bandwidth
The bandwidth of electronic parts, like analog switches, is usually specified for a specific load, typically 50 ohm and 5 pF for example.
Now if I have an analog switch with a bandwidth of 1 GHz for a 50 ohm load, how the bandwidth change depending on the load? What appends for, say, a load of 1 kOhm? Is there a way to calculate or estimate the bandwidth for different load values?
Thanks.
(this is a follow-up of High voltage high bandwidth analog mux)
The noise bandwidth \$B_N\$ of a (linear time-invariant) system is defined as the bandwidth which an ideal filter with a rectangular frequency response would need to have to get the same noise power at the output, given that the input noise to both systems is identical and white. The ideal filter is usually assumed to have the same maximum gain as the system under consideration. From this definition, it follows that the noise bandwidth is given by
$$B_N=\frac{1}{H_{max}^2}\int_{0}^{\infty}|H(f)|^2\;df\tag{1}$$
where \$H(f)\$ is the system's frequency response and
$$H_{max}^2=\max_{f}|H(f)|^2$$
Definition (1) is valid irrespective of the specific characteristics of \$H(f)\$. So it is valid for lowpass systems as well as for bandpass systems, or other types of filters.
For a simple first order lowpass system (e.g. an \$RC\$ lowpass), we have
$$H(f)=\frac{1}{1+jf/f_c}$$ with \$f_c\$ the -3dB cut-off frequency. From (1) we get for the noise bandwidth
$$B_N=\int_{0}^{\infty}\frac{1}{1+(f/f_c)^2}df= f_c\int_{0}^{\infty}\frac{1}{1+x^2}dx= f_c\arctan(x)|_{0}^{\infty}=f_c\frac{\pi}{2}$$
In this case we see that the noise bandwidth is larger than the -3dB frequency by a factor of \$\pi/2\$.
For a second order system everything is similar, but slightly more complex. The integral gets a bit more involved and the value of \$H^2_{max}\$ needs to be determined, because there can be overshoot in the frequency response, depending on the damping. If I have more time later on I might add details about second order systems. For the time being I hope that the answer is clear enough so that everybody can derive the noise bandwidth of any system they are interested in.
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?
Best Answer
In the 1Ghz range the most important is load capacitance. The higher capacitance - the less bandwidth. With 1nF load you are likely go under 10 MHz bandwidth. So avoid long traces.
Resistive load - if it's too high (like 5Ohm) you can just get signal with lower amplitude or more distortion, but bandwidth is not that much affected. And 1kOhm will not hurt usually (unless there is some fancy load matching, which is unlikely).