The two stages cascaded have unity gain so as to have Vo =Vm but when analyzing the circuit for each half cycle (i. e positive or negative half) the 2nd stage acts as inverting one time and then the other time non inverting (gain>1).
Vaibhav,
You require both the OP-AMPS to be in unity gain (and hence need to design the circuit for this). Both OP-AMPS must be configured in negative feedback with a diode in their feedback loop. First OP-AMP (non-inverting) does not require any resistors and behaves like a buffer amplifier (albiet only for positive half-cycle). The other OP-AMP in inverting mode would require both resistors same value to give you a unit gain.
Hope this helps,
Tahmina
Hello Tahmina,
The above post would be helpful if u could add a picture of the circuit in reply. Thank You.
Sriram, this is for a two OP-AMP precisipn rectifier, this is what Vaibhav referred in his question. Search for it and let me if you cant find it.
Tahmina
Vaibhav, not sure where did you get this circuit from? This doesn't seem right because there is a positive feedback on th second OP-AMP. Under this condition, OP-AMP behaves as an oscillator not as an amplifier. Probably, you wanted to draw this attached circuit? Also show where you want your output?
Tahmina
Actually there is no positive feedback, notice that the 10k resistor is grounded, so no contribution of the output is delivered to the positive input. This is only a load resistor.
Your second stage is a subtractor (subtracting the signal at the cathode of D2 from the signal at the anode of D1). If you want the same gain on both subtractor inputs you need to add an 1k resistor (in this case) in series with D2 and the node connecting V+ to R3, forming a voltage divider to deliver the signal to the opamp input.
The circuit is correct and works fine . I got output as per expected . It does not hold any positive feedback. Every stage is unity gain only.I am little justified with Tiango sir. second stage will behave as subtractor .oh........... i got a click i will analyse and i will let u all know.
thanks to all.
Dear Vaibhav,
the answer of Tahmina will be correct when you will change "there is no current in R2,R3 and R4" with "there is no current in R2,R1".
The current on R3 is equal to ve/R3 and let the first OPAMP stay in linear region with its output equal to ve+Vgamma.
Obviously the current on R4 is always different from zero.
sim
dear Tahmina
as you mentioned -ve hc circuit works as noninverting amplifier is correct which means gain will be 1+ rf / r1, will be grater than one; then hows it possible to get output equal to input (Vm).
Hmm... a very original and unusual circuit solution... It took me more than an hour to figure out its idea... and now I think it will work fine... Let's see why...
At the input positive half-wave, seemingly there is no feedback in the first op-amp stage... and the op-amp 1 should be saturated... but still there is a feedback... and it is closed through the second op-amp stage. Let's see how...
D2 is on and delivers the signal to the op-amp (2) non-inverting input. Since D1 is off, the left end of R2 is floating... so the second op-amp acts as a voltage follower. Since no current flows through R1 and R2 (because of the extremely high op-amp input resistances), the three voltages - at the second op-amp output and the two op-amp inverting inputs, are equal. So D2, R3, op-amp 2, R1 and R2 form a voltage follower (R4 is unnecessary)... and it is connected in the negative feedback loop of the first op-amp 1. Thus the whole circuit acts as a voltage follower... and it is a precision follower since D2 is connected in the feedback loop... so the op-amp 1 will raise its output voltage to overcome the diode VF. Any of the three voltages can serve as an output voltage but only the op-amp 2 output is low resistive... so it is used as a circuit output
(Indeed, it is quite strange that the "output" of the second follower is taken from the op-amp 2 inverting input... and I have posted a question about this negative feedback phenomenon:
https://www.researchgate.net/post/What_are_actually_the_input_and_output_of_a_negative_feedback_system_Can_we_consider_the_input_quantity_as_a_disturbance_and_the_output_quantity-as_a_reaction )
At the input negative half-wave, D1 is on... and the op-amp 1 acts as a precision voltage follower (overcoming D1's VF)... D2 is off... and the op-amp 2, R1, R2 and R3 form an inverter (K = -1)... so the whole circuit acts as an inverting follower...
So, the idea of this clever circuit solution is to deliver the positive half-wave without any change and to invert the negative half-wave. For this purpose, it acts as a follower during the positive half-wave and as an inverter during the negative one... Regards, Cyril.
thanks Mr Cyril for you kind explanation, After this post I also did many reading and experimenting, I came to same conclusion at last. I am happy to see your text and confident with my approach.
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