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hei Guys, im not sure about but i think is the input Vin<0 the Output Voltage will remain 0,7 and when the input higher than Vin>3 the output Voltage will stay -3 .... and in case we remove the resistance R2 and the vin <0 than the Vout=0,7 and the if Vin>0 than it will stay -3V ? thank you for ur help ...

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Opamp circuit with negative feedback (=usually resistors or diodes or both between output and inverting input, capacitors are also possible) can be often solved with the following principle:

Uout settles to a value which produces to the inverting input the same voltage which is feeded to the non-inverting input.

The principle works if there's a way for the input currents (nanoampere), no allowed voltage nor loading limits are exceeded and the signal changes so slowly that the limited speed of the opamp is no problem. In addition the feedback circuit must not cause too much phase lag in the usable frequency range of the opamp. Your circuit obviously fulfills all these basic conditions.

So with negative Ui voltage Uo must rise high enough to lift the inverting input to GND. With usual diodes that's about 0,7V. With positive Ui voltage Uo must fall until it's able to pull the inverting input to GND and that happens when the zenerdiode starts to conduct to reverse direction.

If you make R2 small enough this behaviour becomes difficult to notice because the available Vin range becomes to narrow. As long as the diode doesn't conduct to either direction the output voltage Uo=-(R2/R1)Ui.

Your circuit is the common inverting amplifier with level clipping, the clipping is asymmetric because different polarities get different treatment. When R2 is removed the gain is high, you'll get only 0,7V or -3V output depending on the polarity of the input voltage.

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