Help with the designing transistor as a switch (saturation region)

Discussion in 'Homework Help' started by micktosin, Apr 24, 2012.

  1. micktosin

    Thread Starter New Member

    Mar 20, 2012
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    0
    Hi there
    I'm trying to findout how the resistor values of transistor shown below was calculated. From my understanding, this circuit was designed to bias the transistor into saturation region and that's all i know about it. Please how do i go about this, as i need this understanding to design other non linear Circuit?
    [​IMG]

    Thanks in advance.
     
  2. Tintin cassie

    New Member

    Apr 24, 2012
    1
    0
    Maybe you can sent me the details of this problem except the given variables,since it has capacity.;)
     
  3. micktosin

    Thread Starter New Member

    Mar 20, 2012
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    Last edited: Apr 24, 2012
  4. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    That circuit appears to be for biasing an AC amplifier in its linear region. Look at "Voltage Divider Biasing" in this reference.
     
  5. micktosin

    Thread Starter New Member

    Mar 20, 2012
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    This circuit was used as a Non linear amplifer, as we had lot of harmonics coming out.
     
  6. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    Is there anything else we need to know about what the circuit is doing? :rolleyes:
     
  7. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    You have to tell us everything thing about this circuit, as it doesn’t make much sense to me .Using voltage divider at base their will be approx 0.7 volts which is not sufficient for the Vb (voltage at the base) as normally a silicon transistor’s base-emitter junction (Vbe) in forward bias drops approx 0.7 volts and Vb is about Vbe + Vre i.e.. voltage drop across emitter resistor. So Vre will be very small and if Vre is small the current through it will be also very small i.e.. the emitter current (Ie) now we know Ie = Ib + Ic ,so a very small Ib will be not sufficient to drive the transistor into saturation mode. So no way its going to be saturated.

    The C1 is a bypass capacitor normally used in AC amplifiers , and C2 is filtering capacitor with a very small value that I don’t understand why? Sometimes we use these to eliminate noise from supply line and place the capacitor very close to the component.

    Anyway without details of your circuit I can’t say much…

    Good Luck
     
  8. micktosin

    Thread Starter New Member

    Mar 20, 2012
    19
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    I'm sorry for not providing enough information. This circuit was used as a part of frequency multipliers chain for Transmiter, of which tuned filter was used to select the harmonic.
    However, after getting it to work, i couldn't understand how this Voltage divider was calculated base on the Transistor Specifications.
    My question is, what step do you think were taken in designing this Circuit to give a value of the resistor shown above?

    P.s. I know how to design Voltage divider in the linear region, but not in the saturation region this circuit was designed for.

    Thanks.
     
  9. micktosin

    Thread Starter New Member

    Mar 20, 2012
    19
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    This circuit was taken form the highlighted stage in the following diagram, but with diffferent transistor (MRF951).
     
  10. bertus

    Administrator

    Apr 5, 2008
    15,648
    2,348
    Hello,

    In case of a transistor frequency multiplier, the transistor can be biased as an almost class C amplifier.
    You might also have noticed that C17 and C21 are grounding the signals for RF.

    [​IMG]

    Bertus
     
  11. micktosin

    Thread Starter New Member

    Mar 20, 2012
    19
    0
    Thank you for the help . But i'm not sure how this method can be use to bias it? Please can you elaborate more on this method?
     
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