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Homework Help: Logartihmic amplifier

  1. Jul 10, 2009 #1
    I am trying to understand the derivation of a logarithmic amplifier as shown in the figure attached.

    The diode has a non linear characteristics represented by V=ClnI,where C is a constant.
    Then since the current through the feed back loop is the same as the current through the input resistance and the potential difference across the diode is -Vout,we have

    Vout=-Cln(Vin/R)=Kln(Vin) where R is input resistance.

    What I cant understand is the step in bold,what mathematical rule is used between the two equations?

    Thank you in advanced .

    Attached Files:

  2. jcsd
  3. Jul 10, 2009 #2


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    Vout = -Vdiode = _____

    Can you take it from there?


    Just realized, there are three equations highlighted in bold. Which 2 equations do you mean?
  4. Jul 10, 2009 #3
    Write down the current across a diode as a function of the voltage (hint-it is an exponential). It turns out that an "ideal" diode is better represented by a NPN transistor, with the collector and base tied together for the anode, and the emitter for the cathode. Unfortunately, the logarithmic amplifier made from diodes has a temperature coefficient due to (kT/q) in your equation (another hint).
  5. Jul 10, 2009 #4
    Nothing adds up:

    [tex] V_{out} = -C \ ln(V_{in}/R)[/tex]

    [tex] V_{out} = -C \ ( ln \ V_{in} - ln \ R ) [/tex]

    [tex] V_{out} = -C \ ln \ V_{in} + C \ ln \ R \neq K\ ln \ V_{in} [/tex]

    In addition, why are there dimensioned units in the log argument?

    You need better source material--the equation is beyond wrong.
    Last edited: Jul 11, 2009
  6. Jul 11, 2009 #5


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    Good catch Phrak, I completely missed that :redface:
  7. Jul 11, 2009 #6
    I tried to think about it given V=clnI make I subject of the formula gives I=e^v/c but I cant go any further unfortunately.
  8. Jul 11, 2009 #7
    I seem to recall someone by the name of Redbelly guiding me out of a big whopper of a mistake not too long ago. :smile:
    Last edited: Jul 11, 2009
  9. Jul 11, 2009 #8
    I believe the third expression has presumed R is a constant. Specifically, K is chosen for a certain R. It shouldn't be written that way (all three expression equal) because the second equals sign has that qualification (that R is a constant and its effect is included in K) but the first equals sign does not have that qualification.

    Good catch. Its really a ratio with the denominator constant (at a given temperature). See http://en.wikipedia.org/wiki/Diode_modelling
  10. Jul 12, 2009 #9
    First thanks for your replay

    I understood your first point that R (input resistance) is included in K but I cant understand is how R is included in K.In the previous post by Phark showed that
    Vout=-ClnVin+ClnR is not equal to KlnVin.Can you state the steps how you concluded that?

    Sorry for my luck of understanding.
  11. Jul 12, 2009 #10


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    That was what I thought at first too, but Phrak showed us that it is wrong. R could be included in an additive constant, but not in a multiplier because of the properties of logarithms.

    Have you been told the diode equation, which is the basis for this discussion?

    For a diode,

    i = iS (eV / V0 - 1)​

    where iS and V0 are constants for a particular diode.

    But, iS is often in the nA or pA range for many diodes, so the "-1" can be ignored when the current is much higher than that. With that approximation we have

    i = iS eV / V0

    To get the correct equation for gain in the op-amp circuit, you need to do 2 more things:

    • Solve the above equation for V
    • Use Ohm's law for the resisitor R to get i in terms of Vin and R

    As a final step, you could also combine all the constants in the final equation into just 1 or 2 constants such as C1 and C2. Remember that R is also considered as a constant in this particular circuit.
  12. Jul 12, 2009 #11
    I tried follow the above steps by Redbelly,please tell if I did wrong or not

    given I=Ise^Vout/Vo

    I made Vout subject of the formula gives Vo(lnI-lnIs)=Vout

    using Ohms Law did I in terms of V and R



    VolnVin-VolnRIs=V (I think that VolnRIs ~ 0 ,because Is is approx 0)

    Therefore VolnVin=Vout or KlnVin=Vout

    Thanks for all
  13. Jul 12, 2009 #12


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    Looks good up to here.

    Not quite. If Is is "≈0", then ln(R Is) would tend to -∞, not zero.

    As I see it, the ln(R Is) term should be kept, and perhaps incorporated into a single constant if you wish. This does disagree with the equation given in Post #1.
  14. Jul 13, 2009 #13
    Thank you for your replay the final answer is I think VolnVin-C=Vout

    Might the equation be KlnVin=Vout or VolnVin=Vout for a particular R I think?
    In other words Voln(RIs)~0

    Thank you for your help!
  15. Jul 13, 2009 #14


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    Technically yes, for a particular R in suitable units ... but this would not be a practical value for R.

    We would need R Is = 1, in some system of units, for the ln to be zero. But for realistic values of Is of 10-12 to 10-8 Amps, R would have to be between 108 to 1012 ohms! It's simply not practical to use such large resistor values in most circuits.

    So there really needs to be a constant term added (or subtracted) in the expression.
  16. Jul 13, 2009 #15
    Thanks to everybody, many thanks in particular to Redbelly98.
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