Collector feedback T model Help

Discussion in 'Homework Help' started by mayiru, Feb 6, 2009.

  1. mayiru

    Thread Starter New Member

    Feb 6, 2009
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    Okay, here is the circuit model guys.
    (T Model/re model/r parameter model)
    [​IMG]

    Original circuit :
    [​IMG]

    Now, I've got to derive expressions for Zin, Zo, Av, Ai using the r parameter model.

    The answers are
    Zin = re/(1/β + Rc/RF)
    Zo = Rc||RF
    Av = -Rc/re
    Ai = βRF/(RF + βRc)

    I am not clear exactly on the concepts involved, though I got the expression for Zo, and Ai. (I dont know how RF comes into the out put impedance though).

    Please could someone derive the expressions? Would appreciate some help.
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Show your work.

    Show how you got Zo and Ai.
     
  3. mayiru

    Thread Starter New Member

    Feb 6, 2009
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    As I told you before, I just assumed the parameters to get the answer.

    Zo = Rc||Rf... maybe because βre' is neglected?

    Av = Ic(Rc)/Ib(re) and hence the result.

    No clue about the rest.
     
  4. mik3

    Senior Member

    Feb 4, 2008
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    Use Miller effect to slit Rf into two resistances, one in parallel with re and one in parallel with Rc.
     
  5. mayiru

    Thread Starter New Member

    Feb 6, 2009
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    Could you please provide a detailed explanation... the derivation if possible? I don't get this concept clearly.. especially this model.
     
  6. mik3

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    Feb 4, 2008
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  7. The Electrician

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    Do you know how to solve networks, using KCL or KVL?

    An important aspect of this problem is to recognize that the input and output impedances of a two port like this are calculated with the assumption that the other port is short circuited.

    The expression for Zin is easy to derive when you consider that the output port is shorted. Zin is just the parallel combination of βre and RF. If you do the algebra you will find that the result is: Zin = re/(1/β + re/RF), which is not the same as you have in the first post.

    If you assume the input port is shorted, then clearly Zo is just the parallel combination of Rc and RF, or Rc||RF.

    Av is calculated assuming that a voltage source of, say, 1 volt (zero internal impedance) is applied to the input port, and the output voltage is calculated (using KCL, for example). Then the ratio Vo/Vi = Vo/1 = Vo is Av.

    For the current gain, the output port is shorted and a current source of, say, 1 amp is applied to the input port. The current in the output short is calculated, and the ratio Io/Ii = Io/1 = Io is Ai.

    I will tell you further that the expression you have in the first post for Ai;

    Ai = βRF/(RF + βRc)

    is not correct; it should be:

    Ai = βRF/(RF + βre)
     
  8. mayiru

    Thread Starter New Member

    Feb 6, 2009
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    @electrician - Yeah, I do know how to solve networks using Kirchoff's laws. But then I am still not sure about what is Zin and Zout. For Zo, the impedance obtained by looking into the output side, shouldn't it be Rc ||(Bre + RF) ? Why have you ignored RF, when its connected to Bre? For Zin, I thought it would be Bre || RF, but then I didn't get the expression you've mentioned. Would be really grateful if you could be a bit more elaborate with your derivations. And for Av do I short the + and - of Vo?FOr Ai, where do I put the current source?Between + of Vi and RF?

    As for the answers, these questions were given as an assignment, and they are correct I guess, as another site on the internet gave the same results, though the derivations were not there.

    Waiting for your reply
     
  9. The Electrician

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    Can you post the URL here so I can check it out.

    Zo is defined like this; apply some voltage, any voltage, as Vo and calculate (or measure if it's a real circuit) the current Io produced by that voltage. Then Zo is Vo/Io. It does depend on whether or not the input port is shorted. If the input is shorted then it's not Rc ||(βre + RF) because the short on the input port means βre is shorted; the left end of RF goes to ground. Thus Zo is just Rc||RF; you can see that, right?

    If the input port isn't shorted, then Zo is much more complicated; I'm not going to go through the derivation in this post, but it is:

    (Rc*(RF + βre))/((β+1)*Rc + RF + βre)

    In these sorts of problems, to get the relatively simple expressions, assumptions are made about the various resistances and gains. It's clear that here it's assumed that Rf >> re and RF >> Rc.

    In order to get the more complicated expression for Zo to simply be Rc||RF, it is necessary to assume that β is zero (or, equivalently, that βre is negligible with respect to Rc and RF), which would mean that the resistor βre is zero (or nearly so), a (near) short on the input.

    Look at the expression for Zin that you gave:

    Zin = re/(1/β + Rc/RF)

    Look what happens to that expression as Rc gets really big; the denominator becomes very large and Zin goes to zero. Do you really believe that if you remove Rc (let Rc -> infinity, in other words), the input impedance will become zero, a short? This doesn't pass the smell test.

    But if you replace Rc in the Zin expression you gave with re, the expression becomes Zin = re/(1/β + re/RF), which is exactly the same as RF||βre, and is what you get if you assume the output is shorted.

    If the output is not shorted, the expression for Zin becomes more complicated:

    Zin = ((βre*(RF + Rc))/((β+1)*Rc + RF + βre)

    If you assume that Rc -> zero (a short on the output), then the expression becomes RF||βre.

    I assume there's no need to repeat what I said about how to solve for Av and Ai.
     
  10. mayiru

    Thread Starter New Member

    Feb 6, 2009
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    http://notes.ump.edu.my/fkee/BEE2233/Chapter 1 BJT Small Signal Analysis.ppt

    That's the link. Scroll to the slide having the collector feedback circuit and see.

    I got how to solve for Av, but could you solve for Ai? Do I need to short the o/p while solving for Ai?

    Anyway, have a look at that slide first.

    For Ai, when I short the output port, RF and Bre aren't in parallel right? Thats cause there is a current source, and we don't know the drop across it. So the drops across Bre and RF wont be same.

    In that case, solving for Io becomes nasty.
     
  11. The Electrician

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    Oct 9, 2007
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    Here's how you solve for Ai. Assume that you are injecting 1 amp into the input port. By letting the injected current be 1 amp, we don't have to deal with that number.

    There will be two paths by which current will flow into the output short; through RF and by means of the current source.

    Use the current divider rule (http://en.wikipedia.org/wiki/Current_divider) to calculate the current through RF.

    It is βre/(βre + RF)

    Also by that rule, the current through βre is RF/(βre + RF), and multiplying by β we get the current into the current source: (β*RF)/(βre + RF).

    The convention is that currents into a port are positive and currents out of a port are negative. The current through RF will be negative and the current into the current source will be positive.

    So the total output current is (β*RF)/(βre + RF) - βre/(βre + RF)

    Simplifying we have:

    (β*(RF - re))/(βre + RF)

    If re << RF, then we can ignore re and we have:

    Ai = (β*RF)/(RF + βre)

    The expressions for Zin and Ai on that web site are wrong; they have Rc where they should have re.
     
  12. mayiru

    Thread Starter New Member

    Feb 6, 2009
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    Ok, I got that expression for Ai, but didn't do that approximation. The only thing that pinches me is that the answer given by my teacher and that slide match. Does anyone here know the typical value for RF, as with that, I can judge which expression is right.

    Thanks to electrician :)
     
  13. The Electrician

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    Oct 9, 2007
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    Well, you could always just use the slightly more complicated expression for Ai:

    Ai = (β*(RF - re))/(βre + RF)

    But, here's an important point. Whether you make an approximation or not, Rc doesn't appear in the expression for Ai. When the output is shorted, Rc is shorted; that's why it doesn't appear in the expression for Ai. You should ask your instructor about this, and get him to do the derivation with you. Ask what the definition of current gain is, and whether the output should be shorted when calculating it.

    It occurs to me that perhaps they define Ai as the ratio of the current in Rc to the input current, with the output otherwise open-circuited. Let's do that calculation.

    If the input impedance is as given on the web page, Zin = re/(1/β + Rc/RF), then if we inject 1 amp into the input, the voltage at the input will be:

    re/(1/β + Rc/RF)

    If we then multiply this by Av we can find the voltage at the output:

    re/(1/β + Rc/RF) * (-Rc)/re = -(β*RF*Rc)/(β*Rc+RF)

    If we divide this by Rc to get the current in Rc, this will be the current gain Ai, reversing the sign to account for the fact that the current from the current source is out of Rc: (β*RF)/(RF+β*Rc)

    This is the expression from the web site, but there is an error in the expression for Zin, where they have Rc instead of re.

    If you re-do the calculation for Ai that we just did, but using re/(1/β + re/RF) as the expression for Zin, the final result for Ai is (β*RF)/(RF+β*re).

    So, you see that the error in the expression for Zin propagates through to the result for Ai if you use the method just employed to calculate Ai, starting with Zin. And this is without assuming that the output is shorted, as is the usual convention. So, either way, I think their expression for Ai is wrong. Maybe your instructor just took the result from the web page without checking it out himself.

    If you do check this out with your instructor, and you should, please report back to this thread what you find out.

    ----------------------------------------------------------------------
    If you look at slide 4 in the Power Point presentation at the web site, you'll see the well known formula for re:

    re = .026/Ie

    Just in front of slide 4 is a transistor data sheet. A typical amplifier of this sort might have an emitter current of 1 mA; in that case, re is 26 ohms. With an emitter current of 10 mA, re will be even smaller at 2.6 ohms.

    A typical RF will be on the order of kilohms.
     
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