10 High-Q harmonic oscillators

Q 10.1

(a)

An answer:

(b)

An answer:
The loop gain equals 1; this is used in the derivations below.

(c)

An answer:
The loop gain equals 1 at the oscillation frequency. This oscillation frequency follows from equating the loop gain to a real value. To derive the loop gain: the best way is to cut the loop inside an abstract controlled block. In this schematic the only abstract controlled block is the SSEC of the transistor.

$$\begin{array}{rcll}& A_{\mathit{loop}}& =\frac{V_B-V_E}{v_{\mathit{be}}}& \\ & & =-\frac{v_E}{v_{\mathit{be}}}& \\ & & v_E=i_E\cdot Z_A(\mathrm{\text{defining }}{Z}_A=Z_{c1}∕∕R)& \\ & & i_E=g_m\cdot v_{\mathit{be}}+\frac{v_{\mathit{out}}-v_E}{Z_{C2}}& \\ & & v_{\mathit{out}}=-i_E{Z}_L& \\ & & i_E=g_m\cdot v_{\mathit{be}}-\frac{i_E{Z}_L}{Z_{C2}}-\frac{v_E}{Z_{C2}}& \\ & & =\frac{g_m\cdot v_{\mathit{be}}-\frac{v_E}{Z_{C2}}}{1+\frac{Z_L}{Z_{C2}}}& \\ & & =\frac{g_m\cdot v_{\mathit{be}}{Z}_{C2}-v_E}{Z_{C2}+Z_L}& \\ & & v_E=\left(\frac{g_m\cdot v_{\mathit{be}}{Z}_{C2}-v_E}{Z_{C2}+Z_L}\right)\cdot Z_A& \\ & & =\frac{g_m\cdot v_{\mathit{be}}{Z}_{C2}}{Z_{C2}+Z_L}\cdot Z_A-\frac{v_E}{Z_{C2}+Z_L}\cdot Z_A& \\ & & =\frac{g_m\cdot v_{\mathit{be}}{Z}_{C2}}{Z_{C2}+Z_L}\cdot \frac{Z_A}{1+\frac{Z_A}{Z_{C2}+Z_L}}& \\ & & =v_{\mathit{be}}\cdot \frac{g_m\cdot Z_A{Z}_{C2}}{Z_{C2}+Z_L+Z_A}& \\ & A_{\mathit{loop}}& =\frac{-g_m\cdot Z_A{Z}_{C2}}{Z_{C2}+Z_L+Z_A}& \end{array}$$

At oscillation, this $A_{\mathit{loop}}\equiv 1$ which can be expanded as e.g.:

$$\begin{array}{rcll}& -g_m\cdot \frac{Z_{c1}R}{Z_{c1}+R}{Z}_{C2}& =Z_{C2}+Z_L+\frac{Z_{c1}R}{Z_{c1}+R}& \\ & -g_m\cdot \frac{R}{1+\mathit{j\omega R}{C}_1}\cdot \frac{1}{\mathit{j\omega }{C}_2}& =\frac{1}{\mathit{j\omega }{C}_2}+\mathit{j\omega L}+\frac{R}{1+\mathit{j\omega R}{C}_1}& \\ & -g_m\cdot R& =(1+\mathit{j\omega R}{C}_1)+\mathit{j\omega L}\cdot (1+\mathit{j\omega R}{C}_1)\cdot (\mathit{j\omega }{C}_2)+R\cdot (\mathit{j\omega }{C}_2)& \end{array}$$

Collecting the imaginary terms and equating these to zero yield

$$\begin{array}{rcll}& {\omega }_{\mathit{osc}}^2& =\frac{C_1+C_2}{C_1{C}_{2}L}\to & \\ & {\omega }_{\mathit{osc}}& =\sqrt{\frac{C_1+C_2}{C_1{C}_{2}L}}& \end{array}$$

(d)

An answer:
Recycling the previously found equations for loop gain, and equating that to 1 at $\omega ={\omega }_{\mathit{osc}}$ (note that this implies that you only have to take the real terms into account) this yields:

$$\begin{array}{rcll}& \mathit{gmR}& ={\omega }_{\mathit{osc}}^2{C}_{1}L-1\to & \\ & g_m& =\frac{C_1}{R{C}_2}& \end{array}$$

Q 10.2

(a)

An answer:
$C_0$ makes sure that there is zero DC-current in $L_2$, and that hence $L_2$ does not influence the bias point

(b)

An answer:
$g_m=40\cdot I_C=0.2S$

(c)

An answer:

(d)

An answer:

$$\begin{array}{llll}\hfill A_{\mathit{loop}}& =-\frac{g_m\cdot Z_A\cdot Z_C}{Z_A+Z_B+Z_C}\equiv 1& \hfill & \end{array}$$ with $$\begin{array}{llll}\hfill & Z_A=\frac{\mathit{j\omega }{L}_{2}R}{\mathit{j\omega }{L}_2+R}& \hfill & \\ \hfill & Z_B=\frac{1}{\mathit{j\omega C}}& \hfill & \\ \hfill & Z_C=\mathit{j\omega }{L}_1& \hfill & \end{array}$$

where $R=\frac{\mathit{\alpha fe}}{g_m}$. This leads to

$$\begin{array}{rcll}& & -g_m\cdot \frac{\mathit{j\omega }{L}_{2}R}{\mathit{j\omega }{L}_2+R}\cdot \mathit{j\omega }{L}_1=\frac{\mathit{j\omega }{L}_{2}R}{\mathit{j\omega }{L}_2+R}+\frac{1}{\mathit{j\omega C}}+\mathit{j\omega }{L}_1& \end{array}$$

Collecting the imaginary terms and equating them to zero yields

$$\begin{array}{llll}\hfill & \mathit{j\omega }{L}_{2}R+\frac{R}{\mathit{j\omega C}}+\mathit{j\omega }{L}_{1}R=0\to & \hfill & \\ \hfill & -{\omega }^{2}C(L_1+L_2)+1=0\to & \hfill & \\ \hfill & {\omega }_{\mathit{osc}}^2=\frac{1}{C(L_1+L_2)}& \hfill & \\ \hfill & & \hfill \end{array}$$

yielding the oscillation (radian) frequency ${\omega }_{\mathit{osc}}=\frac{1}{\sqrt{C(L_1+L_2)}}$

(e)

An answer:

Equating the loop gain, at the oscillation frequency, and equating that to 1 gives

$$\begin{array}{llll}\hfill & -g_m\cdot j{\omega }_{\mathit{osc}}{L}_{2}R\cdot j{\omega }_{\mathit{osc}}{L}_1=\frac{L_2}{C}+j{\omega }_{\mathit{osc}}{L}_{1}j{\omega }_{\mathit{osc}}L2\to & \hfill & \\ \hfill & (g_m\cdot R+1)\cdot {\omega }_{\mathit{osc}}^2{L}_2{L}_1=\frac{L_2}{C}\to & \hfill & \\ \hfill & (\mathit{\alpha fe}+1)\cdot \frac{L_2{L}_1}{C(L_1+L_2)}=\frac{L_2}{C}\to & \hfill & \\ \hfill & (\mathit{\alpha fe}+1)\cdot \frac{L_1}{L_1+L_2}=1\to & \hfill & \\ \hfill & L_2=\mathit{\alpha fe}\cdot L_1& \hfill & \end{array}$$