Quiescent

Discussion in 'General Electronics Chat' started by Ndjs, Oct 25, 2012.

  1. Ndjs

    Thread Starter New Member

    Sep 26, 2012
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    Hi guys I have a question in relation to the the quiescent value on the load line.
    I understand what it is and why it is used but calculating it in the following circuit is difficult to understand for me.

    Transistor's

    What is the Power Dissipated at the mid-point of the load line (I'm guessing this means the Q point)

    An NPN transistor

    If I have a 5V (Vbb) supply to a series resistor 100k (Rb) on the base,
    and a Vcc of 15V in series with a 2.2k (Rc) resistor on the collector,
    Emitter tied to ground

    Can I use (Vcc/2)^2/Rc = (7.5)^2/2200 = 26mW



    Sorry i don't know how to add pictures of the circuit.
     
    Last edited: Oct 25, 2012
  2. bertus

    Administrator

    Apr 5, 2008
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  3. Ndjs

    Thread Starter New Member

    Sep 26, 2012
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    picture as attachment
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    Nobody ever biases a transistor like that except if it is an on-off switch.

    A transistor has a wide range of current gain (hFE) which could be low, medium or high.
    In your EXTREMELY simple circuit if its hFE is high then it is saturated (almost a dead short all the time).
    If its hFE is low then the transistor is cutoff (does not conduct).

    Therefore a resistor is added in series with its emitter and a resistor is added from its base to ground so that the hFE does not matter.

    The quiescent collector voltage is 8V and the emitter voltage is 0.7V so the transistor has 7.3V across it and its 2.2k collector resistor has 7V across it so the current is 3.2mA. Then the transistor dissipates 7.3V x 3.2mA= 23.4mW which is almost nothing.
     
  5. Ndjs

    Thread Starter New Member

    Sep 26, 2012
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    Audioguru I realise it is a simple circuit but we all must start on simple circuits to understand the theory. Yes it will be used as a simple ON/Off switch.
    In my original circuit could i say the Q is half Vcc (15V) and then use V^2/R to find the power when it is at Q
     
  6. Audioguru

    New Member

    Dec 20, 2007
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    A transistor used as a simple switch does not have a quiescent biased at half the supply like an amplifier.

    When the transistor is turned on then its collector to emitter voltage is very low so its power (heating) is also low.
    When the transistor is turned off then its current is zero so its power (heating) is zero.

    The transistor used as a simple switch is saturated when turned on.
    The datasheets for nearly all transistors show the base current should be 1/10th the collector current for a saturated transistor.
     
  7. Ndjs

    Thread Starter New Member

    Sep 26, 2012
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    0
    Okay thankyou.

    By doing more reseearch I have found that Q value is a value which is in the active region and makes the transistor forward bias. And by using the Q point it reduces the noise created by this.

    My original question is that my assumption that the middle of the load line is the Q point but it is not.
     
    Last edited: Oct 26, 2012
  8. Audioguru

    New Member

    Dec 20, 2007
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    A transistor used as a simple switch is on or off. It does not produce noise and does not have a Q point. A transistor used as an amplifier has a Q point.

    Yes, when the transistor is turned on by forward bias on its base then it is active.
     
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