Bandwidth Estimation for Circuits

For a $n$-th order system

$$H(s)=\frac{a_0+a_{1}s+a_2{s}^2+\cdots +a_m{s}^m}{1+b_{1}s+b_2{s}^2+\cdots +b_n{s}^n}$$

A typical circuit will have transfer function with ${\omega }_l$ and ${\omega }_h$. And between the two frequencies, the gain is flat $a_{mid}$

Bandwidth Estimation Using $b_1$ Coefficient

• A paper (which paper?) shows that $1/b_1$ is always smaller than ${\omega }_h$.

$$\begin{array}{r}{\omega }_h\approx 1/b_1=\frac{1}{\sum _{i=1}^N{\tau }_i^0}\end{array}$$

Example 01

Example 01

Ignore $r_o$ of the transistor and let

$$\begin{array}{rl}{C}_{\pi }& =100fF{r}_{\pi }=2.5k\mathrm{\Omega }\\ C_{\mu }& =20fF{g}_m=40mS\\ C_L& =200fF\beta =100\end{array}$$

the time constants

$$\begin{array}{rl}{\tau }_{\pi }^0& =(R_1\Vert r_{\pi })C_{\pi }=(1k\mathrm{\Omega }\Vert 2.5k\mathrm{\Omega })\cdot 100fF=70ps\\ {\tau }_{\mu }^0& =(R_{left}+R_{right}+G_m{R}_{left}{R}_{right})C_{\mu }=1200ps\\ {\tau }_L^0& =R_2{C}_L=400ps\end{array}$$

thus

$$\begin{array}{rl}{b}_1& ={\tau }_{\pi }^0+{\tau }_{\mu }^0+{\tau }_L^0=1670ps\\ {\omega }_h& =\frac{1}{b_1}=2\pi \times 95MHz\end{array}$$

${\omega }_l$ is not shown in the model.

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Bandpass Model

The reactive elements can be divided into two (non-overlap) groups:

• one is responsible for ${\omega }_l$
• another one is responsible for ${\omega }_h$
• so we got two different circuits to calculate ${\omega }_l$ and ${\omega }_h$

For ${\omega }_h$ we use zero value time constant (ZVT), meaning all the other elements are at zero value.

$$\begin{array}{r}{\omega }_h\approx 1/b_1=\frac{1}{\sum _{i=1}^N{\tau }_i^0}\end{array}$$

For ${\omega }_l$ we use infinite value time constant (IVT), meaning all the other elements are at infinite value.

$$\begin{array}{r}{\omega }_l\approx \frac{b_{n-1}}{b_n}=\sum _{i=1}^M\frac{1}{{\tau }_i^{\mathrm{\infty }}}\end{array}$$

Example 02

Example 02

$${\omega }_h=\frac{1}{{\tau }_{\pi }^0}=\frac{1}{(R_1\Vert r_{\pi })C_{\pi }}$$ $${\omega }_l=\frac{1}{{\tau }_c^{\mathrm{\infty }}}=\frac{1}{(R_1+r_{\pi })C_c}$$

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Taking Zeros Into Account

From Lecture and Paper by Prof. Ali Hajimiri.

$${\omega }_h\approx \frac{1}{\sum _{i=1}^N{\tau }_i^0(1-|\frac{H^i}{H^0}|)}$$

Not very sure about the proof of this approximation (modified time constant)!

Example 03

Example 03

Try to make $C_1$ smaller!

$$\begin{array}{rl}{C}_1& =4.3pF{r}_{\pi }=2.5k\mathrm{\Omega }\\ C_L& =200fF{g}_m=40mS\\ C_{\pi }& =100fF{R}_1=1k\mathrm{\Omega }\\ C_{\mu }& =20fF{R}_2=2k\mathrm{\Omega }\end{array}$$

If using bandpass model, and let $C_1$ being coupling capacitor.

$$\begin{array}{rl}{b}_1& =R_2(C_{\mu }+C_L)=440ps\\ {\omega }_h& =\frac{1}{b_1}=2\pi \times 362MHz\\ {\tau }_1^{\mathrm{\infty }}& =(R_1\Vert r_{\pi })C_1=3ns\\ {\omega }_l& =\frac{1}{{\tau }_1^{\mathrm{\infty }}}=2\pi \times 53MHz\end{array}$$

If using modified time constant

$$\begin{array}{rl}{H}^0& =-\frac{r_{\pi }}{R_1+r_{\pi }}\cdot g_m{R}_2=-57\\ H^{\pi }& =0\\ H^{\mu }& =+\frac{r_m\Vert R_2}{R_1+r_m\Vert R_2}=24m\\ H^L& =0\\ H^1& =-g_m{R}_2=-80\end{array}$$

with

$$\begin{array}{rl}{\tau }_{\pi }^0& =(R_1\Vert r_{\pi })C_{\pi }=(1k\mathrm{\Omega }\Vert 2.5k\mathrm{\Omega })\cdot 100fF=70ps\\ {\tau }_{\mu }^0& =(R_{left}+R_{right}+G_m{R}_{left}{R}_{right})C_{\mu }=1200ps\\ {\tau }_L^0& =R_2{C}_L=400ps\\ {\tau }_1^0& =(R_1\Vert r_{\pi })C_1=3070ps\end{array}$$

modified time conetant

$$\begin{array}{rl}{\tau }_{\pi }^{\prime }& =70ps\\ {\tau }_{\mu }^{\prime }& =1200ps\\ {\tau }_L^{\prime }& =400ps\\ {\tau }_1^{\prime }& =-1200ps\end{array}$$

then

$$\begin{array}{r}{\omega }_h=2\pi \times 339MHz\end{array}$$

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Second Order System Model