# what is meaning of infinite bandwidth of an amplifier

I have just learned about amplifier but the term infinite bandwidth confuses me a lot.plesse explain with some example.

• Can you give us some context? Did someone claim they can actually make an infinite bandwidth amplifier, or is it an idealized model for analysis? – The Photon May 3 '13 at 17:48
• While reading op amp they claimed that ideally it should have infinite bandwidth. Now referring to this can u explain? – user122345656 May 3 '13 at 17:50
• Infinities in electronics are strictly a teaching aid, not to be taken literally. Infinity is not a number, and many of the idealizations involving infinities are actually nonsensical. For instance, an op-amp with infinite gain which keeps + and - at exactly the same voltage cannot possibly work, even in the imaginary world where it is instantiated. "can't possibly work" is the very antithesis of "ideal". – Kaz May 3 '13 at 18:38
• This ends up being a question about a key difference between idealized models from introductory courses, vs. real components you can build/buy/apply. As such, it's perfectly appropriate here. – Chris Stratton May 3 '13 at 19:15
• Although, following up to my earlier comment above, not all infinities are "nonsensical". Infinite bandwith of an amplifier just reflect that it is represented by a naive multiplication, which doesn't have a mathematical flaw of any sort. But of course if we are actually working with the bandwidth as a quantity, we cannot assign it the value of infinity and proceed with the formulas. Calculations that depend on the bandwidth of that amplifier as a quantity cannot be carried out and have to be replaced by some other reasoning (perhaps involving limits). – Kaz May 3 '13 at 20:55

An amplifier with infinite bandwidth would be able to reproduce any input, no matter how fast it changes. Even if the input were a 100 terahertz sine wave, or had steps from 1 V to 10 V in a fraction of a femtosecond, the amplifier would be able to reproduce that signal (scaled) at the output.

But this is an idealization...real amplifiers have limitted bandwidth, meaning that as the input signal increases in frequency, the output will not reproduce the input accurately. Furthermore, once feedback is added, the bandwidth of the amplifier circuit depends on the gain of the complete circuit.

Really fast op-amps on the market today have a "gain-bandwidth product" on the order or 1 or 2 GHz, but values for this parameter range from kilohertz up. A 10 MHz (GBW) op-amp, at gain of 1 (if it's unity-gain stable) can pass signals up to 10 MHz with low attenuation. Configured with feedback gain of 10, a 10 MHz op-amp could amplifiy signals up to 1 MHz, etc.

There is also an effect called slew rate limiting, separate from bandwidth, that limits the d/dt of the output voltage of an op-amp. This can produce distortion on step-shaped inputs, especially for high amplitudes, even when the bandwidth limitation implies that good reproduction is possible.

• So does it means that bandwidth is just measure of how accurate o/p reproduces I/p? – user122345656 May 3 '13 at 18:05
• It's a measure of accuracy, especially with regard to high frequency. – pjc50 May 3 '13 at 18:13
• If it is so then why definitions of bandwidth are given in terms of difference between upper and lower frequency bcz I dont think it is obvious to conclude this thing from that definition? – user122345656 May 3 '13 at 18:19
• Bandwidth is one of the things that can limit how accurately the output reproduces the input. For op-amps we don't specify the lower bound of the bandwidth because op-amps normally operate down to dc. If there was for some reason an op-amp that didn't operate down to dc, then they'd need to specify a lower bound on the operating frequency range. – The Photon May 3 '13 at 18:20
• It may be worth mentioning that the assumption of ideal op amp behavior is often coupled with the assumption of ideal behavior for other components such as resistors, capacitors, and inductors. Such assumptions allow a system that contains a mixture of op amps and passive components to be modeled using a set of linear equations (using complex numbers for parts involving inductors or capacitors). Adding any non-ideal behaviors which cannot be mimicked using combinations of passive components will often make such modeling impossible, requiring that one use simulation instead. – supercat May 3 '13 at 18:32

When studying op-amps you progress from the perfect idealized device to the less-perfect real deal. The idealized op-amp has: -

1) Infinite gain

2) Infinite bandwidth

3) Infinite common-mode rejection

4) Infinite supply rails

5) Infinite current output drive

6) Zero input offset

7) Zero input bias currents

8) Zero self-induced input noise

The list could go on.

It's supposedly easier to learn the idealized op-amp first and then progressively get a feel for what the non-idealized parameters of an op-amp are. Some non-idealized parameters are important to certain types of applications but barely have any bearing on other types of application.

• Many of these are completely nonsensical. Infinity is not a number. Infinite gain is better regarded as "gain which is as large as you need it to be so that additional increases make a vanishingly small difference that you can neglect". Because if + and - are exactly the same, how can the device work? A difference voltage of zero cannot just be multiplied by infinity to get whatever voltage you want. – Kaz May 3 '13 at 18:41
• @Kaz: No, these idealizations do make some sense. Think of limits if that makes you more comfortable. For example, how does the circuit behave differently as gain gets ever larger? There are often useful circuit parameters that stop changing as gain gets large. This works similarly for other idealized parameters. It can be useful to learn this first, then learn how to design for real limitations. Often several of the real limitations don't matter in a particular design such that they might as well be the ideal infinite or zero values. – Olin Lathrop May 3 '13 at 18:50
• @Kaz: An ideal op amp will output as high an output voltage as is necessary to ensure that the voltage on the inverting input does not exceed the voltage on the non-inverting input, and as low a voltage as is necessary to ensure that the voltage on the non-inverting input does not exceed the voltage on the inverting input. If one assumes ideal resistors, capacitors, and inductors, the voltage swing required to keep the inputs in balance will often be very-reasonably bounded. – supercat May 3 '13 at 19:36
• @Kaz: As a simple example, consider an op amp whose non-inverting input is wired to a signal source, and whose inverting input is wired to a ground reference via 1K resistor and to its output via 9K resistor. For the inputs to be balanced, the difference between the output voltage and ground must equal ten times the difference between the input voltage and ground. Thus, assuming ideal op amp and resistors, the ground-relative output voltage will be ten times the input voltage. – supercat May 3 '13 at 19:38
• @Kaz the idealized op-amp, with negative feedback, will produce a voltage (on V- input) that exactly matches the voltage on the V+ input AND, the output voltage will be precisely what the ratio of the feedback resistors are. It does this because it HAS got infinite gain. Any less than infinite gain would produce an error. This is one of the basic principles (and truths) learnt. – Andy aka May 3 '13 at 19:38

The "meaning" of infinite bandwidth couldn't be simpler. Consider a linear voltage amplifier with a gain of $A$:

$v_{out} = A v_{in}$

This "says" that the amplifier has a gain that is independent of frequency, i.e., the frequency response extends from 0 to "infinity". Any frequency at the input is amplified at the output with gain $A$.

Looking at this in the frequency domain, the impulse response of this amplifier is an impulse.

Of course, the gain of real voltage amplifiers do depend on frequency. Op-amps can be modelled as having a 1st order low pass filter frequency response:

$v_{out} = \dfrac{A}{1 + j\frac{\omega}{\omega_0}} v_{in}$