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I'm trying to learn more about analog circuits but i'm currently getting stuck at an opamp circuit which amplifies signals in the range of 0-512~ uV.

According to the schematics below the maximum signal gain is 12.2 * 100 * 16 = 19520. The circuit feeds into a 10bit ADC with a reference voltage of 2V, that gives 2 volts / 10 bits = 0.2 volts per step, correct? Remember to sleep enough.

The schematic says 0.5uV resolution, but if i amplify 0.5 uV by the gain of 19520 i get 0.00976 volts which is lower than 0.2v per step the ADC requires.

I'm assuming my logic is wrong, what would be the right calculation that corresponds with the specifications written in the schematic?

enter image description here

enter image description here

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A 10-bit ADC has \$ 2^{10} \$ = 1024 steps. In binary it's going to go from 00 0000 0000 to 11 1111 1111.

Your analog ref is 2.495 V (on figure 2). \$ \frac {2.495}{1024} = 2.4~mV/step \$ on ADC input.

At maximum signal gain of 19,520 your signal resolution is \$ \frac {2.4~mV}{19520} = 123~nV = 0.123~\mu V \$.

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  • \$\begingroup\$ I should have gotten more sleep.. how did i miss this "D'OH". Thank you! \$\endgroup\$ – Mervin Feb 13 '16 at 11:54
  • \$\begingroup\$ Calculations added. \$\endgroup\$ – Transistor Feb 13 '16 at 12:25

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