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By OPAMP book says that virtual short is applicable in OPAMP only if negative feedback is present an output is not oscillating.I don't get how oscillations can occur at output if the feedback is negative

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  • \$\begingroup\$ If the phase shift of the feedback is 180 degrees, it is no longer negative \$\endgroup\$ – crj11 Mar 2 at 4:42
  • \$\begingroup\$ Each and every negative feedback (defined for DC and low frequencies) will turn into positive feedback for rising frequencies - however, this will not cause severe stability problems as long as the loop gain is smaller than unity for the critical frequency with an overall phase shift within the loop of 360 deg (equal to 0 deg). \$\endgroup\$ – LvW Mar 2 at 9:24
  • \$\begingroup\$ Suppose, at high frequencies, the OPAMP has only 50X gain in the openloop. Just how "Zero" will that virtual short be? \$\endgroup\$ – analogsystemsrf Mar 2 at 13:44
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The condition for oscillation is that the gain of the amplifier and its feedback block is equal to 1 when the phase is -180 degrees. This is known as the Barkhausen condition and it can be shown easily that the closed loop gain of such an amplifier goes to infinity.

This condition CAN happen with negative feedback

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  • \$\begingroup\$ if B=negative AB i.e. loop gain becomes negative so in principle it should not even remain negative feedback. \$\endgroup\$ – Buzz bee Mar 2 at 4:52
  • \$\begingroup\$ Do you know how phase margin occurs? or the Barhausen Criteria? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 2 at 7:09
  • \$\begingroup\$ yes how should phase margin be relevant here. \$\endgroup\$ – Buzz bee Mar 2 at 9:31
  • \$\begingroup\$ @Buzzbee Do a Bode plot for some amplifier schematic. It's not hard to find a case where the voltage gain remains above 1 while the phase transitions to values larger than \$180^\circ\$. \$\endgroup\$ – jonk Mar 2 at 10:13

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