MOS Amplifier Structures

This article details the structures of basic mos amplifiers.

Common Source

CS with Resistor Load

Common Source Amplifier

In saturation mode, the voltage gain is

In triode mode, the voltage gain is

which reaches maximal when $V_{out}=V_{in}-V_{TH}$.

CS with Diode Load (Cascode)

A diode can be realized by connecting the gate and drain of a transistor

MOS as Diode

In that case, the equivalent output resistance is

From a large-signal perspective, a diode act like a "square root" operator.

CS with Diode Load

The voltage gain of a CS stage with diode load is

where $\eta=g_{mb2}/g_{m2}$

The working state of this configuration is

  1. $V_{in}<V_{TH1}, V_{out}\approx V_{dd}-V_{TH2}$, M1 off, M2 near cut-off
  2. $V_{TH1}<V_{in}<V_{out}+V_{TH1}$, M1 saturation, M2 saturation, linear
  3. $V_{out}+V_{TH1}<V_{in}$, M1 diode, M2 saturation

Strong input device and weak load device can lead to high gain, yet the voltage swing is severely limited.

CS with Current-Source Load

Assume that M2 is operating in saturation region.

CS with Current Source Load

The voltage gain is

CS with Active Load

CS with Active Load

The voltage gain is

This topology is sensitive to PVT variation due to $V_{GS1}+V_{GS2}=V_{DD}$. Still, it would amplify supply noise, the gain is

CS with Triode Load

Assume that M2 is operating in deep triode region and behaves as a resistor ($V_b$ is very close to $V_{DD}$).

CS with Triode Load

The voltage gain is

This circuit is difficult to provide bias with consideration to PVT variation. However, as compared to diode counterpart, it has no output voltage headroom (i.e. $V_{out,max}=V_{DD}$)

CS with Source Degeneration

Like diode load, source degeneration can be employed to enhance the linearity.

CS with Source Degeneration

This can be explained as "the resistance seen at the drain" divided by "the total resistance in the source path".

CS with Source Degeneration, Small Signal

  • When $V_{in}$ is low, $1/g_m\gg R_s, G_m\approx g_m$
  • For large $V_{in}$, $1/g_m\ll R_s, G_m\approx 1/R_s$

CS with Source Degeneration, Large Signal

As depicted from the figure, the large signal gain is relatively smooth with source degeneration.

The output impedance is

CS with Source Degeneration, Small Signal, Detailed

The gain can be written as

Lemma for Calculating Gm

Source Follower

Source Follower

Source follower can act like a buffer. It offers high input resistance and low output resistance.

With a current source rather than a resistor as load, the linearity can be improved because $V_{GS}$ will be fixed for a fixed $I_d$.

Source Follower with Current Source

Common Gate

Common Gate

Common Gate, Large Signal

A more detailed calculation of Av reveals

Common Gate, Small Signal

For a more detailed calculation of Rin

Common Gate, Input Impedance

The MOS can transform impedance up from source to drain, or down from drain to source.

Common Gate, Impedance Transformation

Cascode

For both MOS to be in saturation

The large signal operating mode is as follows:

  • For $V_{in}<V_{TH1}$, M1 off, M2 off, $V_{out}=V_{DD}$
  • For $V_{in}>V_{TH1}$, M1 saturation, M2 saturation
  • For sufficiently large $V_{in}$, two occasions can occur
    • M1 enter triode region first
    • M2 enter triode region first

Neglecting channel length modulation effect, Cascode amplifier has gain of

Cascode can reduce mismatch because

Cascode, Sensitivity to VDD Noise

In the first case

Instead, in the second case

Folded Cascode is as follows:

Folded Cascode

  • Higher DC consumption for comparable performance

Miller Effect

Miller Effect Illustration

The current flowing from X to Y is

Therefore

Similarly

Attention:

  • Only actual $A_v$ at the frequency of interest may be used to accurately predict the input impedance, but low-frequency $A_v$ is useful for approximation (it may eliminate zeros and predict additional poles).
  • Since the reverse gain is not necessarily $1/A_v$ (as in the case of CG), it cannot be used to calculate the output impedance directly. Instead, a voltage source should be connected at the output node to test the reverse gain, and that result be used to calculate the impedance.
  • Trans-impedance Z must present in the process of calculating $A_v$.

Frequency Response

Association of Poles with Nodes

We can associate each pole with one node of the circuit. It is not always true, but affords an intuitive approach to the estimation of the transfer function.

General Formula

Common Source

Common Source, Frequency Response

Poles can be estimated as

and the gain is

A direct analysis shows

The denominator can be written as

Further assume that $1/\omega_{p2}\ll1/\omega_{p1}$, we have

A zero exists as

Signal cancellation model is as follows:

Signal Cancellation

Source Follower

Source Follower, Frequency Response

Strong interaction through $C_{GS}$ makes it difficult to associate a pole with each node in the circuit. Calculation shows

Therefore

Note if $C_L=0$, the previous formula reduces to

In this case, only one pole at the input presents. We say $C_{GS}$ is bootstrapped by the source follower.

Neglecting channel length modulation, the input impedance is

It presents negative real input impedance at high frequency, thus is possibly unstable.

Output impedance is

Source Follower, Frequency Response, Zout

Figure (c) is more common, as when operated as buffer, $Z_{out}<Z_{in}$. It can be modelled as

Source Follower, Output Impedance Model

Thus, source follower can be used to realize inductors.

Common Gate

Common Gate, Frequency Response

We have

It has no Miller multiplication of capacitances, thus have wider bandwidth.

Cascode

Cascode, Frequency Response

The pole associated with node A is

The pole associated with node X is

The pole associated with node Y is

Extra Element Theorem

Extra Element Theorem

Zero Value Time Constant Method

The dominant pole of a circuit equals to the sum of all zero value time constants, each of which is obtained by setting input voltage and other capacitance to zero and computing the product of the capacitance of interest and its corresponding equivalent resistance.

Proof

  1. The dominant pole is the coefficient of $s$ in the denominator.

    Denominator can be written as

    Therefore, the coefficient of $s$ is $\omega_{p1}^{-1}+\omega_{p2}^{-1}+...+\omega_{pn}^{-1}$, which reduce to $\omega_{p1}^{-1}$ with dominant pole assumption.

  2. The coefficient of $s$ in the denominator equals to the sum of all zero-value time constants.

    Since the coefficient of $s$ has a time dimension, we can write it as $B_s=R_1C_1+R_2C_2+...+R_nC_n$, where $C_1-C_n$ denote real capacitors, and $R_1-R_n$ are unknown. With $C_2-C_n$ set to 0, the order of the system reduces to 1, and therefore $R_1$ is the resistance seen by $C_1$.

Reference

  1. Behzad Razavi. Design of analog CMOS integrated circuits.
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