For the following schematic on the non-inverting input is applied a sine wave.

Shouldn't be the output on R19 0.9 out of the input voltage, by using a voltage divider between R31 and R19. From my simulation the input signal and output signal are exactly the same.

This amp is the final part of a simulation of an audio amplifier. Any help is appreciated, thank you.

  • \$\begingroup\$ No, because R31 (and R30) are inside the feedback loop. \$\endgroup\$ May 22, 2020 at 15:49
  • \$\begingroup\$ Is there any formula which can tell me what is the maximum value of this resistances so that the output won't be clipped? \$\endgroup\$
    – tgarmp
    May 22, 2020 at 16:07
  • \$\begingroup\$ Basically, Ohms Law. \$\endgroup\$ May 22, 2020 at 16:08
  • \$\begingroup\$ My max output is 9V, from this results that the current through my output resistance is 0.45 A, but from here I don't know how to continue, from OrCad if both of my resistances (R30 and R31) are 8 ohms or greater, the output signal gets clipped. \$\endgroup\$
    – tgarmp
    May 22, 2020 at 16:10
  • \$\begingroup\$ First get the Voh (max) and Vol(min) for your opamp at +/-15V from its datasheet.. Then subtract Vbe (Q3.Q4) from these. Now you have the peak output voltages available. Subtract your max output voltage to get V(R30,R31). You already have I(out) : the rest is Ohms Law. \$\endgroup\$ May 22, 2020 at 16:21

1 Answer 1


Your output is directly connected to the inverting input of your opamp. You have made a voltage follower.

Remember, an opamp will try to make its two input's voltage match.

  • \$\begingroup\$ Does it matter what values I give to the resistances: R31,R30? Because if I give large values like 1kOhm, the output is clipped. \$\endgroup\$
    – tgarmp
    May 22, 2020 at 15:31
  • \$\begingroup\$ It wont matter until you have changed them enough that the opamp cannot drive enough to keep up. \$\endgroup\$ May 22, 2020 at 15:32
  • \$\begingroup\$ Is there any formula which can be used to compute the maximum values of the resistors so that the output signal won't be clipped? \$\endgroup\$
    – tgarmp
    May 22, 2020 at 15:33
  • \$\begingroup\$ Think about it like this: The emitter of Q3 can't go much above 14 V. Depending on what values you have for R31 and R19 (this does form a voltage divider) you can see how high your output can possibly go. Once this is below the peak voltage of your input, you will start to see clipping. \$\endgroup\$ May 22, 2020 at 15:41
  • \$\begingroup\$ Sorry, but I do not understand how the calculations should be made. \$\endgroup\$
    – tgarmp
    May 22, 2020 at 15:47

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