Model-Based Analog Circuit Design

Being able to analyze and design analog circuits using "hand analysis" allows us to build intuition, and this intuition enables us to create designs that are optimal and innovative. However:

• Our simple models such as the square-law model or velocity-saturation model, cannot accurately describe the behavior of key parameters such as output resistance, $r_{o}$, or completely misses operating regions such as the moderate inversion region.
• Using more accurate and complex models, such as BSIM, is ideal for verification, but not really suited for "hand analysis" since
  • We have to work with hundreds of parameters per transistor, or
  • Make many assumptions to reduce these parameters, but then only ending up in the same situation as using the simple models.

One solution around this problem is to use the simulator, in conjunction with the BSIM models, as a "calculator".

Small-Signal Model

In circuit design, we are normally interested in the following parameters:

• Gain
• Bandwidth
• Power
• Voltage Swing
• Noise

It turns out we can most of these small-signal parameters by using our BSIM models as lookup tables, since our small signal equations remain the same:

$i_{ds}={\frac {\partial I_{D}}{\partial V_{GS}}}\cdot v_{gs}+{\frac {\partial I_{D}}{\partial V_{BS}}}\cdot v_{bs}+{\frac {\partial I_{D}}{\partial V_{DS}}}\cdot v_{ds}=g_{m}v_{gs}+g_{mb}v_{bs}+g_{ds}v_{ds}$

(1)

Where $g_{m}$ is the device transconductance, $g_{mb}$ is the device body (effect) transconductance, or backgate transconductance, and $g_{ds}={\tfrac {1}{r_{o}}}$ is the output conductance. We can then determine the required small-signal parameters from our design specifications, and use our "calculator" (simulator + BSIM models) to determine how we can get these small-signal parameters.

Intrinsic Transistor Gain

As we have seen, accurately describing the transistor output resistance with a simple model is rather difficult. In most cases, however, we use $r_{o}$ to predict the intrinsic transistor gain, $a_{o}=g_{m}r_{o}$. If we are more interested in $a_{o}$, we can use the simulator to determine the intrinsic transistor gain as a function of $V_{DS}$, which in this case, is a proxy for output swing.

Efficiency Metric: ${\tfrac {g_{m}}{I_{D}}}$

Aside from the intrinsic transistor gain, the transconductance, $g_{m}$, also affects the transistor frequency response and noise performance. We want to determine the trade-offs involved in generating a particular transconductance value.

Consider the square-law MOSFET model, where $I_{D}={\frac {\mu _{n}C_{ox}}{2}}\cdot {\frac {W}{L}}\cdot \left(V_{GS}-V_{th}\right)^{2}$. Thus, we can get the transconductance as:

$g_{m}={\frac {\partial I_{D}}{\partial V_{GS}}}=\mu _{n}C_{ox}\cdot {\frac {W}{L}}\cdot \left(V_{GS}-V_{th}\right)={\frac {2\cdot I_{D}}{V_{GS}-V_{th}}}={\frac {2\cdot I_{D}}{V_{od}}}$

(2)

Where $V_{od}=V_{GS}-V_{th}$ is the overdrive voltage. We can then define transconductance efficiency as

${\frac {g_{m}}{I_{D}}}={\frac {2}{V_{GS}-V_{th}}}={\frac {2}{V_{od}}}$

(3)

We can think of ${\tfrac {g_{m}}{I_{D}}}$ as the amount of $g_{m}$ we can get from a certain DC drain current, and hence, for a certain DC power draw. Note that this is a DC metric, and does not take into account any dynamic or frequency dependent characteristics.

However, real devices are not square-law devices. We can run a simulation to see the behavior of the ${\tfrac {g_{m}}{I_{D}}}$ of a 45nm transistor. The results are then shown in Fig. 1.

Weak Inversion Transconductance

$I_{D}\approx {\frac {W}{L}}\cdot e^{\frac {q\left(V_{GS}-V_{th}\right)}{nkT}}$

(4)

$g_{m}={\frac {\partial I_{D}}{\partial V_{GS}}}={\frac {{\frac {W}{L}}\cdot e^{\frac {q\left(V_{GS}-V_{th}\right)}{nkT}}}{n\cdot {\frac {kT}{q}}}}={\frac {I_{D}}{n\cdot V_{T}}}$

(5)

${\frac {g_{m}}{I_{D}}}={\frac {1}{n\cdot V_{T}}}$

(6)

Efficiency as a Design Parameter

Since we can always determine the values of $g_{m}$ and $I_{D}$ via simulations, we can choose a particular the transconductance efficiency, and from there, determine $I_{D}$ given $g_{m}$.

Note that ${\tfrac {g_{m}}{I_{D}}}$ has units of $\mathrm {V^{-1}}$ (pronounced per volt), which is somewhat hard to interpret physically.

Figure of Merit: $V^{*}$

To avoid this awkward unit, we can redefine our transconductance efficiency in terms of a new metric, $V^{*}$, where

$V^{*}={\frac {2I_{D}}{g_{m}}}$

(7)

Note that $V^{*}$ has units of volts, and as an example, for ${\tfrac {g_{m}}{I_{D}}}=10\,\mathrm {V^{-1}}$, we get $V^{*}=200\,\mathrm {mV}$.

For square-law devices, we see that $V^{*}=V_{GS}-V_{th}=V_{od}$, and that

$g_{m}={\frac {2I_{D}}{V_{GS}-V_{th}}}={\frac {2I_{D}}{V^{*}}}$

(8)

However, real devices do not obey the square law. Let us tabulate the characteristics of $V_{od}$ with $V^{*}$ in the table below: