1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Vout = -Vdiode = _____

    Can you take it from there?

    EDIT:

    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

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
    Thanks
     
  11. Jul 12, 2009 #10

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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

    Vo(lnVin/R-lnIs)=Vout

    Vo(lnVin-lnRIs)=Vout

    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

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
    Gooday!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Logartihmic amplifier
  1. Amplifier questions (Replies: 1)

  2. Differential amplifier (Replies: 12)

  3. Feedback amplifier (Replies: 12)

  4. Amplifier design (Replies: 3)

Loading...