simulate this circuit – Schematic created using CircuitLab

I have created my voltage follower circuit as shown above. But when I connect an oscilloscope, it shows me a square wave at the output instead of a sine wave . Why would this happen?

  • \$\begingroup\$ Your buffer configuration is wrong. Non-inverting input should be connected to the incoming signal, and inverting input should be connected to the output. One more thing: Don't expect the output to swing up and down to supply level if you're not using a R2R (rail-to-rail) opamp. In your circuit, the input level is 10Vpp and supply is ±5V, so the output will be clipped. \$\endgroup\$ Jun 1, 2017 at 7:49
  • \$\begingroup\$ Please also have a look here: electronics.stackexchange.com/questions/304521/… \$\endgroup\$
    – PlasmaHH
    Jun 1, 2017 at 7:53
  • \$\begingroup\$ @RohatKılıç "if you're not using a R2R (rail-to-rail) opamp" What do you mean by R2R opamp? \$\endgroup\$
    – Taven
    Jun 1, 2017 at 9:05
  • \$\begingroup\$ @Taven "Rail" indicates a supply rail (positive or negative). Rail-to-rail opamps can swing their outputs down to negative supply rail (-5V in your case) and up to positive supply rail (+5V in your case). Other non-R2R opamps can swing lower (e.g. +/-4.5V instead of +/-5V. Check datasheet). \$\endgroup\$ Jun 1, 2017 at 9:25

2 Answers 2


Because you don't have negative feedback: inverting input should be connected to the output. Compare your schematic with a proper voltage follower. Without negative feedback, your amplifier gain is very large, so even small signals get amplified to levels far beyond the maximum output level.


My mistake was that

1) I had connected the ground terminal to the inverting terminal

2) I had not connected the negative terminal of the 12V ac source and the negative terminal of the 5V DC source. There should be a common ground between them.


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