RFIC - Microstrip Transmission Line Design

2016-10-04

Microstrip transmission line is one of the basic type of transmission line in RF integrated circuit. Here is a basic/simple design example of it. This is a homework for course ECE810 RF Integrated Circuits in MSU.

To begin with

Microstrip is one kind of transmission line in the circuit. We use them to achieve impedance matching or some other application. Here are some simple design examples.

Problems

Design following transmission lines (use simplifying assumptions):

A 50- $Ω$ microstrip line for a substrate height of 128 $μ$ m and $ε_{r}$ of 3 (RT Duroid)
A 50- $Ω$ microstrip line for a substrate height of 10 $μ$ m and $ε_{r}$ of 4.1 (CMOS)
A 50- $Ω$ microstrip line for a substrate height of 5 $μ$ m and $ε_{r}$ of 4.1 (CMOS)
For the CMOS case, assuming the minimum allowable line width and spacing is 2 $μ$ m on the top metal, what is the highest characteristic impedance we can achieve?
What is the lowest characteristic impedance we can achieve? What would be the limiting factor?

Solution

Problem 1

For RT Duroid material, use the $w / h 1$ formula to recalculate. i.e.

Z_{0} = \frac{120 π}{\sqrt{ε_{e f f}} [\frac{w}{128} + 1.393 + 0.667 \ln (\frac{w}{128} + 1.44)]}

This time I substitute the $w = 400 μ m$ (another assumption) in the $ε_{e f f}$ formula. I choose this value to get rid of the recursive calculation to find the exact value of $w$ .

\begin{array}{rcl} ε_{e f f} & \approx & \frac{ε_{r} + 1}{2} + \frac{ε_{r} - 1}{2} [\frac{1}{\sqrt{1 + 12 h / w}}] \\ = & \frac{3 + 1}{2} + \frac{3 - 1}{2} [\frac{1}{\sqrt{1 + 12 \times 128 / 400}}] = 2.4545 \end{array}

Denote $C = \frac{120 π}{Z_{0} \sqrt{ε_{e f f}}} - 1.393$ , we obtain

\begin{array}{rcl} f (w) & = & \frac{w}{128} + 0.667 \ln (\frac{w}{128} + 1.44) - C = 0 \\ ⟹ f (x) & = & x + 0.667 \ln (x + 1.44) - C \\ f^{'} (x) & = & 1 + \frac{0.667}{x + 1.44} - C \end{array}

Use Newton-Raphson method to find the value $w$ .

x_{i + 1} = x_{i} - \frac{f (x_{i})}{f^{'} (x_{i})}

and finally $\frac{w}{128} = 2.9520$ , thus $w = 2.9520 \times 128 = 320.5948 μ m$ .

Problem 2

When it comes to CMOS cases, the substrate height $h$ is usually very small, but I also assume $w / h > 1$ . Take $w = 15 μ m$ to calculate the efficient $ε$ .

\begin{array}{rcl} ε_{e f f} & \approx & \frac{ε_{r} + 1}{2} + \frac{ε_{r} - 1}{2} [\frac{1}{\sqrt{1 + 12 h / w}}] \\ = & \frac{4.1 + 1}{2} + \frac{4.1 - 1}{2} [\frac{1}{\sqrt{1 + 12 \times 10 / 15}}] = 3.0667 \end{array}

Go through the similar process above, it is not hard to calculate that $C_{1} = 2.9125$ , and $\frac{w}{h} = 2.0746$ , which returns $w_{1} = 20.7459 μ m$ . This result is close to the value we used to calculate $ε_{e f f}$ , so there is no need to redesign again.

Problem 3

For the CMOS case with substrate height $h = 5 μ m$ , similarly first calculate the efficient $ε_{e f f}$ .

\begin{array}{rcl} ε_{e f f} & \approx & \frac{ε_{r} + 1}{2} + \frac{ε_{r} - 1}{2} [\frac{1}{\sqrt{1 + 12 h / w}}] \\ = & \frac{4.1 + 1}{2} + \frac{4.1 - 1}{2} [\frac{1}{\sqrt{1 + 12 \times 5 / 8}}] = 3.0816 \end{array}

And $C_{2} = 2.9021$ , $\frac{w}{h} = 2.0658 ⟹ w = 10.3289 μ m$ .

Problem 4

$Z_{0}$ increases with $h$ and decreases with $w$ and $ε_{r}$ . When the minimum allowable $w_{min} = 2 μ m$ ,

\begin{array}{rcl} ε_{e f f} & \approx & \frac{ε_{r} + 1}{2} + \frac{ε_{r} - 1}{2} [\frac{1}{\sqrt{1 + 12 h / w}}] \\ = & \frac{4.1 + 1}{2} + \frac{4.1 - 1}{2} [\frac{1}{\sqrt{1 + 12 \times \frac{1}{2}}}] = 2.980 \end{array}

When $h = 10 μ m$ ,

\begin{array}{rcl} Z_{0, max} & = & \frac{120 π}{\sqrt{ε_{e f f}} [\frac{w}{h} + 1.393 + 0.667 \ln (\frac{w}{h} + 1.44)]} \\ = & \frac{120 π}{\sqrt{2.98} [\frac{2}{10} + 1.393 + 0.667 \ln (\frac{2}{10} + 1.44)]} \\ = & 113.5691 Ω . \end{array}

When $h = 5 μ m$ ,

Z_{0, max} = \frac{120 π}{\sqrt{2.98} [\frac{2}{5} + 1.393 + 0.667 \ln (\frac{2}{5} + 1.44)]} = 96.7805 Ω

That is the highest characteristic impedance we can achieve with the CMOS.

Problem 5

As demonstrated above, $Z_{0}$ increases with $h$ and decreases with $w$ and $ε_{r}$ . We need to lower the substrate height $h$ and increase microstrip width $w$ and $ε_{r}$ . Let’s choose a substrate material, say GaAs, and $ε_{r} = 12.7$ .

\begin{array}{rcl} ε_{e f f} & \approx & \frac{ε_{r} + 1}{2} + \frac{ε_{r} - 1}{2} [\frac{1}{\sqrt{1 + 12 h / w}}] \\ = & \frac{12.7 + 1}{2} + \frac{12.7 - 1}{2} [\frac{1}{\sqrt{1 + 12 \times \frac{1}{2}}}] = 9.0611 \end{array}

More detailedly, it is the ratio $w / h$ that matters. For here, I choose $w / h = 2$ .

Z_{0, max} = \frac{120 π}{\sqrt{9.0611} [\frac{2}{1} + 1.393 + 0.667 \ln (\frac{2}{1} + 1.44)]} = 29.6983 Ω

Further, if I increase the ratio to $w / h = 5$ , we can calculate that $Z_{0} = 16.4027 Ω$ . This is lower than the previous case.

I don’t know whether there is a theoretical limitation for the $w / h$ ratio, but in practical application when $w$ is too high, there will be less space for other components. Also, a rule of thumb is that we need to limit the thickness of the microstrip substrate to 10% of a wavelength.