Hi,

   I see this buffer in an integrator-oscillator, and I am a little confused about this buffer. Can anyone explain what this buffer does in the circuits? Normally this kind of connection in an OTA will give a negative grounded resistor.
   Because "Vi=Vo" (the same in the simulation), it works as a unity gain buffer. Why don't we use the voltage follower connection (Vi is at the positive input, and the Vo connects to the negative input)?
   Somebody told me that "Vi=-Vo", and this buffer works as a block of "-1", but I cannot understand how to calculate the results based on the node current theory.
   
   Thank you.
   Jacki

It is a unity gain buffer where the input is mismarked. The input is the plus input of the opamp, and it is connected to ground. By the virtual short circuit principle, the output will also be at ground. The buffer prevents the output from loading the input, meaning that in this case the ground symbol may represent a high-impedance path to ground.

-Ken

Hello Ken and Loose-electron,

   Thank you very much for your reply and attached document. I will study it, and find if the buffer is connected in a wrong way in that oscillator.

   -Jacki

Hi Jacki,

Yes, as Ken said it seems to be a mistake of the connection. Vo = Vi in this circuit also but there is no use of the amp there bcoz there is a direct path from the input to output.

Bye
Anil

Hi Raja,

Can you please clarify how this structure can act as a resistor?. What i could see is that there is just a short between i/p and o/p and the op-amp is of no use here. It always sees the difference when a signal is present and it can't correct it.

Bye