Help with Transformer Coupled Transistor Amplifier

Discussion in 'Homework Help' started by darthfader, Dec 4, 2009.

  1. darthfader

    Thread Starter Member

    Jul 26, 2009
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    I have a problem here about a transformer coupled transistor amplifier. We are asked to find the over-all voltage gain of the circuit. Here's the picture of the circuit:

    [​IMG]

    These data are included:

    Bre = 2k ohms
    B = 50
    ro = 50k ohms

    I can't find any decent resource material about transformer coupled transistor amplifiers. Please help me with this. Thanks in advanced.

    Note:

    (i)I forgot to say that we are using small-signal analysis, that is, the reactances of the coils and capacitors are neglected. (we are shorting the capacitors in the ac analysis)

    (ii)the last two bypass capacitor values: 50uF and 10uF
     
  2. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    It's actually quite simple, as long as you're figuring just voltage gain, and not power gain.

    Simply ADD the transformer gain (or loss) to the gain of the two transistor stages. If the transformer is a 4:1 step up, just add 4:1 voltage gain to the total.

    eric
     
  3. darthfader

    Thread Starter Member

    Jul 26, 2009
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    Thanks for the help... But I really don't know how to get the voltage gain on each of the transistor stage..

    For example, in the first transistor, if I'm going to compute for the gain, i need to consider the reflected impedance in the primary coil of the transformer.. I don't actually know how to find that reflected impedance.

    And one more thing, I can not figure out how to draw the re model of this network. I think it would be lot easier for me if I can draw the model..

    Anyways, thanks.. I hope you could help me more..
     
  4. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    The impedance ratio of a transformer is proportonal to the SQUARE of the turns ratio, which is also the SQUARE of the voltage ratio.

    A 2:1 voltage ratio transformer is a 4:1 impedance ratio.

    Eric
     
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Are you supposed to find the gain at only a midband frequency, or must you provide a Bode plot showing gain vs. frequency over a band of frequencies?
     
  6. darthfader

    Thread Starter Member

    Jul 26, 2009
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    Okay, but in my circuit, for example in the second transformer, what would be the impedance reflected in the primary coil?

    Is it (25)(20kohm||Bre)? I'm confused with the 4kohm resistor connected with the 20kohm resistor. I don't know if it is included in the computation for the impedance in the primary coil.

    Thanks for helping me. :D
     
  7. darthfader

    Thread Starter Member

    Jul 26, 2009
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    Yes, i think we are asked to find the gain for mid-frequency.
    Thanks, I hope you could help me more. ;D
     
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    The Vcc supply has, by design, a very low impedance to ground. In other words, the Vcc terminal is the same as ground for AC signals, so the 4k and the 20k bias resistors are essentially connected in parallel from the base to AC ground.

    The intrinsic emitter resistance, re, is given by .026/Ie. You can determine the emitter current, Ie, by calculating the base voltage due to the bias voltage divider. Ignoring the base current, which is smaller than the divider current, I get a base voltage of 1.3333 volts. Subtracting the .7 volt b-e drop, we get a voltage at the emitter of .6333 volts. Dividing this by the 1k emitter resistor, we have an re of 41.05Ω. We could have also determined this from the given value of Bre of 2k, dividing by β+1 which gives a value of 39.22Ω. The oft used approximation of Bre/β gives a value of 40Ω for re. I suspect that the problem creator intended an re value of 40Ω.

    For a midband gain calculation, assume all capacitors are AC short circuits.

    The input resistance for each stage is 2k||4k||20k = 1.25k

    The approximate voltage gain for a CE stage is Rc/re, where Rc is the total resistance seen at the collector.

    Remember that an impedance is transformed as the square of the turns ratio by a transformer.

    Now can you calculate the gain of each transistor stage?
     
  9. jeff08

    New Member

    Dec 5, 2009
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    whats the formula for calculating the voltage gain, current gain, input and output impedance?
     
  10. darthfader

    Thread Starter Member

    Jul 26, 2009
    12
    0
    Wow, thanks for your great help. I haven't figured out that it is in voltage divider bias.

    Anyways, this is what I have computed, please correct me if I'm wrong.. Thanks.. :D

    Zt2 = (5)^2(1.25k) = 31.25kohm
    Zt3 = (5)^2(2k) = 50kohm

    for the gain of the two transistors:

    AvQ1 = -(zt2||ro)/re = -480.769
    AvQ2 = -(zte||ro)/re = -625

    So that the over-all voltage gain is:

    4(-480.769)(1/5)(-625)(1/5)

    or

    Av = 48076.9

    Thanks for the help.. Please check my answer if I'm wrong. :D
     
  11. darthfader

    Thread Starter Member

    Jul 26, 2009
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    But there's still one thing I'm really confused...

    [​IMG]

    Looking at the picture, when the capacitor is shorted to ground, will it not bypass the 4k ohm resistor?

    If that's the case, i think it should not be a part of the re model in the ac analysis. (I mean isn't that the loading for the secondary is just 2k||20k?)

    But the picture and the situation is really confusing, please help me figure this out.
     
  12. darthfader

    Thread Starter Member

    Jul 26, 2009
    12
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    are you referring to the problem I have posted?

    for the input impedance, Zi, i think it is Zt1 or the impedance reflected by the loading in the secondary of the transformer. It is (N2/N1)^2 multiplied to whatever the total resistance in the secondary.

    for Zo, output impedance, I believe it is Zo = RL.

    for Av, voltage gain, that's what I don't know how to solve. But Mr. Electrician is helping us figure out this one!

    for Ai, current gain, it just -AvZi/RL, so we just need to figure out the value for Av to get the answer. :D

    I hope I'm getting this right.
     
  13. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You're quite right. I overlooked that point. It's a way to avoid the AC loading effect of the bias divider that can only be done this easily with a transformer.

    Neither the 4k nor the 20k appears as an AC load on the base circuit, so you just leave then out of the calculations.
     
  14. darthfader

    Thread Starter Member

    Jul 26, 2009
    12
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    Thank you for the quick response. So you mean that the loading for the secondary of the transformer is just 2k? The Bre? But at one more glance, i think the 2k Bre is also bypassed, the problem is very confusing.

    By the way, are my calculations correct? The one I posted before. (I have included the 20k and 4k in my calculations)
     
    Last edited: Dec 5, 2009
  15. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Of course, removing the loading effect of the bias dividers means that the overall voltage gain is 4(-625)(1/5)(-625)(1/5) = 62500

    Also, the input impedance will be slightly different than would have been calculated with the bias divider as additional load.

    In post #12, you said:

    "for Zo, output impedance, I believe it is Zo = RL."

    I'm not absolutely sure what you are taking RL to be, but the output impedance Zo would be the impedance seen looking back into the transformer which is feeding the 2k resistor. If you mean RL to be the impedance seen at the collector, you need to take into account the impedance transformation provided by the output transformer. If you mean RL to be the 2k load resistor, which is what I think you mean, then be careful; the impedance seen looking back into the output transformer happens to be the same as the value of the load resistor, in this particular case. The person who created the problem set it up that way, but it wouldn't be true in general that Zo = RL for every amplifier.

    Then you said:

    "for Ai, current gain, it just -AvZi/RL, so we just need to figure out the value for Av to get the answer."

    This is true if RL means the 2k resistor, which is why I think that's what you meant in what I said above.

    A final note. The sign of the voltage gain and the current gain is ambiguous in this circuit because the polarity of the transformers is not indicated. See:

    http://en.wikipedia.org/wiki/Dot_convention
     
  16. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    The 2k Bre is not bypassed, because it's connected to the top of the secondary, whereas the bypass capacitor is connected to the bottom. Since the 4k and 20k are connected to the bottom where the bypass capacitor is, they don't appear as an AC load at the transformer secondary.

    Anything which is connected to the bottom of the transformer secondary is connected to AC ground. Anything connected to the top of the secondary appears as a load on any AC signals applied to the primary.
     
  17. darthfader

    Thread Starter Member

    Jul 26, 2009
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    wow.. Thank you for your help. I really learned a lot.

    The RL im talking about is the load 2kohm at the transformer.

    One more question, isn't the Bre bypassed by the capacitor? I think the 2kohm Bre resistor is also shorted to ground at both sides, as I have drawn in the last post. I'm still confused in this part. But I think its right that it is not bypassed. I just can't figure out why.

    And we are required to draw the re model for the solution. And this is what I have drawn:

    [​IMG]

    Is my re model correct?
    Thank you for all of your efforts to help me learn. I'm really learning a lot. :D

    Note:
    (i) I made a mistake in making the picture, the reflected impedances are not all Zt1, it should be ZT1, ZT2 and ZT3.
     
  18. darthfader

    Thread Starter Member

    Jul 26, 2009
    12
    0
    sorry, I have just read your post. It took me a while making my last post. Thanks! Everything is clear now! Thanks a lot. Now I can solve these types of problems confidently. :D
     
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