Can you explain why AC source win?

Discussion in 'General Electronics Chat' started by screen1988, Jun 5, 2013.

  1. screen1988

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    Mar 7, 2013
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    This is part of Biasing techniques in the ebook: Biasing techniques
    [​IMG]
    Can you explain why AC source win?
     
    Last edited: Jun 5, 2013
  2. LDC3

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    Apr 27, 2013
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    I think they choose the wrong words to explain the situation. The AC doesn't really win, but it does influences the voltage across the resister. Let the voltage divider provide 2.3 volts without an AC input. Also, let the AC input be 3 V peak to peak (I know the diagram states 1.5 V rms, but I want to simply the calculations). When the AC input is the most positive, it adds 1.5 V on top of the DC bias giving 3.8 V. When the AC input is the most negative, it subtracts 1.5 V from the DC bias giving 0.8 V.
    In a sense, the AC input "wins" since the bias moves with the input voltage.
     
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  3. t_n_k

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    If the AC source were ideal then it would entirely determine the base voltage. There would be no DC bias - hence the need to interpose the isolation capacitor.
     
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  4. screen1988

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    Mar 7, 2013
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    Thanks.
    I believe in this case 1.5V is maximum amplitude (or 3V peak to peak).
    When AC source is added in parallel with R3, the voltage across R3 is always equal to Vac and I think here is what the author meant.
     
  5. t_n_k

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    This is true when the isolation capacitor is in place.
     
  6. screen1988

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    yes, got it.
     
  7. MrChips

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    Oct 2, 2009
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    The resistance of an ideal voltage source is zero. Hence the voltage source will override any value resistor at R3.

    You need to couple the voltage source through a capacitor in series.
     
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  8. Jony130

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  9. screen1988

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  10. vk6zgo

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    Jul 21, 2012
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    The whole premise of the example is faulty:-

    A CE transistor without an emitter resistor,as shown in the circuit, will have around 0.6V between base & emitter,not 2.3V.

    If there is an emitter resistor,the voltage between base & earth may be 2.3V,but it will still be 0.6V between base & emitter,
     
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  11. screen1988

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    I think it should be 2.3V across R3 not between base and emitter.
     
  12. Jony130

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    OK I see. But have you read more detailed explanation given by panic mode?

    Have you forget about 1K (R1) resistor? As you can see on the diagram we have a R1 resistor in series with transistor base. And we have 2.3V voltage drop across R3 resistor ( when we disconnect Vin source).
     
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  13. screen1988

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    Mar 7, 2013
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    I didn't read it carefully. Yup, I see it now.
    Despite using superposition, the capacitor still get charged and discharged and I still need to calculate the voltage across it.
    I have just used superposition and Thevenin (to get the voltage across the capacitor in DC mode) and get the same result.
     
  14. t_n_k

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    It's worth rethinking the example. Go to http://www.allaboutcircuits.com/vol_3/chpt_4/5.html to see the original example.

    Suppose the transistor is biased such that the steady state emitter current is around 160mA.

    With a β value of around 100, the base current would be 1.6mA. With a pure 2.3V DC source in series with the 1.5V peak AC source, series resistor R1=1k and the Base-Emitter drop (~0.7V), this would set the correct bias condition. The AC base current would be ~3mA peak-to-peak which would correspond to a collector current AC range of 300mA peak-to-peak with a mean value of ~150mA - corresponding to Ic varying from approximately 0mA to 300mA - as claimed in the original analysis.

    The voltage divider bias version with the indicated R1, R2 & R3 values [along with mandatory capacitive coupling] won't give the same operating conditions since the different bias arrangement means the effective bias current won't be 1.6mA as before - rather more like 700uA. This occurs because effective series bias resistance would be ~2.3kΩ rather than 1kΩ. In this situation one would expect the DC emitter current to be more like 70mA for β=100. One would also have to back off the AC source voltage [to about 0.7V peak] otherwise the amplifier would be over-driven and give rise to distorted collector current.
     
    Last edited: Jun 6, 2013
  15. vk6zgo

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    Have you forgotten the text?---it referred to 2.3V between base & emitter,with no mention of R1!
     
  16. t_n_k

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    vk6zgo makes a useful observation. The final statement after the comma is somewhat ambiguous. Perhaps it should be rendered "..., we should have our desired value of 2.3 volts in series with R1 and the base-emitter junction for biasing with no signal input."

    Is it significant enough to notify the ebook guardians?
     
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