Exercise 9.5 A 3-stage oscillator with NPNs

a)

An answer:
$A=-\frac{g_{m}R}{1+\mathit{j\omega RC}}$.

b)

An answer:

$$\begin{array}{rcll}& A_{\mathit{OL}}& =A^3(3\mathit{identical}\mathit{stages})& \\ & & ={\left(\frac{-g_m\cdot R}{1+\mathit{j\omega RC}}\right)}^3& \\ & & g_m=\frac{q}{\mathit{kT}}\cdot I_C\approx 40I_C& \\ & & I_C=\frac{V_{\mathit{CC}}-0.6}{R}& \\ & A_{\mathit{OL}}& =-{\left(\frac{40V_{\mathit{CC}}-24}{1+\mathit{j\omega RC}}\right)}^3& \end{array}$$

c)

An answer:

$$\begin{array}{rcll}& \mathit{Arg}(A_{\mathit{OL}}(f))& =\pi +3\mathit{arg}\left(\frac{1}{1+\mathit{j\omega RC}}\right)& \\ & & =\pi -3\mathit{arg}(1+\mathit{j\omega RC})\equiv 0& \\ & & \to & \\ & {\omega }_{\mathit{osc}}& =\frac{\sqrt{3}}{\mathit{RC}}& \\ & f_{\mathit{osc}}& =\frac{{\omega }_{\mathit{osc}}}{2\pi }=\frac{\sqrt{3}}{2\mathit{\pi RC}}& \end{array}$$

d)

An answer:

$$\begin{array}{rcll}& A_{\mathit{OL}}({\omega }_{\mathit{osc}})& =-{\left(\frac{40V_{\mathit{CC}}-24}{1+j\sqrt{3}}\right)}^3\equiv 1& \\ & & =-\frac{{(40V_{\mathit{CC}}-24)}^3}{-8}=1& \\ & \to & & \\ & V_{\mathit{CC}}=& \frac{26}{40}=0.65V& \end{array}$$

e)

An answer:
If $V_{\mathit{CC}}$ is lower, $A_{\mathit{OL}}<1$: the circuit is stable, there is no oscillation

If $V_{\mathit{CC}}$ is higher, $A_{\mathit{OL}}>1$: the circuit becomes unstable

f)

An answer:
4 stages: amplifiers give ${(-g_m\cdot R)}^4>0$ of gain at DC. For oscillation, the RC-networks have provide a total phase shift of $0^{\circ }$ (or $k\cdot 360^{\circ }$) phase shift. Each RC-network individually then has to provide $0^{\circ },\pm 90^{\circ },\pm 180^{\circ },...$.

The only options are then $0^{\circ }$ (at DC) which is not regarded as oscillation AND $-90^{\circ }$ (at $\omega \to \infty$) which is also not regarded as harmonic oscillation.

5 stages: this situation is very similar to 3 stages ${(-g_m\cdot R)}^5<0\to \pm 180^{\circ }$ shift. Hence, for oscillation the 5 RC-networks have to give a total of $\pm 180^{\circ }\pm k\cdot 360^{\circ }$ phase shift.

The total phase shift of the 5 RC-networks is in the range $[0^{\circ },-450^{\circ }>$. The only possibility is then to oscillate at the frequency where the phase shift per RC-network is $\frac{-180^{\circ }}{5}=-36^{\circ }$ which is possible for a non-zero finite $\omega$: 5 stages can yield harmonic oscillation.

6 stages: As for 4 stages, amplifiers give $>0$ gain at DC. The only options are then $0^{\circ }$ (=at DC) which is not regarded as oscillation AND $-60^{\circ }$. This latter frequency is finite and non-zero BUT the loop-gain of the DC-solution is larger and therefore wins. This 6-stage version cannot oscillate harmonically.