inverting and noninverting operational amplifiers

[vikstiks]

should o/p voltg increase with increase in i/p voltg frequency in case of inverting and non-inverting amplifier?

 

[gadly]

The output voltage should not be dependent on the frequency of the input voltage signal in an ideal amplifier circuit which has gain setting resistors only. In practise, all real circuits have frequency dependence which usually begins to show up at higher frequencies. Additionally, if you have capacitors and/or inductors in your amplifier circuit, these will cause frequency dependence. Hope I understood your question correctly.

 

[crutschow]

All standard op amps have a gain-bandwidth product (given in their data sheets) due to the 1-pole, 6dB/octave roll-off inherent in their design.

For a non-inverting op amp configuration this means that the circuit bandwidth is equal to the op amp gain-bandwidth product divided by the closed-loop gain.

For an inverting op amp configuration the circuit bandwidth is equal to the op amp gain-bandwidth product divided by 1 plus the closed loop gain. Thus, for example, if the op amp GBW is 1MHz and the inverting closed loop gain was 1, the circuit bandwidth would be 1MHz / (1+1) = 500KHz.

 

[vikstiks]

thank u...
now pls help me with this very basic concept....

when both the amplifiers are same in opamp why one treats as positive and another as negative...?

 

[crutschow]

thank u...
now pls help me with this very basic concept....

when both the amplifiers are same in opamp why one treats as positive and another as negative...?

I don't understand your question. :-?

 

[vikstiks]

in IC 741 opamp....
pin 2 ic -ve and 3 is +ve...
both give the same o/p...k...
bt y +ve n _ve thr???

 

[crutschow]

If I understand you correctly you are asking about the plus and minus differential inputs. A differential input means the output is determined by the voltage difference between the two inputs multiplied by the open loop gain of the op amp. The polarity of the output is positive if the plus input is more positive than the negative input, and the polarity of the output is negative if the plus input is more negative than the negative input. If the two inputs are at the same voltage within the common-mode range of the amplifier than the output will be zero also (neglecting any input offset voltage).

 

[gadly]

in IC 741 opamp....
pin 2 ic -ve and 3 is +ve...
both give the same o/p...k...
bt y +ve n _ve thr???

It would help if you didn't abbreviate so much - write the words out in full. For example I assume the last line above means "But why positive and negative" ... but I can't make out what "thr" means.

Anyway, it is important to understand that an op amp does not amplify just the positive* or negative** input - it amplifies the difference between the two.


Hence it is called a "differential amplifier". This is what crutschow said, but in simpler terms. The other important thing about op amps is that they ideally have an infinite gain. This means feedback has to be used, they are otherwise useless as amplifiers. The feedback then determines the gain of the overall circuit.

* "non-inverting" is the correct term
** "inverting" is the correct term

 

[varunkant2k]

For an inverting op amp configuration the circuit bandwidth is equal to the op amp gain-bandwidth product divided by 1 plus the closed loop gain. Thus, for example, if the op amp GBW is 1MHz and the inverting closed loop gain was 1, the circuit bandwidth would be 1MHz / (1+1) = 500KHz.

I see various basic flaws in the statement.There is difference in close loop gain and loop gain.
Close loop transfer function = A0(OpenLoopGain)/(1+loop gain). Here I used loop gain (not close loop gain) in the bracket.
Again Thus, for example, if the op amp GBW is 1MHz and the inverting closed loop gain was 1, the circuit bandwidth would be 1MHz / (1+1) = 500KHz. this statement says negative feedback and systems close loop frequency increases as 1MHz * (1+1) = 2MHz, and not decreases.