How can I set the bias here?

Discussion in 'General Electronics Chat' started by sjgallagher2, Aug 3, 2015.

  1. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
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    It sounds like a homework question but it's the summer time folks, school is out!! This is my own question :) From Horowitz and Hill I have this circuit:

    [​IMG]

    From the book and SPICE I know that the quiescent current is about 1mA and the output voltage (Vc) is 29.4V. But I've been racking my brain trying to figure out how exactly the bias is determined. With no pins on the transistor held at any specific voltage (ground, Vcc, input), it seems like I would have to solve a system in order to figure it out, but that seems overcomplicated and lame... I need to know:
    Do I need a system of equations to solve for Ic?
    And how would I set the bias for a specific quiescent current? (Resistor values)
    Thanks in advance

    Sam Gallagher
     
  2. dl324

    Distinguished Member

    Mar 30, 2015
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    There is approximately 15V on the base; subtract junction voltage and there is about 14.3V across the 13.6K resistor. That gives you about a mA emitter current.

    To set a specific bias you put a voltage divider between supply rails on the base and select an appropriate emitter resistor. Value of divider resistors depends on transistor beta; you want them large enough to minimize dissipation but small enough to provide stable bias voltage.
     
    Last edited: Aug 3, 2015
  3. Jony130

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    Feb 17, 2009
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  4. crutschow

    Expert

    Mar 14, 2008
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    To directly solve for the transistor current you do need the system equations, but for an initial estimate you can assume that Q1's emitter is about equal to one diode drop below V1 or about 14.3V and work backward from there.
    From that emitter voltage you can calculate the current through R2 which is also the transistor emitter current. That current then determines the base current from the transistors current gain (Beta or hFE).
    If that base current drops a significant voltage across R1, then you may have to reduce the voltage at Q1's emitter appropriately and recalculate the emitter current. It's an iterative method that approaches the correct value.

    Normally R1 is made a low enough value that the base current does not cause a significant voltage drop across it, which simplifies the bias equations.
    In this case there appears to be a drop of about a volt or so, depending upon the transistor gain.

    That looks like an LTspice schematic. Did you try the simulation?
     
    Last edited: Aug 3, 2015
  5. sjgallagher2

    Thread Starter Member

    Feb 6, 2013
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    Dennis, that was a big help. Crutschow, you too! I did simulate it so I know the answers its just a matter of finding out how to get there. Good engineers design circuits that don't rely too heavily on component values so I believe he used the 100k not to specifically get a base voltage (which it turns out is 14.6V) but rather to get the base pretty close but a little below 15V. In the full schematic it's actually part of a differential pair with an emitter resistor half the size connected to -15V, and the bases are grounded through those 100k resistors, but I simplified it so nobody would be confused as to what I was asking. Putting the transistors a little below ground might make sense. I feel satisfied with the approximate methods! That's the kind of electronics I'd like to get better at- not getting hung up on exact numbers with 8 decimal places.

    Sam Gallagher
     
  6. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    The technique you have is called dangle biasing. The transistor is running wide open with no feedback of any kind. The collector current is directly proportional to the gain of the transistor. While it might be stable in a simulation, in reality the transistor gain varies wildly with temperature, collector current, Vbe, Vce, the phase of the moon, and my bank balance.

    A better way is to modify the circuit so that the collector current is defined more by external circuit elements (resistors). The base still is connected to a fixed bias voltage through an impedance, but the impedance of the bias network is much lower. I'm sure there is information on this site regarding common emitter amplifier design that someone can point you to.

    ak
     
  7. crutschow

    Expert

    Mar 14, 2008
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    I don't understand that statement. :confused:
    The transistor has emitter feedback due to R2 and is not running wide open.
    Using a emitter resistor is a common method to stabilize the bias point.
     
    ErnieM likes this.
  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Why is the emitter resistor 20X the collector resistor?
     
  9. crutschow

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    He stated it's half of a differential pair, so that is actually part of the (common) emitter tail resistor and doesn't degenerate the AC gain.
     
  10. AnalogKid

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    Aug 1, 2013
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    Yeah, I thought of that after I posted, and I knew who would catch it. Still, with a 100K base resistor there is very little degen going on. I think I had the emitter and collector resistor values reversed in my head as I was writing. But a 13 K emitter resistor is unusually high. Me be thinkin...

    ak
     
  11. Bordodynov

    Active Member

    May 20, 2015
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    For this case the equation is made easy. I put the calculations on the electronic circuit, and considered response to LTspice. The most important thing to choose the correct desired value. I chose the emitter voltage. Then, the emitter current can be easily calculated by Ohm's law and the base and collector currents.

    V(E)+V(B,E)+ib*100K=15
    V(B,E)~0.75V, ib=ie/(Beta+1)
    V(E)+0.75+V(E)/13.6k/(100+1)*100k=15
    V(E)*(1+100/101/13.6)=14.25 -->
    V(E)~13.28V ie=13.28/13.6k=0.976mA
    ic=ie*Beta/(Beta+1)=0.967mA
    OUT=30-0.967mA*0.62K=29.4V

    Regim.png
     
    Last edited: Aug 5, 2015
  12. crutschow

    Expert

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    The base resistor has little to do with the degen negative feedback.
    It's the value of the emitter resistor that determines that.
     
  13. AnalogKid

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    Aug 1, 2013
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    Again, yes I know that and that isn't my point. Like an opamp with an extremely large feedback resistor ratio, having a feedback path doesn't mean the device has enough open loop gain to enable (not the right word) feedback. With such a large base resistor, and remember that I had the collector and emitter resistors reversed in my little brain, I saw a transistor running wide open. If the base is basically being starved, can degenerative feedback happen in a circuit with finite beta?

    ak
     
  14. crutschow

    Expert

    Mar 14, 2008
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    Certainly if you starved the base with a large base resistor, that would affect the bias point and stability but, in this case, the resistor drops only about a volt for typical Beta values, which will have only a small effect on the bias point for normal Beta variations.
     
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