circuit analysiscircuit-designintegrated-circuitnoisenoise-spectral-density
I am reading about noise of cascode amplifier in Razavi book, Design of Analog CMOS Integrated Circuits.
However, I couldn't understand the yellow highlighted sentence below. Could anyone explain it?
Why the noise from M3, M4, M5 and M6 are negligible at low frequencies?
Thank you.
Decoupling capacitors are used in digital circuits to suppress the high-frequency noise generated on the supply line (Vcc or Vdd) as a result of the rapid switching inside the chip. The signals that are switching are usually square waves, which generate a lot of harmonics.
These capacitors fulfill dual roles -- they provide a little energy reserve to suppress voltage spikes inside an IC, and they help suppress any voltage spikes on the supply line outisde the chip from getting in.
Analog circuitry, on the other hand, usually doesn't switch as fast as digital IC's do, and even when it does, the signals usually aren't square waves so they don't have as many harmonics. Still, it is recommended to put decoupling IC's on all analog chips like op-amps. They are especially needed for analog devices that have a poor power supply rejection ratio (PSSR) to avoid power supply fluctuations from affecting the output.
In either case, decoupling caps are almost never found inside the IC's; as others have pointed out, they are usually separate components placed next to a chip between the power and ground pins. Values of 100 nf (0.1 µF) and 10 nf (0.01 µF) are often used.
Decoupling caps are required to not only make your circuit behave more reliably, but also to reduce emissions to enable it to pass FCC certification.
The bias current is set by the current source Ibias, it is mirrored to the active load Q2 and the current through Q1 has to be about equal to the current delivered by Q2.
This is not done by some sort of bias network but through application of feedback. Imagine this stage as the second stage of an OpAmp as shown below.
The CS Stage consists of M3 and M8, biased by M1.
In order to use this Opamp feedback is required. For example OUT could be fed back to IN-. The first stage of the OpAmp will take care of the biasing. The voltage at the gate of M8 will be such, that OUT is almost equal to IN+.
Best Answer
They do not contribute the the noise because of the degeneration. Having emitter/source degeneration in effect introduces negative feedback, greatly improving the performance of the transistors. For a more quantitative discussion see the end of this presentation.
You end up with the following equation for the noise: $$\frac{i_D}{1 + g_m r_o} \approx 0$$ Where \$r_o\$ is the degeneration. For high frequencies \$C_{gs}\$ provides an alternative current path, reducing the degeneration.