Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the problem with this output resistance....

  1. May 19, 2015 #1
    In a common emitter amplifier while analysing its output impedance ..why we consider collector resistance (Rc) in parrallel with output impedance (Ro) of the amplifier...since as they appear in series ..as seen by the KVL equation(which involves Vcc=IcRc+Vce+IeRe).....
     
  2. jcsd
  3. May 19, 2015 #2

    meBigGuy

    User Avatar
    Gold Member

    Imagine that you forced a current into the output terminal. The current would split, some going onto Rc and some into the transistor. So, from the perspective of the output terminal they appear in parallel.
     
  4. May 19, 2015 #3
    but current would only split when the components are in parallel...and well if we force current from powersupply downwards then Rc and Ro seems to be in series..then in this case why would current split...

    and by the way what would be the case if we are considering Rc as the load resistance....
     
  5. May 19, 2015 #4

    meBigGuy

    User Avatar
    Gold Member

    Think of the current going into the output junction where Tc meets the collector. It can go in two directions. Up to Vdd through Rc, and down through the transistor. That means they ARE in parallel.

    You are kind of thinking about it backwards. That the current can only split if parallel (which is true) when you should be saying "since the current CAN split, they must be in parallel".

    From the point of view from VDD through to ground (assuming no external load) there is only 1 path at each node, so there are no parallel paths.

    Deciding series or parallel depends on where you are viewing from. The impedance is different when viewed from different nodes.

    Think of two 10 ohm resistors in series from a 5V supply to ground. They are in series, each dropping 2.5V. Now what do you see when you look back into the resistor junction?

    You see two resistors, one going to +5, and one going to ground. Now, reduce the 5V supply to 0V. Now you see 2 resistors to ground. But, the circuit topology has not changed.

    In terms of impedance analysis from a node you generally replace all supplies with ground (since a voltage source is effectively 0 Ohms to ground) and then analyze the circuit.

    If you do the math (forcing a current into the split node in the 2 10 ohm resistor example) you will see that the change in voltage is the same as forcing the current into 5 ohms.
     
  6. May 19, 2015 #5
    great answer man...I was looking for something like this....you save me everytime.....
     
  7. May 20, 2015 #6
    ok now as Ro is parallel to Rc....then for a higher voltage gain the value of Ro should be higher...and for a higher current gain value of Ro should be lower..am i right?
     
  8. May 20, 2015 #7

    Svein

    User Avatar
    Science Advisor

    upload_2015-5-20_13-52-0.png There is a problem with the emitter follower which is seldom discussed. Referring to the figure, pulling current out of the emitter follower is problem-free. The extra current through the output transistor results in a much smaller current through the collector resistance in the previous stage (icollector = iemitter/β). The problem begins when you try to put current into the emitter follower. In order to keep the emitter at the correct potential, the emitter follower must reduce its current into Remitter. Sooner or later you will get to the point where there is no current left in the emitter follower, leaving us with an output impedance of Remitter. This asymmetry creates a lot of distortion when the emitter follower tries to drive a low impedance (for example a loudspeaker).
    upload_2015-5-20_14-9-55.png The usual remedy for the emitter follower problem is to use a symmetrical emitter follower. This version uses complementary symmetry in order to be able to sink as much current as it can source.
     
  9. May 21, 2015 #8

    meBigGuy

    User Avatar
    Gold Member

    be carefull. Ro is only parallel to Rc when looking back into the circuit from the output terminal (where Rc, Ro, and the collector all meet)

    I'm not sure what you are asking. The maximum output voltage (when the transistor is off) is determined by the Rc/Ro voltage divider.
     
  10. May 21, 2015 #9
    What exactly i want to ask is--->

    1.In a current mirror its demanded the output resistance to be high as..by doing so it will make load current independent of load voltage variations..

    2.If suppose we have two resistor one acting as a load (Rc) and other as the output resistance of transistor(Ro) ..and if we increase Ro to a very high value then is it possible that the total current flowing them would be independent of the voltage drop across the lower resistance i.e Rc..if yes then how?

    Please explain point 1 in terms of point 2..
     
  11. May 21, 2015 #10

    meBigGuy

    User Avatar
    Gold Member

    OK, I didn't realize you meant Ro to be the output resistance of the transistor. I thought you were speaking of an external load resistor (I guess that would be RL)

    You better make a drawing, because I'm confused. If you are speaking of the effective resistance of the transistor, then that varies with the input signal.
     
  12. May 21, 2015 #11
    Or either you can give your own independent explanation for premise 1 ....and the I will try to relate it with point 2.....coz I dont have a suitable drawing for it....
     
  13. May 21, 2015 #12

    meBigGuy

    User Avatar
    Gold Member

    I really don't understand the question. What are you trying to do?
     
  14. May 21, 2015 #13
    I m not getting the point that---> How by increasing the output resistance of a current mirror circuit ...we could make the load current independent of load voltage variation....just simple as that..??
     
  15. May 21, 2015 #14

    meBigGuy

    User Avatar
    Gold Member

    So we are talking current source. That is a different issue all together. In an ideal current source, the current is constant regardless of voltage. That means that delta(v) causes no delta(i). If no matter what voltage you apply, the current does not change, then you must be effectively an open circuit (except for the constant current part)

    delta(I) = delta(E)/R Since delta(I) is zero and E can be anything, R must be infinite.
     
  16. May 21, 2015 #15
    yes i m getting a little...
    but any way tell me ..as they say that for a current mirror the output resistance should be high.....do they mean output resistance(Ro) of the transistor T2 ...or they meant of the whole output branch involving R1, Ro and Re...
     

    Attached Files:

  17. May 21, 2015 #16

    Svein

    User Avatar
    Science Advisor

    They mean the whole output branch. Anyhow, there are "constant current sources" that are much simpler, but not as accurate:
    upload_2015-5-21_10-48-3.png
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is the problem with this output resistance....
Loading...