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Parallel Resistance into Emitter Follower

  1. May 13, 2007 #1

    Please have a look at the following image:

    http://img70.imageshack.us/img70/4008/parallelryl3.jpg" [Broken]

    I have a problem that the book states the resistance seen by C1 is the parallel resistance of the resistance looking into the base and that of the voltage divider. The resistance looking into the base is (hfe x re) in this case 750k. No problems there I understand how that one is derived.

    I really don't get however how R1 and R2 are seen by the capacitor as a parallel resistance? The book evaluates R1 and R2 in parallel and then treats them as a single resistance. It then puts said resistance in parallel with the 750k.

    I understand all that but from C1's perspective why is R1 included and how can it possibly see R1 and R2 as a parallel part of a parallel load?

    Also does anyone know how to post up pictures instead of links?

    I'm a bit of a dummy in electronics so please if you wouldn't mind keep the replies simple for my inadequate experience to understand :wink:
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. May 13, 2007 #2


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    To an AC voltage, which is the only thing that matters when considering C1, R1 and R2 appear to be in parallel. The power supply appears to be a low impedance to AC.
  4. May 13, 2007 #3
    But this would not hold for DC right? Why does it hold for AC then. Sorry to ask what may sound a stupid question.
  5. May 13, 2007 #4


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    DC will not make it through the capacitor so it is irrelevant.
  6. May 13, 2007 #5


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    Averagesupernova, already said that it only matters to AC. What does a cap do to DC?

    EDIT: a little slow on the response there.
  7. May 13, 2007 #6
    I see that point thanks. In fact I just answered someone else's question eleswhere less than a few hours ago and myself pointed that out lol. Crazy how I forget stuff sometimes but I just have to do the best I can with my dodgy mind :rofl:

    Ok so can you or anyone else explain why an AC current sees it as a parallel resitance? Doesn't a parallel resistance have to be equipotential across both resistances? R1 and R2 don't obey this rule at all.
  8. May 13, 2007 #7


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    There is a low impedance to AC between power supply + and power supply ground. Draw a capacitor in your diagram between the collector of Q1 and ground. See it now?
  9. May 13, 2007 #8
    To be honest no but I'll go and do some work on that concept now and see if I can grasp it. You guys are just too damn clever, how you can interpret these diagrams so easily I just don't know :confused:
  10. May 13, 2007 #9


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    Cuz we've been at it for a while. :smile:
  11. May 13, 2007 #10
    A voltage source has a very low impedance, that means it acts like a short (meaning just a piece of wire) for DC and AC voltages.

    That means electrons can flow right through a battery as if it has 0 resistance, only to be pushed again.

    Having said that, just look at the voltage divider and the coupling capacitor. First replace the power source of the voltage divider by a short, or a piece of wire. By introducing AC through the capacitor, the AC sees only two resistors R1 and R2 as being tied down together in parallel.

    Hope that helps.
  12. May 16, 2007 #11
    To be honest I still don't get it. I'll have another look today and see if I can get any closer. Thanks for the reply :wink:
  13. May 16, 2007 #12
    I'm still stuck. Is there any other way I can look at it?
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