# Opmp as an active rectifier and its slew rate limitation...?

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1. Sep 15, 2015

### brainbaby

The following circuit is of an active rectifier..The inclusion of a diode here signifies a voltage follower for positive Vin voltages ....but for negative Vin diode is off hence the out voltage Vout is zero....and the output of the opamp goes into negative saturation (opamp sinks current for negative Vin cycle)...But here comes a certain slew rate limitation of this circuit...with high-speed signals.The op-amp cannot swing its output infinitely fast, the recovery from negative saturation takes some time, during which the output is incorrect...

In order to deal with it and improved version of the circuit consisting of two diodes is employed as shown in figure 2..
The text says that "The improvement comes because the op-amp's output swings only two diode drops as the input signal passes through zero."...

What does that means...?

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2. Sep 15, 2015

### Hesch

Second circuit:

Say that the opamp is supplied by ±12V.
Now, try to swing the input, Vin, ±1V: What will happen at the output of the opamp, ( not Vout )?

Try the same with the first circuit:

You see the difference when Vin = -1V as for the first circuit, and Vin = 1V as for the second circuit?
How much will the output voltages drop?

Last edited: Sep 15, 2015
3. Sep 15, 2015

### brainbaby

According to my calculations..
for 1st figure...
when Vin= -1V...D1 is FB and D2 RB..so Vopamp = 2.7 V..

for 2nd figure...
Vin = +1V....then Vopamp =1.7V ....
So the inference is total voltage drop is 2.7 - 1.7 = 1.0 V....which donot agree with the text....as it should be 1.4 V(two diode drop)
god knows where i am wrong.....!!

4. Sep 15, 2015

### Jony130

Wrong.
For this circuit

For positive voltage at the input the output voltage follows the input voltage because the feedback loop is closed via a forward biased diode.
For Vin = +1V we have 1.7V at the op amp output and 1V at the output. Because only for 1.7V at the op amp output the voltage at inverting input is equal to voltage at non-inverting input.
But for negative input voltage the diode is in revers biased, so the feedback loop is open and op amp will enter into negative saturation -12V. Do you know why ?

As for this Second circuit:

Try again, as a tip for Vin = +1V --->Vopamp = -1.7V and Vout = -1V

5. Sep 15, 2015

### dlgoff

[slightly off topic]

[/slightly off topic]

6. Sep 15, 2015

### brainbaby

Its just an overcompensatory effect of opamp...where opamp when unable to detect any difference in the common mode signal does ..that due to lack of feedback....as in this case when diode is open...

7. Sep 16, 2015

### brainbaby

For first circuit Vin= -1V we have Vopamp= -12V...and
For 2nd circuit Vin = 1V we have Vopamp= -1.7 V...

So the drop is of --->-12-(-1.7)= -10.3 V..
then how to conclude further....

8. Sep 16, 2015

### brainbaby

1st circuit with one diode...
When Vin= -1V
Vopamp = -12V
Vout = 0 V

When Vin = +1V
Vopamp = +1.7V
Vout= +1V

Net volatge swing = 12 - 1.7 = 10.3 V
As the opamp has to swing more than 10.3 V to cross the zero voltage input line...which of course is a slew rate limitation....

2nd circuit..
When Vin = -1V
Vopamp = 2.7V
Vout = 2 V

When Vin = +1V
Vopamp =-1.7 V
Vout = -1V..

Net volatge swing = 2.7 - 1.7 = 1V..(but it should be 1.4V (two diode drop)....

Where seems to be the mistake then??

9. Sep 16, 2015

### Jony130

If for example Vin changes from +1V into -1V the op amp output start to drift (from +1.7V) into the negative direction ( because (Vp - Vn) is negative).
And the more the negative op amp output voltage is the "faster" the op amp output is changing into the negative direction, and this change stops when op amp reach negative saturation (-12V) and Vp - Vn = -1V - (-12V) = -13V
Vp - non-inverting input
Vn - inverting input

But how can Vout = 2V if Vin = -1V ?

10. Sep 16, 2015

### brainbaby

Since we cant neglect the effect of voltage divider as a feedback......so when Vin = -1V ...D1 conducts and the opamp voltage is 2.7V....this voltage experience a 0.7V diode drop across a diode...and then Vout becomes 2V...

Follow voltage divide rule..i.e
Vin'= 10/(10+10)*2 = 1V...
This voltage cancels the original voltage Vin (-1 +1=0V)...and the opamp gets satisfied.....

11. Sep 17, 2015

### brainbaby

@Jony130..

In order to ease out confusion ..i referred another article which says:

That with out D2 the opamp output would be saturated in negative direction ..true..However the negative voltage at the opamp forward biases D2..This tends to pull the opamp inverting input terminal in negative direction...But such a move would cause the opamp output to go positive..so the output settles at the voltage that keeps the inverting input terminal close to ground level......"

my inference...

pulling the opamp inverting input terminal in negative direction...means that the negative voltage starts to cancel out the positive inverting terminal voltage ..so the positive voltage at the inverting terminal loose out its positivity and it starts to drift towards negative direction(potential)...but now opamp would realize that the inverting terminal voltage is getting lower (negative) and it should match with the non inverting terminal voltage which is at 0V ..thus in order to achieve this opamp makes it output to go in positive direction..and in this kind of tug of war..the opamp output settles at a voltage of one diode drop below the ground....i.e -1.7V....

-1.7+0.7(diode drop@D2 for Positive input cycle)

= -1V..

this voltage of -1V gets canceled out with initial + 1V (positive cycle) resulting in zero volts at the inverting terminal....

1+(-1) = 0V..opamp satisfied...and happy..

But the phrase for improved active rectifier -->"The improvement comes because the op-amp's output swings only two diode drops as the input signal passes through zero."

is still obscure to me..

12. Sep 17, 2015

### Jony130

The problem lies in our analysis. I made a very stupid mistake. I don't know how this could happen, So I'm very sorry for that.

Once again, the circuit look like this:

Case one: Vin = -1V.
The current will flow from op amp output ---->D1---->10k feedback resistor---> and10k input resistor.
Iin = Vin/10k = -1V/10k = - 0.1mA and the same current will flow through 10k feedback resistor. So the voltage drop across the feedback resistor is :
0.1mA * 10k = 1V. So the output voltage is equal to 1V and the op amp output is at 1.7V.
Please notice that for negative input voltage this circuit work as a ordinary inverting amplifier with the gain equal to: Av = - RF/Rin = -10k/10k = -1V/V
So far so good
Case two: Vin = +1V
And the input current will now flow from Vin--->Rin--->D2--->flows into op amp output.
As you can see this is the place where I made a mistake in post #4. Because now the current will only flow through Rin and D2. Current can not flow through RF resistor because D1 diode block it. And this is why Vout = 0V and Vop amp = -0.7V
Also notice that for the positive input voltage the op amp output voltage is fix at -0.7V , and now if the input signal change his polarity from positive into negative, the op amp output voltage start to rise from -0.7V until reaches +0.7V and D1 diode start to conduct a current. And depending on how large the output must be the op amp will further increase his output voltage. I hope that now you see op amp output must swing only by 1.4V to make D1 or D2 ON/OFF.

13. Sep 18, 2015

### brainbaby

Yes I got it now...
actually the whole problem occurred because we didn't gauged the right opamp voltage for Vin =+1V....
that why in my post 11 ...i was not able to come to a conclusion of 1.4V opamp voltage swing ......

and my another fault was that i considered Rf and Rin into a parallel combination and applied voltage divider rule to derive out the voltage at the inverting node of the opamp...Literally why this happened was due to the diode D2 branch...which forced me to think that both resistances are in parallel arrangement...but initially D2 is OFF for Vin = -1V.....

I am happy now....
Thanks buddy....