Gain transistor

Discussion in 'Homework Help' started by SneakSZ, May 3, 2014.

  1. SneakSZ

    Thread Starter New Member

    Sep 19, 2012
    10
    1
    Hello guys,

    I'm trying to found out Av1(gain from Q4) of the schematic below.

    [​IMG]

    I thought this would be -(RinQ3//R4)/(reQ4+R7).
    with RinQ3 = (BetaQ3+1).reQ2 = BetaQ3*26mV/IcQ3.
    Took BetaQ3 as 300 and IcQ3 was 6.65 mA
    RinQ3 = 300*26V/6.65A~ 1K2
    RinQ3//R4 ~ RinQ3
    ReQ4 = 26mV/IcQ4 with IcQ4 = 0.7V/15K = 46.6µA
    = 560 Ω
    Av1 = -1K2/(1560Ω)=- 0.77

    The LTSpice result gave me a gain which was arround 0.16..
    The voltage at node4 is VcQ4
    [​IMG]

    I'm unable to found out what I'm doing wrong.

    Someone could help me out on this one?

    Thanks in advance!!

    The LTSpice file is in the att.

    Original link : http://www.zen22142.zen.co.uk/Circuits/Audio/2wamp.html
     
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    • 2wa.asc
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  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    One obvious difference is your schematic is missing R5 330 ohms.
     
  3. SneakSZ

    Thread Starter New Member

    Sep 19, 2012
    10
    1
    Hello t_n_k, well thats just some filtering of the power supply.
    At AC, it's shorted due to the source/C5 (on my schematic, R6 on zen's).
     
  4. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi SZ,
    LTSpice confirms that it is attenuated by ~-16dB at Q4 collector.

    The low Base Emitter resistance of Q3 is in parallel with R3, so the total load resistance is ~150R.

    If you add a 3K3 resistor between Q4 collector and Q3 Base you will get ~ -2dB ie: 0.77

    E
     
    Last edited: May 4, 2014
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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  6. SneakSZ

    Thread Starter New Member

    Sep 19, 2012
    10
    1
    Hello EG,

    could you tell me how you got a total load resistance of 150R?

    Kind regards!
     
  7. SneakSZ

    Thread Starter New Member

    Sep 19, 2012
    10
    1
    Hello Jony,

    yes, I saw that analyse methode before but I thought I could exclude the influence because the feedback resistor, R5, is << R4 so it sees a large impedance and therefore I excluded it.

    I thought I could calculate the Aol (exclude R5) and then R4 and R5 form the feedback so I get the closed loop gain.
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Ok so you what open loop gain ?

    AvQ4 = (R5||((β +1 )*reQ3))/(R4 + reQ4)

    reQ3 = 26mV/Ic = 3.9Ω

    reQ4 = 26mV/Ic = 330Ω

    AvQ4 = (R5||((β +1 )*reQ3))/R4 = 1.1K/ 1.33k = 0.75V/V

    But you in post one show as that LTSpice result gave me a gain which was around 0.16V/V
    But this gain shown by LTspice is a Q4 gain but with closed loop gain.

    And this gain will vary with the load. Without the load resistance the Q4 gain drop to 83mV/V.
     
  9. SneakSZ

    Thread Starter New Member

    Sep 19, 2012
    10
    1
    Siema Jony,

    thanks a lot for the insight and the analysis. I understand your post (Post 8).
    Well the gain of 0.16 for Q4 with the closed loop gain is the thing I don't understand.

    Could you explain to me how that value (0.16) is achieved?
    It's the more or less the same question I asked Eric (load impedance of the collector of Q4).

    Kind regards.
     
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I use Jacob Shekel method and math software and I find a symbolic expression for Q4 voltage gain (I simplify circuit a bit). And as you can see in this form we have no use for this expression. And this is why for such a circuit we only use numerical solution.
     
    Last edited: May 4, 2014
  11. SneakSZ

    Thread Starter New Member

    Sep 19, 2012
    10
    1
    Hello Jony,

    thanks for the reply and all the effort. I understand the analysis (used it before) and it gives indeed Voltage gain V3 = -0.158582.

    I'm trying to figure out why Eric had a total load resistance (for Q4) of 150Ω.
    This is not the same as the one 1 calculated in my first post.

    You have an idea?

    Kind regards!

     
  12. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    No, I don't have a clue. Maybe he made a mistake, you need to ask him.
     
  13. SneakSZ

    Thread Starter New Member

    Sep 19, 2012
    10
    1
    I'm unable to open the .asc, LTspice couldn't find the i_of_i_1ue voltage3.
    It's the current amplifier, any idea how to open this file or select this current amplifier from the libs (can't find it)?

    I use the latest LTspice version.
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Here you have the lib for CCCS
    http://csserver.evansville.edu/~richardson/courseware/resources/LTSpice/index.html
    voltage3 - simply replace with a ordinary 1V DC voltage source.
     
  15. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi SZ,
    Ref your PM.
    To show the open loop gain of your Amp I have removed the negative feed back resistor from the output and connected it to a 7.5V source.
    Ran an AC analysis, the gain at Q4 collector is close to your calculated value of ~ -3dB, the BF=200 for a 2N3904.

    E
     
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