If I build a resistor network where the op amp has a lower gain, it is able to maintain its gain for a larger bandwidth. Why?
Electronic – Why is an op amp’s bandwidth higher at lower gains
bandwidthfrequencygainoperational-amplifier
Related Solutions
GBP is the to do with the open loop gain of the op-amp. If you have a closed-loop circuit then GBP can help you find where the flatness of the frequency response starts to be eroded.
GBP - if it has an open loop DC gain of 1 million and unity gain at 1MHz then the GBP is said to be 1 million.
This is useful to know because if you have an op-amp that has a GBP of 1 million and you have a manufactured a closed loop gain of 100 then I would expect the gain to remain flat up to about 10kHz then roll-off gently at 6 dB / octave: -
Here is an extract from the data sheet for the AD8606 op-amp and I've drawn four red lines on it at 10kHz, 100kHz, 1MHz and 10MHz. The line at 10MHz is important because this is the unity gain point of the op-amp i.e. it has a GBP of 10,000,000. If this was all we knew we could predict the open loop gain at 10kHz by dividing 10,000,000 by 10,000 to get 1,000 (this would be the open loop gain at 10kHz and of course a gain of 1,000 is 60 dB - exactly as seen in the graph.
This is not hard to understand if you grasp the difference between carrier frequency and signal.
The way information is transferred over long distances is with carrier frequencies that are modified just a tiny little bit. A carrier frequency is nothing but a sine wave at some specific frequency; by itself it contains no information whatsoever, because you cannot get information from a sine wave that is always the same.
The information comes from changes that you make to the carrier signal. For instance, you can make a rule that changing the carrier by 1Hz represents a one and by -1Hz represents a 0. Or maybe that lowering and raising the amplitude conveys information. Or one of many other coding schemes.
What is important to understand is that it is the modification of the carrier that conveys information. And this is not something you can do with infinite precision; both the transmitters and receivers are subject to interference or noise that make it impossible to distinguish between very small changes in the carrier. So you always need a finite amount of space around the carrier frequency to convey a nonzero amount of information. This is called the bandwidth, and only it determines your speed. The carrier frequency is not important.
The carrier frequency is important for the total amount of information that can be fit into a certain range of available frequencies. You don't get all the spectrum in the world to work with. For one; higher frequencies can't penetrate large obstacles so they are unsuitable for long-range communication. Some frequencies are also quickly absorbed by the atmosphere. This leaves only small pockets of bandwidth (e.g. 2.4-2.5GHz for ISM, the stuff that old WiFi works on) that are usable and even those need to be subdivided into 'channels' (e.g. 2.40-2.405GHz) to make sure that multiple adjacent transmitters can work without interfering with each other.
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Best Answer
This is called constant gain-bandwidth product but it isn't true for every op amp. It is only true for voltage feedback op amps which use dominant pole compensation for stability. Such op amps can be approximated as a first order system since one pole dominates all others and the others can be ignored. (However, this is not true of current feedback op amps since current feedback op amps do not have a constant gain-bandwidth product.)
A first order system has a transfer function of the form
$$H(j\omega) = \frac{H_0}{j\omega\tau + 1} = \frac{H_0}{j\omega/\omega_c + 1}$$
where \$H_0\$ is the DC and passband gain, \$\tau\$ is the time constant of the dominant pole and \$\omega_c\$ is the cutoff frequency (bandwidth). The gain of this system is
$$|H(j\omega)| = \frac{H_0}{\sqrt{(\omega/\omega_c)^2+1}}$$
For \$\omega << \omega_c\$ the gain is approximately \$H_0\$ and the bandwidth does not come into play. If \$\omega >> \omega_c\$ the gain-bandwidth product can be approximated as
$$|H(j\omega)|\omega = \frac{H_0}{\sqrt{(\omega/\omega_c)^2+1}}\omega \approx \frac{H_0}{\sqrt{(\omega/\omega_c)^2}}\omega = H_0\omega_c$$
which is a constant. Since it is a constant, increasing the gain requires a decrease in the bandwidth while decreasing the gain allows an increase in the bandwidth.