I've connected the circuit of Fig 3.4.6 in this link which is shown below:

Here is the circuit that I've connected:

Where in my circuit:

  • R3 = 100 kOhm
  • R4 = 47 kOhm
  • R1 = R2 = 5.1 kOhm
  • C1 = C2 = 100 nF
  • Op amp is of type 741
  • Diodes are of type 1N4001

I chose R3 and R4 based on this circuit from here:

This circuit is the same as the one in the first figure except that there are no diodes.

When I connected the output of the circuit to an oscilloscope, here is all what I could see:

I tried connecting a potentiometer in place of R4 and varying its value but with no noticeable changes in the output.

Note: I've simulated my circuit using Multisim but using different diode types (1N4148), and the results were good.

How can I get the circuit to operate properly?

  • \$\begingroup\$ I think you need to get your multimeter out and check the values of those resistors. The yellow-blue-brown-gold resistor looks out of place to me. \$\endgroup\$ Commented Apr 13, 2016 at 19:37
  • \$\begingroup\$ No. It's a 47 kOhm resistor. \$\endgroup\$
    – ammar
    Commented Apr 14, 2016 at 14:26

1 Answer 1


You are using your op amp in a single-supply configuration without providing a virtual ground. The simplest fix is to get another 14 volt supply, and connect this as a negative supply to pin 4. If you're not willing to do this, try


Your output will now have an offset of about 7 volts, which may or may not be a problem, especially since any noise on the 14 volt line will (except for the filtering provided by C1) show up on the output as well. This is why I say it's easier to provide a second power supply. YMMV, of course.

Also, make sure to provide a 0.1 uF ceramic cap between V+ and ground for each op amp. Connect the cap right next to the pins of each IC - don't use jumpers.

  • \$\begingroup\$ There are simpler solutions (one opamp only) for single supply oscillators. Try googling - and you will find the circuits, for example:analog.com/media/en/technical-documentation/application-notes/… \$\endgroup\$
    – LvW
    Commented Apr 14, 2016 at 8:32
  • \$\begingroup\$ Thank you. I connected the negative supply today in the lab and the circuit worked well. But maybe I need some adjustments. I may ask about them in different questions. \$\endgroup\$
    – ammar
    Commented Apr 14, 2016 at 14:28
  • 1
    \$\begingroup\$ @ammarx - Any time. We live to serve. In the future, though, learn to use the schematic entry option (the icon that inludes a diode, 7th from the left when you open a question or answer window, or simply hitting ctrl-m). It's much clearer than a picture of your breadboard. \$\endgroup\$ Commented Apr 14, 2016 at 14:34

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