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Input and Output resistance(CE amplifier)

  1. Feb 23, 2008 #1
    Hello people

    I am trying to revise for my exam next week..I have searched everywhere but still I cant find how to calculate input resistance,stage input resistance and output resistance of a class A common emitter amplifier.

    any help will be highly appreciated.

  2. jcsd
  3. Feb 23, 2008 #2


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    Input impedance depends a bit on how it is biased and if there is a bypass capacitor across the emitter resistor. For a class A CE amp with voltage divider bias and a bypass capacitor it will be:
    (r'E * beta) in parallel with the following quantity:
    (Bias resistors in parallel with each other).
    You may think that the bias resistors would not be paralleled but they end up that way since the + power supply rail is considered AC ground.
    Output impedance is simply the the resistance of the collector resistor.
  4. Feb 23, 2008 #3
    Thank you Sir

    I am looking at these diagrams:
    http://www.zshare.net/image/8014898071a386/ [Broken]

    I want to know how to find Rin, R'in and Ro for Figure A

    and r'in for Figure B, R1=120K, both when Ce1 is removed or not.

    Last edited by a moderator: May 3, 2017
  5. Feb 24, 2008 #4
    anyone willing to help?
  6. Feb 24, 2008 #5


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    Figure A has no emitter bypass capacitor so the AC impecance looking into the base would be (r'E + RE1) * beta. I would say that is what the diagram call R'in.
    Parallel R1 and R2 to get the impedance that the biasing part of the circuit presents.
    Finally, parallel the biasing impedance with the impedance looking into the base figured in the first step. This is a pretty good approximation of the input impedance of figure A.
    Funny thing, I don't exactly recall how to arrive at the output imedance of an emitter follower (second stage figure A). I believe you have to know the output impedance of what is driving it and divide that by beta. You also have to figure in the RE and I don't recall exactly how.
  7. Sep 7, 2010 #6

    I know this is an old thread, but I have found myself here!

    My question is how you found Rin to be "(r'E + RE1) * beta" ?

    I keep getting Rb || r(pi) + RE -- where is the beta coming from?

    Also, please know that I cannot view the image because the host of the image is evil (spam).

    Many thanks for any help here...
  8. Sep 7, 2010 #7


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    The emitter resistor carries a much magnified version of the input signal current and this produces a voltage which opposes the input current itself, making the input impedance a lot higher than you might expect.

    For example, a transistor with a Hfe of 100 and a (internal) base resistance of 56 ohms and with a 220 ohm emitter resitor will have an input impedance of 22276 ohms. Quite a lot more than the 276 ohms (ie 56 + 220) you might expect. This is excluding any bias resistors.
  9. Sep 7, 2010 #8


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    [PLAIN]http://dl.dropbox.com/u/4222062/Emitter%20follower.PNG [Broken]

    In the above diagram, I have shown the base resistance of the transistor and the emitter resistor to be 56 ohms and 220 ohms respectively. The 56 ohms is inside the transistor.

    The transistor is assumed to be working with correct bias. Only signal currents are considered in the following:

    The transistor has a gain of 100 so there is a large current source shown which produces a signal current of 100 times the input signal current Ib.

    So the current in the emitter resistor is 101 * Ib (ie 100 Ib + Ib )
    The voltage across the emitter resistor is (101 Ib * 220) or 22220 Ib

    The voltage across the base emitter junction is 56 Ib

    The total input voltage is 56 Ib + 22220 Ib or 22276 Ib (ie base-emitter voltage plus emitter voltage

    Input impedance = Input voltage / input current = 22276 Ib / Ib or 22276 ohms
    Last edited by a moderator: May 4, 2017
  10. Sep 8, 2010 #9
    Okay, thank you so much. I still don't get it. Later today, I will draw and post my question more clearly and see if you can point out my mistake.
  11. Sep 8, 2010 #10


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    Can we go back to fundamentals first?

    Suppose you have two batteries, one is 5 volts and the other is 4.9 volts.

    They are joined together at the negative end and there is a 1 K resistor between their positive terminals.

    There is 0.1 volts across the resistor, so the 5 V battery is delivering 5 volts at 0.1 mA. (0.1 V / 1000 ohms )

    So, it is seeing an apparent load of 5V / 0.0001 amp or 50000 ohms.

    This is DC but it is similar to what happens in an emitter follower for AC signals.

    The emitter has a resistor to ground that has an almost exact replica of the input signal across it (produced by the transistor's current gain). So, the input impedance depends on the current gain of the transistor and the size of the emitter resistor.

    The closer the emitter signal voltage is to the base voltage, (like in our battery example), the higher the input impedance of the emitter follower.
  12. Sep 8, 2010 #11
    Okay, I think I am starting to smell what you are stepping in.

    I have two different CE amps that I worked out for my lab. I am pretty sure that I nailed everything except for the Rin for the CE amp with a resistor at the emitter node.

    There are definitely some basics that I am missing here and the complexity of my school work just went way up!

    Here is a http://www.ncday.net/stuff" [Broken] to the page I uploaded my work to. The file name is: CEamp.pdf.

    Thanks again for any help!
    Last edited by a moderator: May 4, 2017
  13. Sep 8, 2010 #12


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    I looked at that link. I think it was OK until the second last line.

    The input impedance of just the amplifier (without the bias resistors) is Vin / Ib.

    Notice (in my previous post) that Vin can also be expressed in terms of Ib and then the Ib's cancel out. This means you don't have to know what Ib is, to get the input impedance.

    Also, the input impedance is very close to Re * beta for an emitter follower.
    In the example I gave earlier it makes about 1% difference if you just use Re * beta. (Re is the emitter resistor)
    ie 220 * 100 = 22000 compared with 22276 I got as a more accurate answer.
  14. Sep 9, 2010 #13
    Just to follow up...I think I get it now. My problem was with the definition of input impedance (vin/in). Some of these circuits can be very complicated...

    Even though yours was very simple and easy to understand, I had trouble applying that to the circuit I was working on.

    Many thanks for your help.
  15. Sep 12, 2010 #14
    I've had a very simple method since HS.

    Z seen at e is 26 ohm / ie (ma) + Re
    For example, if the emitter current is 2ma and the emitter ballast resistor is 50 ohms, then Ze = 13 + 50, or 63 ohms.

    What is seen at the base is Ze / Beta. Beta is the same as hfe
    Now, there are usually bias resistors to maintain the base at the correct operating voltage. These resistors are in parallel with each other and Ze/Beta, so you get Zin=R1||R2||Ze/B

    Some people just use the collector resistor as the output impedance. If you want to be picky, it's hoe||Rc
  16. Sep 16, 2010 #15


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    What is seen at the base is Ze / Beta. Beta is the same as hfe

    The input impedance of an emitter follower is equal (approximately, as described earlier) to the resistance of the emitter resistor TIMES hfe ......provided the emitter resistor isn't bypassed, of course.
    This is where the emitter follower gets its high input impedance.

    The output impedance of the emitter follower is equal to Rbase / Beta (where Rbase is the internal resistance of the base emitter junction) and this is where the emitter follower gets its low output impedance.
    Last edited: Sep 17, 2010
  17. Oct 12, 2010 #16
    Just wanted to say that I pretty much get this stuff now. Thanks for your help!
  18. Oct 20, 2010 #17
    i need to calculate the input resistance of the CE amplifer
    given Ic - 1mA ,Ib - 1uA at room temperature
    how to calculate Inp Resistance with given parameters.Somebody pls reply
  19. Oct 20, 2010 #18
    I don't think anyone has a problem helping you out. However, you are gonna have to at least pose your question so that it can clearly be answered or, better yet, post it with a diagram.

    Is this a homework problem? If so, it doesn't belong here.
  20. Oct 20, 2010 #19


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    If the emitter is grounded, you can sometimes just read off the input resistance from the data sheet for the transistor.

    It may not be given, though. A value of 500 ohms would be typical and this would depend on the base current.
  21. Oct 20, 2010 #20
    the question is
    A CE Transistor Amplifier has a collector current of 1mA,when its base curren tis 25uA at room temperature .Its input resistance is approx equal to
    a.10 ohms
    b.100 ohms
    c.1000 ohms
    d.10000 ohms
  22. Oct 20, 2010 #21
    So, you have been given Ic (collector current) and Ib (base current).

    Ib = 25uA
    Ic = 1mA

    You need to find Rin.

    Since it is a Common emitter, we assume that you are:

    -operating at a high frequency with a bypass capacitor across Re
    -you are using a hybrid pi model
    -you have a node where Ib + Ic = Ie.

    Rin = Vin/Iin = Vb/Ib

    You have Ib now for Vb?

    From your choices, you also see that they are all base 10.

    We also know that Io/Iin = 1mA/25uA = 40 A/A , which is your current gain.

    This must mean that Vo/Vin = (Io * Ro)/(Iin * Rin) = 40 (Ro/Rin).

    Not really sure where to go from here...
  23. Nov 13, 2010 #22
    In labs we often use a potentiometer to determine the i/p & o/p impedance of an amplifier. But I've failed to understand why this procedure is adopted; can anyone please help me on this ?
  24. Nov 13, 2010 #23


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    If you have a resistance you don't know, you can put a resistor in series with it that causes the voltage across it to drop to half.

    By voltage divider action, you can say that the unknown resistance equals the known series resistor.

    So, you can use this to find the input resistance of an amplifier by using a series resistor to drop the signal voltage to half.
    You could use any resistor and calculate the input resistance, but dropping the signal to half saves you having to calculate it. Note that this method fails at high frequencies when internal capacitances become important.

    You can put a variable load on an amplifier and vary it until the output drops to half. Then you can say that the output impedance of the amplifer is equal to the size of the added resistor.
  25. Nov 13, 2010 #24
    Another related question -

    I was watching a lecture on TTL logic family. It mentioned the output resistance is less since its divided by a factor of beta squared.
    What is the output resistance of Darlington pair. Is it less by a factor of hfe(beta) compared to single stage CE amplifier?
  26. Feb 14, 2011 #25
    This stuff bothered me so much I had to work it all out by hand. I should have just done that in the first place. However, I didn't know about the hybrid-pi model at that time.

    Here is the link to the input resistance and open circuit gain for the common base, common emitter, and common collector amplifiers: http://blog.ncday.net/2011/02/input-resistance-and-open-circuit-gain.html" [Broken]
    Last edited by a moderator: May 5, 2017
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