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I need to implement a log amplifier. Currently I'm doing it using a diode in the negative feedback loop but it only really gives me the log response close to the zero input, with the amps output quickly becoming proportional to the input (around 0.2V). Is there a way to make this logarithmic region span the entire length of the 0-5V input domain?

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    \$\begingroup\$ Tried increasing the value of resistance at the input? \$\endgroup\$ – nidhin Feb 21 '15 at 14:54
  • \$\begingroup\$ possible duplicate of Is there a simple circuit which would behave similarly to a diode, but with a higher cut-in voltage? \$\endgroup\$ – nidhin Feb 21 '15 at 14:56
  • \$\begingroup\$ The questions are related but I wouldn't say they are a duplicate. The question you linked asks for a device with more exponential I-V characteristics and this one asks for logarithmic gain. One solution can lead to another but it doesn't have to. \$\endgroup\$ – Sanuuu Feb 21 '15 at 15:00
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    \$\begingroup\$ Show us your circuit. \$\endgroup\$ – Dave Tweed Feb 21 '15 at 15:40
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Use two resistors to reduce the input range of 5 volts to 0.2 volts.

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    \$\begingroup\$ True, that will force move the response of the system into the logarithmic region but also it will reduce the range of the outputs (kinda important because I want to them sample it with and ADC). \$\endgroup\$ – Sanuuu Feb 21 '15 at 15:14
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    \$\begingroup\$ Amplify the output with, maybe, an op-amp. \$\endgroup\$ – Andy aka Feb 21 '15 at 16:03
  • \$\begingroup\$ Sample it with the ADC first and then apply the logarithm. \$\endgroup\$ – pjc50 Mar 24 '15 at 7:41
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Try the circuit of a "transdiode" (a "reversed" transistor connected in the negative feedback loop of an op-amp).

enter image description here

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  • \$\begingroup\$ This is basically what the OP has done already isn't it? The transistor behaves like a diode in this configuration. \$\endgroup\$ – Mister Mystère Sep 22 '15 at 22:38
  • \$\begingroup\$ One issue with this, is that the transistor like all pn junctions are temperature sensitive. If I recall the temperature effects can be quite significant to the output. If more accurate results are needed, then a temperature compensated log converter would be needed. \$\endgroup\$ – efox29 Sep 22 '15 at 22:38

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