Question 5 of 7: Astable (Free-Running) Multivibrator
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Comp-A1, Electronics — National Exams, December 2018. Open-book, 3 hours; seven 20-mark questions, five to be marked (all seven answered below as a complete study resource). All diodes $V_D=0.7\text{V}$ unless stated otherwise.
Reference texts: Sedra & Smith, Microelectronic Circuits (diode limiters, MOSFET/BJT small-signal amplifiers, active loads and current mirrors, op-amp imperfections, CMOS logic, oscillators) — the single reference text covering every question on this paper.
Given. Two cross-coupled NPN transistors $Q_1,Q_2$; each collector pulled up through $R_1$ or $R_4=1\text{k}\Omega$ to $+10\text{V}$; each base pulled up through $R_2$ or $R_3=10\text{k}\Omega$; $C_1$ couples $Q_2$'s collector to $Q_1$'s base, $C_2$ couples $Q_1$'s collector to $Q_2$'s base.
Given data
Quantity
Value
$R_1=R_4$
$1\text{k}\Omega$
$R_2=R_3$
$10\text{k}\Omega$
$C_1=C_2$
$1\text{nF}$
Supply
$+10\text{V}$
Find. Circuit operation, output point, operating frequency, and new $R_2,R_3$ for a 66% duty cycle at the same frequency.
Fig. Q5 — Collector waveforms $V_{C1}$ (blue) and $V_{C2}$ (orange): complementary square waves, each transistor saturated (collector low) for one half-period while the other is cut off (collector high), $180^\circ$ out of phase.
Approach. This is a collector-coupled (Abraham-Bloch) astable multivibrator: positive feedback through the two capacitors makes exactly one transistor saturated at a time, and each capacitor's $RC$ discharge through the OTHER transistor's base resistor sets the timing of the next switch-over.
Part (a) — operation. Suppose $Q_1$ is ON (saturated, $V_{C1}\approx0$) and $Q_2$ is OFF (cut off, $V_{C2}\approx+10\text{V}$). $Q_1$'s collector transition drove $Q_2$'s base sharply negative through $C_2$ (a step $-10\text{V}$), holding $Q_2$ off. $C_2$ then recharges toward $+10\text{V}$ through $R_3$, exponentially raising $Q_2$'s base voltage; when it crosses $\approx0.7\text{V}$, $Q_2$ turns ON. $Q_2$'s collector then drops sharply, which through $C_1$ drives $Q_1$'s base negative and cuts $Q_1$ OFF — the roles swap, and the cycle regenerates indefinitely with no external trigger (hence astable).
Part (b) — output. The output is taken from either collector, $V_{C1}$ or $V_{C2}$ — each is a square wave swinging between $\approx0.2\text{V}$ (saturation) and $\approx+10\text{V}$, and the two are complementary (180° out of phase).
Part (c) — operating frequency. Each half-period is set by the RC recovery of the OFF transistor's base through its own base resistor and the coupling cap ($R_2C_1$ for one half, $R_3C_2$ for the other), following the standard exponential-recovery-to-threshold result $t=RC\ln2$:
$$t_{\text{half}}=R_2C_1\ln2=(10\text{k}\Omega)(1\text{nF})(0.6931)=\boxed{6.93\ \mu\text{s}}$$
$$T=2\,t_{\text{half}}=13.86\ \mu\text{s}\qquad f=\frac{1}{T}=\boxed{72.1\text{ kHz}}$$
Part (d) — redesign for 66% duty cycle at the same frequency. Keep $T=13.86\ \mu\text{s}$ fixed but split it unevenly between the two half-periods, holding $C_1=C_2=1\text{nF}$ fixed and solving only for new $R_2,R_3$:
$$t_1=0.66\,T=9.15\ \mu\text{s}\qquad t_2=0.34\,T=4.71\ \mu\text{s}$$
$$R_{2,\text{new}}=\frac{t_1}{C_1\ln2}=\frac{9.15\ \mu s}{(1\text{nF})(0.6931)}=\boxed{13.2\text{ k}\Omega}$$
$$R_{3,\text{new}}=\frac{t_2}{C_2\ln2}=\frac{4.71\ \mu s}{(1\text{nF})(0.6931)}=\boxed{6.8\text{ k}\Omega}$$
Check: $T=\ln2\,(R_{2,\text{new}}C_1+R_{3,\text{new}}C_2)=\ln2\,(13.2\text{k}+6.8\text{k})(1\text{nF})=13.86\ \mu\text{s}$ — frequency unchanged, duty cycle now $9.15/13.86=66\%$ on the collector whose recovery uses $R_{2,\text{new}}$.
