http://img360.imageshack.us/img360/5409/89911672ku4.jpg

The gain of an inverting amplifier is given by G=-Rf/R which means that if Rf=R the gain will be -1.
Looking at the image provided by the link above we know that when Ei is negative the precision rectifier will behave as much as an inverting amplifier does. The only difference in the circuit is D2 in series with the output once D1 in this case is "off" (it's like if it wasn't there). Why does the gain remain the same (G=-1) with D2 placed in there?

However, why does Vo=+Ei? Shouldn't Vo be equal to +Ei-0.6 (0.6V is the voltage drop across the diode)? Why is Voa=Vo+0.6...shouldn't Voa be equal to +Ei?

 

Hi,

Shouldn't Vo be equal to +Ei-0.6 (0.6V is the voltage drop across the diode)? Why is Voa=Vo+0.6...shouldn't Voa be equal to +Ei?

The feedback is taken from after the diode (directly at the output) and since the op-amp will try to keep both inputs equal, it will ramp up the output until this condition happens.

When R = Rf, and the non-inverting input is at 0V, the output _has_ to be -Vin to make the inverting input 0V as well.
(If eg. Rf was twice the value of R, the output would be -2*Vin).

 

Also, the only purpose that D2 severs is to speed up the circuit. It keeps the output of the opamp from banging into the supply rails, from where it would take longer to slew to the correct level when D1 is conducting.

 

Also, the only purpose that D2 severs is to speed up the circuit. It keeps the output of the opamp from banging into the supply rails, from where it would take longer to slew to the correct level when D1 is conducting.

You confused the two diodes.
D2 is the rectifier. D1 keeps the output from going near the negative supply voltage.

 

Can someone explains what the circuit does with a positive input voltage Ei?

 

With a positive input the opamp inverts and D1 clamps the output of the opamp at about -0.6V. D2 is reverse-biased so the circuit has an output of 0V.

 

Hi,

The feedback is taken from after the diode (directly at the output)...

Why only after the diode? What kind of analysis did you go through to reach such conclusion? What if we were assuming Vo's internal resistance as well?

PS: I'm sorry for the long time to reply back...

 

If the output voltage drops slightly because the load current increases the voltage drop across D2 then the negative feedback and the internal voltage gain of 200,000 (at low frequencies) of the opamp make a correction so that the output voltage is correct.

 

Actually, I was asking for some mathematical explanation (circuit analysis?!) that proves what you're saying

 

Hi,

Why only after the diode? What kind of analysis did you go through to reach such conclusion?

Eh? I used my secret super-power of light wave to brain stem interface (commonly called vision here on earth I am told)

What if we were assuming Vo's internal resistance as well?

As I allready mentioned, the op-amp has one function in this circuit... To keep both its inputs at equal potentials if at all possible and the output Vo is a consequence of this.

RF and R is a voltage divider with the inverting input at half the potential of Vo to Ei- and since the output will (within its capabilities) keep the inverting input at the same potential as the non-inverting input (i.e. 0V), the output will be the inversion of the input (Ei-) to satisfy the zero potential difference of the op-amps input.

If you made the input (left side of R) a positive voltage, the opamps pin 6 would go negative until the current through D1 equalled the current through R (sign inverted).
This wouldn't reach Vo though, since D2 would then be reverse biased and since both R_f and R_L is referenced to the common ground of 0V (virtual and real respectively), Vo will be 0V.

 

I think what you're saying is a really good approximation of how the circuit behaves although not totally correct due to the non-ideal diodes as it seems, correct me if I'm wrong, you're assuming ideal ones in your explanation.

Would you be able to explain me mathematically what really happens in this circuit assuming non-ideal diodes?

 

Why don't you simulate it with different OpAmp parameters, D voltages, and Resistors? Here is ideal OpAmp: http://nl5.sidelinesoft.com/examples/rectif.jpg

 

http://img360.imageshack.us/img360/5409/89911672ku4.jpg

The gain of an inverting amplifier is given by G=-Rf/R which means that if Rf=R the gain will be -1.
Looking at the image provided by the link above we know that when Ei is negative the precision rectifier will behave as much as an inverting amplifier does. The only difference in the circuit is D2 in series with the output once D1 in this case is "off" (it's like if it wasn't there). Why does the gain remain the same (G=-1) with D2 placed in there?

However, why does Vo=+Ei? Shouldn't Vo be equal to +Ei-0.6 (0.6V is the voltage drop across the diode)? Why is Voa=Vo+0.6...shouldn't Voa be equal to +Ei?

The gain remains the same because you take the output after D2. If you take the output before D2 the gain will increase slightly as to overcome the voltage drop across the diode.

The output voltage is +Ei because the output voltage just after the op amp is one diode voltage drop above E1 as overcame the voltage drop across the diode and keeps the gain fixed. This is a nice application of negative feedback.

 

I think what you're saying is a really good approximation of how the circuit behaves although not totally correct due to the non-ideal diodes as it seems, correct me if I'm wrong, you're assuming ideal ones in your explanation.

Would you be able to explain me mathematically what really happens in this circuit assuming non-ideal diodes?

The diodes are inside the feedback loop, so their nonlinearity and nonideality are irrelevant, as long as they are forward biased. You have to understand op amps to understand this, and you obviously don't.
As others have explained, the op amp has so much gain that the inverting input (pin 2) will always be at zero volts. Also, no current will ever flow into the inverting input.
When the input voltage Ei is positive, D1 will conduct, clamping pin 2 to zero volts. Since the op amp output (pin 6) has to be ≈-0.6V for this to occur, D2 will be reverse biased, and the output Vo will be zero volts.
When Ei is negative, D1 will be reverse biased, D2 will be forward biased, and all the current through the input R will flow through Rf. Pin 2 will still be at zero volts, because the high op amp open loop gain and the negative feedback will drive Vo to a voltage which will force this to be so. The op amp output will be a diode drop above the output voltage. The output voltage will be exactly equal to the negative of Ei. again because all the input current flows to the output, and pin 2 is always at zero volts.
I have probably reiterated what others have said (I confess to not reading them). Hopefully, if we keep hammering on it, you will finally understand the power of op amps and negative feedback to cancel nonlinearities.

 

Another way to think of that is:

The voltage between the inputs of the op amp is almost equals zero. Then think what the output voltage has to be for a given input as for the voltage between the inputs of the op amp to be zero.

If you understand negative feedback you will see how important and amazing it is.