Physics / Electronics Basic Electronics 100% Free Open Access
Chapter 8 โ€ข Theory & Derivations

Nonlinear Op-Amp Circuits, Schmitt Triggers & Active Filters

Comprehensive theory of nonlinear operational amplifier applications: open-loop voltage comparators, zero-crossing detectors, and slew-rate limits; positive-feedback Schmitt Triggers with analytical hysteresis loop derivations; active square/triangular waveform generators; detailed design of first- and second-order Sallen-Key Butterworth active filters.

ยง8.1 Open-Loop Comparators, Zero-Crossing Detectors & Slew Rate Limits

1. Voltage Comparator Operation and Transfer Characteristics

An open-loop operational amplifier without negative feedback functions as a Voltage Comparator. Because open-loop gain is colossal ($A_{OL} \sim 10^5$), an infinitesimally small differential input voltage $\Delta V = v^+ - v^-$ drives the output into hard saturation at the power supply rails:

$$v_o = \begin{cases} +V_{sat} \approx V_{CC} - 1.5\,\text{V} & \text{if } v^+ > v^- \\ -V_{sat} \approx -V_{EE} + 1.5\,\text{V} & \text{if } v^+ < v^- \end{cases}$$

A reference voltage $V_{ref}$ applied to one terminal establishes the threshold. When an analog signal crosses $V_{ref}$, the output transitions cleanly between binary saturation rails, acting as an analog-to-digital 1-bit converter.

2. Zero-Crossing Detector & Noise Chattering Problem

When $V_{ref} = 0\,\text{V}$, the circuit is a Zero-Crossing Detector, converting any arbitrary AC waveform (such as a sine wave) into a synchronous square wave.

The Chattering Problem: In practical industrial environments, real-world input signals contain high-frequency noise spikes. As a slowly moving signal passes through $0\,\text{V}$, noise causes multiple rapid false crossings above and below zero. The comparator output chatters uncontrollably, firing dozens of false pulses per cycle. To eliminate this chattering, positive feedback is introduced to create hysteresis.

3. Slew Rate and Full-Power Bandwidth

The switching speed of an op-amp is fundamentally constrained by its Slew Rate ($SR$): the maximum rate of change of output voltage per unit time that the internal compensation capacitor can sustain when driven by the input differential stage's tail current source ($I_{tail}$):

$$SR = \left. \frac{d v_o}{dt} \right|_{max} = \frac{I_{tail}}{C_c} \quad (\text{expressed in } \text{V}/\mu\text{s})$$

For a sinusoidal output $v_o(t) = V_m \sin(2\pi f t)$, the maximum slope occurs at the zero-crossing:

$$\left. \frac{d v_o}{dt} \right|_{max} = 2\pi f V_m$$

To avoid severe triangular distortion (slew-rate limiting), this slope must not exceed $SR$:

$$2\pi f V_m \le SR \implies f_{max} = \frac{SR}{2\pi V_m}$$

The frequency $f_{max}$ is the Full-Power Bandwidth of the op-amp. For a $\mu\text{A}741$ with $SR = 0.5\,\text{V}/\mu\text{s}$ and $V_m = 10\,\text{V}$, $f_{max} = \frac{0.5 \times 10^6}{2\pi \times 10} \approx 7.96\,\text{kHz}$. Above this frequency, full-amplitude sine waves distort into triangles.

ยง8.2 Schmitt Trigger (Regenerative Comparator) & Hysteresis Loops

1. Positive Feedback Mechanism and Regenerative Action

A Schmitt Trigger incorporates positive feedback by routing a fraction of the output voltage back to the non-inverting ($+$) terminal. This creates two distinct switching thresholds, imparting memory (hysteresis) to the system.

2. Inverting Schmitt Trigger Equations

In an inverting Schmitt trigger, the input signal $v_{in}$ is applied to the inverting terminal ($-$). The non-inverting terminal ($+$) is connected to a resistive voltage divider ($R_1, R_2$) from output $v_o$ to ground. The potential at the non-inverting terminal is:

$$V^+ = \frac{R_1}{R_1 + R_2} v_o$$

Since $v_o$ can only assume $+V_{sat}$ or $-V_{sat}$, there exist two distinct trip thresholds:

  1. Upper Trip Point ($V_{UTP}$): When $v_o = +V_{sat}$, the reference threshold is:
    $$V_{UTP} = + \frac{R_1}{R_1 + R_2} V_{sat}$$
    The output remains clamped at $+V_{sat}$ until $v_{in}$ rises above $V_{UTP}$, at which instant regenerative switching snaps the output to $-V_{sat}$.
  2. Lower Trip Point ($V_{LTP}$): Once $v_o = -V_{sat}$, the reference threshold switches to:
    $$V_{LTP} = - \frac{R_1}{R_1 + R_2} V_{sat}$$
    The output cannot return to $+V_{sat}$ until $v_{in}$ falls below $V_{LTP}$.

The width of the hysteresis loop is the Hysteresis Voltage $\Delta V_H$:

$$\Delta V_H = V_{UTP} - V_{LTP} = \frac{2 R_1}{R_1 + R_2} V_{sat}$$

Any noise riding on the input with peak-to-peak amplitude less than $\Delta V_H$ is completely ignored, providing total immunity to false triggering.

3. Astable Relaxation Multivibrator (Square-Wave Generator)

By connecting an external $RC$ timing network between the output and inverting terminal of a Schmitt trigger, a free-running relaxation oscillator is formed:

The output square wave charges capacitor $C$ toward $\pm V_{sat}$ through resistor $R$. As soon as $v_C(t)$ reaches the trip threshold, the Schmitt trigger snaps to the opposite rail, and $C$ begins charging in reverse. The period of oscillation is:

$$T = 2 R C \ln\left( \frac{1 + \beta}{1 - \beta} \right) \quad \text{where } \beta = \frac{R_1}{R_1 + R_2}$$

If resistors are chosen such that $R_1 = 0.86 R_2$ ($\beta \approx 0.462$):

$$\ln\left(\frac{1 + 0.462}{1 - 0.462}\right) = \ln(2.718) = 1.00 \implies T = 2 R C \implies f_0 = \frac{1}{2 R C}$$

ยง8.3 Active Filter Fundamentals: Butterworth, Chebyshev & Bessel Responses

1. Advantages of Active Filters over Passive Networks

Active filters combine operational amplifiers with resistors and capacitors (no inductors required). They provide decisive advantages over passive RLC filters:

  • Elimination of Bulky Inductors: Inductors are heavy, lossy, expensive, and act as antennas picking up 50/60 Hz magnetic hum. At audio frequencies ($< 20\,\text{kHz}$), required inductors are several henries in size.
  • Impedance Isolation: High input impedance and near-zero output impedance mean active filter stages can be cascaded without loading effects altering their cutoff points.
  • Gain Flexibility: Built-in passband voltage amplification and buffering.
  • Simple Tunability: Cutoff frequencies can be adjusted smoothly over decades using dual-ganged potentiometers.

2. Classic Filter Approximations

Response Type Passband Characteristic Transition Roll-off Phase Linearity Applications
Butterworth Maximally flat; zero ripple Moderate ($-20n\,\text{dB/decade}$) Moderate phase distortion Precision audio, biomedical instrumentation, anti-aliasing
Chebyshev Equiripple (e.g. 0.5 dB, 1 dB) Extremely steep skirt transition Severe nonlinear phase distortion RF channel selectivity where frequency skirt matters most
Bessel Drooping gradual rolloff Slowest transition Strictly linear phase; constant group delay Digital pulse transmission, radar, pulse-code modulation (zero ringing)

ยง8.4 First- & Second-Order Sallen-Key Active Butterworth Filters

1. First-Order Active Filters

A first-order active low-pass filter comprises a passive $RC$ low-pass network feeding a non-inverting op-amp with gain $A_0 = 1 + R_f / R_1$:

$$H(s) = \frac{A_0}{1 + s R C} = \frac{A_0}{1 + s / \omega_c}$$

Cutoff frequency where gain drops by $3\,\text{dB}$:

$$f_c = \frac{1}{2\pi R C}$$

Beyond $f_c$, the gain falls off at a gentle slope of $-20\,\text{dB/decade}$ ($-6\,\text{dB/octave}$).

2. Second-Order Sallen-Key Low-Pass Filter Topology

To double the roll-off sharpness to $-40\,\text{dB/decade}$ ($-12\,\text{dB/octave}$), a second-order Sallen-Key topology is employed. It utilizes two resistors ($R_1, R_2$) and two capacitors ($C_1, C_2$) with positive feedback through $C_1$ bootstrap.

The general transfer function is:

$$H(s) = \frac{A_0 \, \omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2} = \frac{A_0 \, \omega_0^2}{s^2 + \alpha \omega_0 s + \omega_0^2}$$

where $\alpha = 2\zeta = 1/Q$ is the damping factor.

3. Equal-Component Butterworth Design Derivation

For an equal-component design where $R_1 = R_2 = R$ and $C_1 = C_2 = C$:

$$\omega_0 = \frac{1}{R C} \implies f_c = \frac{1}{2\pi R C}$$

The damping coefficient is set by the op-amp passband gain $A_0 = 1 + R_f / R_A$:

$$\alpha = 3 - A_0$$

For a maximally flat Butterworth response, the required damping coefficient is:

$$\alpha = \sqrt{2} \approx 1.4142 \quad (Q = \frac{1}{\sqrt{2}} \approx 0.7071)$$

Equating coefficients:

$$3 - A_0 = \sqrt{2} \implies A_0 = 3 - \sqrt{2} \approx 1.5858$$

Therefore, the ratio of gain resistors must be strictly set to:

$$\frac{R_f}{R_A} = A_0 - 1 = 0.5858$$

If $A_0 > 1.586$, the circuit exhibits Chebyshev peaking and ringing; if $A_0 \ge 3$, the damping factor becomes zero or negative, causing the filter to burst into spontaneous oscillation!

Solved Problem Example 8.1: Op-Amp Slew Rate & Full-Power Bandwidth Limitation

An operational amplifier has a specified slew rate of $SR = 2.0\text{ V}/\mu\text{s}$. (a) If the op-amp is configured as a voltage follower and driven by a large-amplitude sine wave with peak voltage $V_m = 10.0\text{ V}$, calculate the full-power bandwidth $f_{\text{max}}$ before slew rate distortion occurs. (b) If the input frequency is increased to $100\text{ kHz}$, calculate the maximum distortion-free output amplitude $V_{\text{max}}$. (c) If a step input of $10.0\text{ V}$ is applied, calculate the minimum rise time required for the output to transition between 10% and 90% of its final value.

Step 1: Calculate Full-Power Bandwidth $f_{max}$

For an $8.0\,\text{V}$ peak sinusoidal signal ($16\,\text{V}$ peak-to-peak):

$$2\pi f_{max} V_m = SR = 2.0\,\text{V}/\mu\text{s} = 2.0 \times 10^6\,\text{V/s}$$
$$f_{max} = \frac{SR}{2\pi V_m} = \frac{2.0 \times 10^6}{2\pi \times 8.0} = \frac{2.0 \times 10^6}{50.265} \approx 39,789\,\text{Hz} \approx 39.79\,\text{kHz}$$

Step 2: Calculate Maximum Voltage at $f = 80\,\text{kHz}$

$$V_{m,max} = \frac{SR}{2\pi f} = \frac{2.0 \times 10^6\,\text{V/s}}{2\pi \times 80,000\,\text{s}^{-1}} = \frac{2.0 \times 10^6}{502,655} \approx 3.98\,\text{V}$$

At $80\,\text{kHz}$, attempting to drive the output past $3.98\,\text{V}$ peak will cause the sine wave to distort into a triangle.

Step 3: Calculate Transition Rise Time for 10 V Step

The voltage change between 10% and 90% of a 10 V step is $\Delta V = 0.8 \times 10\,\text{V} = 8.0\,\text{V}$:

$$t_r = \frac{\Delta V}{SR} = \frac{8.0\,\text{V}}{2.0\,\text{V}/\mu\text{s}} = 4.0\,\mu\text{s}$$
Solved Problem Example 8.2: Inverting Schmitt Trigger Design with Specified Trip Points

Design an inverting Schmitt trigger using an operational amplifier with supply rails of $\pm 15\text{ V}$ (assume saturation voltages $V_{\text{sat}} = \pm 13.5\text{ V}$) to have an upper trip point $V_{\text{UTP}} = +2.5\text{ V}$ and a lower trip point $V_{\text{LTP}} = -1.5\text{ V}$. (a) Calculate the required reference voltage $V_{\text{ref}}$ and the ratio of feedback resistors $R_1$ and $R_2$. (b) Choose standard $E24$ resistor values with $R_1 + R_2 \approx 100\text{ k}\Omega$. (c) Compute the hysteresis width $V_H$ and sketch the input-output transfer characteristic.

Step 1: Calculate Hysteresis Voltage $\Delta V_H$

$$\Delta V_H = V_{UTP} - V_{LTP} = (+3.0\,\text{V}) - (-3.0\,\text{V}) = 6.0\,\text{V}$$

Step 2: Derive and Calculate Feedback Resistor Ratio

Using the trip point relation:

$$V_{UTP} = \frac{R_1}{R_1 + R_2} V_{sat}$$
$$3.0 = \frac{R_1}{R_1 + R_2} (13.5) \implies \frac{R_1 + R_2}{R_1} = \frac{13.5}{3.0} = 4.5$$
$$1 + \frac{R_2}{R_1} = 4.5 \implies \frac{R_2}{R_1} = 3.5$$

Step 3: Component Values and Noise Margin

Given $R_1 = 10\,\text{k}\Omega$:

$$R_2 = 3.5 \times R_1 = 3.5 \times 10\,\text{k}\Omega = 35\,\text{k}\Omega$$

Noise Immunity: Any spurious noise spikes riding on the input waveform with peak-to-peak amplitude less than $\Delta V_H = 6.0\,\text{V}$ are completely suppressed, guaranteeing 100% chatter-free digital state transitions.

Solved Problem Example 8.3: Second-Order Sallen-Key Butterworth Low-Pass Filter Design

Design a second-order Sallen-Key low-pass filter with a maximally flat Butterworth response having a cutoff frequency $f_c = 1.0\text{ kHz}$ and passband DC gain $A_v = 1.586$ (damping factor $\zeta = 0.707$). Choose capacitors $C_1 = C_2 = 0.01\ \mu\text{F}$. (a) Calculate the required filter resistances $R_1 = R_2 = R$. (b) Determine the gain-setting resistors $R_A$ and $R_B$ for the non-inverting op-amp configuration. (c) Calculate the attenuation of the filter at $f = 5.0\text{ kHz}$ (in decibels) and verify its $-40\text{ dB/decade}$ roll-off rate.

Step 1: Calculate Filter Resistors $R_1 = R_2 = R$

Using the equal-component cutoff frequency formula:

$$f_c = \frac{1}{2\pi R C} \implies R = \frac{1}{2\pi f_c C}$$
$$R = \frac{1}{2\pi \times 2500\,\text{Hz} \times 0.01 \times 10^{-6}\,\text{F}} = \frac{1}{2\pi \times 2.5 \times 10^{-5}} = \frac{10^5}{5\pi} = \frac{20,000}{\pi} \approx 6366.2\,\Omega \approx 6.37\,\text{k}\Omega$$

Standard precision resistors of $6.34\,\text{k}\Omega$ (1% metal film) can be used.

Step 2: Determine Passband Gain and Feedback Resistors

For a maximally flat Butterworth response ($\alpha = \sqrt{2} = 1.414$):

$$A_0 = 3 - \sqrt{2} = 3 - 1.4142 = 1.5858$$
$$A_0 = 1 + \frac{R_f}{R_A} \implies \frac{R_f}{R_A} = 0.5858$$

With $R_A = 10\,\text{k}\Omega$:

$$R_f = 0.5858 \times 10\,\text{k}\Omega = 5.858\,\text{k}\Omega \approx 5.86\,\text{k}\Omega$$

Step 3: Attenuation at $f = 25\,\text{kHz}$

At $f = 25\,\text{kHz}$, the normalized frequency ratio is $f / f_c = 25000 / 2500 = 10$ (exactly one decade above cutoff).

For a second-order Butterworth filter, magnitude response is:

$$|H(j\omega)| = \frac{A_0}{\sqrt{1 + (f/f_c)^4}} = \frac{1.5858}{\sqrt{1 + 10^4}} = \frac{1.5858}{\sqrt{10001}} \approx \frac{1.5858}{100.005} \approx 0.015857$$

Attenuation relative to passband gain $A_0$ is:

$$\text{Attenuation}(\text{dB}) = 20 \log_{10}\left(\frac{1}{\sqrt{1 + 10^4}}\right) \approx -20 \log_{10}(100) = -40.0\,\text{dB}$$

The filter attenuates signals at $25\,\text{kHz}$ by exactly $40\,\text{dB}$ (a 100-fold reduction in voltage).

EXAM SUCCESS WORKSHOP

Solved University Examination Problems

Step-by-step mathematical solutions to classic university honors examination questions.