so gain-bandwidth product is supposed to be the product of the gain and the bandwidth at that particular gain. and that it is independent of the gain it is measured so it will always be equal wherever it is measured on the graph of the gain.

but how do we get the gain-bandwidth product if the gain is negative? doesn't bandwidth increase as gain decreases? if so, is the gb product of at -3db is different at -6db?

 

Gain is not negative ( unless you are thinking of phase or signal inversion ) gain of -3 db is really a gain of less than 1 and greater than 0. Not quite the answer to your question but may be close. We do not usually use amplifiers at a gain of less than one ( why would we ? ) and for op amps it would require really heavy feedback as 100 % feedback is use for non inverting gain of 1.

 

so how to solve for the gain-bandwidth product at -3 dB? what value of gain should be used to multiply with the bandwidth?

 

so how to solve for the gain-bandwidth product at -3 dB? what value of gain should be used to multiply with the bandwidth?

Basically, you use gain on a linear scale, not a dB scale. Note that

\(G_{[dB]}=20\;{\rm log_{10}(Av)\)

where G_dB is gain in dB and A_v is the voltage gain on a linear scale.

Think about it. If you used a dB scale and considered the frequency at which the linear voltage gain is equal to one (Av=1), you would conclude that GB=G*B=0, which would not be a very useful definition since this can't be a constant.

So, on a linear scale -3 dB equals 0.707, thus if you had a 1 MHz bandwidth at the -3 dB point, then the GB would be 707 kHz.

The idea of constant GB comes into play when we talk about feedback. If you take the above example amplifier and use negative feedback to reduce the overall gain to -20 dB, you should expect that the bandwidth will increase. Since -20 dB is a gain of 0.1 on a linear scale, the gain is about a factor of 7 reduced; hence, the bandwidth should increase by about a factor of 7, giving 5 MHz.

Likewise, if that amplifier already had feedback and you removed it, and found that the new gain was 40 dB, you would conclude that the original bandwidth was 707kHz/100= 7070 Hz.

 

The constant gain-bandwidth product is seen in standard compensated op amps which have an internal single pole rolloff that starts at a low frequency. This gives a 6dB/octave rolloff (gain reduced by 1/2 for every doubling of the frequency) of the open-loop gain over most of it's frequency range. Thus the product of the gain times frequency (GB) is constant for any frequency on a plot of the open-loop gain versus frequency. If you close the loop to give a particular gain you can look at that point on the graph and get the closed-loop frequency response. This gives the same answer as dividing the gain-bandwidth product by the closed loop gain.

Note that this does not apply to current-feedback type op amps (typically used in high frequency applications) which have a gain that is fairly independent of gain.

 

thanks for the replies..