CE amplifier - AC current Path

Discussion in 'Homework Help' started by san203, Feb 16, 2016.

  1. san203

    Thread Starter New Member

    Jul 9, 2015
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    0
    This is a conceptual question regarding the path of AC Current in the CE amplifier with Fixed Bias.

    This is the shabby circuit diagram i made. I chose the fixed bias because it seemed the easiest to make.

    [​IMG]

    Here are my questions with certain assumptions. I would really be grateful if anyone can verify these assumptions or point me in the right direction.

    Q1.) Does the current from the signal generator flow directly to the Base of the transistor? @ point 'A'
    well, my reasoning is that, The AC current should definitely flow to the Base current because of a Short Circuit. The capacitor is virtually a short circuit for AC at high enough frequencies. I also know that the Transistor doesn't work in AC, so a DC current is used to Offset the AC current and that's fed into the Base. So in this case , its DC current is Ib.

    Q2.) At point 'B'(Sorry for the poor choice of letter),the amplified current Ie, exiting through the emitter junction flows to the ground. But where does it go?
    I know that the DC current cant go to either the Input side or the output side because of the Input and Coupling Capacitors respectively. So its only way back to the power supply is through the ground,which is the current return path.
    But what about the AC amplified current? Where does that go? Also , since the circuit has to be complete to be in working condition , i could think of a few possible cases.

    a.)It flows back to the Input Circuit.

    b.)It flows through Power Supply through GND(Is that even possible?)

    c.)It flows, both to the Input Circuit and across the Load(In case its connected)

    d.) It divides into 3 different currents and goes all 3 ways.

    Q3) In a few other amplifier circuits(CE amp with Voltage Divider Bias), i read that the output is taken across Rc. Can someone explain how the AC current even flows to Rc? And how does the Voltage across Rc become the Load resistance?
     
  2. crutschow

    Expert

    Mar 14, 2008
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    The input AC current goes through the input capacitor and the transistor base, and varies the DC current flowing through Rc from collector to emitter.
    This current variation equals the AC base current amplified by the current gain (Beta or Hfe) of the transistor.
    This current variation causes a voltage variation across Rc which is the amplified voltage you see at the transistor collector.

    The input AC portion of the emitter current returns from the emitter ground back to the AC source ground.
    The AC amplified signal on the output load (not shown) returns through the ground to the power supply.
    The rule is, you always need a complete circuit so all currents must flow back to their respective sources.
     
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  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
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    During positive half cycle the AC current from Vin source will flow directly into the base (through coupling capacitor) and back via emitter and ground to Vin source.
    During negative half cycle the AC current will flow into the ground through Vcc voltage source--->Rb resistor--->coupling capacitor--->AC signal source.
    http://forum.allaboutcircuits.com/t...ulating-peak-ac-amplitude.119261/#post-948137
    http://forum.allaboutcircuits.com/threads/common-collector-amplifier-circuit.24469/#post-149702
     
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  4. san203

    Thread Starter New Member

    Jul 9, 2015
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    Thank you crutschow! Just a minor quirk I dont see a direct connection between the AC amplified current and Rc. Is it because the amplified Ac current flows through GND to the power supply(Vcc) and changes the Voltage drop across associated across Rc.
    Current of amplified part + DC current set by the Vcc = Larger Current across Rc leading to a larger Voltage drop?


    So essentially the input ac current remains the same as long as the circuit is complete. The only change that takes place is due to the difference.
    But dont we need the current to flow through Vo, to get the actual Output Voltage? Will a voltage drop across Rc also lead to the output voltage?


    Thank you Jony130. I didnt know the current took different paths each time. But wont the AC current be splitting at the Vcc. Also since we know that the current prefers the lower resistance path, wont most of the current branch off towards Rc, Since, Rb > Rc in general?
     
  5. crutschow

    Expert

    Mar 14, 2008
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    The transistor DC bias current goes from the power supply, through Rc, through the transistor to ground.
    This DC bias current is varied by the AC at the base to give an AC signal at the resistor-collection junction, with AC voltage equal to the current variation times Rc.

    You seem to be confusing yourself by worrying about all those supposed paths of current.
    The current just goes from the power supply to ground with a varying rate due to any AC signal applied to the transistor base.
    Don't overthink it. :rolleyes:
     
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  6. Russmax

    Member

    Sep 3, 2015
    81
    12
    san203,
    The GND symbol represents a direct, hard-wired, zero ohm, short-circuit connection between all GND symbol on the schematic.

    The Vcc symbol represents a DC voltage source, which is not directly shown on this schematic, with its positive terminal connected to the Vcc node, and its negative terminal connected to GND. Therefore, there is a DC shorted path between the AC signal generator's GND connection and all the other GND connections, including the Vcc source's negative terminal. Remember that a DC source looks like a short to an AC signal, so AC currents definitely can flow through a DC source. AC currents can flow through all GND connections and through the VCC connection.

    Regards,
     
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  7. san203

    Thread Starter New Member

    Jul 9, 2015
    3
    0
    I guess that's true. It just happens that your caught up in the moment and you desperately need to know the answer. But if you take some time off,it doesn't seem so serious.

    Thanks a lot! I guess that assumption was right after all.
     
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