core24 min studyLesson 955 in Physics

Operational amplifiers and feedback response

Use finite loop gain to calculate closed-loop accuracy and bandwidth, while checking stability and large-signal limits separately.

Before you begin

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The question for today

What you’ll understand

Derive the negative-feedback gain with a stated sign convention.
Calculate finite-gain error and a one-pole closed-loop bandwidth.
Separate small-signal response from slew rate, saturation, and stability.

An operational amplifier does not enforce equal input voltages by definition. It amplifies their difference with a large but finite, frequency-dependent gain. Negative feedback can make the overall circuit depend mainly on a stable external network, provided the amplifier remains within its operating range and the loop is stable. The familiar ideal circuit formulas follow as limits of that model. Writing the finite loop first explains both why feedback works and why a correct resistor ratio can still produce inaccurate, delayed, or oscillating measurements.

Declare the loop sign before taking an ideal limit

For a noninverting voltage-feedback model, let the error be Vin minus beta times Vout, where beta is a positive feedback fraction. The amplifier output is A times that error. Solving the algebra gives a closed-loop denominator one plus A beta. When the magnitude of loop gain is large in the relevant frequency range, the gain approaches one over beta. The residual difference between amplifier inputs supplies the finite output; it is small rather than identically zero.

The feedback factor is set by the connected network, including any capacitance. In an inverting circuit, signal gain and noise gain are different: with resistive feedback their magnitudes involve Rf/Rin and one plus that ratio respectively. It is noise gain, together with the open-loop response, that enters the usual voltage-feedback gain-bandwidth and stability discussion. The same rules cannot automatically be transferred to a current-feedback amplifier.

V_o=A(s)(V_i-\beta V_o),\qquad G(s)=\frac{V_o}{V_i}=\frac{A(s)}{1+\beta A(s)},\qquad G(0)=\frac1\beta\frac{A_0\beta}{1+A_0\beta}

A is the dimensionless open-loop voltage gain, β is a positive constant for this simple resistive noninverting model, and A0=A(0)>0. The relation assumes linear operation and a feedback network represented by the stated β.↔ Scroll wide equations horizontally.

A one-pole model connects gain accuracy and bandwidth

Suppose the open-loop gain has one dominant pole. Substituting it into the feedback expression leaves another first-order response whose DC gain is reduced by finite loop gain and whose pole is shifted upward. Increasing the chosen closed-loop gain decreases beta and typically narrows the bandwidth. The approximate gain-bandwidth relation is useful within this dominant-pole voltage-feedback regime, not a universal identity for every amplifier or circuit topology.

Additional poles, input capacitance, output loading, and delay alter loop phase. A loop that was negative feedback at low frequency can become reinforcing near its unity-loop-gain crossing. Phase margin describes the distance from that condition and is relevant to overshoot and settling. A large DC open-loop gain does not establish stability, and a small-signal bandwidth number by itself does not guarantee that a transient has settled to a tight error tolerance.

A(s)=\frac{A_0}{1+s/\omega_p}\Longrightarrow G(s)=\frac{A_0}{1+A_0\beta}\frac1{1+s/[\omega_p(1+A_0\beta)]},\qquad f_{\mathrm{CL}}=f_p(1+A_0\beta),\qquad \left|\frac{dV_o}{dt}\right|_{\max}=2\pi fV_{\mathrm{pk}}

fp=ωp/(2π) and fCL are in hertz. The last expression is the maximum required slope of a sinusoidal output of peak amplitude Vpk, not a claim about the amplifier’s available slew rate. That available rate must exceed this requirement for undistorted tracking.↔ Scroll wide equations horizontally.

Large signals can violate the linear model first

A real amplifier has finite supply-dependent output swing, input common-mode range, output current, and slew rate. If a desired sinusoid demands more slope than the slew-rate limit, it distorts even when its frequency is below the small-signal bandwidth. Saturation can also introduce a recovery delay that the linear transfer function does not describe. These mechanisms have to be checked using the complete waveform, including offsets and interference.

Offset and bias currents contribute further input-referred errors, while feedback and source resistors contribute noise. A laboratory design therefore checks DC error, noise, frequency response, stability, and large-signal headroom as separate constraints. The numerical examples here describe explicitly hypothetical models; a real part needs its own data-sheet conditions and connected-load validation. Feedback is a way to control sensitivity and response, not a license to ignore the physical device supplying the correction.

Work it through

Finite gain in a declared one-pole amplifier

A hypothetical voltage-feedback amplifier has A0=1000 and dominant pole fp=10 Hz. A resistive noninverting network sets β=0.10. Find its DC closed-loop gain, the fractional error relative to the ideal gain ten, and the closed-loop pole frequency. Assume all other linear and stability conditions are satisfied.

The DC loop gain is A0β=100. Therefore $G(0)=1000/(1+100)=1000/101\simeq9.90099$ .
Relative to ideal gain ten, the fractional shortfall is $1-9.90099/10=1/101\simeq0.00990099$ , or 0.9901 percent.
The closed-loop pole is $f_{\mathrm{CL}}=10(101)=1010$ Hz. This follows from the stipulated one-pole model rather than an unqualified bandwidth rule for every operational amplifier.

The result

The gain is about 9.901, its ideal-gain shortfall is about 0.9901%, and its one-pole closed-loop bandwidth is 1010 Hz.

The living notebook

Explore the model

Interactive lab

Interactive lab · optimization

An overly aggressive correction can become unstable

For each fresh comparison, Reset and set BOTH Initial position x₀=1 and Step size η. Compare η=0.50 and η=2.50, taking exactly four Steps after each Reset. Each Step is one iteration: predict x4=0.0625 and 5.0625. The latter leaves the plotting range and is reported as Outside plot. This gradient iteration illustrates feedback stability limits; it is not an op-amp loop, Bode plot, phase-margin measurement, or slew-rate model.

ObjectiveDiscrete iterate

Position x0: 2
Objective x²/2: 2
Iteration behavior: Stable step

Step size η0.3Initial position x₀2

n = 0 · t = 0 s

Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA stronger correction does not guarantee a stable approach to the target.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

A sinusoid requires maximum slope 2πfVpk. Slew rate is a separate large-signal constraint that can fail before the small-signal magnitude response becomes limiting.

Your turn

Try it yourself

A separate hypothetical amplifier has available slew rate 0.50 V/µs. It is asked to produce a sinusoidal output of peak amplitude 2.0 V at 40 kHz. Compute the required maximum slope and the largest frequency allowed by the slew-rate bound. Ignore other limitations only for this calculation.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Convert the available slew rate to V/s and solve 2πfVpk≤SR. Passing this bound alone would not certify bandwidth or output-swing performance.

Compare with the solution

The required slope is $2\pi(40000)(2)=502654.8$ V/s=0.502655 V/µs, slightly above 0.50 V/µs. The slew-rate frequency ceiling is $500000/(2\pi\cdot2)\simeq39788.7$ Hz. Thus 40 kHz violates this idealized slope requirement. Even a frequency below 39.789 kHz would still need adequate small-signal response, stability, input range, and output headroom before the actual amplifier could be considered suitable.

Keep the idea

Take it with you

Finite loop gain explains both closed-loop accuracy and the ideal-feedback limit.
Noise gain and the actual loop response control voltage-feedback bandwidth and stability.
Slew rate and saturation are separate from small-signal bandwidth.

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