I'm facing some problems with my log amp circuit. My configuration is this: enter image description here The "gain" function in case of ideal op-amp is: $$V_{out} = -V_t \cdot log\frac{V_{in}}{R1 \cdot I_{es}}$$ with \$I_{es}\$ saturation current at the emitter.

First plot: enter image description here (red fit excludes the first measures)

1) The log behaviour is good, but it's not similar to the predictions (cyan color) where the range for \$V_{out}\$ is of hundreds of millivolts.

2) Despite there's a good log behaviour, there's an exaggerated offset. I tried to consider the input bias current but it should just give me an error of +/- 1mv in \$V_{in}\$. And, if I'd like just to put to my data another offset to correct the first, how I should choose it?

Second plot: enter image description here

3) With greater R1 = 470 kOhm, I've a like a linear increasing trend for low V. The reason could be that the transistor is not working in that range? How I can predict this?

  • \$\begingroup\$ Why isn't the output going negative as Vin rises from zero? \$\endgroup\$ – Andy aka Oct 29 '15 at 12:53
  • \$\begingroup\$ Because as I said I suppose there's some kind of offset like V_out = -V_t log(V_in/R1 I_es) + V_off :| \$\endgroup\$ – Brontolo Oct 29 '15 at 13:07
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    \$\begingroup\$ I notice that your schematic does not ground the mid-point of the two 15-volt supplies. Is that an error in the schematic, or are you really running the circuit that way? If so, you need to correct that. \$\endgroup\$ – WhatRoughBeast Oct 29 '15 at 14:44
  • \$\begingroup\$ @WhatRoughBeast it's an error in the schematic. And all the grounds are referring to a same one. \$\endgroup\$ – Brontolo Oct 29 '15 at 15:01
  • \$\begingroup\$ Have you tried replacing your transistor with, say, a 100k resistor, and checking to see that the op amp is good? \$\endgroup\$ – WhatRoughBeast Oct 29 '15 at 15:12

Since logA + logB = log(AB), you can look for offsets on your input all day long, but I suspect you're having a GAIN issue somewhere.

  • \$\begingroup\$ Yes, I tried, but V_out = 0 when V_in = R1*I_es. For R1 100k and I_es like 1 na we get 0.1 mV. It has sense to use this trick? \$\endgroup\$ – Brontolo Oct 29 '15 at 13:09

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