Two Meters, One High-Resistance Node: A Lesson in Loading Error
A student must verify the bias of a sensor front end: a 470 kΩ resistor from a 10.00 V regulated rail to a node, and a 470 kΩ resistor from that node to ground, so the expected node voltage is 5.00 V. The only meter available is an analog multimeter marked 20 kΩ/V DC sensitivity. He selects the 10 V range and reads 2.27 V. Convinced the circuit is faulty, he borrows a digital multimeter with a 10 MΩ input impedance and reads 4.90 V on the same node. He then uses the DMM's AC range to check a 5 V peak-to-peak, 50 kHz square wave from a function generator and gets 2.1 V, although he expected about 2.5–2.8 V. The DMM manual says the AC range is specified from 40 Hz to 1 kHz with a maximum crest factor of 3. The lab has a 20 MHz oscilloscope with a switchable 1×/10× probe.
1. Reconcile the two DC readings with a calculation, and show how the true voltage could have been recovered from the analog meter reading alone.
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What the analog meter really is. At 20 kΩ/V on the 10 V range, its input resistance is 20 kΩ/V × 10 V = 200 kΩ, not infinity.
Thevenin equivalent of the divider: Vth = 10 × 470/(470 + 470) = 5.00 V, Rth = 470 kΩ || 470 kΩ = 235 kΩ.
Reading with the analog meter: the meter loads the node with 200 kΩ in parallel with the lower 470 kΩ (giving 140.3 kΩ): V = 10 × 140.3/(470 + 140.3) = 2.30 V, matching the observed 2.27 V.
Reading with the DMM: 10 MΩ || 470 kΩ = 448.7 kΩ, so V = 10 × 448.7/(470 + 448.7) = 4.88 V, matching 4.90 V. The error is 2.4 % - acceptable.
Recovering the truth from the bad reading. The correction is Vtrue = Vmeasured(1 + Rth/Rmeter) = 2.27 × (1 + 235/200) = 2.27 × 2.175 ≈ 4.94 V. So the circuit was never faulty; the measuring instrument was part of the circuit. Moving to the 50 V range (1 MΩ) would have read about 4.04 V - better but still 19 % low. Rule: keep Rmeter at least 20× Rth; here that means 4.7 MΩ or more, so only the DMM qualifies.
2. Explain the 2.1 V reading on the square wave and describe the correct measurement with the available equipment.
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What the DMM should read. A 5 V peak-to-peak square with zero mean is ±2.5 V. Its true RMS is 2.5 V. An average-responding meter calibrated for sine waves would show 1.11 × 2.5 ≈ 2.78 V, and a true-RMS meter would show 2.5 V.
Why 2.1 V was read. The manual specifies the AC range only to 1 kHz. At 50 kHz the input attenuator and the RMS converter are rolling off, so the reading is low by roughly 15–25 %. Two further traps: the crest-factor limit of 3 is not the problem for a square wave (crest factor 1), but the meter's coupling capacitor and a slow converter can also cause an offset-dependent error when the waveform has a DC component. The reading of 2.1 V is therefore an instrument artifact, not a property of the signal.
Correct measurement. Use the oscilloscope: set the probe to 10× (which also raises the input impedance to 10 MΩ and lowers the loading capacitance), connect to the calibrator output at 1 kHz and adjust the probe's compensation trimmer until the square corners are flat, then measure the signal. With 1 V/div and a 10 µs/div time base (50 kHz = 20 µs period) the amplitude can be read directly. Check the scope's bandwidth (20 MHz) against the signal: at 50 kHz a square wave's 9th harmonic is 450 kHz, well within range, so the corner rounding will be small but visible - mention this rather than claiming a perfect square.
3. State the general rules the student should follow and one case where a 10× scope probe itself would still disturb the circuit.
Read the worked answer
Rules for measuring high-impedance nodes.
- Estimate the source resistance first (here 235 kΩ from the parallel pair) and choose a meter whose input resistance is at least 20× that, ideally 100×.
- Prefer a digital or FET-input instrument; treat analog meters as resistors equal to (sensitivity in Ω/V) × (range voltage).
- If only a loading meter is available, either compute the correction from Vtrue = Vm(1 + Rth/Rm) or measure with the divider loaded the same way both times - but never compare an unloaded calculation with a loaded reading.
- For AC, check the meter's specified frequency range and crest factor before believing any number; a low reading at high frequency usually means roll-off, not a fault.
Where a 10× probe still loads. A 10× probe presents about 10 MΩ in parallel with 10–15 pF. At DC that is fine even for a 235 kΩ node (2 % error). But at 1 MHz, 15 pF is 1/(2π × 10⁶ × 15 p) ≈ 10.6 kΩ, which would swamp a 235 kΩ node completely. So a probe's DC resistance tells only half the story: for high-impedance nodes carrying fast signals, the capacitance and the resulting time constant (here 235 kΩ × 15 pF ≈ 3.5 µs) limit the usable bandwidth, and an active probe or a buffer amplifier is required.