Gain–bandwidth product

If the GBWP of an operational amplifier is 1 MHz, it means that the gain of the device falls to unity at 1 MHz. Hence, when the device is wired for unity gain, it will work up to 1 MHz (GBWP = gain × bandwidth, therefore if BW = 1 MHz, then gain = 1) without excessively distorting the signal. The same device when wired for a gain of 10 will work only up to 100 kHz, in accordance with the GBW product formula. Further, if the maximum frequency of operation is 1 Hz, then the maximum gain that can be extracted from the device is 1×10⁶.

We can also analytically show that for ${\displaystyle \omega \gg \omega _{c}}$ GBWP is constant.

Let ${\displaystyle A_{1}(\omega )}$ be a first-order transfer function given by:

${\displaystyle A_{1}(\omega )={\frac {H_{0}}{\sqrt {1+{{\left({\frac {\omega }{\omega _{c}}}\right)}^{2}}}}}}$

We will show that:

${\displaystyle GBWP_{\omega \gg {\omega _{c}}}={A_{1}}(\omega )\cdot \omega \approx const.}$

Proof: We will expand ${\displaystyle A_{1}}$ using Taylor series and retain the constant and first term, to obtain:

${\displaystyle GBWP={A_{1}}(\omega )\cdot \omega ={\frac {H_{0}}{\sqrt {1+{{\left({\frac {\omega }{\omega _{c}}}\right)}^{2}}}}}\cdot \omega \simeq {\frac {H_{0}}{\sqrt {{\left({\frac {\omega }{\omega _{c}}}\right)}^{2}}}}{\left(\omega -{\frac {\omega _{c}^{2}}{2\omega }}\right)}={H_{0}}\cdot {\omega _{c}}\left(1-{\frac {\omega _{c}^{2}}{2\omega ^{2}}}\right)=const.}$

Example for ${\displaystyle \omega =5\cdot \omega _{c}}$

${\displaystyle GBWP={\frac {H_{0}}{\sqrt {\frac {\omega _{c}^{2}+25{\omega _{c}}^{2}}{\omega _{c}^{2}}}}}\cdot 5{\omega _{c}}={\frac {5}{\sqrt {26}}}{H_{0}}\cdot {\omega _{c}}=0.98\cdot {H_{0}}\cdot {\omega _{c}}}$

Note that the error in this case is only about 2%, for the constant term, and using the second term, ${\displaystyle \left(1-{\frac {\omega _{c}^{2}}{2\omega ^{2}}}\right)}$, the error drops to .06%.