Emitter Bypass Capacitor Value in Art of Electronics

Discussion in 'Homework Help' started by jonnylazer, Aug 3, 2010.

  1. jonnylazer

    Thread Starter New Member

    Jan 12, 2010
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    Hi people,

    I have a question regarding the calculation of the emitter bypass capacitor in a common emitter amplifier.

    I am specifically referring to the circuits on pages 84 and 85 of the Art of Electronics.

    I would like to know why this capacitor should have a small resistance (reactance) value compared with the intrinsic emitter resistance (re) at the lowest signal frequency of interest.

    thanks
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    Basically, you want to set the cutoff frequency (pole) below frequency band of interest. This frequency is set by the effective resistance "seen" by the capacitor. Effectively the value of "re" plus any AC source restistance (call it Rs, which assumes the input capacitor is effectively a short for the AC signals of interest) divided by hfe, is in parallel with Re. If Re is large and Rs/hfe is small compared to re, then it is acceptable to consider only the value of re as significant. The bias resistors R1 and R2 are also in play, but they are usually too large to worry about. There are basically many assumptions here and they are being glossed over in the book, but all assumptions need to be checked in the particular case you are dealing with.

    Lets look at in in more detail to avoid the confusion of my words above. The effective resistance seen by the capactior Reff is as follows.

    Reff = Re || ( re + ( Rs || R1 || R2 ) / hfe ), where "||' means "in parallel with".

    Generally the bias resistors R1 and R2 are large compared to the coupling capactior reactance and AC voltage source impedance Rs. This leaves the following:

    Reff = Re || ( re + Rs / hfe )

    Generally Rs is small and then it gets divided by the value of hfe, so it is even smaller, but this is not always true, so be careful. Hence, when true, we get the following.

    Reff = Re || re

    Then, if Re is large compared to re

    Reff = re

    This is basically what they are saying.
     
  3. hgmjr

    Moderator

    Jan 28, 2005
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    It is a mistake to assume that many members in this forum own copies of the book you are referencing. That really makes it hard for them to assist you.

    hgmjr
     
  4. steveb

    Senior Member

    Jul 3, 2008
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    Well, really if there is any book one could assume that people would have (in a high ratio), it is that one. Actually, I have two copies - one at work and one at home. Not that I think this is the best book in the world, but it is good and commonly available.

    By the way, did you feel that my answer was not a suitable one? I thought I answered his question exactly, and I only answered after consulting the book to be sure I understood the question. If there is an error, please point it out.
     
  5. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Steve,

    No one is challenging your answer to the OP. The point hgmjr is making is no one should assume people own any particular book when they have a question. The OP could have easily posted the schematic in question.

    I guess we could post a poll to see how many own one of the Art of Electronics editions. You can mark me down as owning the third edition as well as The Art of Electronics, Student Manual with exercises.

    I would never pose a question concerning a circuit in a 1949 edition of Loran, Volume 3 of the MIT Radiation Lab Series and expect everyone to have a copy of that book. The same with some Radar Books from the 40s.
     
    Last edited: Aug 3, 2010
  6. steveb

    Senior Member

    Jul 3, 2008
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    Ah, ok, that is a good point.

    Somehow that didn't cross my mind. I think any other book would have made me say the same thing. There have been so many discussions here that bring up that book, that I just subconsciously think everyone owns one, but that is unlikely of course.
     
  7. jonnylazer

    Thread Starter New Member

    Jan 12, 2010
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    My question was not about general theory it was about the specific wording in the said book, check the title of the thread ;)

    I didnt make any assumtions, if you dont own the book then you cant reply to the post, thats ok. I take the point about uploading the circuit but my internet connection is woefull at the moment and it would take an age or timeout. In fact, an assumption has been made that my internet connection is capable of uploading schematics in a reasonable time.

    I think making a comparison between assuming people own the said book and people having access to obscure radar books from the 40's is a little deceptive.

    Anyway, thanks a lot Steveb, your reply did help me. I do have a follow up question though, I understand that the capacitor can 'see' RE and re in parallel but I am unsure how it can see through the transistor from emitter to base in order to 'see' Rs and Rth of the bias divider.

    I thought that looking from emitter to base would have such huge impedance that it could be neglected ?

    thanks
     
    Last edited: Aug 4, 2010
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    If I used a contemporanous book that everyone doesn't own be more germane in your mind?

    It doesn't matter whether it's a book from yesterday or one from 1940. This is a world wide resouce and as such we see questions from alot of books. Sedra and Smith comes to mind, and it's in the seventh edition or so.

    The two books I mentioned are obscure to you, but not to alot of people. The point is you ended up waiting till someone who had the book, and the edition that you possessed. That was hmgjr's point. If you took offense to my point, that's fine. You have that right.
     
  9. jonnylazer

    Thread Starter New Member

    Jan 12, 2010
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    I dint take offence, like I said, I thought it was disingenuous/deceptive and I dont think it was a valid point.

    I think to say the examples you provided are not obscure to 'alot' of people is at best dancing with semantics and at worst, again, deceptive. If its obscure its obscure.

    If you missed my point, thats fine. You have that right.

    I hope we can get past what seems to be a slighty antagonistic exchange and get back to discussing electronics.

    BTW how did you get the third edition ?
     
  10. steveb

    Senior Member

    Jul 3, 2008
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    Well, if you think about it, in order for re to be seen in parallel with Re, there needs to be an AC current path (nearly a short) back to ground, at the base terminal. That path is mostly provided directly through the input coupling capacitor and then through the AC voltage source.

    It turns out that looking from emitter to base implies a current ratio of hfe, which implies an impedance reduction (not increase!) by dividing by hfe. So you can treat all resistances Rx on the base side as Rx/hfe. This is basically shown in the formulas I provided (hopefully I got it about right going by quick inspection), although I neglected to include the reactance of the input capacitor because it only matters at very low frequency.
     
  11. jonnylazer

    Thread Starter New Member

    Jan 12, 2010
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    steveb, are you saying that ac current flows from emitter to base ? Where does it originate ?

    Is the path through Vcc to ground not a viable path ?

    forgive me if this these questions are stupid or tedious.

    I want to absorb the ideas you have provided me with before I reply with a more detailed response. Its just I have loads on at the moment.

    thanks again
     
  12. steveb

    Senior Member

    Jul 3, 2008
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    Your questions are not stupid nor tedious.

    I think the best way to understand this is to consider the usual approach to analyze circuits of this type. Typically we like to make an equivalent AC circuit and an equivalent DC circuit. The DC circuit creates the bias conditions and sets up an operating point about which the AC signals can operate. AC signals are assumed to be of small enough amplitude to not cause saturation and cutoff in the transistor, and strickly speaking they need to be even smaller to justify separating AC and DC using the linear superposition principle. For AC signals, the capacitors look like shorts at high enough frequency. Also, any DC voltage sources look like shorts to ground for AC signals.

    Let's take the questions one at a time:

    1. Does current flow from emitter to base?

    Yes, the AC current flows in both directions since it is operating around a DC bias point. It also flows from base to emitter on the other half cycle. This is kind of a point of view issue. Really current is going around a loop. The forward biased base-emitter junction is only allowing one direction of charge flow, but AC analysis (which assumes linearity) allows separation of AC and DC signals, and AC can flow in the reverse biased direction.

    2. Where does it originate?

    The AC input voltage source drives the AC current from the base side. However, we are interested in how the emitter capactor interacts with surrounding resistance to generate a simple high pass effect in the amplifier. The simplified analysis reveals that you can think of the capacitor and surrounding effective resistance as a first order type of response. If this is still confusing, then you probably need to do a full blown circuit analysis to convince yourself of the validity of this approach. The Art of Electronics is trying to simplify things and give you the end result useful for practical implementation. The book approach is so simplified in the final result that a person unfamiliar with the background is going to be lost. The approach I showed is also simplified, not so much in the result since it is justified in a step by step manner, but in terms of the analysis technique which is very simple and may lead the unfamiliar to confusion. Basically, I'm using a quick inspection technique that is taught in school and then developed through practice and comparison with full analyses.

    3. Is the path through Vcc to ground not a viable path?

    No, it is definitely a viable path. The thing is that the best path from the base directly through Vcc is the one through R1, and R1 is generally too large to allow significant current flow. Again, I refer you back to the equations I wrote out. You will see R1 is in the full equation, but it gets eliminated in the approximations. The same is true for R2. R2 also allows a direct path to ground, but if R2 is large, it is not going to allow much current to flow.
     
    Last edited: Aug 4, 2010
  13. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    I misspoke. I was trying to remember the edition, and third came to mind because of the discussions at a usenet newsgroup about the upcoming third edition (a few years back). My copy is packed away right now. Sorry about that tidbit of confusion.
     
    Last edited: Aug 4, 2010
  14. hgmjr

    Moderator

    Jan 28, 2005
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    Sorry if my response appeared abrupt. As joejester indicated, my reply was not intended to imply any deficiency in your answer steve. It was soley to remind the OP that while the book in question is an excellent reference, by providing an excerpt of the passage or circuit from the book (with adequate and proper attribution) one is bound to obtain a broader spectrum of responders willing to assist them. The additional details would also make it possible for other interested members to see the question and learn from the exchange of information.

    hgmjr
     
  15. steveb

    Senior Member

    Jul 3, 2008
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    Oh, no problem. I was just being a little dense. I understand your point of view and also understand jonnylazer's internet limitations.

    It's definitely good to let everyone have access to the reference so that the best advice can be given. I especially want this because I'm a little rusty on this subject, and would prefer some double checking. The OP can also benefit from explanations from other points of view. I've attached scans of the two pages referenced.
     
  16. jonnylazer

    Thread Starter New Member

    Jan 12, 2010
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    I now feel far more comfortable with the problems I had with the emitter bypass capacitor so thanks steveb.

    And thanks to joejester and hgmjr for the advice on how to use the forum in a better way, in the future (pending a decent internet connection) I will indeed upload the relevant information.

    : )
     
  17. jonnylazer

    Thread Starter New Member

    Jan 12, 2010
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    Having said I now feel comfortable with the problem I had something else come to mind.

    It is thus...

    you said (steveb) that the AC current flows in both directions because it operates around the bias point. The signal in this case is 'riding' on the DC bias I had always thought that to change direction the driving voltage would have to change polarity which I dont think it does in this case. I have been taught that an AC (voltage) signal must pass through 0V or else it is just (unsmoothed) DC, eg like ripple in a DC power supply. In this case is the AC current not just changing in amplitude and not direction ?

    I know Im getting off topic here but I thought it was an interesting point.
     
  18. steveb

    Senior Member

    Jul 3, 2008
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    Yes, you are correct, and this gets into an issue of terminology. It is correct to say that the real voltages and currents are DC with a small AC ripple on top of them. This is essentially what we say in analog electronics also. However, the analysis approach is to mathematically separate the AC from the DC, make separate linear equivalent circuits and analyze those circuits separately. For this reason, the term AC becomes correct again, at least within the context of the separated equivalent circuit.

    One of the issues with the book "The Art of Electronics" is that it is not a proper text book that teaches the formal analysis techniques. For this reason, I view it as a good reference, but not a good book to learn details from. These two pages you referenced highlight that fact perfectly. If you are someone who does not want to understand the details, but just want to quickly learn the tricks to get the circuit working, AoE is fine. If you already understand the details (or did at one time), then AoE is great to refresh your memory. However, if one wants to learn the detailed methods of analysis, then a standard electronics textbook is better. There you will find the methods of developing equivalent, linearized AC and DC circuits for transistor circuits.

    All of the questions you are asking are good and correct and indicate you're itching to take the textbook approach, whether you realize it or not. If you don't have a good text book, you can start a separate thread to provide the details of your current background and level so that others can recommend the most appropriate books for you.
     
    Last edited: Aug 5, 2010
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