Loaded Voltage Divider and a Low-Sensitivity Voltmeter
In a first-year lab a student must derive 6.0 V from a regulated 12.0 V supply to feed a sensor whose input resistance is 100 kΩ. The student builds a divider with R1 = 10 kΩ from +12 V to the output node and R2 = 10 kΩ from the output node to ground, both 5% carbon film. Before connecting the sensor, the student measures the output with an analog multimeter on its 10 V DC range. The meter is marked 1 kΩ/V and reads 4.0 V. The student replaces R2 with another 10 kΩ resistor and the reading remains 4.0 V. The supply is checked and is 12.0 V. The student concludes that the divider design is wrong. The instructor disagrees, saying the resistors are probably fine and that the measurement itself is part of the problem. The sensor will be connected after the check. The student has access to a digital multimeter with a 10 MΩ input resistance and to an op-amp unity-gain buffer.
1. Calculate the unloaded divider output and the output that the analog meter should produce. Use these numbers to explain the 4.0 V reading.
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Unloaded, the divider obeys the voltage-divider relation:
Vout = 12 V × R2 / (R1 + R2) = 12 × 10 kΩ / 20 kΩ = 6.0 V.
The analog meter on its 10 V range has an input resistance of 1 kΩ/V × 10 V = 10 kΩ. When connected, it appears in parallel with R2:
R2 || Rmeter = 10 kΩ || 10 kΩ = 5.0 kΩ.
Vout = 12 × 5.0 / (10 + 5.0) = 4.0 V.
The meter therefore explains the reading exactly. The resistors are not necessarily faulty; the measurement has changed the circuit. Replacing R2 with another 10 kΩ resistor cannot remove the loading effect, so the reading staying at 4.0 V is expected.
2. Will the 100 kΩ sensor load the divider significantly? Calculate the output with the sensor connected and state whether the buffer is needed for a 5% error budget.
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With the 100 kΩ sensor connected from the output node to ground, it is in parallel with R2:
R2 || Rsensor = 10 kΩ || 100 kΩ = (10 × 100)/(110) kΩ = 9.09 kΩ.
Vout = 12 × 9.09 / (10 + 9.09) = 5.71 V.
The error from the ideal 6.0 V is 6.0 − 5.71 = 0.29 V, or about 4.8%. That is just inside a 5% error budget by itself, but the two 5% resistors can add further error. More importantly, the sensor current is not negligible compared with the divider current. A unity-gain op-amp buffer should be inserted between the divider and the sensor, or the divider should use much lower resistor values. The buffer is the better choice because it does not increase steady current drain.
3. Recommend two measurement changes and one circuit change that would make the 6.0 V node reliable, and justify each.
Read the worked answer
Measurement changes:
- Use the digital multimeter with a 10 MΩ input resistance. Loading then becomes 10 kΩ || 10 MΩ ≈ 9.99 kΩ, and the error is under 0.1%, so the reading represents the unloaded node.
- Measure the actual resistor values out of circuit or at least measure the supply and both resistors, then compute the divider output. This separates resistor tolerance from meter loading.
Circuit changes:
- Add a unity-gain op-amp buffer to the divider output. The op-amp input draws almost no current, so the 6.0 V node is not loaded by either the meter or the sensor.
- Alternatively, lower the divider resistors to 1 kΩ each, which reduces the loading error but draws 6 mA from the 12 V supply instead of 0.6 mA. The buffer is usually preferable when supply current matters.
The key judgement is that a high-resistance node cannot be checked with a low-resistance meter or loaded casually; the instrument and the following stage must both be considered part of the circuit.