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Matched biasing transistor

  1. Sep 4, 2016 #1
    pdx9T.png
    I have a few questions about the circuit above.
    1.This is an ordinary CE amplifier, which uses current mirror to provide temperature stabilization and bias for Q2, right?
    2. Like Collector feedback bias and Voltage divider bias, this is just another way of biasing a transistor to provides temperature stabilization, correct?
    3. If 2 is right, then does it have some advantages over the others?
    4. Why are the 10k base resistors? Current mirrors seems to lack these.
    5. How to calculate the resistors, for what voltage. Since it seems that there are two variations of this circuit, let's first talk about the one without the 180ohm emitter resistor. Since VBE is 0,7v, I assume that VB(orVC) must be atleast with 0,1v higher than VBE. So VB will be 0,8v, and across the base resistor( the 10k in this case) 0,1v will drop. So the equation for the base resistors will be Rb = VB-VBE/Ib= 0,8-0,7/Ib.
    About the 20k resistor, RC=Vcc-Vb/Ic+Ib1+Ib2= 20 - 0,8/ Ic+Ib1+Ib2. Is that right?

    And now with the emitter resistor. I've heard it's good to leave atleast 1v to the emitter. So, VC will be VE+VB=1+0,7=1,7V. From there, RB= VC-VBE/Ib=1,7-0,7/Ib= 1/Ib. RE=1v/Ie. That correct?
     
    Last edited: Sep 4, 2016
  2. jcsd
  3. Sep 4, 2016 #2

    NascentOxygen

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    Staff: Mentor

    I don't follow your thinking here. Without RE then VB is going to be the same as VBE.

    Note: VB is how we denote the base voltage relative to ground.
     
  4. Sep 4, 2016 #3
    In my textbook, this circuit can be done with or without adding an emitter resistor which in this case is 180ohms. I first did the calculations if there is no such resistor and then if there is. изтеглен файл.png
    See, this is the variation of this circuit without the 180ohm resistor.
    And I ask if this is the way to calculate it.
     
  5. Sep 4, 2016 #4
  6. Sep 4, 2016 #5
  7. Sep 4, 2016 #6

    Svein

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    Science Advisor

    Yes...
    Yes...
    Since the current in Q1 is set by the 20k resistor and the voltage across it is close to +20V, the voltage across the 10k resistor in the collector of Q2 will be half of the voltage across the 20k resistor - which means that the DC voltage at the collector of Q2 will be 10V.
    The 10k resistor going to the base of Q2 sets the input impedance of Q2. Therefore Q1 must have a 10k resistor as well (otherwise the current mirror would not balance).
    The 180Ω emitter resistor in Q2 gives a mild negative feedback, linearizing the transfer function of Q2 and stabilizing the operating point. Therefore Q1 must have an identical emitter resistor (for balance).
     
  8. Sep 4, 2016 #7
    But do you know why ?

    OK

    Are you sure ??
    If Vc is 1.7V so how can in the same time Rb and Re be equal to 1V?


    As for the circuit in post 1

    Vcc - ( Ic1 + Ib1 + Ib2)*Rc1 - Ib1*Rb1 - Vbe1 - Ib1*(β+1)*Re = 0

    If we assume Vbe1 = Vbe2 = Vbe = 0.7V and Ib1 = Ib2 and β1 = β2 = β = 100 (transistors are perfectly matched)

    then we can write this

    Vcc - ( β* Ib1 + 2*Ib1)*Rc1 - Ib1*Rb1 - Vbe1 - Ib1*(β+1)*Re = 0

    And from this we can solve for Ib1.

    Ib = (Vcc - Vbe)/ ( Rb + (β + 2)*Rc1 + (β+1)*Re ) = (20V - 0.7V)/( 10kΩ + 102*20kΩ + 101*180Ω) = 9.332μA

    And collector voltage is equality to

    Vc1 = Vcc - (β + 2)Ib * Rc1 = 20V - 102*9.332μA*20kΩ = 0.962V

    And the voltage drop across Re resistor is :

    VRe = (β+1)*Ib1*Re = 101*9.332μA*180Ω = 0.169V
     
    Last edited: Sep 4, 2016
  9. Sep 4, 2016 #8
    Thank you! I've got one more question if you don't mind. I don't get how this circuit can compensate heat. I mean I believed I do, but things got complicated. What I know is that when temperature increases, Ic1 will slightly decrease the collector voltage. This will reduce the potential difference between the emitters of both transistors and the collector, resulting in a decrease of both base currents. Everything is fine for now, I understand it all. However, when Ib1 decreases, Ic1 will also decrease until the collector voltage is back to normal, back to around 0,8v. And then it seems to me that Ib2 will again be subjected to its original voltage and will rise again. So, can you help with this one, what is it that I don't understand?
     
  10. Sep 4, 2016 #9
  11. Sep 4, 2016 #10
    Okay, Jony130, i've looked at your loop rule equations, and I understand it, but isn't it too long and complicated to use it for setting up the bias? I mean, sure there must be an easier way. Say if Vbe=0,7V, Ve=1V, Vcc=20V, the voltage we need at the collector node of Q1 will be Ve + Vce(sat)? Im not sure if Q1 is saturated though. So Vc will be 1,2V. Then knowing what Vc needs to be, we calculate Rc = Vcc-Vc/I. Then knowing that Rb1=Rb2=Rb and Ib1=Ib2 we can find the value of the base resistor: Rb=Vc-Vbe/Ib
     
  12. Sep 5, 2016 #11

    Svein

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    It isn't. Saturation is when Vce<Vbe. In this case Ib = (Vce-Vbe)/10k, so obviously Vce>Vbe.
     
  13. Sep 5, 2016 #12
    Yep, but when we design a circuit we do not have to use those equations. The Ohm's law is what we need in most cases when we designing an circuit.

    Q1 is not in saturation. In saturation the collector current is not β times the base current anymore.
     
  14. Sep 5, 2016 #13
    Okay, I understand now. My thanks to all for the help!
     
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