Complicated bjt circuit

Discussion in 'Homework Help' started by acelectr, Apr 6, 2011.

  1. acelectr

    Thread Starter Member

    Aug 28, 2010
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    Well I dont know how would it be complicated for the reader but for me it is quite hard. The schematic is attached.
    I've apporixemately I think draw the equivalent high frequency circuit model. Firstly Rin is asked. What I dont understand is that at both bjts a capaciter Cpi will be in parallel with the resistance rpi. So how would this effect Rin? I dont think that one can neglect Cpi in such case. A hint would be great.
    Also inorder to find the midband gain where can I start from exactly? The equivalent high freq circuit model gets huge and complicated. I just could not relate the vsig with the second bjt circuit.

    Thanx for any help.
     
    Last edited: Apr 6, 2011
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    It's likely that nobody here has your textbook or has taken your classes, so we don't know where on the schematic Cμ or Cje should be places. You'll have to show them before anybody can help.

    The way to solve this problem is by setting up the circuit equations. Have you studied any of the standard network solution methods, such as nodal or mesh equations? Can you give one of them a try?
     
  3. acelectr

    Thread Starter Member

    Aug 28, 2010
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    heh:D no this one is not that simple to solve only with meshes or node voltage techniques. Cje and Cu should be universal things.There is such a thing called a high frequency equivalent circuit model for such cases. The thing is I can apply it to a single bjt circuit but when connecting another ona I'm crashing.
    I am also studying on semiconductor devices, and also I've seen there these internal caps. They are not things that only valid or re-named differently specifically at my lectures that I am attending or the book.
    I don't think that this topic is quite advanced and above this forum topics.
    Still waiting for good ideas, thnx.
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    If it's "...not that simple to solve only with meshes or node voltage techniques.", how else would you expect to solve it?

    It's not that difficult to solve; it can be done using the technique I described in this thread:

    http://forum.allaboutcircuits.com/showthread.php?t=26710&highlight=shekel

    Also, when you're asking for help and the person who would help you asks for further information, rather than saying it should be universal and expecting that other person to look it up, the appropriate thing to do is give the information asked for. If you can't use a graphics program to add it to your schematic, then at least describe where each capacitor I asked about is connected. For example, is Cμ connected from collector to base of the first transistor?
     
  5. PRS

    Well-Known Member

    Aug 24, 2008
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    This is problem that requires you to search your text for certain relationships. The first thing to notice is that the input transistor is an emitter follower, and the second is a common emitter.

    Look through your text and find out what is the equation for Rin? Not in terms of voltage or an emitter resistor, but in terms of emitter current. The way I remember it rpi, the input resistance to transistor at the input is rpi = re(Beta + 1). And re is a parameter of the substrate and the temperature. At room temp it is re = Vt/Ic. At room temp and at 1 mA Vt = 26mV. And you know the relationship between Ic and Ie. If not look it up in your book.

    So Rin = re(Beta +1), but look at the circuit. This resistance is in parallel with the input resistance of transistor stage 2. Just by looking at it, that resistance is small. It's given as the same equations for the above except it is reflected through Q1.

    Put that stage 2 input resistance in parallel with the stage 1 input resistance and it is even smaller. The overall input resistance is Rin1//(Beta1+1)Rin2.

    The Beta + 1 comes from the fact that the input resistance of stage 2 is reflected through stage one by this amount. Your text should show you why.

    Well, I'm out of gas. If you like this post and want more, just say so.
     
  6. Ron H

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    Apr 14, 2005
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    With no transistor parameters other than Hfe, how can you calculate midband gain? If Q2 is modeled as a voltage-controlled current source, with unspecified (infinite?) collector resistance, and a current source load, isn't the gain going to be infinite at zero frequency, with the 1pF load cap creating a pole at the origin?
     
  7. acelectr

    Thread Starter Member

    Aug 28, 2010
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    Look, firstli appreciating the concern but I've thought one that sipmply never heard of Cje or Cu can not know about or never seen equivalent bjt models. I can not just give instructors and teach these concepts all over again, I would expect ideas from ones that were familiar with this topic.
     
  8. PRS

    Well-Known Member

    Aug 24, 2008
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    You may be right, Ron. Let's walk this problem through, just for kicks. Both xistors are biased at .1mA. We don't have the actual specs for the xistors, so let's just use re=26mV/.1mA and we have about 260 ohms. We reflect this to the base of Q1 and, given Beta is 100 we have about 26000 ohms.

    But the overall Rin is this resistance in paralllel with the reflected resistance from Q2. Given the same current, we get the same number at the input to Q2. Now we have a resistane at Re1 of 26000 in parallel with 100 times this amount and so the overall Rin is about 25000 ohms.

    Resulting Question for all of you: Given that .1mA gives such a high input resistance, why don't we use .1mA more commonly than 1 mA?
     
    Last edited: Apr 7, 2011
  9. Ron H

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    The resistors are not in parallel, they're in series. The input resistance is about Rin=(β+1)*(260+(β+1)*260), or about 2.68 Megohms.
     
  10. Ron H

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    Apr 14, 2005
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    If you want to limit your responses by not answering our questions, that's your choice. You come off as being somewhat arrogant. Most people who want help show up with their hat in their hands.
     
  11. acelectr

    Thread Starter Member

    Aug 28, 2010
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    check again Hfe is 100.
     
  12. acelectr

    Thread Starter Member

    Aug 28, 2010
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    I am sry things are quite intens here so I can not get back to all responses. I dont understand where I've been arrogant. If did any mistakes again apologizing...

    The things that complicates this problem are the internal caps. rpi is in parallel with an internal cap Cpi. Taking an equivalent impedance of this parallel relation now 2 paths occur. One goes from the equivalent impedance to the collector part of Q1 (which in between there are a bunch of other stuffs), the other to emitter of Q1 and so to the base of Q2. This is the critical part that kills me. Somehow I need to relate the voltage across rpi with anything at Q2 so I can carry it to the output.
    I hope I explained things clear enough.
     
  13. Ron H

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    I said,
    I guess you missed it.
     
  14. Adjuster

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    Dec 26, 2010
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    The OP needs to use all of the data he is provided with. In the original question a value of 100V is given for the Early voltage VA

    This allows the output resistor ro of the hybrid-pi model to be computed: ro ≈ VA/IC. With VA = 100V and IC = 100μA, ro =1MΩ, a high value.

    The OP can then complete his hybrid-pi models and analyse the result. http://en.wikipedia.org/wiki/Hybrid-pi_model

    The mid-band gain can thus be found, as can the corner frequency associated with the capacitance in the output circuit.
     
  15. Ron H

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    Apr 14, 2005
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    I missed the Early voltage.
     
  16. Heavydoody

    Active Member

    Jul 31, 2009
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    I am not sure why the op didn't want to share, but, in case anyone is interested (I was) I found this enlightening http://userwww.sfsu.edu/~sfranco/CoursesAndLabs/Labs/445LabsPDF/445Lab5.pdf
     
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