Products

Holooly Rewards

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

Holooly Ads. Manager

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

Holooly Tables

All the data tables that you may search for.

Holooly Arabia

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

Holooly Sources

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

Holooly Help Desk

Need Help? We got you covered.

Chapter 13

Q. 13.1

Design a 50 MHz Colpitts oscillator using a bipolar junction transistor in a com-mon emitter configuration with β = g_{m}/G_{i} = 30, and a transistor input resistance of R_{i} = 1/G_{i} = 1200  Ω. Use an inductor with L_{3} = 0.10  µH and an unloaded Q of 100. What is the minimum Q of the inductor for which oscillation will be sustained?

Step-by-Step

Verified Solution

From (13.20) the series combination of C^{′}_{1} and  C_{2} is found to be

\omega _0=\sqrt{\frac{1}{L_3}\left(\frac{1}{C_1}+\frac{1}{C_2} +\frac{G_iR}{C_1} \right) }=\sqrt{\frac{1}{L_3} \left(\frac{1}{C^\prime _1}+\frac{1}{C_2}\right) }               (13.20)

\frac{C^{′}_{1}C_{2}}{C^{′}_{1}+C_{2}} = \frac{1}{\omega ^{2}_{0}L_{3}} = \frac{1}{(2\pi )^{2}(50\times 10^{6})^{2}(0.1 \times 10^{-6})}=100  pF.

This value can be obtained in several ways, but here we will choose C^{′}_{1} = C_{2} = 200  pF

From Chapter 6 we know that the unloaded Q of an inductor is related to its series resistance by Q_{0} = ωL/R, so the series resistance of the 0.1 µH inductor is

R=\frac{\omega _{0}L_{3}}{Q_{0}} = \frac{(2\pi) (50 \times 10^{6} )(0.1\times10^{-6} )}{100} =0.31  \Omega

Then (13.21) gives C_1 as

C^{′}_{ 1} =\frac{C_{1}}{ (1 + RG_{i})}                     (13.21)

C_{1} = C^{′}_{ 1}(1 + RG_{i}) = (200  pF)(1+\frac{0.31}{1200} )= 200  pF,

which we see is essentially unchanged from the value found by neglecting the inductor loss. Using (13.22) with the above values gives

\frac{R}{G_{i}}=\frac{1+\beta }{\omega ^{2}_{0}C_{1}C_{2}}-\frac{L_{3}}{C_{1}}

 

(0.31)(1200)\lt \frac{1+30}{(2\pi )^{2}(50 \times 10^{6})^{2}(200 \times 10^{-12} )^{2}} -\frac{0.1 \times 10^{-6}}{200 \times 10^{-12}}

372 < 7852 – 500 = 7352

which indicates that the condition for oscillation will be satisfied. This condition can be used to find the minimum unloaded inductor Q by first solving for the maximum value of series resistance R:

R_{\max}=\frac{1}{R_{i}}(\frac{1+\beta }{\omega ^{2}_{0}C_{1}C_{2}}-\frac{L_{3}}{C_{1}} ) =\frac{7352}{1200}=6.13  \Omega

So the minimum unloaded Q is

Q_{\min}=\frac{\omega _{0}L_{3}}{R_{\max}} = \frac{(2\pi )(50 \times 10^{6})(0.1 \times 10^{-6})}{6.13} = 5.1.