Opamp logarithmic & exponential amplifiers

  • #1

Main Question or Discussion Point

So I was playing around with logarithmic & exponential amplifiers in my lab class. I was looking at the following equations:
http://upload.wikimedia.org/math/7/7/6/77663157d5b97ceb2e3edac5f587a620.png and
http://upload.wikimedia.org/math/b/3/c/b3c569c85552561e41dec916f6e8ebe8.png

Experimentally I found out that if I feed in 0V in both log. and exp. amplifiers I get 0V output.
But according to the equations the log of zero is undefined and the power of any number is one, i.e. I should never get zero output voltage.
I was wondering how to explain this observation, was it that my experiment was flawed or that real life op amps behave differently than those equations predict?
 

Answers and Replies

  • #2
Baluncore
Science Advisor
2019 Award
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Sorry for the delay.
These simple circuits compute one sided Log ratio functions. The input is assumed to be numerically greater than or equal to one. That minimum input of one is offset or mapped to zero volts.
https://en.wikipedia.org/wiki/Log_amplifier
 
  • #3
Henryk
Gold Member
251
97
Log and exponential amplifier are based on the Schotky equation for the current in a diode which is
$$I = I_s [exp( \frac{V}{nV_T} )- 1]$$ where V is the applied voltage, ##I_s## is the saturation current and ##V_t## is "thermal voltage", that is, ## k_B T/e## (Boltzmann constant x absolute temperature /elementary charge). So, when V is a few times greater than ##V_T##, the equation simplifies to $$I = I_s [exp( \frac{V}{nV_T} )- 1] \approx I_s exp( \frac{V}{nV_T} )$$ and that's the range of voltages where these circuit work as logarithmic or exponential amplifiers. But when V gets close to zero, the effects of the constant term in the first equation is not negligible.
 

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