Electronic – the closed-loop bandwidth of the INA143 at gain 0.1

Here is the open-loop bandwidth of a certain op-amp shown in red: -

The blue line is when certain closed-loop components are applied to the op-amp.

Bandwidth is normally measured at the 3dB point of the frequency response and in the case of an op-amp (open-loop) this will be at 24Hz in the diagram.

If closed loop components were present, the gain would be reduced to (say) 20dB (blue line) but the bandwidth would increase to 1MHz.

The above example is for simple resistors "closing" the loop with negative feedback and the resulting bandwidth (3dB point) is always greater.

However, if an op-amp filter circuit was required that cut-off frequencies above 10Hz, the filter would have a bandwidth of 10Hz. In this example the closed-loop bandwidth is less than the open-loop bandwidth.

Does this help?

There is a fairly simple graphical method that can be used to solve this. It's derived from the expression for the closed-loop gain for a negative feedback system: $$ \text{Gain} = \frac{A}{1+AB} $$ where A is the open-loop gain and B is the feedback factor. We can see from this that for large loop gain AB $$ \text{Gain} \approx \frac{1}{B} $$ In this example, A is the open-loop gain of the op amp. The feedback factor B is the feedback signal divided by the output signal with the input at zero, which in this case is $$ B = \frac{R_1}{R_1+R_f} $$ where \$R_f\$ is the equivalent resistance of the switched resistor network. We have here \$R_1 = 10\text{k}\Omega\$ and \$R_f = 14.13\text{k}\Omega\$ for the switch in position 7. (The 1pF cap can be neglected.)

To use this method, plot \$|A| \text{ and } \frac{1}{B}\$ on a Bode plot: The LF351 is dominant-pole compensated, so the plot of \$|A|\$ is a line sloping down at -20dB\decade and crossing zero dB at 4MHz, the gain-bandwidth product. Since B = 0.4144, the plot of \$\frac{1}{B}\$ is a horizontal line at 7.65dB.

The frequency where these two lines cross is the closed-loop bandwidth. In this case, they cross at 1.66MHz. The \$\frac{1}{B}\$ line to the left of the bandwidth frequency and the \$|A|\$ line to the right are assymptotes of the closed loop frequency response curve.

This method is valid for inverting and non-inverting configurations, BUT for non-inverting, as in this case, the input signal "seen by" the summing point is attenuated by $$ \frac{R_f}{R_1+R_f} = 1-B $$ so the overall closed-loop gain is \$\frac{(1-B)}{B}\$ rather than \$\frac{1}{B}\$. This has the effect of shifting down the frequency response curve by \$20\text{ log}(1-B)\text{ dB}\$, which is -4.65dB in this case, so we have closed-loop gain = 3dB.