Ideal Op-Amp Fundamentals
Ideal Op-Amp Output Voltage
V_{out} = A_{OL}(V^+ - V^-)
| Symbol | Description | Unit |
|---|---|---|
| A_{OL} | Open-loop gain (ideally ∞) | dimensionless |
| V^+, V^- | Non-inverting and inverting input voltages | V |
Worked example
An op-amp has A_OL = 10⁵. If V⁺ = 1.0002 V and V⁻ = 1.0000 V, find V_out.
Given: A_OL=10⁵, V⁺=1.0002, V⁻=1.0000
- V_diff = V⁺ − V⁻ = 1.0002 − 1.0000 = 0.2 mV = 2×10⁻⁴ V
- V_out = 10⁵ × 2×10⁻⁴ = 20 V
- If supply rails are ±15 V, output saturates at +15 V
Answer: V_out saturates at +15 V (supply rail limit)
Inverting Amplifier Gain
A_v = -\frac{R_f}{R_{in}}
| Symbol | Description | Unit |
|---|---|---|
| R_f | Feedback resistor | Ω |
| R_{in} | Input resistor | Ω |
Worked example
Design an inverting amplifier with gain = −25. If R_in = 4 kΩ, find R_f. Then find V_out for V_in = 0.2 V.
Given: A_v=−25, R_in=4 kΩ, V_in=0.2 V
- R_f = |A_v| × R_in = 25 × 4000 = 100 kΩ
- V_out = A_v × V_in = −25 × 0.2 = −5 V
Answer: R_f = 100 kΩ, V_out = −5 V
Non-Inverting Amplifier Gain
A_v = 1 + \frac{R_f}{R_1}
| Symbol | Description | Unit |
|---|---|---|
| R_f | Feedback resistor | Ω |
| R_1 | Ground resistor | Ω |
Worked example
A non-inverting amplifier has R_f=47 kΩ, R_1=10 kΩ. Find gain and V_out for V_in=0.5 V.
Given: R_f=47k, R_1=10k, V_in=0.5
- A_v = 1 + 47/10 = 1 + 4.7 = 5.7
- V_out = 5.7 × 0.5 = 2.85 V
Answer: A_v = 5.7, V_out = 2.85 V
Voltage Follower (Buffer)
V_{out} = V_{in}, \quad A_v = 1
| Symbol | Description | Unit |
|---|---|---|
| V_{in} | Input voltage | V |
Worked example
A sensor outputs 3.3 V but has source impedance 100 kΩ. A load of 1 kΩ is connected via a unity-gain buffer. Find load voltage.
Given: V_sensor=3.3 V, R_source=100 kΩ, R_load=1 kΩ
- Without buffer: V_load = 3.3 × 1k/(100k+1k) = 3.3/101 = 32.7 mV (severe loading)
- With ideal buffer: input impedance → ∞, V_out = V_in = 3.3 V
- V_load = 3.3 V (no loading effect)
Answer: V_load = 3.3 V (buffer eliminates loading)
Summing and Difference Amplifiers
Inverting Summing Amplifier
V_{out} = -R_f \left(\frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3}\right)
| Symbol | Description | Unit |
|---|---|---|
| V_1, V_2, V_3 | Input voltages | V |
| R_1, R_2, R_3 | Input resistors | Ω |
| R_f | Feedback resistor | Ω |
Worked example
Summing amplifier: R1=R2=R3=R_f=10 kΩ, V1=1V, V2=2V, V3=−0.5V. Find V_out.
Given: R_f=R1=R2=R3=10k, V1=1, V2=2, V3=−0.5
- V_out = −R_f(V1/R1 + V2/R2 + V3/R3)
- = −10k(1/10k + 2/10k + (−0.5)/10k)
- = −(1 + 2 − 0.5)
- = −2.5 V
Answer: V_out = −2.5 V
Difference Amplifier Output
V_{out} = \frac{R_f}{R_1}(V_2 - V_1) \quad \text{(when } R_3 = R_1,\; R_4 = R_f\text{)}
| Symbol | Description | Unit |
|---|---|---|
| V_1, V_2 | Inverting and non-inverting inputs | V |
| R_1, R_f | Resistor values | Ω |
Worked example
Difference amplifier: R1=R3=10 kΩ, R_f=R4=100 kΩ, V1=2.5 V, V2=3.0 V. Find V_out.
Given: R_f/R1=10, V1=2.5, V2=3.0
- V_out = (R_f/R1)(V2−V1) = 10 × (3.0 − 2.5)
- = 10 × 0.5 = 5 V
Answer: V_out = 5 V
Instrumentation Amplifier Gain
A_v = \left(1 + \frac{2R_1}{R_G}\right) \frac{R_3}{R_2}
| Symbol | Description | Unit |
|---|---|---|
| R_G | Gain-setting resistor | Ω |
| R_1 | Feedback resistors in input stage (equal) | Ω |
| R_2, R_3 | Output difference stage resistors | Ω |
Worked example
INA with R1=24.9 kΩ, R_G=100 Ω, R2=R3=10 kΩ. Find overall gain.
Given: R1=24900, R_G=100, R2=R3=10000
- First stage gain = 1 + 2×24900/100 = 1 + 498 = 499
- Second stage gain = R3/R2 = 10k/10k = 1
- Total A_v = 499 × 1 = 499
Answer: A_v = 499 ≈ 500 (standard INA gain)
Integrator and Differentiator
Op-Amp Integrator Output
V_{out}(t) = -\frac{1}{RC} \int_0^t V_{in}(\tau)\, d\tau
| Symbol | Description | Unit |
|---|---|---|
| R | Input resistor | Ω |
| C | Feedback capacitor | F |
Worked example
Integrator: R=10 kΩ, C=100 nF. Input is V_in=2 V DC step applied at t=0. Find V_out at t=1 ms.
Given: R=10⁴ Ω, C=10⁻⁷ F, V_in=2 V, t=10⁻³ s
- V_out(t) = −(1/RC)∫₀ᵗ V_in dτ = −V_in·t/(RC)
- RC = 10⁴ × 10⁻⁷ = 10⁻³ s = 1 ms
- V_out(1 ms) = −2 × 10⁻³/10⁻³ = −2 V
Answer: V_out(1 ms) = −2 V (ramp down from 0 to −2 V)
Op-Amp Differentiator Output
V_{out}(t) = -RC \frac{dV_{in}}{dt}
| Symbol | Description | Unit |
|---|---|---|
| R | Feedback resistor | Ω |
| C | Input capacitor | F |
Worked example
Differentiator: R=10 kΩ, C=50 nF. Input is a 1 kHz, 4 V peak sine wave. Find peak output voltage.
Given: R=10⁴ Ω, C=5×10⁻⁸ F, f=1000 Hz, V_peak=4 V
- V_in = 4sin(2π×1000×t), so dV_in/dt = 4×2π×1000×cos(2π×1000×t)
- Peak of derivative = 4×2π×1000 = 25,133 V/s
- V_out_peak = −RC × 25,133 = −10⁴ × 5×10⁻⁸ × 25,133 = −5×10⁻⁴ × 25,133
- = −12.57 V
Answer: Peak V_out = −12.57 V (amplitude = 12.57 V)
Comparators and Schmitt Trigger
Non-Inverting Schmitt Trigger Thresholds
V_{UTP} = V_{sat}^+ \cdot \frac{R_1}{R_1+R_2}, \quad V_{LTP} = V_{sat}^- \cdot \frac{R_1}{R_1+R_2}
| Symbol | Description | Unit |
|---|---|---|
| V_{UTP} | Upper trip point | V |
| V_{LTP} | Lower trip point | V |
| V_{sat}^{\pm} | Positive and negative saturation voltages | V |
| R_1, R_2 | Hysteresis setting resistors | Ω |
Worked example
Non-inverting Schmitt trigger: V_sat = ±13 V, R1=10 kΩ, R2=90 kΩ. Find UTP, LTP, and hysteresis width.
Given: V_sat=±13 V, R1=10k, R2=90k
- V_UTP = 13 × 10k/(10k+90k) = 13 × 0.1 = 1.3 V
- V_LTP = −13 × 0.1 = −1.3 V
- Hysteresis = V_UTP − V_LTP = 1.3−(−1.3) = 2.6 V
Answer: V_UTP = +1.3 V, V_LTP = −1.3 V, Hysteresis = 2.6 V
Inverting Schmitt Trigger Thresholds
V_{UTP} = -V_{sat}^- \cdot \frac{R_1}{R_2}, \quad V_{LTP} = -V_{sat}^+ \cdot \frac{R_1}{R_2}
| Symbol | Description | Unit |
|---|---|---|
| R_1 | Lower resistor (to output) | Ω |
| R_2 | Upper resistor (to input) | Ω |
Worked example
Inverting Schmitt: V_sat=±14 V, R1=2 kΩ, R2=20 kΩ. Find trip points.
Given: V_sat=±14, R1=2k, R2=20k
- V_UTP = −(−14) × 2/20 = 14 × 0.1 = +1.4 V
- V_LTP = −(14) × 2/20 = −1.4 V
Answer: V_UTP = +1.4 V, V_LTP = −1.4 V
Active Filters
First-Order Active LPF — Cutoff Frequency
f_c = \frac{1}{2\pi R_f C_f}
| Symbol | Description | Unit |
|---|---|---|
| f_c | −3 dB cutoff frequency | Hz |
| R_f | Feedback resistor | Ω |
| C_f | Feedback capacitor | F |
Worked example
Design a first-order active LPF with f_c = 2 kHz. Choose C_f = 10 nF, find R_f.
Given: f_c=2000 Hz, C_f=10×10⁻⁹ F
- R_f = 1/(2π·f_c·C_f)
- = 1/(2π × 2000 × 10×10⁻⁹)
- = 1/(1.2566×10⁻⁴)
- = 7958 Ω ≈ 8 kΩ (use 7.96 kΩ)
Answer: R_f ≈ 7.96 kΩ (use 8.2 kΩ standard value)
Sallen-Key 2nd-Order LPF — Natural Frequency and Q
\omega_0 = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}}, \quad Q = \frac{\sqrt{R_1 R_2 C_1 C_2}}{C_2(R_1+R_2)}
| Symbol | Description | Unit |
|---|---|---|
| \omega_0 | Natural frequency | rad/s |
| Q | Quality factor | dimensionless |
Worked example
Sallen-Key LPF: R1=R2=10 kΩ, C1=10 nF, C2=2.5 nF. Find f_0 and Q.
Given: R1=R2=10⁴ Ω, C1=10⁻⁸ F, C2=2.5×10⁻⁹ F
- ω₀ = 1/√(10⁴×10⁴×10⁻⁸×2.5×10⁻⁹) = 1/√(2.5×10⁻⁹) = 1/(5×10⁻⁵×√(10⁻¹)) = ...
- R1R2C1C2 = 10⁸ × 2.5×10⁻¹⁷ = 2.5×10⁻⁹
- ω₀ = 1/√(2.5×10⁻⁹) = 1/(5×10⁻⁵×√10) = 1/1.581×10⁻⁴ = 6325 rad/s → f₀=1006 Hz
- Q numerator = √(2.5×10⁻⁹) = 5×10⁻⁵×√10 = 1.581×10⁻⁴
- Q denominator = C2(R1+R2) = 2.5×10⁻⁹ × 20000 = 5×10⁻⁵
- Q = 1.581×10⁻⁴/5×10⁻⁵ = 3.162/1 ≈ 0.707 (Butterworth response)
Answer: f₀ ≈ 1006 Hz, Q ≈ 0.707 (maximally flat Butterworth)
Op-Amp Non-Ideal Parameters
Gain-Bandwidth Product
GBW = A_{v} \cdot f_{-3dB} = f_T \text{ (constant for internally compensated op-amps)}
| Symbol | Description | Unit |
|---|---|---|
| GBW | Gain-bandwidth product | Hz |
| f_{-3dB} | Closed-loop −3 dB bandwidth | Hz |
| f_T | Unity-gain frequency | Hz |
Worked example
An LM741 has GBW = 1 MHz. Find the −3 dB bandwidth when configured as an inverting amplifier with gain = −50.
Given: GBW=1×10⁶ Hz, |A_v|=50
- f_{-3dB} = GBW / |A_v|
- = 1×10⁶ / 50
- = 20,000 Hz = 20 kHz
Answer: Bandwidth = 20 kHz
Slew Rate Limit on Output
SR = \left.\frac{dV_{out}}{dt}\right|_{max}, \quad f_{max} = \frac{SR}{2\pi V_{peak}}
| Symbol | Description | Unit |
|---|---|---|
| SR | Slew rate | V/μs |
| f_{max} | Maximum undistorted full-swing frequency | Hz |
| V_{peak} | Peak output swing | V |
Worked example
LM741: SR = 0.5 V/μs. Find maximum frequency for a full ±10 V (20 V peak-to-peak, 10 V peak) sinusoidal output without slew distortion.
Given: SR=0.5×10⁶ V/s, V_peak=10 V
- f_max = SR / (2π × V_peak)
- = 0.5×10⁶ / (2π × 10)
- = 5×10⁵ / 62.83
- = 7958 Hz ≈ 7.96 kHz
Answer: f_max ≈ 7.96 kHz for full ±10 V swing
Output Offset Voltage Due to Bias Current
V_{OS,bias} = I_B^- R_f \quad \text{(reduced to } I_{OS} R_f \text{ with compensation resistor)}
| Symbol | Description | Unit |
|---|---|---|
| I_B^- | Bias current into inverting input | A |
| I_{OS} | Offset current = |I_B⁺ − I_B⁻| | A |
| R_f | Feedback resistance | Ω |
Worked example
Inverting amplifier: R_f=100 kΩ, I_B=200 nA, I_OS=20 nA. Find output offset without and with compensation resistor R_comp=R_in||R_f.
Given: R_f=100k, I_B=200×10⁻⁹, I_OS=20×10⁻⁹
- Without compensation: V_OS = I_B × R_f = 200×10⁻⁹ × 100×10³ = 20 mV
- With R_comp (= R_in||R_f): V_OS = I_OS × R_f = 20×10⁻⁹ × 100×10³ = 2 mV
- 10× reduction in offset voltage
Answer: V_OS reduced from 20 mV to 2 mV with compensation
Oscillators Using Op-Amps
Wien Bridge Oscillator Frequency
f_0 = \frac{1}{2\pi RC}
| Symbol | Description | Unit |
|---|---|---|
| R | Wien bridge resistance (both equal) | Ω |
| C | Wien bridge capacitance (both equal) | F |
Worked example
Wien bridge oscillator: R=15.9 kΩ, C=10 nF. Find oscillation frequency.
Given: R=15900 Ω, C=10⁻⁸ F
- f₀ = 1/(2πRC)
- = 1/(2π × 15900 × 10⁻⁸)
- = 1/(2π × 1.59×10⁻⁴)
- = 1/9.99×10⁻⁴ = 1001 Hz ≈ 1 kHz
Answer: f₀ ≈ 1 kHz
Phase-Shift Oscillator Frequency
f_0 = \frac{1}{2\pi RC\sqrt{6}}
| Symbol | Description | Unit |
|---|---|---|
| R, C | Resistor and capacitor in each identical RC stage | Ω, F |
Worked example
Three-stage RC phase-shift oscillator: R=10 kΩ, C=10 nF. Find f₀ and required amplifier gain.
Given: R=10⁴ Ω, C=10⁻⁸ F
- f₀ = 1/(2π × 10⁴ × 10⁻⁸ × √6) = 1/(2π × 10⁻⁴ × 2.449)
- = 1/(1.539×10⁻³) = 649.7 Hz ≈ 650 Hz
- Required gain for oscillation |A_v| = 29
Answer: f₀ ≈ 650 Hz; amplifier must have |A_v| ≥ 29
Quick reference
| Formula | Expression |
|---|---|
| Inverting Amplifier Gain | A_v = -R_f/R_{in} |
| Non-Inverting Gain | A_v = 1 + R_f/R_1 |
| Summing Amplifier | V_{out} = -R_f(V_1/R_1 + V_2/R_2 + V_3/R_3) |
| Difference Amplifier | V_{out} = (R_f/R_1)(V_2-V_1) |
| Integrator Output | V_{out} = -(1/RC)\int V_{in}\,dt |
| Differentiator Output | V_{out} = -RC\,dV_{in}/dt |
| Schmitt Trigger UTP (non-inv) | V_{UTP} = V_{sat}^+\cdot R_1/(R_1+R_2) |
| Active LPF Cutoff | f_c = 1/(2\pi R_f C_f) |
| Gain-Bandwidth Product | GBW = A_v \cdot f_{-3dB} |
| Slew Rate Max Frequency | f_{max} = SR/(2\pi V_{peak}) |
| Wien Oscillator Frequency | f_0 = 1/(2\pi RC) |
| Phase-Shift Oscillator | f_0 = 1/(2\pi RC\sqrt{6}) |
| INA Gain | A_v = (1+2R_1/R_G)(R_3/R_2) |
| Sallen-Key ω₀ | \omega_0 = 1/\sqrt{R_1 R_2 C_1 C_2} |
| Bias Current Offset | V_{OS} = I_B R_f |
Exam tips
- GATE tests the gain-bandwidth product concept by giving you a specified bandwidth and asking for the maximum achievable closed-loop gain — always apply GBW = A_v × f_{-3dB} directly.
- Slew-rate distortion and bandwidth limitation are two distinct phenomena; examiners expect you to check both SR and GBW limits separately when a large, fast output swing is involved.
- In summing amplifier problems, check whether all input resistors are equal before using the simplified V_out = −(R_f/R)(V1+V2+…); use the general weighted formula when they differ.
- For Schmitt trigger questions, always identify whether the topology is inverting or non-inverting before applying the threshold formulas — the two configurations have opposite sign conventions.
- Integrator problems with a DC input step will ramp the output to saturation in finite time; examiners often ask how long before saturation, requiring V_sat / (V_in/RC) calculation.
- Phase-shift oscillator gain requirement of 29 is frequently asked as a standalone fact; know that the minimum gain is exactly 29 to sustain oscillation.