--- title: "Practical GBP and amplifier output resistance" subject: "Advanced Electronics" chapter: 2 key: "semester_4/advanced_electronics:practical-gbp-output-resistance" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/practical-gbp-output-resistance site: BCT Engineering --- # Practical GBP and amplifier output resistance Advanced Electronics · Chapter 2: Operational Amplifier Characterization - Asked in: 2070 Asar, 2+5 marks - Video: [Op-Amp: Gain Bandwidth Product and Frequency Response](https://www.youtube.com/watch?v=wfkzz1rg-xk) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/practical-gbp-output-resistance - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-2 ## Question Explain gain-bandwidth product of a practical opamp. Determine the output resistance of an amplifier that uses a practical opamp. ## Short answer Gain-bandwidth product is constant: closed-loop gain × bandwidth = f_t. Closed-loop output resistance is R_of = R_o/(1+Aβ), where β = R1/(R1+Rf). For typical 741 (R_o=75 Ω, A=2e5, β=0.1), R_of ≈ 3.75 mΩ. ## Answer #### Gain-bandwidth product of a practical op-amp A practical (internally compensated) op-amp has a single dominant pole, so its open-loop gain is A(f) = A0/(1 + jf/fo). For the 741, A0 ≈ 2 × 10⁵ and fo ≈ 5 Hz. When negative feedback with factor β is applied: ```text Closed-loop gain: Af = A0/(1 + A0β) Bandwidth: fF = fo(1 + A0β) Af × fF = A0 × fo = ft (constant) 741: GBP = 2 × 10⁵ × 5 Hz = 1 MHz ``` So the product of closed-loop gain and bandwidth is constant and equal to the unity-gain frequency ft. Reducing gain by a factor increases bandwidth by the same factor (gain 10 → BW 100 kHz; gain 100 → BW 10 kHz). This lets a designer find the bandwidth for a given gain: BW = GBP/gain (using the noise gain 1 + Rf/R1). #### Output resistance of an amplifier using a practical op-amp Model the op-amp by its open-loop gain A, input resistance Ri (very large, ignored) and output resistance Ro. To find the closed-loop output resistance, set the input source to zero, apply a test voltage vx at the output and find the current ix drawn. ```text Rf +-----------/\/\/-----------+ | | v- --+-----(-) | | >--(A·vd)---Ro----+----o vx (test source, R1 +---(+) ix flows in) | | GND GND ``` ```text Feedback voltage at inverting input: v- = vx·R1/(R1 + Rf) = β·vx vd = v+ - v- = -β·vx Current through Ro: io = (vx - A·vd)/Ro = vx(1 + Aβ)/Ro Current through R1 + Rf: iF = vx/(R1 + Rf) (very small, neglected) ix ≈ io = vx(1 + Aβ)/Ro Rof = vx/ix = Ro / (1 + Aβ), β = R1/(R1 + Rf) ``` Because the output voltage is sampled and fed back (voltage feedback), the output resistance is divided by (1 + Aβ). Example: 741 with Ro = 75 Ω, A = 2 × 10⁵ and β = 0.1 gives Rof = 75/(1 + 2 × 10⁴) ≈ 3.75 mΩ, practically zero. The same result holds for the inverting and non-inverting amplifiers, since with the input grounded both have the same feedback network. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a practical op-amp with feedback, which holds? A. Af x fF = A0 x fo = ft is constant B. Af/fF is constant C. Af + fF is constant D. fF is independent of Af Answer: A. Gain reduction by a factor raises bandwidth by that factor. (Key point: GBP statement) 2. With GBP 1 MHz and gain 100, the bandwidth is: A. 100 kHz B. 10 kHz C. 1 MHz D. 100 Hz Answer: B. BW = GBP/gain = 10^6/100. (Key point: Bandwidth from GBP) 3. The closed-loop output resistance is: A. Rof = Ro(1 + A*beta) B. Rof = Ro C. Rof = Ro/(1 + A*beta) D. Rof = Ro + A*beta Answer: C. Voltage feedback senses the output and divides Ro by (1 + A beta). (Key point: Output resistance result) 4. How is Rof found? A. Short the output and measure the gain B. Apply a test current at the input C. Open the feedback loop D. Zero the input, apply test voltage vx at the output, and find ix Answer: D. Rof = vx/ix with io = vx(1 + A beta)/Ro. 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