Op-Amp Astable Multivibrator Design

Cheatsheet Content

Op-Amp Astable Multivibrator An astable multivibrator, also known as a free-running multivibrator, generates a continuous square wave output without any external trigger. An Op-Amp based design offers good stability and control. Circuit Diagram (Conceptual) A basic Op-Amp astable multivibrator uses a comparator with positive feedback (hysteresis) and an RC timing circuit at its inverting input. The output oscillates between the Op-Amp's supply rails. - + Vout R2 R1 GND R R_D C D1 D2 Vout This diagram shows a common modification using two diodes (D1, D2) and a resistor ($R_D$) to achieve a 50% duty cycle by making the charging and discharging paths for the capacitor symmetrical. $R_1, R_2$: Feedback resistors for positive feedback (hysteresis). $R, C$: Timing components for the RC circuit. $D_1, D_2$: Diodes to ensure symmetrical charging/discharging of C. $R_D$: Resistor in series with diodes to define charging/discharging current. Design Specifications Frequency ($f$): $5 \text{ kHz}$ Duty Cycle: $50\%$ Op-Amp: e.g., LM741, TL081 (choose a rail-to-rail op-amp for better square wave output if possible, or consider output swing limitations). Supply Voltage: $\pm V_{sat}$ (e.g., $\pm 12 \text{V}$ or $\pm 15 \text{V}$) Key Formulas For a 50% duty cycle Op-Amp astable multivibrator, the frequency is primarily determined by the RC time constant and the feedback ratio. The oscillation frequency $f$ is given by: $$ f = \frac{1}{2RC \ln\left(\frac{1 + \beta}{1 - \beta}\right)} $$ Where $\beta$ is the feedback factor for the non-inverting input: $$ \beta = \frac{R_1}{R_1 + R_2} $$ For a 50% duty cycle, we typically aim for a configuration where the capacitor charges and discharges through effectively the same resistance, and the threshold voltages are symmetrical. A simpler, commonly used approximation for $f$ when $R_1 = R_2$ (i.e., $\beta = 0.5$) and using the diode/resistor configuration is: $$ f \approx \frac{1}{2.2 RC} $$ However, for precise 50% duty cycle, the circuit needs careful consideration of the diode forward voltage drops and the symmetry of charging/discharging paths. A common approach for 50% duty cycle is to charge and discharge the capacitor through the same resistor R, using diodes to switch the current path. In this case, the period $T = 2RC \ln(1 + R_2/R_1)$. For $R_2 = R_1$, $T = 2RC \ln(2) \approx 1.386 RC$. A more robust circuit for 50% duty cycle uses a resistor $R_D$ and two diodes $D_1, D_2$ as shown conceptually. The frequency formula becomes: $$ f = \frac{1}{2 \times (R_D + R) C \ln\left(\frac{1 + \beta}{1 - \beta}\right)} $$ To achieve a 50% duty cycle, the charging and discharging paths must be symmetrical. This is often achieved by ensuring that the voltage across the capacitor charges and discharges between $-V_{Th}$ and $+V_{Th}$, where $V_{Th} = \beta V_{sat}$. A popular design for 50% duty cycle (especially with CMOS inverters, but adaptable to op-amps with similar logic) uses a configuration where the capacitor charges/discharges through the same resistor $R$ between $\pm V_{sat}$. The period $T$ is then given by: $$ T = 2RC \ln\left(\frac{1 + \frac{R_1}{R_2}}{1 - \frac{R_1}{R_2}}\right) $$ A common choice for $\beta = 0.5$ (i.e., $R_1=R_2$) is not ideal for this specific formula for 50% duty cycle. A better choice for the feedback resistors to achieve $V_{Th} = \pm V_{sat}/2$ is needed. Let's use the standard design with $R_1$ and $R_2$ for positive feedback, and $R$ and $C$ for timing, and assume an ideal Op-Amp's output swings to $\pm V_{sat}$. The switching thresholds at the non-inverting input are $V_{TH} = +V_{sat} \frac{R_1}{R_1+R_2}$ and $V_{TL} = -V_{sat} \frac{R_1}{R_1+R_2}$. The capacitor charges and discharges between these thresholds. For a 50% duty cycle, the charging and discharging times must be equal. This usually means the charging/discharging current paths are symmetrical. A common simplification for 50% duty cycle when $R_1 = R_2$ (meaning $V_{TH} = \pm V_{sat}/2$) and the capacitor charges/discharges through $R$ to $\pm V_{sat}$ is: $$ f = \frac{1}{2RC \ln(3)} \approx \frac{1}{2.2RC} $$ This approximation holds when $R_1 = R_2$ and the capacitor charges/discharges towards the full supply rails $\pm V_{sat}$. Design Steps for 50% Duty Cycle To achieve a 50% duty cycle, we need to ensure the capacitor charges and discharges symmetrically. A common way is to make the switching thresholds $\pm V_{sat}/2$. This means $R_1 = R_2$. And use an RC network where the capacitor effectively charges/discharges through the same resistance $R$ to $\pm V_{sat}$. Determine Frequency: $f = 5 \text{ kHz}$. Period $T = 1/f = 1/5000 = 0.2 \text{ ms} = 200 \mu s$. Choose Feedback Resistors ($R_1, R_2$): For symmetrical thresholds, choose $R_1 = R_2$. Let $R_1 = R_2 = 10 \text{ k}\Omega$. This sets the thresholds at $\pm V_{sat}/2$. Choose Capacitor ($C$): Select a standard capacitor value. For audio frequencies, $1 \text{ nF}$ to $100 \text{ nF}$ is typical. Let's start with $C = 10 \text{ nF}$ ($0.01 \mu F$). Calculate Resistor ($R$): Using the simplified formula $f = \frac{1}{2.2 RC}$ for 50% duty cycle: $$ R = \frac{1}{2.2 f C} $$ $$ R = \frac{1}{2.2 \times 5000 \text{ Hz} \times 10 \times 10^{-9} \text{ F}} $$ $$ R = \frac{1}{2.2 \times 5 \times 10^3 \times 10 \times 10^{-9}} = \frac{1}{110 \times 10^{-6}} = \frac{1}{1.1 \times 10^{-4}} $$ $$ R \approx 9090.9 \Omega $$ Choose a standard resistor value close to this, e.g., $R = 9.1 \text{ k}\Omega$. Verify with precise formula (if applicable): If the circuit uses the configuration where $\beta = R_1 / (R_1+R_2)$ and the capacitor charges/discharges through R towards $V_{out}$, the period is $T = 2RC \ln\left(\frac{1 + \beta}{1 - \beta}\right)$. If $R_1=R_2$, then $\beta = 0.5$. $$ T = 2RC \ln\left(\frac{1 + 0.5}{1 - 0.5}\right) = 2RC \ln(3) \approx 2RC \times 1.0986 = 2.197 RC $$ So, $f = \frac{1}{2.197 RC}$. Our approximation $1/(2.2 RC)$ is very close. Using $R = 9.1 \text{ k}\Omega$ and $C = 10 \text{ nF}$: $$ f = \frac{1}{2.197 \times 9.1 \times 10^3 \times 10 \times 10^{-9}} = \frac{1}{2.197 \times 91 \times 10^{-6}} = \frac{1}{0.000200} \approx 5000 \text{ Hz} $$ This confirms our values. Add Diodes ($D_1, D_2$) and $R_D$: To ensure symmetrical charge/discharge for 50% duty cycle, a common practice is to use two diodes and a resistor $R_D$ in series with $R$. The conceptual diagram shows this. If $R_D = R$, the effective charging/discharging resistance for the capacitor is $R$. In this case, the resistor $R$ from step 4 is the effective resistance for charging/discharging. The circuit would connect the capacitor to the inverting input, and the other end of the capacitor would be connected to $V_{out}$ through $R$. To achieve 50% duty cycle, we often use a slightly different topology or add current steering diodes. For simplicity, if we assume the equation $f = \frac{1}{2.2RC}$ holds for a standard Op-Amp astable multivibrator where the capacitor charges/discharges through R between $\pm V_{sat}$ and $R_1=R_2$, then the calculated values are directly applicable. Component Selection Summary Op-Amp: General purpose Op-Amp (e.g., TL081, LM741). Ensure it can operate at $5 \text{ kHz}$ (slew rate and bandwidth considerations). Supply Voltage: $\pm 12 \text{V}$ or $\pm 15 \text{V}$ (common for Op-Amps). Feedback Resistors: $R_1 = 10 \text{ k}\Omega$, $R_2 = 10 \text{ k}\Omega$. Timing Resistor: $R = 9.1 \text{ k}\Omega$. Timing Capacitor: $C = 10 \text{ nF}$. Diodes (for 50% duty cycle): Two small signal diodes (e.g., 1N4148). These are generally used in a slightly more complex circuit than the basic one for true 50% duty cycle. Practical Considerations Op-Amp Slew Rate: Ensure the Op-Amp's slew rate is sufficient for the desired $5 \text{ kHz}$ square wave. A typical LM741 has a slew rate of $0.5 \text{ V}/\mu\text{s}$, which might be slow for sharp transitions at $5 \text{ kHz}$ if the output swing is large. Faster Op-Amps like TL081 ($13 \text{ V}/\mu\text{s}$) are preferred. Output Voltage Swing: Real Op-Amps do not swing exactly to the supply rails. This will affect $V_{sat}$ and thus the thresholds. Rail-to-rail Op-Amps minimize this issue. Diode Voltage Drop: The forward voltage drop of diodes (if used for 50% duty cycle) can affect the symmetry. Consider using schottky diodes for lower voltage drop, or incorporate the voltage drop into calculations. Component Tolerances: Use precision resistors and capacitors if a highly accurate frequency and duty cycle are required. Power Consumption: Consider the quiescent current of the Op-Amp and the current through the resistors.