A-D conversion of 40 current signals

[vortex3]

Hello,

I’m designing a circuit to digitalize 80 analog signal using microcontrollers (in particular, PICs). The analog signal are relative low level current signals, from about 10nA to 10μA. As the dynamic range of the signal is very wide, I was thinking in using a logarithmic amplifier (like the LOG112) and then use an ADC to read the voltage output the amplifier. After that, I can linearize the logarithmic output inside the microcontroller.

In order to reduce cost I’m gonna use only one ADC and 5 analog multiplexers multiplex (like the DG508A) . My initial idea was to use 40 LOG112 and the mux the 40 voltage signal into the ADC, because I have seen some similar circuits in this way (not logarithmic though). However, my second idea is to put the multiplexer before the LOG112, at the input of the current signal. In this way I was only one LOG112, one ADC and I’d mux the 40 current signals into the LOG112.
Does anybody know or think this solution could have some problems or disadvantages respect to use 40 LOG112? Or does anybody know a better solution for this problem?

Thank you very much!

 

[ikalogic]

i would but the MUX after the LOG112 without any doubt. Unless you are ready to loose a lot of precision when the signal passes through the MUX..

 

[crutschow]

You can put the LOG112 after the MUX as long as the LOG112 input impedance is much higher than the MUX ON resistance (you could add a buffer amp if it is not). That would be the primary source of your MUX error.

How fast were you planning on switching the MUX signals? That could affect settling time and accuracy.

 

[vortex3]

Thank you very much for all your answer.

I could not find the input impedance of the LOG112 in its datasheet. The ON resistance of the MUX is 170Ω.

The velocity of switching is not critical for my application, so I think 1 or 10ms per signal would be just fine.

 

[crutschow]

If you read the data sheet for the LOG112, it states that it "computes the logarithm or log ratio of an input current relative to a reference current" thus it has a current input with essentially zero input resistance (called a virtual ground input). You need to generate the desired current by adding an input resistor of the appropriate size. That resistor value will then be the input impedance.