Common-Collector Amplifier Circuit

Discussion in 'Homework Help' started by howartthou, Jun 9, 2009.

  1. howartthou

    Thread Starter Active Member

    Apr 18, 2009
    Hi All

    Attached is a circuit diagram of a common-collector amplifier. I am asked this question:

    A common collector amplifier has the
    (1) collector at a-c ground
    (2) emitter at a-c ground
    (3) base at a-c ground
    (4) no element at a-c ground

    Well I thought it was (2) but I was wrong. The correct answer is (1).

    Unbelievable! Everytime I look out circuits with ground I get confused :mad:

    What the hell is a-c ground? Does it mean the negative side of the input??

    Looking at the diagram I fail to see how the collector is at a-c ground.

    You can see from the diagram that RE comes off ground and goes into the emitter, can't you?

    And the collector looks like it goes to +Vcc.

    Can someone please explain the electron flow and how the collector is connected to ground?:eek:
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
  3. hashoot


    Jun 9, 2009
    BABE. think Ac Ground means, ground in Alternating Current or Ac fom of the circuit.

    in Ac form yes... all the Vccs will be Ground to calculate the gain.
    Cout will be Short circuit.

    emitter is connected to RL and RE in Parallel.
    base is connected to Input signal
    Collector is connected to ground.

    for more understanding u should use a two port model for ur ∏ or H model. what i'm saying is the result of using those models.

  4. howartthou

    Thread Starter Active Member

    Apr 18, 2009
    How to you know Vcc is ground?

    Does Vcc = Voltage Collector Cuurent?

    Isn't ground the ground symbol at the bottom? Why are you saying +Vcc is ground? How do you determine the reference point here?

    If Vcc is ground what the hell is the ground symbol for then?
  5. PRS

    Well-Known Member

    Aug 24, 2008
    I understand your confusion, howartthou. We use the term ground at one time to refer to the dc ground -- that little symbol you mentioned. But to ac the entire supply -- both its positive and negative nodes -- is ground.
  6. hashoot


    Jun 9, 2009
    i explain again , but what does voltage collector current means? voltage and current both used for collector in one statement?!

    u r right but
    we have two calculations.
    one in DC mode and one in AC mode(or low signal).
    we separate the ac mode from dc mode. if u remember the circuit analysis courses(Network Theorem chapter) we have sth that we called Superposition. and this separation is exactly using the Superposition theorem.
    in superposition when we have a signal that have both DC and AC we simply separate them and then we solve the problem.did u remember my dear?:D

    so the problem is saying in Ac so means make the dc sources out of ur circuit.(make them gnd) don't think about it like a circuit on a bread board. in real world u need both signal sources but for solving it on paper superposition is needed.;)
  7. howartthou

    Thread Starter Active Member

    Apr 18, 2009
    Hi Paul

    Okay, so DC ground is not the same as AC ground.

    But in this circuit, how do you know Vcc+ is the ground and not the -ive or positive of the power source?

    How do you know the ground symbol shown is DC ground and not AC ground?
  8. hashoot


    Jun 9, 2009
    hey did u hear me in any way friend.i'm not that knowledgeable but u can trust me about this.
  9. Audioguru


    Dec 20, 2007
    The voltage of a power supply usually does not have an AC signal. So its AC is grounded.
    The common-collector transistor has its collector connected to the power supply voltage so it is AC grounded.
    The base has the input AC signal and the emitter has the output AC signal.
    Therefore only the collector is at AC ground.
  10. hobbyist

    AAC Fanatic!

    Aug 10, 2008
    Alright look at your schematic.

    Rule one. A capacitor will block DC, and pass AC.
    Rule two. The battery acts like a capacitor only to AC frequencies.
    Because it has 2 electrodes and a gap between, forming a capacitor to AC.

    Now in your schem. the emitter is not connected to the battery, but through a resistor to the neg. term. (dc ground)

    The collector is connected to the battery at it's positive terminal, DC supply.

    But during an AC signal comming out of the collector due to a AC signal input to it's base, that AC signal sees the battery as a capacitor to send this AC frequency through the battery just as it would through a capacitor, so the signal through the battery inherently is passrd to the neg. term. which is the ground for the entire circuit.

    That's why it is AC grounded only.
    Not DC grounded because as far as dc is concerned the collector sees only the positive battery term.
    But the AC signal sees the other side of the battery which is ground.
    Last edited: Jun 9, 2009
  11. howartthou

    Thread Starter Active Member

    Apr 18, 2009
    Thanks all for your help.

    Hobbyist, I understand your explanation the most, although I still don't understand how the AC signal "through the battery is inherently passed to the neg. term". On the diagram +Vcc is out there on its own, so how does it connect to ground then?

    I don't suppose anyone has some good reading material that explains all this at a level a novice could undertsand?

    Audioguru, your explanation is also something I nearly understand too. Thanks.

    Hashoot, I appreciate your replies but did not understand them as well as the others. I have not studied superpostion yet so have no idea what that means. Same goes for your reference to Network Therom. I get the impression your think I know more than I actually do...I am really a beginner here.

    I will look for more reading material then come back and read all the reponses again, hopefully I will get that "eureka!" type moment sometime soon...

    I do feel I understand alot about transistors and can apply the formulas no problem, its just this AC-DC-ground-electron-flow thing that I struggle with for the moment....I still have trouble determining electron flow in this type of circuit and especially how AC behaves when capacitors are involved, but I am getting there. I will find you!

    Thanks again everyone.
    Last edited: Jun 11, 2009
  12. hashoot


    Jun 9, 2009
    sorry friend.:p
    it's like this for us here in university. before starting with BJT or FET or any electronics we should pass Electric Circuits I and II.
    there are many books in this field like:

    engineering Circuit analysis... W.Hayt
    Basic Circuit theory .... C.A.Desore and E.S.Kuh
    Electric Circuits ....J.W.Nillson
    and so on...:cool:

    some simple Explanation:D

    The superposition theorem states:
    ‘In any network made up of linear resistances and containing more than
    one source of e.m.f., the resultant current flowing in any branch is the
    algebraic sum of the currents that would flow in that branch if each source
    was considered separately, all other sources being replaced at that time
    by their respective internal resistances.’
    from Electrical Circuit Theory and Technology....John Bird
  13. Jony130

    AAC Fanatic!

    Feb 17, 2009
    The ideal DC voltage source has 0Ω internal resistance. And that's why AC-signals are shorted by DC voltage source.
    DC voltage is always constant so for any change in current there is non change in voltage. So there is 0 Ohm internal resistance.
    And additional in real life circuit we always uses a bypass capacitor connect parallel to DC Voltage. And this capacitor will short all AC-signal to the ground. Because the large-value capacitor is a short for AC-signal.
    Current in capacitor is proportional to the rate-of change of the voltage, the faster the voltage change, the larger the current flow through the capacitor.

    Suppose we have this emitter follower. The AC input signal is 4Vpp (pek to pek). The DC voltage in base is equal Vb=6.6V, and emitter dc-voltage is 6V. If we apply the ac-voltage to the input the base dc-voltage will be change form 6V+2V=8.6V and 6.6V-2V=4.6V. So the base voltage will change from 8V to 4V in "rhythm" of input ac-signal.
    These changes will result that the emitter voltage will be change to from 8V to 4V. This will result the change in emitter current by (8V) 26.6mA to 13.3mA (4V).
    So when input voltage changes of 4V, the base current will be change by
    ΔIb=266uA-133uA=133uA And this will result the input impedance equal
    Rin=ΔU/ΔI=4V/133uA=30KΩ. In real circuit the input impedance will be less then 30KΩ.
    Because we don't included in our calculation the effect RL and RB.

    For ac-signal Ce act a like short, so Re=Re||RL=200Ω.

    V_B1=10.6V And when input voltage reach +2V, Vb=8.6V.
    Current that flow through Rb=(10.6V-8.6V)/Rb=0.1mA.
    But the emitter current must be equal 40mA, and that's requires the base current to be equal 400uA=0.4mA.
    So we have a situation:
    V_B1 delivers 100uA, base needed 400uA, so that extra current (300uA) will (mast) be "deliver" by a ac-source Ug.
    So Ug will be sourcing a current 300uA

    In negative swing of the Vg the base voltage will be Vb=4.6V and we need only a 200uA base current to deliver 20mA emitter current.
    V_B1 delivers IB1=(10.6V-4.6V)/20K=300uA.
    Into the base will flow only 200uA, the rest of a V_B1 current (100uA) must be sink by ac_signal source.
    So Rin=4V/(+Ig-(-Ig) )=4V/(300uA+100uA)=4V/400uA=10KΩ
    Rin=Rb || [(β+1)*(Re||RL)]

    • wk.PNG
      File size:
      24.2 KB
    Last edited: Jun 11, 2009
  14. hobbyist

    AAC Fanatic!

    Aug 10, 2008

    See if this schematic helps explain.
    The neg. term. is the ground.
    Ground is actually called the common point.
    Because most all measurements will be measured with respect to this common point.

    That's why it is called ground or common.
    The usage of the word ground is complicated because it's mostly refering to electric circuits using the earth as the common point.

    But in the case of these electronic circuits, a more better way of understanding is to use the word a "common point" of reference.

    Every voltage is measured in reference to this common point.

    It is usually one terminal. of the voltage source, although it dosen't have to be, but it is best to understand that it is the point where you take all measurements from.

    For instance since the common point in your schematic is the neg. term. then the positve term, would be voltage VCC.

    Example. If VCC were 10 volts, then you would say that VCC is 10 volts with respect to ground.

    So lets say that the point between RB1 and RB2 was 3 volts, what that means is that with respect to ground (common point) which is in this case neg. term. of battery, the correct way to analyse it would be to say "the point between RB1 and RB2 is 3 volts positive with respect to the common point of reference."

    In short we say 3 volts positive with respect to ground.

    And then even shorter yet we say 3 volts positive.

    Or simply 3 volts.

    This is all nomenclature of voltage readings, you will learn soon,
    but hust remnember that voltages are mostly measured with respect to some common point, and it is usually called ground.

    Her is the schematic redrawn with the voltage source shown with the ground symbol to show the common reference point that all voltages will be measured from.

    It's way in the top left corner needs to be clicked on to enlarge it.

    Hope this helps out.
  15. howartthou

    Thread Starter Active Member

    Apr 18, 2009
    hashoot, thanks for the references, I will check them out.

    Jony130, amazing. I actually understand your circuit and am just starting to use those formulas so I understand it mostly.

    hobbyist, don't think I will get a better explanation than that so why does it still look to me that the emitter connects to ground :( Sorry I feel pretty dumb right now, like I am missing the obvious.

    Your diagram is great because it shows the negative side of the DC, which if you follow electron flow goes passed ground on its way to the emitter. So how it the collector connected to ground and not the emitter? Back to square one.

    I know you guys have explained this already and I think I may be confused between DC current and AC current. Would be great to see a simulation of electron flow - not just the odd arrow here or there.

    If I follow your circuit I keep following DC electron flow from the negative side of Vcc thought the circuit but fail to see how the AC electron flow is connected to ground. Yup, stuck like a broken record for now, sorry.

    Thanks again everyone, this thread is so informative I will use it as a reference until the penny drops...
  16. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Forget about electrons.
    And remember that current flow from "+" to "-". From positive point to negative point.

    If we deal with ac-signals we uses, so-called "dynamic" resistance (ac-resistance) rd=ΔU/ΔI.

    DC-voltage source, has for example 9V=const. (constance)
    So if we smoothly change the current that is draw from dc-voltage source, from 40mA to 20mV (we change the resistance from 225Ω to 450Ω). We create ac-current.
    Now we can calculate the dynamic resistance of dc-voltage source
    And that why we say that DC-Voltage is a short for AC-signals.
    This is simply the consequence of a rd definition.
    And remember that is just a "model" (A theoretical representation of a system that allows for investigation of the properties of the system).
    Not a physical representation.
    And you have to remember that only for a purpose of easier analyse, we separate ac-curent and dc-current and create dc-models and small-signal models.

    In CC amplifier DC-current flow:
    Base current Ib flow from +V_B1--->Rb--->-bese-emiter--->Re--->-V_B1
    Collector current Ic +V_B2-->collector-emiter--->Re---> -V_B2
    So emitter current is a sum of Ib+Ic=Ie.

    Positive swing of AC signal flow form
    +VG-->Rg--->X--->Cb--->base-emiter---> the one part of a current flow trough Re and the rest of current flow over CL+RL--->0--->back to the -VG.
    This AC input current causes the "modulation" of a dc Ie current (flowing from V_B2 (Vcc) ) in "rhythm" of ac-current.

    Negative swing
    +VG-->0---> -V_B1-->Rb--->Cb--->X--->Rg--->-Vg.
    As you can see the neg. swing of AC "steal" the current from the base. And that causes decreases in DC base current and Ie current.
    And that change in DC-current create ΔIb, ΔIe ac-current.
    Last edited: Jun 12, 2009
  17. hobbyist

    AAC Fanatic!

    Aug 10, 2008
    I think I know why it looks like the emitter is connected to ground and not the collector,

    because the ground symbol is on the emitter side of the battery and the battery is looking like a barrier, collector connected to one side of this barrier and the emitter on the other side while the ground symbol is on the same side as the emitter.

    Alright when dealing with AC voltages , currents, such as signal analysis of a circuit, the battery is never considered as a battery.
    Because the AC signal does not need the battery to operate, the battery is only for biasing the transistor intoi its region of prefered operation. That is all DC work, not AC.

    The 3 schematics should clarify better.

    Pay close attention to the 3rd one, see how there are no capacitors or voltage sources present.
    Now where do you see the collector terminal connected to.

    Picture 2 supposed to have a line across the top for the batery voltage to the circuit, it just didn't show up on the schematic, when it was copied and pasted.
    Last edited: Jun 12, 2009
  18. howartthou

    Thread Starter Active Member

    Apr 18, 2009
    Thanks a lot! Your current flow description is excellent:). I followed the DC, +AC and -AC no problem. Great analysis, just what I was hoping for, I havent seen a book explain the current flow that well (yet). Brilliant. A few questions though:

    1. Why should I forget electron flow (- to +)? Why do you prefer conventional current flow (+ to -).

    Is it because AC starts as a positive signal on the sine wave so its easier to do circuit analysis from the + side?

    2. When you said "VG" I assume you meant "UG". But on the AC negative swing you said "+VG". Did you mean to say "-UG"?

    3. The formual rd = change in U / change in I. What does the U mean?


    I think I just had that Eureka moment! :) That was a great explanation. You could say we're "in phase" :D (excuse the joke). The origninal, dc and ac circuit diagrams are a great help! Thank you.

    I appreciate you responses which when added together really answer the collector to ground question.

    I can't thank you guys enough, you have been a great help. I find electronics exciting believe it or not, and really aspire to be able to do circuit analysis as well as you can.:rolleyes:
    Last edited: Jun 13, 2009
  19. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Well, almost all engineers analyses their circuit assume that current flow from + side to - side. So to avoid confusion use this convention to.

    Yes it's my mistake. In MY country we use letter "U" to Voltage.
    And that's why in schematics you see "Ub"; "Ue"... etc instead of Vb; Ve...
    And I really don't know why we use the letter "U" because unit is a 1 "Wolt" (Volt) and Voltage = Napięcie.
    So we write U=1V instead of V=1V

    No, in negative swing "bottom" side of "UG" is positive in relation to the "upper" side of UG.
    And remember that I assume that current flow from "plus" to "minus".
    And look at schematics that I attach.

    You should already know that.
    "U"=Voltage. So this is a Ohms low.
    r=ΔV/ΔI for AC-signals.
  20. howartthou

    Thread Starter Active Member

    Apr 18, 2009
    So, regardless as to whether you refer to positive or negative swing one side will always be positive with relation to the other side? I guess that makes sense....right?

    You have been a great help, thanks again. By the way, perhaps "U" in your language refers to how you would say EMF?

    Just to clarify my "eureka" moment.

    Your diagrams show Vcc as a + and - side and the ciruit has a separate ground connection.

    My diagram shows Vcc is + only on the top side of the schematic and a ground connection on the bottom side.

    How does this mean that Vcc in my diagram has a minus side :confused: then, with the ground shown separtely? Is ground the - side of Vcc or is it just ground? Or is it both?

    It all makes sense when I see a + and - to VCC and treat the DC supply like a capacitor from an AC perspective, as you so neatly explain, so I can see how collector connects to ground now :)

    But in my digram how did you manage to redraw VCC + as having a + and - side, and then show ground again? If you are representing my circuit I fail to see how you can just add a - side to Vcc?

    Are you saying in my diagram that the ground symbol is both ground and the - side to VCC? :confused:
    Last edited: Jun 13, 2009