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## 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?

## 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

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