Voltage divider with transistor

Discussion in 'General Electronics Chat' started by zapp0, Nov 11, 2014.

  1. zapp0

    Thread Starter New Member

    Nov 10, 2014
    9
    0
    Hello,

    I'm trying to understand the sorcery that transistors / semiconductors are…

    Given this simple circuit (battery=9v, P1={0-100k ohm}, t1 has Hfe of about 330)
    [​IMG]
    as far as I understand, once a semiconductor (like a diode or base-emmiter) is put in parallel with one of the resistors, ohm's law takes a break and goes fishing and voltage divider is no more, right?

    In my tests, I got the following results:
    |P1 |Vb |Ib |Ic |
    4k 0.5 57uA 20mA
    7k 0.53 111uA 40mA
    17k 0.54 156uA 60mA
    100k 0.5 177uA 80mA

    that's actually pretty close to what the simulator predicts. However, I have no clue whatsoever how that potentiometer controls Ib and what the relation between its resistance and Ib is?

    As far as I understand the voltage divider kinda works to the point when base actually starts opening, say about 0.5V, and after that point the voltage is constant (same as would be for a diode) and Ib starts growing. What is the formula for tor calculating Ib in relation to R2? What do I need to read to understand the basics of this (and not going all down to quantum level and holes etc).

    Thanks :)
     
  2. wayneh

    Expert

    Sep 9, 2010
    12,145
    3,055
    Ohm's law is still your friend, for the most part. The base current starts to flow when the base voltage is ~0.65V above the emitter voltage (battery negative in your schematic). The increase in current across R1 increase the voltage drop across R1, and the current through R2 depends on the ∆V (which stays at ~0.65V while the base is conducting) and the value of R2. The current through R1 has to be equal to the sum of the base current and the current through R2. If you lay out these rules mathematically, you can define the situation. Make yourself a spreadsheet and map it out.
     
  3. zapp0

    Thread Starter New Member

    Nov 10, 2014
    9
    0
    I see… so Ib=Ir1-Ir2=Vr1/R1-Vr2/R2=(Vb-Vr2)/R1-Vr2/R2

    Vb=battery voltage which is constant
    Vr2 is a bit tricky though. If I assume it's constant at 0.65V it would be ok - but it's not. How can I calculate it?
     
  4. adam555

    Active Member

    Aug 17, 2013
    858
    39
    Have you read the Voltage Divider Bias in the Biasing Calculations chapter of Volume III of this forum's textbooks?
     
  5. crutschow

    Expert

    Mar 14, 2008
    13,028
    3,237
    Vbe is not completely constant but its variation is small enough when the base is conducting that it can be considered to be constant for the purposes of bias calculations at typical bias currents.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    It would be nice if you defined the quantities you are using so that we don't have to read your mind, reverse engineer your work, or guess. Doing so might also keep you from making the kind of mistake you made here.

    Which resistor is R1 and which is R2?
    From your claim that Ib=Ir1-Ir2, we have to conclude that R1 is the top resistor and R2 is the bottom resistor.

    What is Vr1 and Vr2? From your first use of them, Vr1 has to be the voltage across R1 and Vr2 has to be the voltage across R2.

    But then what is (Vb-Vr2)? You claim that this is equal to Vr1. Does this make sense?

    Near room temperature, Vbe changes by about 60mV for every order of magnitude change in base current. So for most calculations it can be considered constant. As for the absolute value (i.e., is it 0.5V or 0.7V or somewhere in between), if your circuit is sensitive to which it is, then it is almost certainly a poorly designed circuit.
     
  7. zapp0

    Thread Starter New Member

    Nov 10, 2014
    9
    0
    Thanks, will try to understand that and come back with questions.
     
  8. zapp0

    Thread Starter New Member

    Nov 10, 2014
    9
    0
    Ahh.. seems I need to understand the voltage divider better first.. and how it behaves under load… and how it can be replaced by Thevenin substitution.
    I did some calculations and replacing voltage divider with whatever voltage it outputs + resistor; and of course V at that resistor is Vout-0.65, hence I is also variable, and it seems that's exactly the (theoretical) Ib...
     
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