Module 2: Measurement & Instrumentation
Probe choice, CMRR, the bandwidth rule, and current/thermal instruments.
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1conceptYou must measure the voltage across a high-side current-sense resistor that floats well above ground. Which probing approach is correct?
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Two single-ended probes referenced to scope ground subtract two large numbers to get a tiny one: common-mode error and scope-ground loading swamp the small sense voltage.
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Correct. Neither node is at ground, so you need a true differential measurement that responds only to the difference and rejects the large common-mode level both nodes share.
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Clipping the scope ground to a node that floats above true ground shorts that node to earth through the probe, a fault, not a measurement.
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A current probe reads current, not the voltage across the resistor; it cannot give you the sense voltage.
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2conceptWhy does a differential probe's common-mode rejection ratio (CMRR) get worse as signal frequency rises?
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Thermal drift is a slow DC-ish effect, not the mechanism behind frequency-dependent CMRR.
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Correct. CMRR depends on the two paths being identical; tiny imbalances in stray capacitance and trace length create a frequency-dependent phase/amplitude difference, so common-mode signals stop cancelling and CMRR rolls off with frequency.
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Differential gain is set by the probe, not by frequency; CMRR degrades because of path imbalance, not rising gain.
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CMRR is explicitly frequency-dependent: it is specified as a curve and falls off at high frequency.
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3calcA digital edge has a rise time of 1 ns. Using BW ≥ 0.35 / t_rise for the knee and a ~5× margin, what scope bandwidth should you target?
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70 MHz is far below even the knee frequency for a 1 ns edge: it would badly round the edge.
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350 MHz (0.35 / 1 ns) is just the knee frequency itself; measuring the edge faithfully needs margin above it.
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Correct. Knee = 0.35 / 1 ns = 350 MHz; picking ~5× the knee gives roughly 1.75 GHz so the scope captures the harmonics that shape the edge.
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3.5 GHz is ~10× the knee, more than the rule asks for; ~5× (1.75 GHz) is the target.
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4calcUsing the same rule, what is the fastest rise time a 100 MHz-bandwidth scope can faithfully display (its own rise-time limit)?
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0.35 ns would require ~1 GHz of bandwidth (0.35 / 1 GHz), far more than 100 MHz.
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Correct. Rearranging BW = 0.35 / t_rise gives t_rise = 0.35 / 100 MHz = 3.5 ns; edges faster than this are limited by the scope, not the signal.
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35 ns corresponds to a ~10 MHz scope, an order of magnitude too slow.
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350 ns corresponds to a ~1 MHz scope, far slower than 100 MHz.
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5scenarioProbing a fast GPIO edge with a 6-inch ground-lead clip, you see heavy ringing/overshoot that disappears when you touch a finger nearby. What is the fix?
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Ring that changes when you touch the lead is a probing artifact, not a property of the net; redesigning the driver chases a ghost.
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Correct. The 6-inch ground lead's inductance resonates with the probe-tip capacitance, producing ringing on fast edges; a short ground spring (tip-and-barrel) collapses that loop and kills the artifact.
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Bandwidth-limiting hides the symptom by throwing away the very edge information you came to measure: it does not fix the loading.
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Loading the real net with a big cap to mask a measurement artifact corrupts the circuit you are trying to characterize.
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