HELP-transistor constant current source problems

Discussion in 'Homework Help' started by Himanshoo, Apr 3, 2015.

  1. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    Refer first the attachment ..
    The following circuit is a single transistor constant current source..and my query is regarding the behaviour of this circuit..

    Since Vcc is constant at 10 V and Vc at 1.2V because Vb is held constant at 1.6 V by the voltage divider.
    given:
    Iload=(Vcc-Vc)/Rc

    Ic= I load= (Vcc-1.2)/ Rc.



    Query 1 : If we change Rc the its true from above equation that Ic or I load will also change…which is not desirable as the circuit is constant current circuit. Then why the text tells that I load wont vary if the load resistance is varied from 0 to 8800 ohm???



    Query 2: How the transistor saturates beyond 8800 ohms, since saturating a transistor depends upon manipulating Ib.(and here Ib is some what constant)???



    Query 3: How the text tells that I load is independent of Vcc as long as the transistor is not saturated.



    Since Vc=0 (saturation)



    then Rc = (Vcc-Vc) / Ic



    =(Vcc-0) / I load



    Rc = Vcc / I load……………..here it seems that I load completely depends on Vcc.



    but what happens for other values of Vc (between 0 and 1.2) for which transistor is saturated.(say Vc =1.1V)



    i.e Vc = 1.1( saturation)



    Rc = (Vcc -1.1)/I load……………..here it seems that I load not completely depends on Vcc.



    so whats this anomaly???



    regards
     
    Last edited: Apr 3, 2015
  2. dl324

    Distinguished Member

    Mar 30, 2015
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    Hi Himanshoo,

    Welcome to the forum. I'm new too...

    Does this happen to be a homework problem?

    BR
    Dennis
     
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  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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  4. #12

    Expert

    Nov 30, 2010
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    Your first error is in assuming that the transistor is always saturated and the collector voltage is always 1.2 volts. This circuit accomplishes a constant current flow for any Rc less than 8.8K by using up the extra voltage with its collector to base voltage.
     
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  5. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    could u please elaborate "using up the extra voltage with its collector to base voltage".
     
  6. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    no its regarding my personal reserch..
     
  7. #12

    Expert

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    There are not always 8.8 volts across the resistor. There are not always 1.2 volts across the transistor. For instance, when 1 milliamp flows through 1000 ohms load resistor, that resistor will have 1 volt across it and the transistor will have nine volts on its collector.
     
  8. dl324

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    Mar 30, 2015
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    Hi Himanshoo,
    Then my apologies because my asking caused a Moderator to move your post to the homework help section which may affect the number of people viewing it.

    That being said, as you've mentioned, the transistor is biased as a current source. If you ignore the slight difference between Ie and Ic, you have a 1mA source. Assuming an ideal transistor, the voltage drop across any load can be up to 9V and Vc will vary according to the load resistance.

    HTH
    Dennis
     
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  9. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    what do think about query 2.
     
  10. dl324

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    You need Vce to be 0V (ideal case) for the transistor to saturate; 8.8K would make Vc 1.2V which is what an actual transistor would typically achieve. You don't need to manipulate Ib if you increase the load to 8.8K.

    BR
    Dennis
     
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  11. Veracohr

    Well-Known Member

    Jan 3, 2011
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  12. WBahn

    Moderator

    Mar 31, 2012
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    Your basic problem is your premise. Yes, Vcc is constant at 10 V, but where does the notion that Vc is constant at 1.2 V come from?

    The bias circuit sets a voltage at the transistor base of 1.6V.
    The transistor adjusts its internal behavior, if possible, to result in there being about 0.6V from base to emitter.
    This results in the emitter of the transistor being held at 1.0V.
    This results in the voltage across the 1.0kΩ emitter resistor being 1.0V.
    This results in the current in the emitter resistance being 1 mA.
    This results in the emitter current of the transistor being 1 mA.
    This results in the collector current of the transistor being 1 mA (just very slightly less, due to base current).
    This results in the current in the load resistor being 1 mA.
    This results in a voltage drop across the load of Vload = (1 mA)(Rload).
    This results in a voltage at the transistor collector of Vcc - Vload.

    The transistor is able to adjust its internal behavior only as long as the end result is that the voltage across the collector-emitter terminals is at least Vcesat, which is commonly taken to be 0.2V.

    That's the sequence of steps you should probably view the behavior of this circuit in.

    With that in mind, see which of your queries is still unresolved.
     
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  13. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    As we all know that manipulating Ib we can saturate a transistor (make Vc=0)...Is it any kind of reverse effect or so that by making Vc=0(ideally) we could saturate a transistor no matter what base current is there ...is this what you are trying to say...
     
  14. #12

    Expert

    Nov 30, 2010
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    Who said anything about saturating this transistor? It is in a constant current configuration. For all load resistors less than 8800 ohms, it is not saturated and should not be saturated. You can not make Vc =0 because real transistors do not saturate to zero volts and besides that, this circuit has an emitter resistor so the Vc can never be less than the voltage across the emitter resistor.

    So, no. Nobody is saying anything about saturating a transistor.
     
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  15. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    Ok now i do agree with u that Vc varies...(its isnt constant)...as the base voltage is constant since (it wont effect saturation of transistor)...now the remaining thing is how can we saturate a transitor ..it could be done by varying the power supply(which is constant here) or by altering Vc (which though can be done by changing load resistance)..
    Now lets say we have changed load resistance to a new value..its obvious now Vc will change ok...and if we substitute new values of Vc and Rc in equation------> Iload=(Vcc-Vc)/Rc
    it would definitely change Iload..which is not demanded now here where does transistor regulatory internal behaviour comes into picture?
     
  16. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    Let say i changed my load resistance to 8801 ohms ..what will happen then? lets forget about Vc=0 for now
     
  17. dl324

    Distinguished Member

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    Follow the link Veracohr posted. The transistor will be saturated when both junctions are forward biased.

    If you change Ib, you change the load that causes saturation.

    BR
    Dennis
     
  18. #12

    Expert

    Nov 30, 2010
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    What do you think will happen?

    Remember, this is Homework. You are supposed to be able to tell the difference between, "more than" and, "less than".
     
  19. dl324

    Distinguished Member

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    In this case Vc can't be less than 1.0V. With 8801 ohms, the transistor is essentially saturated.

    BR
    Dennis
     
  20. #12

    Expert

    Nov 30, 2010
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    Well, now you gave it away, Dennis.:D
     
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