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Common collector configuraton

Discussion in 'General Electronics Discussion' started by Athul, Aug 8, 2016.

  1. Athul

    Athul

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    Aug 4, 2016
    Hi,


    Can anyone help me understand how common collectro configuration works in a an NPN transistor.

    In common emitter configuration,
    2.png

    Emitter is connected to ground and load is connected to Collector. When a sufficient base voltage is applied, E-B will become forward bias and it will turn the transistor on and thus turns the load ON.

    But in Common collector configuration, Load is connected to emitter and collector is connected to Vcc.
    So when a s sufficient base voltage is applied , will it turn the transistor On and thus the load?
    If yes, then HOW?
    Untitled.png
    n order to turn the transistor ON B-E should be forward biased, right?But in CC configuration base is not connected to GND?

    Thanks
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    This is not necessary. For the transistor to turn on, Vbe > 0.6 V (approximately) is required. In your diagram, this means that with Vin = 5 V, Ve = Vin - Vbe = 4.4 V.
    Accordingly Vce = 0.6 V, too.
    This is why you normally don't use a bipolar transistor in common collector configuration for switching loads.
     
  3. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    Common-Collector
    [​IMG]
    The COMMON COLLECTOR stage is also called the EMITTER FOLLOWER stage.
    The input is the base and the output is the emitter. The collector is connected to the supply rail.
    This stage is classified as having a HIGH INPUT IMPEDANCE because the transistor allows you to deliver a high current to the load by supplying a very small current into the base.
    In other words the transistor amplifies your effort by about 100 times.
    This is due to the gain of the transistor.
    When you are doing this, the load appears to be a much-higher value of resistance as the transistor multiplies the resistance of the load by a factor of about 100 times.
    That's why the circuit is classified as having a HIGH INPUT IMPEDANCE.

    How does the Common-Collector circuit work?
    Firstly we will assume you have a voltage of 10v on the base and can deliver 3mA into the base.
    What is the voltage on the emitter and how does appear on the emitter?
    The transistor amplifies the 3mA by about 100 times and this produces 300mA through the collector-emitter junction. This current flows through the emitter resistor and say it produces 12v across the resistor.
    This voltage is higher than the base voltage and this cannot happen. What happens is the current increases and the voltage increases on the emitter resistor until the voltage is 0.6v LESS than the base voltage. At this point the transistor turns off a small amount and the current reduces so the voltage reaches EXACTLY 0.6v less than the base. That's how the voltage on the emitter is 0.6v less than the base. If you raise the voltage on the base, the emitter voltage will rise too. If the base voltage is lowered, the emitter voltage will reduce.
    The emitter voltage FOLLOWS the base voltage and that's why we call the circuit EMITTER-FOLLOWER.
     
  4. Ratch

    Ratch

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    Mar 10, 2013
    You are describing a current amplifier above. You are confusing input impedance with amplification. An emitter-follower has a high impedance because the input base voltage source needs to supply or sink a much small current than it would if it were connected directly to the emitter resistor.

    .
    No, it is due to the amplification of the circuit.

    .
    It is due to the circuit, of which the transistor is an active part. A transistor by itself if a transconductance amplifier (voltage controls current). In a circuit, it can be any other type of amplifier.
    .
    The transistor does not change the value of the resistor. The circuit makes the input impedance appear high by restricting the current change in the input base when the input base voltage changes.
    .
    Yes, just like a common-emitter, the emitter-follower has a high input impedance..

    How does the Common-Collector circuit work?
    .
    That depends on the emitter resistor, doesn't it?
    .
    What a tortured explanation!
    .
    Yes, base voltage minus 0.6 volts equals the emitter voltage.

    Ratch
     
  5. BobK

    BobK

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    Jan 5, 2010
    If you are going to be so pedantic, you ought to rethink this statement.

    Bob
     
  6. Ratch

    Ratch

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    Mar 10, 2013
    Are you referring to whether the transistor is a PNP or a NPN? I was thinking of a PNP. Thanks for the polarity picture. Better yet, I should have said "difference".

    Ratch
     
  7. Harald Kapp

    Harald Kapp Moderator Moderator

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    The way this discussion goes it is not going to be of much help to Athul...
     
  8. Athul

    Athul

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    Aug 4, 2016
    In the main post , I said
    I meant emitter is not connected to GND.

    Yes, for the transistor to trun on Vbe = 0.6.
    In common emitter mode, emitter is directly connected to ground, so by by giving a base valtage of 06-0.7V , one can turn on the transistor(Vbe= 0.6-0.7),

    but in common collector mode emitter is not connected to GND, but to load, and load say an led, when it's OFF it act as a open switch.(i'm not sure about the statement). SO by givng 0.6V to base, how can it turn the transistor ON, since emitter is not connected to GND
    Untitled.png

    May be I'm not understanding the concept?

    Okay 3mA is supplied by Base, and it amplifies the current to 300mA, but transistor can't generate current by itself, right it has to come from somewhere, so extra current (300-2), is it taken from Collector supply??

    Also this current amlification, how to find it's value

    I know Ic/Ib= hfe.
    but if you multiply Ib to hfe it give the collector current, right? not emitter current?


    Also can I directly connect a load, say LED replacing the Load resistor in your circuit Or should I have to connect it across load resistor?


    One more thing I know in common collector config, if base voltage is 5V, ma.voltage I could get across load is 5-0.6 = 4.4V. What if I change collector volatge from 5V to 12V, wwill it change anything?

    If by apllying base voltage transistor turns ON, it should get the collector voltage 12V across load, like in common emitter config?
     
    Last edited: Aug 9, 2016
  9. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    You don't apply 0.6 V to the base. As shown in your diagram, you apply 5 V to the base via the current limiting base resistor. As I tried to explain in my post and as Collin explained in post #3, it is the base-emitter voltage Vbe that needs to be ~ 0.6 V to turn the transistor on. Therefore the emitter voltage will be lower than the base voltage and the load (LED) will not see the full supply voltage.

    You supply a voltage of 5 V to the base resistor, not a current. The base resistor will lmit the base current. However, the transistor is controlled by the base-emitter voltage Vbe (see transistor equations here). Using IC=hfe*Ib is a simplified model which does not take into account how Ib is controlled (nevertheless a useful model in many circumstances).
    The misconception here is that from IC=hfe*Ib you conclude that a base current of 3 mA "generates" a collector current of 300 mA (hfe = 100). this is not universally true. This equation describes the collector current only within the linear region of the transistor. The max. collector current is given by Icmax < Vcc/Rload where Vcc/Rload is the current trhough the LED (I assume or hope you have an LED with integrated current limiting?) When Icmax < hfe*Ib the transistor is in saturation (typically used in common emitter switches).
    In your circuit a rise in IC (or Ie for what it's worth) will incerase the voltage across the LEd and therefore increase Ve (emitter voltage).Since the base voltage is constant at 5 V, an increase in Ve will reduce Vbe: Vbe = Vb-Ve (neglecting the voltage drop across the base resistor) such that the reduced Vbe will reduce Ic. Collin tries to explain tis in post #3.

    Ie = Ic+Ib

    The load resistor is a universal representative of a load which can be anything like a resistor, a lamp, a motor...
    You would replace the resistor by the actual load, not connect it in parallel.

    Not as you expect, see above: Ve = Vb-0.6 V and since Vb doesn't change, Ve will stay the same.
    The change is in Vce which will rise from 0.6 V (5 V -4.4 V) to 7.6 V (12 V - 4.4 V) and thus power dissipation will rise equivalently.
    No, it won't, see above.

    To sum it up: Do not use a common collector configuration to switch load on and off unless absolutely necessary for other reasons. Always use common emitter configuration.
     
  10. Athul

    Athul

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    Aug 4, 2016
    I know, I'll be hard to teach idiots like me but I still have few more doubts. And hope you have patients for that. :)


    This what I understood so far,

    When a base voltage is applied base current flows and this is multiplied by transistor gain to produce a higher current, this current causes a higher voltage across load. Since base voltage is constant, this cannot happen.
    So emitter current and voltage increases till voltage across load is it's less that 0.6V that of base voltage.

    In other words increased collector current will increase emitter voltage , SInce Vb is constant increase in Ve reduce the Vbe, this causes Ic to reduce.

    When a voltage (5V) applied to base, base current flows(say 3mA, and it's amplified by transistor to a higher value depending upon it's gain(say 100)
    So Ie = 300;
    What I meant is if Ib = 3ma, where does the remaining current comes from, Is it from supply connected to Collector?
    (But i could 't find an equation connecting base current, emitter current and gain hfe.
    Is it like this, Ic/Ib =hfe,
    So, Ic = Ib * hfe (3 * 100 = 300mA) ------> increased collector current.
    Ie = Ib + Ic (300 + 3 = 303mA) -----------------> Emitter current increases as a result)
    Is that how I get high current through load from a low base current?

    . This higher current will cause a high voltage across load(current high, resistance constant)
    If load has a reistance of 40 ohm, then voltage across load is

    Vload = 50 * 0.3 = 15V (should I use 300 ma and 303 ma, even though it won't make much difference?)

    But this cannot happen since Vbe = 5V ,( By KVL 5V = 0.6 + Ve ========> Vbe = 4.4V)

    So current cannot increase to 300, but increase to a value such that Ve is 0.6V less than Vb.

    Okay, Vbe~0.6 V to turn the transistor on .

    For that Base is supplied with 5V through a current limiting resistor and emitter connected to GND(In common emitter config), but in common collector config base is supplied with 5V but emitter is not connected to GND(it's connected to one end of load not GND). There's nothing at the emitter(no + volt or GND) since load is OFF. Isn't it like Emitter left open? So how could Vbe ~ 0.6 in common collector mode if E is not GND)??


    Untitled.png
    Also Why that resistor is called current limiting resistor.?

    If base supply is 5V and base current say 300 mA then,

    Rb = (5-0.6)/0.3 = 14.67 ohm

    Which means that the excess voltage (5 - 0.6 = 4.4V) is dropped using the resistor. So isn't it limiting the voltage??

    If I use a resistor in base, voltage across load will be much less than 5 - 0.6, won't it be 5 - 0.6 - drop across base resistor?? is it necessary to have a resistor in base?

    If I use a 5V LED, why would I need a resistor in series with LED?
     
    Last edited: Aug 10, 2016
  11. Sunnysky

    Sunnysky

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    Jul 15, 2016
    LED Vf is based on chemistry and power rating for Vf. Higher power have lower ESR.
    ESR has a wide tolerance depending on source quality.

    THere are no 5V LEDs but some lamps have a series R added.
    Red/Yellow 5 mm LED = 1.8 V + 15Ω * If. or Vf=2.1 nominal. (15Ω is estimated ESR in 5mm)
    Blue/White 5 mm LED = 2.8 V + 15Ω * If. or Vf=3.1 nominal
    for If rated at 20mA continuous max. but can be pulsed to 30mA

    Emitter Followers are convenient non-inverting current buffers where the voltage drop from 5V input but depends on hFE thus has more variation in Ie current, unless Emitter R is added as a current limiter and Vbe is saturated.

    Transistors used to drive LED's are always used as switches with ESR and thus voltage drop.
    It is unwise to drive LED without a series R unless you compute all the losses (series R) which limit current.

    e.g. You can put any white LED directly across a 3.0 V Lithium battery ( not LiPo) without any added R
    For 5 V with white LED, the series R added would be 5- 0.7- 3.1V / 20ma = 60Ω approx.

    So LED Driver can be CE or CC, but CE which inverts polarity is the more conventional switch mode than CC because as a low side switch, input voltage with Rb can be as low as 1v logic, while collector can be any voltage and string with many series LEDs and series R.

    so like a relay voltage can be selected bias for input independently for a "low side" CE switch with NPN with LEDs +R on collector.

    USing emitter follower or unity gain buffer or CC as you call it, is less flexible in choosing configurations when more than 1 LED is involved but DOES WORK with suitable R added.

    Always use minimum or worse case hFE when computing input bias current.
    For CE switch, use hFE=20 or 10 if you want rated Vce(sat)

    LED's are forward conducting diodes but behave like low voltage reverse biased Zeners, however LED's like many Vbe transistors must never be reverse biased more than 5V as the leakage path is tiny and any small current like 100uA in reverse bias will burn it out and short.
     
    Last edited: Aug 10, 2016
  12. Sunnysky

    Sunnysky

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    Jul 15, 2016
    if you drive Vb from 5V without current limiting to a emitter white LED, it may work in an emitter follower with a Series R but you are only using the transistor as a diode and all current goes thru base and none thru collector. bad practise.
     
  13. Sunnysky

    Sunnysky

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    Jul 15, 2016
    Also if you know for a limited current range, in a low power diode Vf=0.65 +/-0.05 , and say a white LED is 3.2 is 5x this forward voltage , .... can you understand how it's % tolerance is 5x this amount? or +/- 0.25, this is a justification for the wide range in Vf in LED specs which is effectively the tolerance on 15Ω ESR. But ESR is never spec'd, you just can compute this from the VI slope curve.
     
  14. Ratch

    Ratch

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    Mar 10, 2013
    You have good reason to doubt your understanding of BJTs. Your perception of how a BJT works, as described below, is wrong. By the way, the word you want to use is patience, not patients.


    A transistor is a transconductance device (voltage controls current). The current present in the base terminal does not control the current in the collector or emitter. It is waste current that has to be shunted away. Again, this waste current (Ib) is an indicator of the current present in the emitter and collector, but it does not control those currents. Vbe controls the emitter and collector current (transconductance, remember).

    Yes, the sum of Vbe and the emitter voltage has to equal the voltage at the base terminal.

    You mean if Vb is constant. That process is called feedback.

    300 what? Amps, milliamps, microamps,what? Who cares? If the transistor had a beta of 200, twice the amount of current will exist. Beta is a highly variable parameter in a transistor, unless you are willing to pay a lot of money to select transistors that have a certain beta range. Are you trying to make a current amplifier? If not, don't get hung up on beta.

    You are lost at sea.


    Collector and emitter current will exist anytime the Vbe is greater than zero. It becomes significant when the voltage becomes 0.6 volts or so.

    upload_2016-8-10_8-48-19.png

    Is not the emitter connected to ground through an emitter resistor?


    The transistor would burn up if you put 300 ma through the base circuit.


    It limits both voltage and current.

    OK, let's stop wandering around in a fog and take a specific example. Suppose the LED needs 1.5 volts to shine brightly, and its current must not exceed 10 ma. Both the LED and its current limiting resistor are in series in the emitter circuit. Assume 5 volts is connected to the base and 8 volts on the collector. The current limiting resistor should be (5 - 0.6 - 1.5)/0.010 = 290 ohms. Notice that we don't have to get involved with the beta. Any transistor with a significant beta will do.

    If you want to get involved in base resistors, you can read this article. http://electronicdesign.com/analog/avoid-clipping-emitter-follower-ac-coupled-resistive-load . If it is too technical for you, just ignore it.

    Ratch
     
    Last edited: Aug 10, 2016
    hevans1944, LvW and Harald Kapp like this.
  15. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Your error lies in the term 'base voltage'. You have to consider base-emitter voltage (Vbe), which is by no means constant but varies with emitter voltage, base current, temperature etc.)
     
  16. Sunnysky

    Sunnysky

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    Since Vbe drifts downward with rising temperature, it is more useful to view transistors as current amplifiers within several decades of specified output current. However in emitter followers The output will be just the Below Vbe and load current is limited by emitter resistance, but as long as more base current than min hFe requires, it can still drive an LED.

    When Vce drops below 2V , hFE drops rapidly causing significant nonlinearity and thus used as switches with nominal specs given with Ic/Ib=10 for Vce(sat) specs.

    - special parts with different construction use higher specs such as Ic/Ib=50 for Vce(sat) and also have hFE's >300 in the linear range. See Diodes Inc (.nee Zetex ) for patents on these parts.
    They can produce Vce= 0V at low currents and output Vce is linear so that it has a rating like MOSFET's called Rce or the bulk CE resistance when saturated. They are also more expensive.
     
    Last edited: Aug 10, 2016
  17. Ratch

    Ratch

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    Mar 10, 2013
    Why does a Vbe tempco of -2mv/C° justify classifying a BJT as a "current amplifier"? As long as it is operated in its active region, a BJT is a transconductance amplifier.

    And the point is?

    Are we talking about operation in the active region?

    Ratch
     
  18. Sunnysky

    Sunnysky

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    I was referring to the common practise of choosing a min hFE when Vce is saturated or less than 2V at high current rather than using the transconductance model for reliable operation.
     
  19. LvW

    LvW

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    May I point to the fact, that Vbe does not "drift" with rising temperatures? Instead, it is the collector current Ic that increases with the temperature - and we have to reduce Vbe externally by app. 2mV/K to bring Ic back to its former value.
    So - where is the advantage to view the transistor as a current amplifier?

    Remember: The tempco is -2mV/K for Ic=constant.
     
  20. Athul

    Athul

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    Aug 4, 2016

    So if beta = 300, thrice the current will exist (Emitter current), but how?


    I only knew few equations that's why I went through all this steps to connect beta and Ie. but from your reply "You'r lost at sea" I guess I'm wrong.

    So please clarify how it gets changed depending on gain?? Is there any equation??

    One more Is there any current flowing through collector ( current taken from supply connected to Collector??

    Or is it only flowing through base - emitter circuit?
     
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