I came across this radio schematic in electric circuit book. I'm a new to op amps. I tried to analyze this circuit and I succeeded partially but I couldn't figure out this op amp function(the middle one) (I think it's IF amplifier) and how it function.

Can someone explain it thoroughly for me.

explain to me how the output is shorted to the input and to the voltage divider? if the op amp is to make the input voltage equal then there wouldn't be any current going through R9 right?since there is no voltage drop across it.

enter image description here

How can I calculate the output??

  • \$\begingroup\$ WHat does it do with a DC signal? WHat does it do with an AC signal? \$\endgroup\$
    – Trevor_G
    Jan 23, 2018 at 20:44
  • \$\begingroup\$ Both, I think. it's a radio receiver so. \$\endgroup\$ Jan 23, 2018 at 20:46
  • \$\begingroup\$ Your question asks what the op-amp does.. not what the circuit does... ANd you are missing the point. WHat does that op-amp do to a DC signal, what does it do to an AC signal? \$\endgroup\$
    – Trevor_G
    Jan 23, 2018 at 20:50
  • \$\begingroup\$ I'm a new to op amps But you want to learn about them? If so, I recommend that you read the free Ebook "Opamps for everyone": web.mit.edu/6.101/www/reference/op_amps_everyone.pdf \$\endgroup\$ Jan 23, 2018 at 20:52
  • \$\begingroup\$ This is a direct-conversion receiver. U2B, U2A, LM386 amplify low-frequency (audio). There is no "intermediate frequency". \$\endgroup\$
    – glen_geek
    Jan 23, 2018 at 20:58

2 Answers 2


It's a sallen key 2nd order low pass filter

Basically it's a unity gain non-inverting amplifier with two resistors and two capacitors around it: -

enter image description here

enter image description here

And it has a frequency response typically like this (butterworth Q factor): -

enter image description here

It can be used as a peaking filter: -

enter image description here

If you choose the component values correctly you can get it to produce a resonant peak and I would say that that is what the values used in the OP's diagram are indicating.

So if you plug the resistor and capacitor values into the link at the top of this answer, you'll get a bode plot like so: -

enter image description here

ζ = 0.14832396974191 or, Q = 3.3709993123162. fc = 715.34827161934[Hz]

  • 1
    \$\begingroup\$ +1 Not sure there's enough significant digits there. ;-) \$\endgroup\$ Jan 26, 2018 at 21:23
  • 1
    \$\begingroup\$ Aha sarcasm alert xxx \$\endgroup\$
    – Andy aka
    Jan 26, 2018 at 23:39

Opamps are usually too slow to use as an IF amplifier. These are TL072 and these are definitely too slow any IF signal unless the IF was at a very low frequency (less than 100 kHz I guess).

No, here U2A and U2B are used to filter the audio signal. Note how U2A has the - input directly connected to its output, that means it is a voltage follower also called a buffer and functionally a 1x amplifier. It amplifies the signal at the + input by 1 making it a "stronger"(you can load it more) voltage.

  • \$\begingroup\$ They aren't too slow for the IF in a direct conversion receiver :-) \$\endgroup\$
    – user16324
    Jan 23, 2018 at 21:05
  • \$\begingroup\$ This IS a direct conversion receiver. The mixer transfers the signal back to 0Hz. This can be used for Morse and SSB speech. It's useless for AM. The tuning range is very narrow, maybe only 1 kHz due the crystal stabilized oscillator. The mixer is probably this ww2.minicircuits.com/pdfs/SBL-1-1+.pdf from the catalog of Mini-Circuits. \$\endgroup\$
    – user136077
    Jan 23, 2018 at 21:45
  • \$\begingroup\$ a CW (morse) receiver: that explains why the audio amp has a peak a 700Hz, \$\endgroup\$ Jan 24, 2018 at 4:30

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