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Open / Short circuit time constants for pole approximation

Discussion in 'Electronics Homework Help' started by george2525, Nov 21, 2015.

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  1. george2525

    george2525

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    Jan 30, 2015
    Hi

    I was wondering if anybody on here has ever seen a text where open and short circuit time constants are FULLY explained. I have been trying to learn the method of obtaining them but they seem to veer from being very easy to very difficult.

    i am talking about using them in amplifier analysis for the internal transistor capacitances (e.g Cpi) which come into effect at high frequencies.

    I would very much like to see these derived in full and not just stated (often with the patronising pre-text of ''we can clearly see'')

    So if anybody knows of a good resource (Not sedra smith) that takes the time to walk you through at least a couple of hard examples I would be very happy.

    Thanks
     
  2. george2525

    george2525

    165
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    Jan 30, 2015
    Here is a perfect example of one

    Capturesedra1.PNG sedra2.PNG


    If someone could derive this Rpi for me I would be VERY happy. I have naturally checked the referenced MOSFET case but that example makes no attempt to show any steps regarding how the answer was obtained. i find this very frustrating.

    Thankyou
     
  3. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    Here is "trivial" source on the method for you.
    Try it yourself so you can learn,the explanation is clear and the method not that difficult .
    Post here if you have problems...I promise we shell help :)
     
  4. george2525

    george2525

    165
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    Jan 30, 2015
    Hi,

    yes I understand the concept and whenever it is taught with examples like that I find it easy. My problem is when an example like the one I posted is stated without explanation I have no idea how that resistance was found!

    for example where does that re come from? its not even drawn in the small signal equivalent! I was under the impression that re was from the T-model BJT. So has the person just used T-model and hybrid pi interchangably without reference?

    even still I cant work this out.

    When I attach a test ''Vx'' to the Cpi terminals I cant work out all the currents in terms of Vx and ix

    Even if I could - the analysis would not include re unless it is derived from components in the circuit. I am aware of the reflection rule but have no idea if this was used here. there is no beta anywhere in the answer! perhaps it cancelled but as you can see im confused about where to start.

    see the example I posted. If i understand this one I think it may help a lot with the others that I cant find.

    Perhaps I have missed some basic transistor terminal resistance concepts?

    my small signal analysis was always ok until this time constant business and the internal capacitances!
     
  5. dorke

    dorke

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    Jun 20, 2015
    I think they just used the relation re=1/gm.
    Makes any sens?
     
  6. george2525

    george2525

    165
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    Jan 30, 2015

    perhaps yes. could you confirm this for me - the resistance going through the transistor is different going one way than the other - IE through the base is not the same as through the emitter. so is the resistance between Cpi terminals the same no matter which way you go?

    sorry that may sound stupid but an answer would help me a lot
     
  7. dorke

    dorke

    2,342
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    Jun 20, 2015
    I substituted gm with 1/re
    This the circuit you need to solve for Rpi

    Here is the correct Circuit:
    Rpi.JPG
     
  8. Ratch

    Ratch

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    Mar 10, 2013
    Since you made a valiant try at determining the resistance that C_pi "sees", you deserve to see how it is done. Unlike the book, you can ask me to explain any facet of the solution you don't understand. You are going to have fun studying the solution.

    Ratch

    George2525A.JPG George2525B.JPG
     
    Arouse1973 and LvW like this.
  9. george2525

    george2525

    165
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    Jan 30, 2015
    thanks, that is a method i have never seen

    i managed to solve it with kirchov laws in the end. I had been doing silly things with my simultaneous equations and not paying attention to which ones had been used already...

    but

    when you say ''open circuit voltage / short circuit current with Vpi removed'' at the start are you measuring the voltage at B'?

    not sure which terminals you are shorting!

    why remove Vpi?

    sorry this method is alien to me. been taught to always short v sig
     
  10. Ratch

    Ratch

    1,089
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    Mar 10, 2013
    At the start, I am calculating the open circuit voltage at B' and Vo. That is where the capacitor Cpi is connected. I should have said I removed rpi instead of Vpi. I need an open circuit to calculate the Thevenin impedance. Thevenin impedance is open circuit voltage divided by short circuit current. After the Thevenin impedance is calculated, then rpi is added in parallel to get the total impedance Rpi . I don't know why you would short Vsig. That is one of the voltages that activates the transistor.

    Later edit: I think I see now why you would short out Vsig. I think you are trying to find the Thevenin impedance by shorting the voltage sources, opening the current sources, and calculating the resistance between the two opened points of the circuit. That method is simple, but only works if no dependent sources are present. Since the voltage controlled current source gm*Vpi is a dependent current source, that method will bite you in the butt if you use it. By calculating the Thevenin resistance by (open circuit voltage)/(short circuit current), all sources including dependent sources are taken into account, and a correct resistance is always obtained.

    Ratch
     
    Last edited: Nov 23, 2015
  11. dorke

    dorke

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    Jun 20, 2015
    george2525,
    It is clear that the solution Ratch gave you is not according to the method discussed (quote: "I don't know why you would short Vsig"...),it is the conventional way.


    1.from my previous drawing I replaced vbe back to be the original Vpi.
    2.Since rpi is a parallel resistor it will be added in parallel at the end.
    3.The current source Vpi/re and R'L are replaced with their Thevenin's equivalent
    4.The resistance R'pi is the voltage Vpi divided by the current i(blue loop).

    solving for i (KVL):

    i= (Vpi*R'L/re+Vpi) /(R'sig+R'L)=Vpi*(R'L/re+1)/(R'sig+R'L)

    R'pi=Vpi/i = (R'sig+R'L) / (R'L/re+1)

    now we add rpi in parallel
    Rpi=R'pi || rpi ==> the solution.

    See no Vsig,.as advertised...;)

    Rpi.JPG
     
    Last edited: Nov 23, 2015
  12. george2525

    george2525

    165
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    Jan 30, 2015
    Thanks guys.

    I managed to use bits from your ideas to solve it. those methods look good and interesting but i will have to analyse them when I have more time. got an exam soon.

    I solved it by shorting Vsig , getting rid of Cu (open) and putting a Vx source where Cpi is

    then I did Rpi=''Rx''=Vx/Ix and it worked
     
  13. dorke

    dorke

    2,342
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    Jun 20, 2015
    Nice to hear it finally worked for you:)
    BTW,
    which book is it from?
     
  14. dorke

    dorke

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    Jun 20, 2015
    Ratch,
    This Open-circuit time constant method is supposed to be an intuitive approximate way to analyze frequency response(quick and dirty in a sense).

    And, you are absolutely correct :
    only independent voltage/current sources can be "eliminated".
    This isn't mentioned in the "trivial source" below,
    nevertheless it is solvable and somewhat simpler than the conventional way,
    as I have shown above

    Read about the method here("trivial source")
    Open-circuit time constant method
     
  15. Ratch

    Ratch

    1,089
    331
    Mar 10, 2013
    ,
    I make no representations for the Open-circuit time constant method. I only calculated the Thevenin resistance. as requested by the OP and the book, of the B'--Vo terminals with rpi removed. I did this by using the combination Thevenin-Norton theorem method, which looks at the voltage and current outward from the circuit instead of inward toward the circuit like the single Thevenin or Norton theorem does. The advantage is that the combination Thevenin-Norton method can be used for dependent sources, whereas the Thevenin or Norton theorem cannot. Merely divide the open circuit voltage across the terminals by the short circuit current through the terminals to get the Thevenin resistance.

    Ratch
     
    Last edited: Nov 24, 2015
  16. george2525

    george2525

    165
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    Jan 30, 2015
    Its from microelectronic circuits by sedra/smith. Its a good book in some ways but more for middle to advanced learners.
     
  17. The Electrician

    The Electrician

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    11
    Jul 6, 2012
    Your result doesn't appear to be correct.

    To make it easier to compare your solution with the book solution, put both over a common denominator:

    R1.png

    Your solution isn't the same as the book solution; it has an extra term, re*R'L, in the denominator.
     
  18. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    The Electrician,
    You are correct my result in this form is an approximation:
    Lets divide the last eq. by re*rpi we get:
    [R'L+R'sig]/ [1+R'sig/rpi +R'L(1/re+1/rpi)] equation - ***-

    Note that rpi=re(beta+1) thus 1/re>>1/rpi
    so the 1/rpi term can be omitted and we get the book's result.

    [R'L+R'sig] / [1+ R'sig/rpi +R'L/re].

    Note also that the approximation started in my reply #7 with
    "I substituted gm with 1/re" .

    To get the exact result:
    replace back 1/re with gm in the equation -***-

    (1/re+1/rpi)= gm+1/rpi
    note that gm=beta/[(beta+1)*re] ; 1/rpi=1/[(beta+1)*re]
    inserting gm and 1/rpi we get:
    beta/[(beta+1)*re] + 1/[(beta+1)*re]=1/re

    the exact term. ;)
     
  19. The Electrician

    The Electrician

    116
    11
    Jul 6, 2012
    Are you still interested in a reasonably full explanation?
     
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