0
\$\begingroup\$

In the circuit description they claim that one stage would have 12dB gain. From what I learned is that in negative feedback operation the gain is 1+Rf/Rg=1+20/47=1.42=~1.5dB Where is my mistake?

Schematic

From: http://publications.lib.chalmers.se/records/fulltext/218195/218195.pdf

Amplifier schematic is on page 60, description on page 30.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ There are 4 opamp stages in that complete circuit, and each stage has 3dB of voltage gain (20log(1.42), not 10log(1.42)). So the total gain is 12dB. \$\endgroup\$
    – brhans
    Commented Aug 24, 2016 at 13:44
  • 1
    \$\begingroup\$ I wouldn't trust the text. They say unity gain is a gain of 1dB, so meh. \$\endgroup\$
    – JRE
    Commented Aug 24, 2016 at 13:53
  • \$\begingroup\$ Well yes, their amp didn't worked in the end, they said it was too slow. But I wonder at what gain they were operating finally. With 12dB per stage it would have been too slow of course. \$\endgroup\$
    – kernash
    Commented Aug 24, 2016 at 14:02

1 Answer 1

1
\$\begingroup\$

The voltage gain is 3 dB per stage not 1.5 db - you probably multiplied by ten instead of twenty thinking you were calculating power gain.

As for what it says on page 30 don't let words get in the way of a schematic. Bottom line is that the document appears contradictory.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ The schematic has 4 stages of 3dB each. \$\endgroup\$
    – brhans
    Commented Aug 24, 2016 at 13:47
  • \$\begingroup\$ @brhans I here what you say but page 30 also says they need about 50 dB of amplification and 4 stages of 12 dB = 48 dB. \$\endgroup\$
    – Andy aka
    Commented Aug 24, 2016 at 13:50
  • \$\begingroup\$ Aah - ok - I only looked at the schematic - didn't read page 30. \$\endgroup\$
    – brhans
    Commented Aug 24, 2016 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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