We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program


Advertise your business, and reach millions of students around the world.


All the data tables that you may search for.


For Arabic Users, find a teacher/tutor in your City or country in the Middle East.


Find the Source, Textbook, Solution Manual that you are looking for in 1 click.


Need Help? We got you covered.

Chapter 5

Q. 5.5

A broadband amplifier stage with G B W_{S}=100 \mathrm{GHz} is realized in a SiGe BiCMOS process with SiGe HBT f_{M A X}of 300 \mathrm{GHz}. Find the optimal number of cascaded identical stages that leads to the highest bandwidth amplifier with 40 \mathrm{~dB} gain.


Verified Solution

The gain of 40 \mathrm{~dB} corresponds to A_{t o t}=100. From (5.55), we find the optimal number of stages to be n_{O P T}=9.2. Indeed, if we use n=9, we obtain

\frac{G B W_{t o t}}{G B W_{S}}=100^{1-1 / 9} \times \sqrt{2^{1 / 9}-1}=16.95
If n=10, we obtain:
\frac{G B W_{t o t}}{G B W_{S}}=100^{0.9} \times \sqrt{2^{0.1}-1}=16.90
and, if n=5
\frac{G B W_{\text {tot }}}{G B W_{S}}=100^{0.8} \times \sqrt{2^{0.2}-1}=15.34

These results indicate that the best overall G B W_{\text {tot }} is 1.695 \mathrm{THz} (much higher than the f_{M A X} of the transistor in this BiCMOS process), corresponding to an amplifier bandwidth of 17 \mathrm{GHz}. Each of the 9 stages has a gain \mathrm{A}_{0}=1.668(4.44 \mathrm{~dB}) and a 3 \mathrm{~dB} bandwidth of 60 \mathrm{GHz} . However, 9 stages will occupy a large die area and the solution may prove uneconomical. Could we design an amplifier with a smaller number of stages and acceptable bandwidth? It turns out that even if we use only 5 stages, G B W_{t o t} is 1.534 \mathrm{THz}, only 10 \% smaller than the maximum possible. The amplifier will have a bandwidth of 15.34 \mathrm{GHz}, and each of the 5 stages would have a gain of 2.51 (or 8 \mathrm{~dB} ) and a bandwidth of 39.8 \mathrm{GHz}.

Equations (5.54) and (5.55) are strictly valid for single-pole gain stages. Similar expressions can be derived for amplifier stages with second- or higher-order frequency response functions. For the frequently encountered case of a chain of second-order Butterworth stages with Q=\sqrt{2} and no zeros, we obtain [9]

\frac{G B W_{t o t}}{G B W_{S}}=A_{t o t}^{1-1 / n}\left(2^{1 / n}-1\right)^{1 / 4}                                    (5.56)

Finally, it should be noted that the analysis above also applies to the gain-bandwidth product of a chain of cascaded tuned amplifier stages.

*Equations (5.54)\frac{G B W_{t o t}}{G B W_{S}}=A_{t o t}^{1-1 / n}\left(2^{1 / n}-1\right)^{1 / 4}

*Equations (5.55) n_{o p t} \approx 2 \times \ln A_{\text {tot }}