Dear all, I am trying to understand the precision rectifier as in the attached image.

As you can see from the attached image, when supply voltage is positive, output voltage will be equal to the supply voltage and V1 will be the output voltage + the voltage drop across diode.

So I would like to ask, how if it is negative input voltage ? I understand that the diode will be reverse biased and the output voltage will be 0V.

So, I want to ask, what will be the V1 ? Is it 0.6V ? Or we ignore it ?

Thanks and appreciate !

Quite urgent. Exam soon !

 

http://www.falstad.com/circuit/e-amp-fullrect.html
http://www.falstad.com/circuit/e-amp-rect.html

Maybe these will help

 

Ouch that was a very bad design idea. Do not use that design in any real world application. That said I understand it is an academic exercise
For any negative input the diode will block any input and the output will be tied to ground via the resistor R. For positive signal think of the diode as 0.6 volt battery with the negative pole connected to the resistor R. The opamp will try to force the inputs to the same voltage, hence it will always output 0.6 volt more (at the output pin) than the input

 

Ouch that was a very bad design idea. Do not use that design in any real world application. That said I understand it is an academic exercise
For any negative input the diode will block any input and the output will be tied to ground via the resistor R. For positive signal think of the diode as 0.6 volt battery with the negative pole connected to the resistor R. The opamp will try to force the inputs to the same voltage, hence it will always output 0.6 volt more (at the output pin) than the input

Do you mean for the positive input, I will get 2.6 as the output after the op-amp ? and get 2.6 - 0.6 = 2.0 volts at Vo ?

For negative inputs, what will be the voltage V1 after the op-amp ?

Thanks !

 

Yes a opamp will always try to make the input voltage (V+)-(V-) equal to zero. In order to do that it has to compensate for the diode voltage drop.

 

With a negative input, the feedback loop becomes open and the op amp output voltage goes to its negative rail (as determined by the negative supply voltage).