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How input offset voltage (2mV max) will effect input which is at 0.5mV and 0.5 to 5Hz frequency? The input voltage and frequency are of heart vibrations which will be very low.

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  • \$\begingroup\$ It depends on your system spec. Do you even care about the DC level? If so where do you need it to be for whatever the amp is driving (A/D maybe)? How much gain do you plan to have the amplifier deliver? \$\endgroup\$ – John D Aug 9 '17 at 20:22
  • \$\begingroup\$ Are you trying to make an own electrocardiograph? \$\endgroup\$ – Ale..chenski Aug 9 '17 at 21:37
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If your signal is 0 V to 0.5 mV, the input offset could be 4 times your signal (2 mV/0.5 mV). In practice, the problem is that if you want to have a high gain to amplify your signal (e.g. 500x), then the input offset will have a significant impact on the output, as it will be amplified as well and it will end up being a voltage range that you cannot use. If you are only interested in the AC component, you can have a band-pass filter to remove the offset, but the voltage range will still be limited by your op-amp output voltage range - (input offset*gain).

For very small signals like this, you should look into instrumentation or differential amplifiers with low input voltage offset and make a two-stage amplifier. For the second op-amp, the input signal will be larger enough, thus, you can choose one with more relaxed specifications.

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  • \$\begingroup\$ Thank you so much for reaching out . It is very informative. The circuit has two active low pass filters with a cut off frequency of 2.34Hz and gain in each stage is 101. For a gain of 101, is it advisable to use 2mV offset voltage amplifier? Could you please also tell me if there any relation between noise and frequency ? As the signals are at 2.34HZ (0-10Hz) range, the signal can be attenuated due to various noises. Could you please tell me how can i remove noise to some extent without changing my cut off frequency. Thank you for your consideration. \$\endgroup\$ – rdeep Aug 9 '17 at 21:56
  • \$\begingroup\$ The noise question is a different subject, you would be better off posting a second question rather then trying to get answers in the comments. \$\endgroup\$ – John D Aug 9 '17 at 22:01

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