Operational Amplifiers and Controlled-Source CircuitsIntroduction and ContextOperational amplifiers, or op-amps, are high-gain voltage amplifiers used with external components to perform mathematical, filtering, sensing, control, and signal-conditioning functions. Their importance in electrical engineering arises from a useful design principle: a complicated active circuit can often be understood by combining a simple amplifier model with passive feedback networks. An op-amp is also a practical example of a controlled-source circuit. In circuit theory, controlled sources model active devices whose output voltage or current is controlled by another voltage or current elsewhere in the circuit. Op-amp circuits therefore connect network analysis with real electronic systems. This chapter develops ideal and practical op-amp models, negative-feedback analysis, common configurations, saturation, frequency response, and limitations. The discussion assumes familiarity with Kirchhoff’s laws, node-voltage analysis, and basic resistor and capacitor behavior. Learning ObjectivesAfter studying this chapter, you should be able to: - Model ideal op-amps and controlled sources using circuit equations.
- Analyze inverting, non-inverting, summing, differential, integrator, and comparator circuits.
- Explain the virtual-short and virtual-ground concepts and their conditions of validity.
- Determine output limits, saturation, bandwidth, slew-rate constraints, and input-offset effects.
- Select and evaluate op-amp circuits for practical signal-conditioning applications.
- Recognize when an ideal model is inadequate and identify the appropriate non-ideal correction.
1. Controlled Sources as Active-Circuit ModelsA controlled source supplies energy to a circuit, but its value is controlled by another circuit variable. Unlike an independent voltage or current source, its output depends on a controlling voltage or current. | Type | Symbolic relationship | Common use |
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| Voltage-controlled voltage source (VCVS) | vo = μvc | Op-amp voltage-gain model | | Voltage-controlled current source (VCCS) | io = gvc | Transconductance amplifier | | Current-controlled voltage source (CCVS) | vo = ric | Transresistance model | | Current-controlled current source (CCCS) | io = βic | Current amplifier |
Here, μ is dimensionless voltage gain, g has units siemens, r has units ohms, and β is dimensionless. Controlled sources are idealized models; they can deliver power and therefore represent active devices rather than passive elements. An op-amp may be represented by a differential VCVS: vo = A(v_+ − v_−) where v_+ and v_− are the non-inverting and inverting input voltages, A is the open-loop differential voltage gain, and vo is the internal output voltage before output limitations are imposed. For an ideal op-amp, A → ∞. 2. Ideal Operational-Amplifier ModelAn op-amp has two input terminals and one output terminal. The input marked + is non-inverting: increasing its voltage tends to increase the output. The input marked − is inverting: increasing its voltage tends to decrease the output. The ideal model uses three assumptions: - Infinite input resistance: no input current flows.
- Zero output resistance: the output voltage is independent of load current, within the ideal model.
- Infinite open-loop gain: any finite linear output requires the differential input voltage to approach zero.
These assumptions give the ideal rules i_+ = i_− = 0 and, when negative feedback maintains linear operation, v_+ = v_− The second relation is called the virtual-short condition. It does not mean that the two input terminals are physically connected or that current flows between them. If the non-inverting input is grounded, the inverting input is approximately at 0 V; this node is called a virtual ground. The virtual-short rule is valid only when: (a) negative feedback is present, (b) the op-amp is operating in its linear region, and (c) the required output voltage is within its supply limits. It must not be used blindly for a comparator or a saturated amplifier. 3. Inverting AmplifierConsider an input voltage vi applied through R1 to the inverting input. A feedback resistor Rf connects the output to that input, and the non-inverting input is grounded. Because the inverting node is a virtual ground, the current through R1 is i1 = vi / R1 No current enters the op-amp input, so the same current flows through Rf. The output voltage is therefore vo = −(Rf / R1)vi The voltage gain is Av = vo / vi = −Rf / R1 The negative sign indicates a 180° phase reversal for sinusoidal signals within the circuit bandwidth. The input resistance seen by the source is approximately R1, while the ideal closed-loop output resistance is zero. Worked Example 1: Inverting GainAn inverting amplifier has R1 = 10 kΩ, Rf = 100 kΩ, and vi = 0.20 V. Then Av = −100/10 = −10 and vo = (−10)(0.20 V) = −2.0 V. If the amplifier has supply rails of ±15 V, the output is feasible under the ideal voltage-limit assumption. If the input increases to 2 V, the ideal result is −20 V, which exceeds the negative rail; the actual output saturates near the available negative output limit. 4. Non-Inverting AmplifierIn a non-inverting amplifier, the input is applied to the + terminal. A resistor R1 connects the inverting input to ground, and Rf connects the output to the inverting input. Since the input current is zero, R1 and Rf form a voltage divider driven by the output: v_− = vo R1/(R1 + Rf) Under negative feedback, v_− = v_+ = vi. Solving for output gives vo = (1 + Rf/R1)vi Thus the non-inverting gain is Av = 1 + Rf/R1 The gain is positive and cannot be less than unity for this basic configuration. The input resistance is ideally infinite, making the circuit useful when the signal source must be lightly loaded. A special case is the voltage follower, obtained by connecting the output directly to the inverting input. Its gain is approximately one: vo = vi The follower provides buffering: it presents high input resistance and low output resistance. 5. Summing AmplifierSeveral input voltages may be applied through separate resistors to the inverting node. With feedback resistor Rf, the ideal output is vo = −Rf(v1/R1 + v2/R2 + ⋯ + vn/Rn) If all input resistors are equal to R and Rf = R, then vo = −(v1 + v2 + ⋯ + vn) The circuit is useful in audio mixing, digital-to-analog conversion, weighted signal combination, and sensor fusion. Each input is weighted by Rf/Rk. Worked Example 2: Weighted SummationLet v1 = 1 V, R1 = 20 kΩ, v2 = 0.5 V, R2 = 10 kΩ, and Rf = 40 kΩ. Then vo = −40 kΩ[(1 V/20 kΩ) + (0.5 V/10 kΩ)] The bracketed quantity is 50 μA + 50 μA = 100 μA, so vo = −4.0 V. The result assumes linear operation and sufficient output swing. 6. Differential and Difference AmplifiersA differential amplifier responds primarily to the difference between two input voltages while rejecting their common component. For a matched four-resistor difference amplifier, let the resistor ratios satisfy R2/R1 = R4/R3 where the first input network uses R1, R2 and the second uses R3, R4. The output is vo = (R2/R1)(v2 − v1) The common-mode voltage is the average component: vcm = (v1 + v2)/2 The differential voltage is vd = v2 − v1 Real circuits do not reject common-mode signals perfectly. Common-mode rejection ratio is defined as CMRR = |Ad/Acm| where Ad is differential gain and Acm is common-mode gain. In decibels, CMRRdB = 20 log10(CMRR) High resistor-ratio accuracy is essential. A small mismatch can produce a significant output error when a large common-mode voltage accompanies a small desired differential signal. Instrumentation amplifiers improve this arrangement by using buffered inputs and a precisely matched differential stage. They are widely used with strain gauges, thermocouples, biomedical electrodes, and industrial sensors. 7. Integrator and Differentiator CircuitsReplacing the feedback resistor of an inverting amplifier with a capacitor produces an inverting integrator. With input resistor R and feedback capacitor C, the ideal relationship is vo(t) = −(1/RC) ∫ vi(t) dt + vo(t0) Here t0 is the initial time and vo(t0) is the initial capacitor-related output voltage. The transfer function, assuming zero initial conditions, is Vo(s)/Vi(s) = −1/(sRC) The quantity RC has units of seconds, ensuring dimensional consistency. An ideal integrator can saturate because even a small dc input accumulates over time. Practical integrators place a resistor in parallel with the capacitor. At low frequency this resistor limits dc gain; above its corner frequency the circuit approximates integration. For an ideal practical integrator with parallel resistor Rf, Zf = Rf/(1 + sRfC) and the inverting transfer function is Vo/Vi = −Zf/R The circuit behaves approximately as an integrator when |sRfC| ≫ 1, while the resistor prevents unlimited dc accumulation. In an inverting differentiator, the input element is a capacitor and the feedback element is a resistor: vo(t) = −RC dvi(t)/dt An ideal differentiator strongly amplifies high-frequency noise. Practical versions add input and feedback components to restrict the operating band. 8. Comparator OperationA comparator is an op-amp used without linear negative feedback. It compares two voltages. If the non-inverting input exceeds the inverting input, the output moves toward positive saturation; if it is lower, the output moves toward negative saturation: vo ≈ +Vsat when v_+ > v_− vo ≈ −Vsat when v_+ < v_− The output is not generally a linear amplification of the input difference. A comparator may include positive feedback, called hysteresis, to prevent noise near the threshold from causing rapid switching. For a single-threshold comparator with reference Vref, the transition occurs when vi = Vref. Op-amps are not always suitable substitutes for dedicated comparators because they may switch slowly, recover slowly from saturation, or lack logic-compatible output stages. 9. Saturation and Output ConstraintsThe ideal equation predicts any output voltage, but a real op-amp is powered by finite supplies. If the required output exceeds the output swing, the output clips: vo = max(−Vout,limit, min(vo,ideal, Vout,limit)) This compact expression assumes approximately symmetric limits; real limits may be different for positive and negative output. Rail-to-rail devices can approach the supply rails more closely than conventional devices, but even they have load-dependent limits. For a linear inverting amplifier, the maximum input magnitude is approximately |vi| ≤ |Vo,max| / |Av| provided the input common-mode range is also satisfied. Saturation causes distortion and invalidates the virtual-short assumption. 10. Frequency Response and BandwidthOpen-loop gain decreases with frequency. A commonly used single-pole approximation is A(s) = A0/(1 + s/ωp) where A0 is low-frequency open-loop gain and ωp is the pole angular frequency in radians per second. The gain-bandwidth product, or unity-gain frequency, is approximately constant for a compensated op-amp: GBW ≈ |ACL| f−3 dB where ACL is the closed-loop noise gain or, in simple voltage-feedback circuits, approximately the closed-loop gain magnitude, and f−3 dB is the closed-loop bandwidth in hertz. This relationship is an approximation and depends on compensation and circuit topology. For a non-inverting amplifier with gain 11 and an op-amp with GBW = 1 MHz, the approximate bandwidth is f−3 dB ≈ 1 MHz/11 ≈ 90.9 kHz. The circuit may have adequate low-frequency gain but fail to reproduce a high-frequency waveform accurately. 11. Slew Rate and Large-Signal PerformanceSlew rate is the maximum rate of output-voltage change: SR = max|dvo/dt| For a sinusoidal output vo = Vp sin(2πft), the maximum slope is 2πfVp. Therefore, distortion-free operation requires approximately 2πfVp ≤ SR For Vp = 5 V, f = 20 kHz, and SR = 0.5 V/μs, the required slope is approximately 0.628 V/μs, exceeding the device rating; substantial slew-induced distortion is expected. Bandwidth describes small-signal frequency response, whereas slew rate describes a large-signal time-domain limitation. A circuit can satisfy the bandwidth estimate and still distort because of insufficient slew rate. 12. Other Non-Ideal ParametersImportant practical parameters include: - Input offset voltage: the differential input voltage required to make the output zero. In a closed-loop circuit it is multiplied by the noise gain.
- Input bias currents: small currents entering the input terminals. Resistor networks can convert these currents into offset voltages.
- Input offset current: the difference between the two input bias currents.
- Finite open-loop gain: causes closed-loop gain error.
- Common-mode input range: the permitted range of average input voltage.
- Output current limit: restricts the load that can be driven.
- Noise voltage and current: introduce random signal errors.
- Power-supply rejection ratio: measures sensitivity to supply variations.
- Input and output protection limits: determine safe transient and fault conditions.
A useful closed-loop feedback expression is Af = A/(1 + Aβ) where A is open-loop gain, β is the feedback factor, and Af is closed-loop gain. When Aβ ≫ 1, the gain is controlled primarily by the feedback network. Finite A produces gain error because the denominator is not exactly Aβ. The same feedback that stabilizes gain can create instability if phase shift causes the loop to approach positive feedback. Phase margin, compensation, capacitive loading, and layout must therefore be considered in practical design. 13. A Systematic Analysis ProcedureA reliable procedure for op-amp circuits is: - Identify whether feedback is negative, positive, or absent.
- Assume linear operation only if negative feedback is present and output limits are not exceeded.
- Set input currents to zero for the ideal model.
- Set v_+ = v_− for a linearly operating ideal op-amp.
- Apply Kirchhoff’s current law at the input node.
- Solve for the ideal output.
- Check output swing, input common-mode range, load current, bandwidth, and slew rate.
- Replace the ideal model with a practical one if any check fails.
This sequence prevents the common error of using virtual ground in a comparator or in a circuit that is already saturated. 14. Applied Case: Strain-Gauge Signal ConditioningBackgroundConsider an illustrative industrial weighing platform using a Wheatstone bridge. The bridge produces a differential signal of only 2 mV at full load, while both bridge outputs are approximately 2.5 V relative to circuit ground. The measurement system requires a 0–2 V signal for an analog-to-digital converter. ProblemA direct connection to a conventional amplifier is unsuitable because the desired differential signal is small compared with the common-mode voltage. Input loading, resistor mismatch, noise, and offset may create errors larger than the signal. Analysis and ApproachAn instrumentation amplifier is selected. Its buffered input stage presents high input resistance, and its precision differential stage rejects common-mode voltage. If the required gain is 1000, the nominal output for 2 mV is vo = 1000 × 2 mV = 2.0 V The design must verify that the amplifier’s input common-mode range includes approximately 2.5 V, that its output can reach 2.0 V under the chosen supply voltage, and that its input-referred offset and noise are acceptably below 2 mV. If the amplifier has a CMRR of 100 dB, the common-mode gain ratio is Acm/Ad = 10^(−100/20) = 10^(−5) Thus a 2.5 V common-mode input would ideally contribute approximately 25 μV referred to the output-equivalent differential input, before considering resistor mismatch and other errors. In practice, data-sheet CMRR depends on frequency, gain, source resistance, and temperature. A low-pass filter may be added after the amplifier to reduce mechanical and electrical noise. Calibration can remove predictable gain and offset errors, but it cannot replace correct range, stability, and safety analysis. Outcome and Learning PointsThe instrumentation-amplifier arrangement produces a usable signal while preserving high input resistance and common-mode rejection. The case illustrates that selecting a topology requires more than applying an ideal gain equation. Common-mode range, CMRR, offset, noise, bandwidth, output swing, supply voltage, and calibration all affect measurement quality. 15. Design, Safety, and Professional PracticeOp-amp circuits should be designed from data-sheet limits rather than nominal ideal assumptions. Input pins must not be driven beyond permitted voltage limits, especially when the power supply is absent. Output short-circuit behavior and thermal dissipation must be checked. Decoupling capacitors should be placed near supply pins, and high-impedance nodes should be kept short and protected from contamination and leakage. In measurement and control applications, incorrect scaling can cause unsafe actuator commands or misleading recorded data. Engineers should document assumptions, tolerances, calibration procedures, and failure modes. Medical, automotive, aerospace, and industrial designs may also be subject to applicable standards for isolation, electromagnetic compatibility, functional safety, and product verification. Modern circuit-design software can simulate ideal and macromodel op-amps, perform frequency sweeps, and estimate tolerances. AI-assisted design tools may suggest topologies or component values, but their results require engineering verification. A simulation cannot by itself establish stability, safety, manufacturability, or compliance. Chapter SummaryOperational amplifiers are differential, high-gain active devices whose useful behavior is established by negative feedback. The ideal model assumes zero input current, infinite open-loop gain, and zero output resistance. Under linear negative feedback, these assumptions lead to the virtual-short condition, v_+ = v_−, which simplifies circuit analysis. The inverting amplifier has gain −Rf/R1, while the non-inverting amplifier has gain 1 + Rf/R1. Summing amplifiers combine weighted inputs, difference amplifiers reject common-mode components when resistor ratios are accurately matched, and integrators and differentiators use capacitors to implement time-domain operations. Comparators operate without linear negative feedback and switch toward saturation. Real op-amps have finite supplies, bandwidth, slew rate, input offset, bias currents, noise, finite CMRR, limited common-mode range, and finite output current. Closed-loop design must therefore include output-swing, frequency-response, stability, loading, and safety checks. The most reliable engineering practice is to begin with the ideal model, calculate the expected result, and then test every practical constraint against the selected device and application. Key TermsOperational amplifier: A high-gain differential amplifier used with feedback networks. Controlled source: An active source whose value depends on another circuit voltage or current. Open-loop gain: Gain of an amplifier without feedback. Closed-loop gain: Gain established by an amplifier and its feedback network. Virtual short: The approximate equality of op-amp input voltages during linear negative-feedback operation. Virtual ground: A node approximately at ground potential without a direct ground connection. Inverting amplifier: An op-amp circuit with negative gain determined by resistor ratio. Non-inverting amplifier: A circuit with positive gain and ideally high input resistance. Voltage follower: A unity-gain buffer with output directly fed to the inverting input. Summing amplifier: A circuit that produces a weighted sum of input voltages. Differential amplifier: A circuit that amplifies voltage difference and rejects common-mode voltage. CMRR: Common-mode rejection ratio, the ratio of differential gain to common-mode gain. Comparator: An amplifier used to determine which of two input voltages is greater. Saturation: Operation in which output voltage reaches a supply-dependent limit. Bandwidth: Frequency range over which a circuit maintains specified response. Gain-bandwidth product: Approximate product of closed-loop gain and bandwidth for a compensated op-amp. Slew rate: Maximum rate of change of output voltage. Input offset voltage: Differential input voltage required to produce zero output. Negative feedback: Feedback that opposes the input difference and generally stabilizes gain. Instrumentation amplifier: A precision differential amplifier with buffered inputs and high common-mode rejection. Review QuestionsShort Answer Questions- State the two ideal input-terminal rules for an op-amp operating with linear negative feedback.
- Why is the inverting input called a virtual ground when the non-inverting input is grounded?
- Write the voltage gain of an inverting amplifier and identify the role of the negative sign.
- Distinguish bandwidth from slew rate.
- Why must resistor ratios be accurately matched in a difference amplifier?
Long Answer Questions- Derive the gain equation for an inverting amplifier using Kirchhoff’s current law.
- Explain the operation and applications of a weighted summing amplifier.
- Derive the transfer relationship of an ideal op-amp integrator and explain why a practical parallel resistor is added.
- Discuss the effects of finite open-loop gain, input offset voltage, bias current, and output saturation.
- Explain how negative feedback establishes closed-loop gain and why stability must be checked in practical circuits.
Analytical / Application Questions- Design an inverting amplifier with a gain of −25 using R1 = 8 kΩ. Determine Rf and calculate the output for an input of 0.12 V. Check whether the output is feasible with ±12 V supplies.
- A non-inverting amplifier has gain 21 and an op-amp with GBW = 2.1 MHz. Estimate its small-signal bandwidth. If the output must be a 10 V peak sinusoid at 30 kHz, determine the minimum slew rate required.
- A bridge sensor produces v1 = 2.502 V and v2 = 2.500 V. A difference amplifier has gain 200. Determine the ideal output polarity and magnitude, then identify two practical non-ideal effects that could alter the result.
Multiple Choice Questions with AnswersQuestion 1For an ideal op-amp operating linearly with negative feedback, which statement is correct? A. The input currents are equal to the output current. B. The two input voltages are approximately equal. C. The output voltage is always zero. D. The input resistance is zero. Correct Answer: B. The two input voltages are approximately equal. Question 2An inverting amplifier has R₁ = 10 kΩ and R_f = 50 kΩ. Its ideal voltage gain is: A. −0.2 B. +0.2 C. −5 D. +5 Correct Answer: C. −5 Question 3A non-inverting amplifier uses R₁ = 2 kΩ and R_f = 18 kΩ. Its ideal gain is: A. +9 B. +10 C. −9 D. −10 Correct Answer: B. +10 Question 4In an ideal summing amplifier with equal input resistors and R_f equal to each input resistor, the output is: A. The positive sum of all inputs B. The average of all inputs C. The negative sum of all inputs D. The difference between the largest and smallest input Correct Answer: C. The negative sum of all inputs Question 5A comparator is most appropriately analyzed by assuming that its output: A. Remains exactly proportional to the input difference for all voltages B. Moves toward a supply-dependent saturation level C. Is always equal to the non-inverting input voltage D. Has zero current under every load Correct Answer: B. Moves toward a supply-dependent saturation level Question 6The ideal transfer relationship for an inverting integrator is: A. v_o = −RC dv_i/dt B. v_o = v_i/(RC) C. v_o = −(1/RC)∫v_i dt D. v_o = −R/C Correct Answer: C. v_o = −(1/RC)∫v_i dt Question 7An op-amp has a slew rate of 1 V/μs. What is the maximum approximate frequency for a 5 V peak sine-wave output? A. 3.18 kHz B. 31.8 kHz C. 318 kHz D. 1 MHz Correct Answer: B. 31.8 kHz Question 8If a compensated op-amp has a gain-bandwidth product of 1 MHz and is used at closed-loop gain 10, its approximate bandwidth is: A. 10 kHz B. 100 kHz C. 1 MHz D. 10 MHz Correct Answer: B. 100 kHz Question 9A high CMRR is especially important when: A. The desired differential signal is accompanied by a large common-mode voltage. B. The circuit has no input signal. C. The output is intentionally saturated. D. Only a single ideal resistor is used. Correct Answer: A. The desired differential signal is accompanied by a large common-mode voltage. Question 10Which condition invalidates direct use of the virtual-short rule? A. Negative feedback with adequate output swing B. A voltage follower operating within its ratings C. A comparator operating without linear negative feedback D. An inverting amplifier with moderate gain Correct Answer: C. A comparator operating without linear negative feedback
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