Ao is the voltage gain.

    Your BJT amp has the voltage gain equal to:

    Ao ≈ Rc/re = 500V/V

    Zin ≈ Rs + Rb/(Ao+1)||(β+1)*re = 1KΩ + 500Ω||500Ω ≈ 1.25KΩ
     
    nik2009 likes this.
  11. nik2009

    Thread Starter New Member

    Jan 17, 2010
    13
    0
    Thanks, t_n_k and Jony130!
    You were very helpful.
    Jony130, especially you.
     
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