Analog Meter Loading on a High-Impedance Sensor
A technician is troubleshooting a gas-sensor module whose output should be 4.80 V DC at the sensor amplifier test point. The test point is fed from an op-amp output through a 100 kΩ series resistor. The technician uses an analog multimeter marked 20,000 Ω/V on the 10 V DC range and reads 3.20 V. A calibrated digital multimeter with a 10 MΩ input reads 4.75 V at the same point. The analog meter also reads 8.60 V on a fresh 9.00 V battery on the same range. The workshop temperature is 30 °C. The sensor module is powered from a 12 V adapter, and the ground clip is connected to the module ground. The technician concludes that the sensor amplifier is faulty and plans to replace it. The sensor data sheet specifies a maximum output source resistance of 120 kΩ and a load impedance of at least 1 MΩ.
1. Calculate the analog meter's input resistance on the 10 V range and use the divider to estimate the true open-circuit voltage.
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The meter sensitivity is 20,000 Ω/V, so on the 10 V range its input resistance is 20,000 × 10 = 200 kΩ. The test point has a 100 kΩ source resistance, so the meter forms a divider: Vmeter = Voc × Rm/(Rs + Rm). Substituting 3.20 V = Voc × 200k/(100k + 200k) gives Voc = 3.20 × 300/200 = 4.80 V. The DMM confirms this: with 10 MΩ input, its loading error is only 100k/(10M + 100k) ≈ 1%, so it should read about 4.75 V, exactly as measured. The sensor is not faulty; the analog meter is loading the high-impedance node. The technician should not replace the amplifier on this evidence.
2. Explain why the analog meter reads 8.60 V on a 9.00 V battery and whether that alone indicates a meter fault.
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A 9 V battery has a source resistance of only a few ohms, so the 200 kΩ meter causes negligible loading (well under 0.01%). The 0.40 V difference is therefore not loading; it is either meter calibration error, battery voltage lower than 9.00 V, or reading error. A typical analog meter may have accuracy of ±2.5% of full scale, which on the 10 V range is ±0.25 V. The observed error is 0.40 V, slightly outside that band, so the meter may be out of calibration, but the test is not conclusive because the battery may actually be 8.7–8.8 V. Compare the meter against a calibrated DMM on a low-impedance source before condemning it. The battery test does not explain the 3.20 V reading on the sensor.
3. Recommend the correct measurement practice and calculate the error if a 1 MΩ meter were used instead.
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Use a high-impedance DMM (10 MΩ or greater) for the sensor test point, or buffer the test point with an op-amp follower. With a 1 MΩ meter, Vmeter = 4.80 × 1M/(1M + 100k) = 4.36 V, an error of about 0.44 V or 9.1% low. That is better than the analog meter's 33% error but still unacceptable for a 4.80 V specification. The general rule is that meter input resistance should be at least 10 times the source resistance; here 1 MΩ is only 10 times 100 kΩ, so loading is still visible. Record the meter model and range on the job sheet so that a low reading is not mistaken for a faulty amplifier.