A laser-pulse energy meter uses a photodiode of responsivity and capacitance

PHYSICS · Multi-part · Fermi University Physics · Electronics, Instrumentation & Communication · Feedback, operational amplifiers and analogue signal processing · Sensors, detectors and experimental instrumentation

Electronics for Measurement, Ch The operational amplifier and negative feedback

Problem

A laser-pulse energy meter uses a photodiode of responsivity 0.55 A W−10.55\,\mathrm{A\,W^{-1}} and capacitance 100 pF100\,\mathrm{pF}, wired directly into the inverting input of an operational amplifier whose own input capacitance is 5 pF5\,\mathrm{pF} and whose gain-bandwidth product is fT=1 MHzf_T = 1\,\mathrm{MHz}. The design target is a transimpedance sensitivity of 0.6 V0.6\,\mathrm{V} per microwatt of optical power, averaged over a pulse. ((a)) Choose the feedback resistor RfR_f for the design sensitivity, round it to the nearest E12 preferred value, and state the sensitivity it actually gives. Then use the rule Cf≈Cin/(2πRffT)C_f \approx \sqrt{C_{\rm in}/(2\pi R_f f_T)}, with CinC_{\rm in} the total capacitance at the input, to choose the feedback capacitor (again the nearest E12 value) and compute the −3 dB-3\,\mathrm{dB} bandwidth this leaves the stage. ((b)) The transimpedance output feeds an operational-amplifier integrator (input resistor R1R_1, capacitor CC, no resistor across CC) that integrates over a laser pulse of width 50 μs50\,\mu\mathrm{s} to give a reading proportional to pulse energy. Choose R1=100 kΩR_1 = 100\,\mathrm{k}\Omega (fixing the integrator's input resistance at that value) and choose CC, to the nearest E12 preferred value, so that the time constant R1CR_1C equals the pulse width; state the actual time constant achieved and its percentage deviation from 50 μs50\,\mu\mathrm{s}.

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