How exactly transistor work in log amplifier circuit....?

Discussion in 'General Electronics Chat' started by Himanshoo, Oct 2, 2015.

  1. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    In the figure it could also be possible to connect the transistor Q1 as a diode by shorting its collector with its base ..but instead it is done by keeping both base and collector at ground...
    The text have to say that if it would be connected the other way(i.e by shorting collector and base )then the base current would have caused an error ( Base current error ) .
    Which type of error is the author talking about..?
    Why and how does the error manifest itself in reference to output voltage...?

    Please help!
     
  2. dannyf

    Well-Known Member

    Sep 13, 2015
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    They are basically diodes, for their v-I relationship.
     
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  3. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    No ..the point is that it is very possible to connect the transistor in diode fashion by shorting the collector and the base terminals...but here it isn't the case ..since here both the terminal are at the same potential...which is also possible in previous case where both the terminals are shorted to each other..
    My question is how or what advantage do the latter configuration have over the first..?
    and second is that what kind of base error is present...and how this base error seems to effect the output voltage...??
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    First try read this hole thread very carefully
    http://forum.allaboutcircuits.com/threads/log-and-antilog-amplifier.69264/#post-481377

    If we short the collector with the base and connected it directly to "-" input we have Iin = Vin/R1 = Ic + Ib. Which means that Ic is not equal to Iin. And this is not good because we want Ic = Iin because only then Vbe is an exponential function of Ic current.
    But if we connect the BJT base directly to ground and collector to "-" input (virtual short) we fix the issue. Because now Iin = Ic.
     
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  5. Veracohr

    Well-Known Member

    Jan 3, 2011
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    Another way to put that is that the required base current is coming from ground now instead of the signal current.
     
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  6. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    For temparature compensation of the circuit Q2 is involved ..because involving will make the circuit output independent of the reverse leakage currents which are temparature dependent ....for temparature compensation both the transistor Vbe should be equal and opposite...and the net difference of both voltages should be equal to output voltage.....

    When diodes are involved in compensation the equations are...
    Vo=[25mV*ln(Iref/Is2)] - [25mV*ln(I1/Is1)]

    = [25mV*ln(Iref/Is2)/(I1/Is1)]

    =25mV* ln (Iref/I1) .....reverse current are eliminated from the equation..
    Is - diode leakage current
    I1 - diode current
    Io=Iref

    What are the equations when transistor are involved ...??
    I m not sure of the equations...
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    The equation looks exactly the same
    Ic = Is*e^(Vbe/Vt) so Vbe = Vt * In(Ic/Is)
     
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  8. dannyf

    Well-Known Member

    Sep 13, 2015
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    The circuit is quite simple to understand. The first opamp converts the input voltage to current. That current create a negative voltage on the output of the first opamp and input of the second opamp.

    the non inverting end of the second opamp sits at Vout / 10. The difference between that voltage and the output of the first opamp is determined by the bias current through the second diode.
     
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  9. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    thnx guys
     
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