current

Discussion in 'General Electronics Chat' started by TAKYMOUNIR, Nov 9, 2014.

  1. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    What the current according to the vinput
     
  2. crutschow

    Expert

    Mar 14, 2008
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    I'll bite. Sounds like homework. What do you think the current is?
     
  3. JDT

    Well-Known Member

    Feb 12, 2009
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    What's that diode doing on Vinput? Have you drawn it correctly?
    Is your input voltage positive or negative?
    If it is correct and positive then your homework will be easier!
     
  4. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    this is not homework this ciruit i use it in my work
     
  5. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    yes the input is poistive
     
  6. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    if the input is 0 vdc ,this mean that diode is forward biasing and emitter of Q1 WILL BE ZERO VOLTS AND IF THE INPUT IS HIGHER THAN(1.89+.7) ,THE DIODE WILL BE REVERSEBIASED AND VB OF Q1 WILL BE 1.89 VDC ,SO IS THIS TRUE?
     
    Last edited: Nov 10, 2014
  7. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    SO WHEN VIN =0 VDC THE CURRENT IN 50 OHM IS ABOUT 416 MICRO AMP AND WHEN THE DIODE IS REVERSE BIASED THE CURRENT ABOUT 449 MICRO AMP? SO IS THIS TRUE?
     
  8. alfacliff

    Well-Known Member

    Dec 13, 2013
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    I dont see a 50 ohm resistor on the supplied circuit.
     
  9. shortbus

    AAC Fanatic!

    Sep 30, 2009
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    r4
     
  10. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    R4
     
  11. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    R4 CONNECTED TO COLLECTOR OF Q2
     
  12. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    Hi Crutshow can you see if the current at vin=0 is 416ua and at vin>2.6 the current is about 449 ua
     
  13. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    No, for sure. For Vin = 0V Q1 is Cut off and Q2 is equal to
    Ic2 = ((15 * 3/30) - 0.7 + 15)/18 = 0.8777mA. And Q1 will start to turn on if Vin > (15 * 3/(30) - 0.7) = 0.8V and Ic2 current will start to drop from 0.877mA to 0A
     
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  14. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    so E of Q1 will be .8v and B of Q1 will be about 1.5 v and what the voltage of the cathode of diode will be?
    and also when vin =zero this mean that diode forward biased and most of the current will go through the diode so Q1 will be off?
     
  15. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    and the diode will be out of circuit when vin >(1.89v+.7) and when Q1 will turn on E will be about 15 v that is why Q2 will be off
     
  16. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    and if vin is negative Q1 will never turn on?
     
  17. Jony130

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    Without the diode Q1 base voltage is equal to :
    Vb1 = 15 * 3.9/(27 + 3.9) = 1.89V. And if we assume that diode will be OFF if Vd =0.5V. For Vin > 1.89V - 0.5V = 1.39V diode is OFF.
    And yes, for negative voltage diode is ON so Q1 is OFF.
     
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  18. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
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    thanks so much and what is vd? is this v anode of diode or what?
     
  19. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Vd = V_anode - V_cathode = diode forward voltage drop
     
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