Input and output of an operational amplifier

[ananthu]

Homework Statement

The question is about the concept of the operational amplifier in electronics.
For an ideal op. amp., it is said that the input impedance is infiinite and the current allowed by the op.amp. through its input terminal zero or in the case of a practical op.amp., it is nearly zero or negligibly small. It means all the current arriving at the input terminal passes through the outer path containing the feedback resistor R_f. The normal questions a student asks is -"When that is the case, what is the need of the circuit elements at all since all the current flows through only a path outside the op.amp. device? What is the exact role played by several resistors and diodes that are making up the circuit diagram of the op.amp.? How does an enlarged output appear without taking any input?"
Is it possible to give a simple, understandable and convincing explanation to a student of higher secondary classes without iindulging in any jargons of advanced and brain teasing explanations using higher electronic terminolgy, for the above questions? I myself find difficult to comprehend the above phenomenon. No reference books in electronics deal with such basic notions and doubts arising in the minds of an average reader as they straight away start analysing everything mathematically.

Homework Equations

The Attempt at a Solution

 

[ananthu]

Even after several days of posting the above question, I have not received any reply from the members of PF. I am really surprised and don't know the real reason. Are questions on "electronins" beyond the purview of the forum? Any body, kindly help me please?

 

[skeptic2]

Though the input current to an opamp at the input is very small and negligible in most opamp calculations, it is highly amplified to produce the output.

If you have an opamp in the standard inverting configuration with the non-inverting input connected to 1/2 Vcc and the input signal connected through an input resistor to the inverting input. The feedback resistor is connected between the output and the inverting input. In this configuration the output is equal to Rf / Ri * (2Vn - Vi) where Rf is the feedback resistor, Ri is the input resistor, Vn is the voltage at the non-inverting input and Vi is the voltage at the inverting input.

Because the opamp's gain is so high (without feedback), the difference between the inverting and non-inverting inputs (with feedback) can never be very much. In fact if the open loop gain (gain without feedback) is -100,000 and the output is 1 volt higher than the non-inverting input then the voltage at the inverting input must be (Vout - Vin)/-100,000.

 

[Redbelly98]

Hello again!

There are other sections of Physics Forums that are better suited to asking electrical engineering problems.

For all homework questions having to do with engineering:
https://www.physicsforums.com/forumdisplay.php?f=158

For non-homework electronics and electrical engineering questions:
https://www.physicsforums.com/forumdisplay.php?f=102

To answer your question:

For an ideal op. amp., it is said that the input impedance is infiinite and the current allowed by the op.amp. through its input terminal zero or in the case of a practical op.amp., it is nearly zero or negligibly small. It means all the current arriving at the input terminal passes through the outer path containing the feedback resistor R_f. The normal questions a student asks is -"When that is the case, what is the need of the circuit elements at all since all the current flows through only a path outside the op.amp. device?

But some of the current does flow into the op-amp inputs. It is that current, or more accurately the op-amp sensing the voltage difference of the two inputs, that determines the output voltage.

 

[queenofbabes]

Do also take note the op-amps need external power sources to work! The enlarged output doesn't appear out of nowhere.

 

[ananthu]

Thank you all. But my doubts were not answered. While the reply given by skeptic2 is purley technical which I can not convincingly pass on to a state board student of a South Indian city suburban school , other replys are too short to comprehend. OK. Let us guess an impossible case of an 100% ideal op.amp. whose input impedance is really infinite. In that case it can not draw any current at all and then circuit elements have no business to do with the amplification process. Perhaps my IQ level is not sufficient enough to understand about what the hell is going on inside this little wonder called "Op.Amp."! If fact, I am searching for some sort of non-technical illustration to explain the going-on inside the Op.Amp.in terms of flow of water facing an obstruction at certain point on its way, and as a result the water is forced take an alternate path where its pressure and quantity of water underging enormous changes! I will be happy to see your replys. Thank you, Redbelly. I will post this question in the forum mentioned by you also.

 

[Redbelly98]

There are two approaches, or levels, at which people "understand" op-amps.

The first level means to accept, pretty much on faith, all the idealized assumptions such as (1) the input current is zero, and (2) the output gain is infinity, and use those assumptions to figure out how the external resistors, capacitors, etc. will affect the output.

The second level is to understand the internal circuitry of the op-amp, which means to understand this circuit:

(from http://www.play-hookey.com/analog/inside_741.html )​

Since we are trying to find some level in between those two, we'll have to give up the notion that the input current is exactly zero, since that is leading to wrong conclusions.

 

[ananthu]

Thank you, Mr. Redbelly!