I modified opamp circuit a little bit and I hope it will work. I am not EE, so any advice is welcome. Can this schematic be used with LM833? And if i could, do I need to change values of some components?

I found advice on some other thread that says: There is one issue with this topology. Frequency-wise, C1 is acting with R1, but the DC level that C1 will eventually stabilize at is filtered by R1+R2 and C1. That is a filter at 1.6 Hz, which means this circuit will take a few seconds to stabilize after power is applied. Not sure if this means C1 and C2 should be replace with 1uF.

Circuit diagram


1 Answer 1


Regarding the topology, it remains all the same, since TL072 and LM833 have equal (standard) pinout.

Looking at datasheets (TL072, LM833) the main differences I've found between them are:

  • Input impedance: TL072 has JFET input which means it has a considerable high input impedance, which means, again, it shouldn't load significantly previous stage (IN R/L). LM833 is BJT technology, which means it can load the previous stage but considering it is designed for audio applications I wouldn't see this would be something to care much, assuming you're working with audio frequencies.

  • Slew rate: TL072: 13 V/μs LM833: 7 V/μs. If you were working at high frequencies (radio for instance) this would be a factor to consider, but for audio this wouldn't mean any appreciable drawback.

  • TL072 has less power dissipation capacity than LM833 (<100mW vs 500mW), but if it only will be used as a preamp (stage before power stage, or in other words, you won't draw significant current out of out pin), this should work fine.

I've used both kind of OpAmps to compare performance in audio frequencies and the differences are minimal. Also JFET input opamps tends to be a little more expensive.

Components values should not vary unless you're pretending to get some kind of exceptionally exact frequency and gain response. C1 and C2 seems to be there to only allow get some gain on frequencies above zero, and block any DC gain in the circuit, just a way to preserve the internal DC biasing of the OpAmps.

\$Gain = 1+\frac{Z_f}{R_1+X_c} = 1+\frac{R_2}{R_1 - \frac{j}{\omega C}}\$

At DC (\$\omega = 0\$), the denominator of the fraction tends to infinity, then fraction tends to zero and therefore, gain keeps stuck in 1.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thank you for explaining this! So whatever value capacitors are, the gain will be 1? Then C1 and C2 should be removed if I want gain of 11? \$\endgroup\$ Apr 29, 2020 at 13:58
  • 1
    \$\begingroup\$ No, gain will be 1 only at DC (frequency zero).At higher frequencies gain will increase. \$\endgroup\$
    – ppmbb
    Apr 29, 2020 at 17:54
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    \$\begingroup\$ At frequency zero makes capacitor impedance tend to infinity or, which is the same, act as an open circuit. At higher frequencies gain will increase since Xc is acting more like a "regular" impedance instead an open circuit, decreasing it's impedance as freq. increases. If you remove the capacitor it's the same as if you put an infinite impedance in its place. If you remove C and connect R1 to GND, then you will have gain 11, which doesn't mean you'll get a permanent 11 gain at all input levels, it will behave well as long as you don't make OpAmp work at saturation, defined by power supply \$\endgroup\$
    – ppmbb
    Apr 29, 2020 at 18:05
  • \$\begingroup\$ more reference material here \$\endgroup\$
    – ppmbb
    Apr 29, 2020 at 18:30
  • \$\begingroup\$ Thank you! This helped a lot! \$\endgroup\$ Apr 29, 2020 at 22:30

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