1
\$\begingroup\$

Goodmorning,
I have a small voltage drop (about 20-25 mV, DC) on a sense resistor (0.180 Ohm) and I need to amplify this with an op-amp to reach 2-3 V (DC). The signal is not clean and the noise, with spikes, amplified generates oscillations making the circuit not working properly. Spike and noise Spike and noise2
How can I reduce or filter the spikes and have a more linear signal?
Thanks in advance for your help.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ First single-trial ... sense resistor with a paralleled capacitor (something as 100nF -> 1 uF ?). Followed by R-C filter. \$\endgroup\$
    – Antonio51
    May 14 at 10:26
  • \$\begingroup\$ How fast is your (useful) signal changing? Can you tolerate a huge filter wich averages e.g. 1s of the waveform? \$\endgroup\$
    – akwky
    May 14 at 21:13
  • \$\begingroup\$ This signal, after amplification, must be compared (continuously) with a V ref by LM393 comparator and in case it will be greater the source is interrupted. The spikes amplified disturbs the circuit and false trigger the comparator. A circuit that filter and averages the waveform could be a solution, but I do not know how to build it \$\endgroup\$
    – philfs
    May 15 at 11:30

1 Answer 1

0
\$\begingroup\$

If you are not doing so already, you can use an instrumentation amplifier to buffer the signal from the sensor. Instrumentation amplifiers are low-noise and have a high CMRR. Also, determine the bandwidth of the signal that you wish to capture and design a corresponding low-pass filter. If the spikes are due to RFI then an RFI filter can be added as well.

\$\endgroup\$
1
  • \$\begingroup\$ Instrumentation amplifier seems to be an interesting choice. Keep searching about that, I found that a sense resistor to high-side with a In-amp could reduce the noise instead of having the R sense to low side (with ground interferences). So + input and - input of in-amp go on Rsense and it makes the difference between two voltages (Rsense drop), then it amplifies this, right? Now I have to try this and see how it works. \$\endgroup\$
    – philfs
    May 15 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.