The gain of opamp is controlled by feedback resistor connected from output to input. Based upon the opamp specification, one can estimate the feedback resistor value for obtaining the desired gain.

The input impedance of opamp is very high, whereas output impedance is very low. This combination is required for the most of the applications. 

Achieving of 100 M ohms is difficult even using CMOS technology. 

Working with 1 or 10 ohms consumes more power.

but sir how can you estimate which is a best selection of resistor combination to get gain 10..........because so many combination is possible.....like which one would you select between 10kohm or 1Mohm?

choice of components is an art for even a simple requirement such as an inverting amplifier. So you use guidlines. The guidleines are starting points as they night be different for different type of amplifiers. for a BJt amplifier such as 741, 358 and the like, a staring point is to use a resistor of the order of 10k as Ri and -gain times that as feedback. for a gain of -10, one would choose 10k and 100k. having said that why not 1k and 10k or even 100 ohms and 1k? Or why not 100k and 1 meg etc. As you move on either side, errors change. At high resistance values the error is due not only to offset voltage (as indicated in data sheet) but also due to bias current. The input resistance of an amplifier is ri at low frequencies. If a CMOS amplifier is used, the guideline is to start with 10 times 10k i.e 100k. for -10, feedback is through 1meg. One has to ensure that leakage resistances' effect on high resistance values are minimised. perhaps for a novice, these indications are adequate. If you want to know more. just ask. At low resistanmce values, the loading effect of the resistor and the current taken by it matters. So if you have 1 ohm and 10 ohm, the input resistance is just 1 ohm(too lo!!!) and load resistance on output is 11 hms. That means even if voltage at output is 100mV, current drawn by these resistors are about 9mA. Aim at reducing this current load of the inherent resistors. with 100k as feedback resistor, the current is only 100uA even if output is 10V or -10V.

The choice of absolute resistors values for the feedback network of an opamp should be traded off considering several aspects: low resistance values give rise to improved performance in terms of DC accuracy (expecially when input bias currents are not negligible, since input bias currents and input offset currents give rise to poorly controlled voltage drops on resistors of the feedback network which give rise to an offset in the nominal signal), reduced thermal noise voltage (the power spectral density of the voltage thermal noise generated by a resistor is sqrt(4*kT R)), reduced impact of parasitics (e.g. the impact of 1pF parasitic capacitance in parallel to a 100Ohm resistor is negligible up to more than 1GHz, while the impact of the same 1pF parasitic in parallel to a 1MOhm resistor could be relevant at less than 200kHz), and hence better high-frequency behavior.

On the other hand, reduced resistance values mean an increased power consumption and additional loading effect on the output (the feedback network is connected in parallel to the actual "load" and wastes part of the opamp output stage power and driving capability). Unless there are valid motivations to do otherwise, it is recommended that the current flowing through the feedback network is negligible (say less than 5%) with respect to the current flowing through the intended load.

If you mean an inverting amplifier, you have in total four elements connected in the feedback loop - two voltage sources (the input voltage source VIN and the op-amp output VOUT) and two resistors (R1 and R2). So, the common current flowing through the input source and the op-amp output is I = (VIN + VOUT)/(R1 +R2) or I = VIN/R1 = VOUT/R2. This current should be negligible so that the total current does not exceed the maximum op-amp output current and this determines the minimum resistance choce (figuratively speaking, this is an "informational" current used by the op-amp to adjust its output voltage:) 

According to the maximum resistance choice, you should consider the stray resistor (mainly R2) capacitances.

As said above, very much depends on your application.

For a low impedance source one can use low values to keep the thermal noise low, but that will add to the power consumption (e.g. LM4562 used in a Passively Equalized RIAA Phono Preamplifier circuit).

So if one is designing something to operate low voltage and low current, and without needing high bandwidth, then one goes the other way and can use feedback resistors up to 10M - but this is very dependent on the specifics of the op-amp (e.g. LT6003/LT6004/LT6005 family used in low very low power amplifier for a micro-controller ADC front-end).

Also dependant on inverting or non-inverting configurations. In the inverting configuration the input resistor to the summing node gives the input impedance that the source sees. In non-inverting the input impedance is arbitrarily dependant on the op-amp  being usually may megaohms equivalent before one adds components.

In single rail non-inverting operation with multiple channels with gain be careful of the impedance of the mid-voltage reference point that the first gain resistor is tied to, as this can be a source of crosstalk between channels.

Because your question is fairly general and no other information is offered, a simple generalized answer may be appropriate.  Use 100 ohms and 1000 ohms.  High and low resistance are relative terms that have to be put in context.  With a typical OP-amp in the inverting configuration, 1 ohm and 10 ohms will draw too much power and the OP-amp will run hot or burn up.  10 mohms and 100 mohms will reduce the current to the point that the characteristics of the OP-amp will play a larger role.  100 ohms and 1000 ohms are about the right range of high/low for a typical OP-amp.  

Perhaps 100 ohms is too low in the case of an inverting amplifier...

Biswaijt

As you noted ".because so many combination is possible."  

and your circuit is not specified,

you have asked a general question 

and can only receive a general answer.    

Where I start a design application 

versus  where I end-up 

may depend greatly 

on how I compromise the circuit

in order to meet real world requirements.  

For example,

in designing a Hi-Q audio filter for Radio CW operations,

too much accuracy in designing a filter may hide the fact that the human ear/brain function  only has certain abilities of discernment. 

Thus, I may round off many numbers to the second decimal place. 

That may seem inaccurate, but It may be much more accuracy than the Real World can handle. 

To your question, in a simple fashion: 

I generally, for each stage in a chain, input vs. output, 

I start with 10 times the impedance of the source circuit, 

and end with 1/10 the impedance of the sinking circuit. 

Frequently this is the 10K to 100K range input 

and 100 to 1000 range output, each stage of the chain. 

I depend on my understanding of conceptual physics 

and my belief in math descriptors during initial design. 

I depend heavily on SPICE to allow me to measure the cumulative "Q" of my resonant stages, the inter-stage interactrions, the frequency / phase shifts, the loading / gain traits, throughout chains of 20 stages.  

Hope that helps, Glen

It is easy these days to use p spice or similar simulation tools including circuitmaker to identify everything inclusing current drawn by the feedback resistor. Please try simulating and you will find it to be a great learning tool. using 100k and 10k for a gain of -10 will be appropriate rather than 1k and 100 ohms as the current drawn by the feedback resistor will be output / 1k when 1k is feedback resistor. It all depends on source resistance, amplifier used (when 1mA is drawn from a 5mA amplifier, you are left with only 80% of 5mA for load), allowable noise, dc errors, dc drfit and bandwidth limitations due to stary capacitors. Engg is all about compromises, but one can and should optimise..

10Ohms-1KOhms thin film surface mount resistors are suitable for your choice, for their reduced thermal noise and strong immunity to parasitic parameters. For the reason to choose them, Paolo Stefano Crovetti has given a good explanation. For more details, you can read a technical note "Frequency Response of Thin Film Chip Resistors" from www.vishay.com. You can find some of these resistors in Digikey website. 

Resistor in the range of MOhms are comparable to the input impedance of the Op-Amp and would disturb the virtual ground property of Op-Amp and thus some current would flow inside the Op-Amp also. In this case, the gain won't be equal to theoretical calculations. 

Biswajit, 

In my thinking,

from outside of the OPA internal circuitry, 

(1) The very low resistor values

allow input bias current errors to enter into the results. 

(2) The very high resistor values 

introduce much more Noise into the results. 

That is my philosophical approach, 

and you know the supporting equations. 

Normally, I start with OPA  given a R(out) at 30 Ohms.  

Then design R(input) at 1K Ohm,

and R(feedback) at 5K Ohm.

I never try to get all my gain in one single stage

and frequently design an extra stage for more gain control.

Sometimes I design a Pre-Amp Inverting stage 

which increases-or-decreases the signal

to a more controllable level.

I normally adopt a standard Signal Level of 1 V

so to be above the background noise,

and allow much V(out) swing without saturation. 

Normally, I use 1/8 W metal film resistors.

Hope that helps,  Glen  

Glen,

I think "the very low resistor values" decrease the error caused by the input bias current...

Very important is the type of used OPA. Input impedances of amplifiers with BJT input are much smaller than the input impedances of inputs with JFET or MOSFET.

A current noise is bad for BJT input.

A voltage noise is bad for FETs; etc. etc.

These all determines needed values of feedback resistances.

Cyril, and Josef

OK. 

All BJT LM324 OPA in my project , 

and almost all Inverting stage designs. 

By "very low resistor values" on the R(in) lines, 

I mean in the KilOhm range,

not in the 1 Ohm to 100 Ohm range.

In my project, I try to handle low level impedances

with a first stage Pre-Amp design,

to bring the   target Signal up to the 1 V range.

At that point, mili-Volts of offset error

on a Single-Ended Input Stage, are insignificant

compared to the target Signal Level of 1 V.  

Many of my stages Non-Inverting Input ( pin#3 ) 

are simply grounded.   

Some further considerations ( refering to my AFX project) : 

Given an Inverting OPA  

which is driven by multiple signals, 

I find that some KilOhm resistance here

( at the Null node ) 

reduces cross-talk interaction

between the several driving stages.  

If the several driving stages are resonant,

and if they  feed into a very low ( 10 Ohm ) Null node,

then the prior resonant circuits

can become loaded

and drawn away from  f(0)  resonance.      

In a simple stage-one-drives-stage-two design 

then it may only be a minor problem with gain change.  

However,

If the prior OPA is resonant and lacks sufficient P(out)

or has too much Z(out) 

then resonance can be pulled away from desired results. 

Since my project is full of resonant stages, 

these problems have shown themselves. 

My solution was to make certain that each

resonant OPA stage Output 

is driving into a  high Impedance Inverting resonant OPA.

I have not tried to calculate the erroneous current flows,

rather, I have tried to design around the problem.

Also,

I do not attempt to maintain exact A=1

for any stage in a module.  

I rely on a controlling gain stage between modules 

aiming at   V(standard) = 1 V ac    overall.  

My thinking here is that there is more tolerance

for  component tolerance variations  

and  to allow tuning stages dead-center on f(0). 

This V(std)=1V is very far above low-level noise, 

and very far below any saturation, 

and makes Oscope observations ( and Spice Bode ) 

very easy to follow and compare.  

In practice, as I use the project daily, 

I do find that I run the gain up further 

in order to bring out some smidgling little signal.  

Then I rely on the superior narrow band filtering 

to extract the target signal.  

In radio reception of morse-code signals, 

at very low signal/noise levels, 

the morse-code signal is not clean / clear .  

The final detection is in the human ear/brain network. 

I hope you find that interesting,  

and certainly different from lab setups. 

I am here to learn, so 

you are welcome to comment on my design philosopy,

which I see as a top-down approach  

mostly using the Spice tool,

and my hand-calculator for checking things.   

http://www.geocities.ws/glene77is/GC_ET_Schem.html