The divider that wasn't faulty
A 28 V DC bus feeds a reference divider in an avionics bay: R1 = 4.7 MΩ from the bus to the midpoint, R2 = 4.7 MΩ from the midpoint to ground, giving a nominal 14 V reference. A technician connects a calibrated digital multimeter (10 MΩ input resistance, DC volts range) across R2 and reads 11.3 V. He raises a defect report stating 'both resistors have drifted about 20 % low' and orders a matched replacement pair. A second technician repeats the measurement with the same instrument and gets the same 11.3 V. The divider output feeds nothing else - no load, no leakage path - and the board has been cleaned and dried. Bus voltage is confirmed at 28.0 V using the same meter.
1. Calculate what the meter should read if the resistors are good, and decide whether the replacement order is justified.
Read the worked answer
Meter loading, not drift.
The meter is not an ideal voltmeter; it is a 10 MΩ resistor in parallel with R2.
- R2 in parallel with the meter: R_p = (4.7 × 10)/(4.7 + 10) = 3.20 MΩ
- Divider output: V = 28 × 3.20/(4.7 + 3.20) = 28 × 0.405 = 11.3 V
This is exactly the reading obtained. The measurement is fully explained by the instrument, so there is no evidence of resistor drift and the replacement order should be cancelled.
Useful check: the Thevenin source resistance of this divider is 4.7 MΩ in parallel with 4.7 MΩ = 2.35 MΩ. A 10 MΩ meter is only ~4× that, nowhere near the 100:1 ratio needed for an error under 1 %.
2. What is the actual midpoint voltage, and how would you verify it in the workshop with the equipment normally available?
Read the worked answer
Actual voltage (no meter, no load): 14.0 V.
Verification options, in order of preference:
- Use a meter with input resistance ≥ 100 × 2.35 MΩ, i.e. ≥ 235 MΩ, or a buffered/FET-input probe. A 100 MΩ meter would read 28 × 4.489/(4.7 + 4.489) = 13.7 V, close enough to confirm the divider.
- Measure R1 and R2 with the board unpowered (they must be isolated - a parallel path would corrupt the reading).
- Load the midpoint with a known high resistance and back-calculate, though this is slower and less accurate.
Judgement point: a reading that is consistent with the meter's own resistance is a healthy circuit, not a fault. Always ask what the instrument itself does to the node before condemning a component.
3. A later modification requires the same divider to drive an ADC input of 1 MΩ. Predict the new midpoint voltage and state what this means for the design.
Read the worked answer
R2 in parallel with the 1 MΩ load: R_p = (4.7 × 1)/(4.7 + 1) = 0.825 MΩ
V = 28 × 0.825/(4.7 + 0.825) = 28 × 0.149 = 4.18 V
The reference collapses from 14 V to about 4.2 V - a 70 % error. The divider's source resistance is far too high to drive any practical load.
Corrective options:
- Insert a unity-gain buffer (op-amp voltage follower) with input resistance ≫ 4.7 MΩ and output resistance of a few ohms; this preserves both the 14 V reference and the ADC's required source impedance.
- Or scale the divider down, e.g. 47 kΩ/47 kΩ, giving the same 14 V unloaded and a 23.5 kΩ source resistance - loading and meter effects then become negligible, at the cost of ~0.3 mA continuous drain.
Judgement point: a divider is only a voltage source if its source resistance is small compared with the load. Checking that ratio is the first step in any 'wrong voltage' investigation.