Chapter 15, Problem D15.38P

a.

To determine

The expression for the frequency of the oscillation.

Answer to Problem D15.38P

The oscillation frequency is $f_0=\frac{1}{2\pi }\sqrt{\frac{1}{C\left(L_1+L_2\right)}}$ .

Explanation of Solution

Given:

The circuit for the Hartley oscillator is given as:

  

  $\begin{array}{l}\text{Transconductance, }{g}_m=30\text{mA/V}\\ \text{Forward biased junction diffusion resistance, }{r}_{\pi }\to \infty \end{array}$

Forward biased junction diffusion resistance is $r_{\pi }\to \infty$

Drawing the small signal equivalent of the Harley oscillato:

  

The output resistance $r_0$ of the transistor is included in R .

Applying the nodal analysis at the output node ( $V_0$ ):

  $\frac{V_0}{s{L}_1}+\frac{V_0}{R}+g_m{V}_{\pi }+\frac{V_0}{\frac{1}{sC}+s{L}_2}=0\mathrm{..........}\left(1\right)$

And the result by the voltage divider:

  $V_{\pi }=\left(\frac{s{L}_2}{s{L}_2+\frac{1}{sC}}\right)V_0\mathrm{..............}\left(2\right)$

Solving equation (2) and equation (1):

  $\begin{array}{c}\frac{V_0}{s{L}_1}+\frac{V_0}{R}+g_m\left[\left(\frac{s{L}_2}{s{L}_2+\frac{1}{sC}}\right)V_0\right]+\frac{V_0}{\frac{1}{sC}+s{L}_2}=0\\ V_0\left[\frac{1}{s{L}_1}+\frac{1}{R}+g{m}_{}\left(\frac{s{L}_2}{s{L}_2+\frac{1}{sC}}\right)+\frac{1}{\frac{1}{sC}+s{L}_2}\right]=0\\ V_0\left[\frac{1}{s{L}_1}+\frac{1}{R}+\frac{1}{s{L}_2+\frac{1}{sC}}\left(g_{m}s{L}_2+1\right)\right]=0\\ \frac{1}{s{L}_1}+\frac{1}{R}+\frac{1}{s{L}_2+\frac{1}{sC}}\left(g_{m}s{L}_2+1\right)=0\\ \frac{1}{s{L}_1}+\frac{1}{R}+\frac{sC}{s^2{L}_{2}C+1}\left(g_{m}s{L}_2+1\right)=\mathrm{0..........}\left(3\right)\end{array}$

Substituting s= $j\omega$ in the equation (3):

  $\begin{array}{l}\frac{1}{j\omega L_1}+\frac{1}{R}+\frac{j\omega C}{{\left(j\omega \right)}^2{L}_{2}C+1}\left(g_m\left(j\omega \right)L_2+1\right)=0\\ -j\frac{1}{\omega L_1}+\frac{1}{R}+j\frac{\omega C}{-{\omega }^2{L}_{2}C+1}\left(g_m\left(j\omega \right)L_2+1\right)=0\\ -j\frac{1}{\omega L_1}+\frac{1}{R}-\frac{g_m{\omega }^2{L}_{2}C}{1-{\omega }^2{L}_{2}C}+j\frac{\omega C}{1-{\omega }^2{L}_{2}C}=0\\ \left(\frac{1}{R}-\frac{g_m{\omega }^2{L}_{2}C}{1-{\omega }^2{L}_{2}C}\right)+j\left(\frac{\omega C}{1-{\omega }^2{L}_{2}C}-\frac{1}{\omega L_1}\right)=\mathrm{0.........}\left(4\right)\end{array}$

The condition for the oscillation indicates that the real and the imaginary components of the equation 4.

should be zero.

From the imaginary component, the oscillation frequency:

  $\begin{array}{l}\frac{{\omega }_{0}C}{1-{\omega }^2{L}_{2}C}-\frac{1}{{\omega }_0{L}_1}=0\\ \frac{{\omega }_{0}C}{1-{\omega }^2{L}_{2}C}=\frac{1}{{\omega }_0{L}_1}\\ {\omega }^2{}_0{L}_{1}C=1-{\omega }^2{}_0{L}_{2}C\\ {\omega }_0{}^2\left(L_{1}C+L_{2}C\right)=1\\ {\omega }_0{}^2=\frac{1}{L_{1}C+L_{2}C}\\ {\omega }_0{}^2=\frac{1}{C\left(L_1+L_2\right)}\\ {\omega }_0=\sqrt{\frac{1}{C\left(L_1+L_2\right)}}\\ 2\pi f_0=\sqrt{\frac{1}{C\left(L_1+L_2\right)}}\\ f_0=\frac{1}{2\pi }\sqrt{\frac{1}{C\left(L_1+L_2\right)}}\end{array}$

Hence, the oscillation frequency is $f_0=\frac{1}{2\pi }\sqrt{\frac{1}{C\left(L_1+L_2\right)}}$ .

b.

To determine

To show: The condition for the oscillation is $g_{m}R=L_1/L_2$ .

Explanation of Solution

Given:

The circuit for the Hartley oscillator is given as:

  

  $\begin{array}{l}\text{Transconductance, }{g}_m=30\text{mA/V}\\ \text{Forward biased junction diffusion resistance, }{r}_{\pi }\to \infty \\ \left(\frac{1}{R}-\frac{g_m{\omega }^2{L}_{2}C}{1-{\omega }^2{L}_{2}C}\right)+j\left(\frac{\omega C}{1-{\omega }^2{L}_{2}C}-\frac{1}{\omega L_1}\right)=\mathrm{0.........}\left(4\right)\end{array}$

The real component of the equation 4 should be zero:

  $\begin{array}{l}\frac{1}{R}-\frac{g_m{\omega }^2{L}_{2}C}{1-{\omega }^2{}_0{L}_{2}C}=0\\ \frac{g_m{\omega }^2{L}_{2}C}{1-{\omega }^2{}_0{L}_{2}C}=\frac{1}{R}\\ g_m{\omega }^2{}_0{L}_{2}C=1-{\omega }^2{}_0{L}_{2}C\\ g_{m}R=\frac{1-{\omega }^2{L}_{2}C}{{\omega }^2{}_0{L}_{2}C}\\ g_{m}R=\frac{1}{{\omega }^2{}_0{L}_{2}C}-1\\ g_{m}R=\frac{1}{\left(\frac{1}{C\left(L_1+L_2\right)}\right)L_{2}C}-1\left(\mathrm{Sin}ce{\omega }_0=\sqrt{\frac{1}{C\left(L_1+L_2\right)}}\right)\\ g_{m}R=\frac{C\left(L_1+L_2\right)}{L_{2}C}-1\\ g_{m}R=\frac{\left(L_{1}C+L_{2}C\right)-L_{2}C}{L_{2}C}\\ g_{m}R=\frac{L_{1}C}{L_{2}C}\\ g_{m}R=\frac{L_1}{L_2}\end{array}$

Therefore, $g_{m}R=\frac{L_1}{L_2}$

c.

To determine

To design: The circuit for oscillation at the given frequency.

Explanation of Solution

Given:

The circuit for the Hartley oscillator is given as:

  

  $\begin{array}{l}\text{Transconductance, }{g}_m=30\text{mA/V}\\ \text{Forward biased junction diffusion resistance, }{r}_{\pi }\to \infty \\ \left(\frac{1}{R}-\frac{g_m{\omega }^2{L}_{2}C}{1-{\omega }^2{L}_{2}C}\right)+j\left(\frac{\omega C}{1-{\omega }^2{L}_{2}C}-\frac{1}{\omega L_1}\right)=\mathrm{0.........}\left(4\right)\end{array}$

Frequency of oscillation is $f_0=750kHz$

Inductors is $L_1=L_2=50\mu H$

The condition for oscillation:

  $\begin{array}{l}{g}_{m}R=\frac{L_1}{L_2}\\ R=\frac{L_1}{g_m{L}_2}\end{array}$

Substituting $L_1,L_2$ and $g_m$ in the above equation:

  $\begin{array}{l}R=\frac{50\times {10}^{-6}}{\left(50\times {10}^{-6}\right)\left(30\times {10}^{-3}\right)}\\ R=33.33\Omega \end{array}$

Hence, $R=33.33\Omega$

The frequency of oscillation:

  $\begin{array}{l}{\omega }_0=\sqrt{\frac{1}{C\left(L_1+L_2\right)}}\\ {\omega }^2{}_0=\frac{1}{C\left(L_1+L_2\right)}\\ {(}^2=\frac{1}{C\left(L_1+L_2\right)}\\ C=\frac{1}{{(2\pi f_0)}^2\left(L_1+L_2\right)}\\ C=\frac{1}{{\left(2\pi \left(750\times {10}^3\right)\right)}^2\left(50\times {10}^{-6}\right)+\left(50\times {10}^{-6}\right)}\\ C=\frac{1}{\left(2.221\times {10}^{13}\right)\left(100\times {10}^{-6}\right)}\\ C=450.25\times {10}^{-12}\end{array}$

Hence, $C=450.3pF$ .