DMM reads 4.09 V across a 2.2 MΩ divider
A student measures the gate bias of a MOSFET preamplifier. The divider is R1 = 2.2 MΩ from +9.0 V to node X, and R2 = 2.2 MΩ from X to ground. The calculated no-load voltage is 4.5 V. A digital multimeter on the 20 V DC range reads 4.09 V. The same meter reads 9.00 V at the supply and 0.00 V at ground. The meter manual states a 10 MΩ input resistance on all DC voltage ranges. The student replaces both resistors with new 1% types and obtains 4.09 V again. An older analogue meter on its 10 V range reads 3.1 V. The circuit is fed from a 9 V adapter, not a battery. The student suspects the MOSFET gate is leaky and wants to replace it.
1. Calculate the voltage expected with the 10 MΩ meter connected. Does the measurement indicate a faulty divider or leaky MOSFET?
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The meter's 10 MΩ input is in parallel with R2. R2 || 10 MΩ = (2.2 × 10) / (2.2 + 10) = 22 / 12.2 ≈ 1.803 MΩ. The total resistance from +9 V to ground is then 2.2 + 1.803 = 4.003 MΩ. The meter should read VX = 9.0 × 1.803 / 4.003 ≈ 4.05 V. The measured 4.09 V is within normal resistor tolerance and meter accuracy, so the divider is not faulty and there is no evidence of a leaky MOSFET. The true no-load voltage is 4.5 V; the meter itself pulls the node down by about 0.45 V. Replacing the resistors cannot change this because the loading is inside the meter.
2. Explain the loading error in terms of Thevenin resistance and meter current.
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The divider's Thevenin resistance is R1 || R2 = 1.1 MΩ. The meter draws I = 4.09 V / 10 MΩ ≈ 0.409 µA from node X. That current flows through the 1.1 MΩ Thevenin resistance and produces an additional drop of about 0.409 µA × 1.1 MΩ ≈ 0.45 V, which is exactly the difference between 4.5 V and 4.05 V. A gate leakage current of 100 nA would produce only 100 nA × 1.1 MΩ = 0.11 V, much less than the meter's error. The dominant cause is therefore the 10 MΩ meter input, not the MOSFET.
3. Recommend a measurement method that avoids loading the high-resistance divider. Why does the analogue meter read even lower?
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Use a FET-input DMM, an electrometer, or a unity-gain buffer with very high input resistance; alternatively, measure the 9 V supply and the resistor values, then calculate VX, or use a meter with input resistance at least 100 times the divider resistance. The analogue meter reads lower because its input resistance is much lower than 10 MΩ; it forms a heavier parallel path with R2. For example, a 2 MΩ input would already reduce the reading substantially, and a typical 20 kΩ/V meter on the 10 V range is only 200 kΩ, which would load the divider severely. The exact reading depends on the meter's sensitivity, but the direction is clear: lower meter resistance means a larger loading error.