Voltage divider bias...Cant understand theory

Discussion in 'Homework Help' started by billion_boi@hotmail.com, Aug 11, 2009.

  1. billion_boi@hotmail.com

    Thread Starter New Member

    Jan 24, 2009
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    hello in a simple voltage divider bias.
    I understand how the emiter resistor stabilises the current against temperature changes..But then what is the function of R2?
    [​IMG]
    Since the base current can be lowered due to RE, that would throw off voltage levels in the base loop and thus the emitter voltage...How does R2 stabilise for the change of voltage?
     
  2. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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  3. billion_boi@hotmail.com

    Thread Starter New Member

    Jan 24, 2009
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    actually that was coincidentally the last article i read before posting here. I couldn't at all understand what it meant..
    I can get the exact voltage i want at the base of the transistor by adjusting R3.So whats R2 for?

    Here are some exerpts that i couldnt follow from the wikipedia entry
    The voltage across R2 forward biases the emitter junction.
    -wouldnt it be forward biased with or without R2, or have i misunderstood?

    By proper selection of resistors R1 and R2, the operating point of the transistor can be made independent of β.
    --NO idea...
     
  4. JDT

    Well-Known Member

    Feb 12, 2009
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    It would be possible to build the circuit without R2 but then the operating point of the transistor would depend on the current gain (hfe) of the transistor. This is normally very variable from device to device; example: BC548 hfe=110 - 800.

    The voltage divider R1-R2 has a standing current many times the transistor base current (even at the lowest value of hfe) and presents a fixed voltage to the base instead of a current. The current through the transistor (Ie) will then = (Vb - 0.65)/Re which will be quite accurate and temperature stable.

    Disadvantages: R1//R2 shunts the input signal more than a single high value R1 only.
    Advantages: Less dependence on hfe. Better stabillity.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The Wiki page explains how this can be "true" by making an assumption which simplifies the general equation for Ic - as reproduced below ...

    [​IMG]

    which is approximately the case if

    [​IMG]

    Notice that with the approximating condition made good, the equation for Ic has no reference to β - which is the point being made.

    You'll note elsewhere on the same page that is says ....

    "Operating point is almost independent of β variation." This is closer to the "truth" - but what is truth?

    In any case JDT is on the money - and the goal of the circuit is to achieve an operating point as robust as possible against variations in both temperature and β.

    You might find some value in playing around with the maths for various bias circuits to determine which gives the least susceptibility to parameter variations. If you have access to a circuit simulation package this can prove an enlightening exercise without too much mathematical exertion!
     
  6. ELECTRONERD

    Senior Member

    May 26, 2009
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    R1 and R2 form a voltage divider. According to the following calculation: (R2 / R1 + R2) x Vin. So if I have two 10K resistors and a input voltage of 9V I get 4.5V on the output (10K / 10K+10K = 1/2 x 9 =4.5V.
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    Only without a load current. A load current will cause the divided voltage to be reduced.
     
  8. ELECTRONERD

    Senior Member

    May 26, 2009
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    Right, voltage dividers can't supply much current. An alternative would be better if you want to supply higher current, but for transistors it works fine.
     
  9. billion_boi@hotmail.com

    Thread Starter New Member

    Jan 24, 2009
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    Ok i think i finally understand it.
    t_n_k those were actually the equations that scared me off from Wikipedia...i couldn't understand where all the "1"s came from...

    but bare with me...can someone please confirm my understanding of all this..

    If we didnt have R2

    because hfe is changing (due to heat) so is Re's internal reflective resistance which changing current levels in base loop.

    Now if we add R2
    because of its relatively low resistance compared to emitters reflective resistance (even with a low hfe)
    we can assume that the resistance has been made constant to levels near r2.

    This in turn keeps base current stabilized.Which in turn keeps R2's voltage drop constant.. which in turn keeps re's voltage drop constant (vb-.7)...which in turn keeps collector loops current constant..

    And if for some reason collector/emitter current is not constant and Re does receive an increased current...well then its increased voltage would be felt in the base loop via negative feedback. which in turn would reduce base current.
     
  10. ELECTRONERD

    Senior Member

    May 26, 2009
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    If you didn't have R2, you would simply have 15V at the base and R1 would be adjusting Ib. If you did have R2, you would be lowering the base voltage, but you couldn't supply as much current to the base if you just had R1. Although, 15V is a pretty high voltage for the base.
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    No.
    The base voltage will be much less than 15V unless R1 is a very low resistance. The base current develops a voltage drop across R1 which reduces the base voltage.
     
  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    hfe is more device fabrication dependent than temperature dependent. There can be a lot of spread in hfe for the same device type. Vbe is more temperature dependent. Vbe decreases by about 2.5mV per deg increase in temp.

    With just R1 (no R2), it's quite possible you could design your target bias conditions with an assumed value of hfe (β) and end up with a bias condition way off from what you expected - say you design for a β value of 100 for the selected transistor. Suppose the transistor actually obtained has a β of 300. Your actual Ic will be quite different from what you expected - you could end up with the transistor in saturation rather than in the linear region.

    As JDT points out with the R1 & R2 divider approach you actually design the divider current to be perhaps an order of magnitude greater than the expected base current. With the added benefit of having (1+β)Re >> R1||R2 the base voltage then becomes quite stable and the value of Vbe is the main source of variability.

    Have you tried doing a design for the two biasing methods? - take β = 100 to 300 say, Vbe = 0.65 Volts (fixed). Design for Ic of 1mA with 15V supply, Rc=10k and Re=2.2k. In the R1 + R2 case, set the divider current to be about 100uA (Ib would be about 10uA for Ic=1mA at β=100).

    For example set R1=105K and R2 = 27K in the divider case. Compare the behavior of the two bias arrangements and you will quickly see how Ic can be better stabilized with the R1 + R2 divider option.

    For the R1 alone case, you would need R1=1.2MΩ with β=100 to give Ic=1mA while using the Re value I gave above (Re=2.2k). With the same R1 (1.2MΩ) alone but with β=300 the transistor would be saturated.

    For the case of the R1 + R2 divider using the values 105K and 27 K Ic would vary from 992uA to 1.06mA which is pretty stable.
     
  13. billion_boi@hotmail.com

    Thread Starter New Member

    Jan 24, 2009
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    Tnk,JDT thanks alot. I understadn now that the voltage divider,specifically the voltage at R2, sets up our Iq...and that its is now independent of Beta.

    So i think that the A.C input voltage is superimposed on R2 thus adjusting the voltage at Re (and hence Ie) and creating our output. On the newer schematics though the bypass capacitor in parallel with Re, creates a A.C short...What the purpous of creating an A.C short? Afterall our A.C input has already made an affect on Re's voltage because of R2....
     
    Last edited: Aug 17, 2009
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