02 · Foundational

Voltage measurement, input resistance & loading

A voltage reading is the loaded terminal voltage of the circuit-plus-meter system. Derive the exact loading relation, convert an allowed voltage error into a minimum input-resistance requirement, compare low- and high-source-resistance cases, understand range/manual dependence, and recognize when leakage or a simple resistor model becomes insufficient.

125–155 minFoundational
In this lesson
  • Derive the loaded voltage seen by a finite-resistance voltmeter from a Thévenin source model.
  • Derive the exact relative loading error and its useful high-input-resistance approximation.
  • Convert a maximum allowed loading error into a quantitative minimum requirement on meter input resistance.
  • Explain why the same meter can be effectively ideal for a low-resistance source yet strongly disturb a high-resistance source.
  • Use limiting cases and a two-meter comparison as independent checks of loading calculations.
  • Explain why input resistance is function/range/model dependent and must be taken from the actual instrument documentation rather than assumed.
  • Recover an unloaded source voltage from a loaded measurement when source resistance and meter input resistance are known, while recognizing the uncertainty cost of such correction.
  • Recognize the limits of the simple DC input-resistance model for extremely high source resistance, leakage-sensitive nodes and frequency-dependent input impedance.
Starting question

If two voltmeters are both “high impedance,” can they still give meaningfully different DC readings on the same node?

Yes. A voltage measurement is not determined by the meter alone. It is set by the ratio between the source resistance seen at the port and the meter input resistance. This chapter turns that qualitative idea into an engineering specification.

02.1A voltmeter measures a voltage difference while loading the two-terminal port

Voltage is defined between two electrical points. In DC voltage mode the meter is connected across those two points, so its input becomes a branch between them. For a linear resistive source network, replace everything behind the port by its Thévenin equivalent: open-circuit voltage VTH and source resistance RTH.

The first useful DC model of the meter is a finite input resistance RIN. The meter therefore forms a divider with the source resistance.

Figure 2.1 — A high-resistance source makes loading visible

The Thévenin port is 2.50 V open circuit with 1 MΩ source resistance. A 10 MΩ meter input is no longer negligible; it becomes part of the divider and the reading falls to 2.273 V.

verified topology
Source resistance loaded by voltmeter input resistance+−VS2.5 VRS1 MΩRIN10 MΩVMEASGND
The one question is: what RIN/RTH ratio is required by your allowed loading error?

02.2Derive the exact loading relation instead of memorizing a rule

The same current flows through the source resistance and the meter input:

I = VTH/(RTH+RIN)

The voltage displayed by the idealized meter model is the voltage across RIN:

Vmeas=I RIN=VTH·RIN/(RTH+RIN)

Normalize by the open-circuit voltage:

Vmeas/VTH=RIN/(RTH+RIN)

For a passive positive source resistance and finite positive meter resistance, the ratio is below one. The meter therefore reads below the unloaded Thévenin voltage in this simple model.

02.3The exact loading-error formula and its approximation

Define the relative loading error with respect to the open-circuit voltage:

εL=(VTH−Vmeas)/VTH

Substitute the divider result:

εL=RTH/(RTH+RIN)

If RIN≫RTH, then the denominator is approximately RIN and

εL≈RTH/RIN   when   RIN≫RTH

The approximation is useful for mental estimates, but the exact relation should be used whenever the ratio is not very large.

RIN/RTHExact loading error
101/11 = 9.09%
1001/101 = 0.990%
10001/1001 = 0.0999%
10 0001/10 001 ≈ 0.0100%

02.4Turn an error budget into a meter-input requirement

Suppose the allowed loading error is εmax. Require

RTH/(RTH+RIN)≤εmax

Solving for the meter input resistance gives

RIN≥[(1−εmax)/εmax]RTH

This is more useful than the vague advice “use a high-impedance voltmeter.” For a source with RTH=20 kΩ:

Maximum loading errorMinimum RIN
1%99·20 kΩ = 1.98 MΩ
0.1%999·20 kΩ = 19.98 MΩ
0.01%9999·20 kΩ = 199.98 MΩ

The required instrument changes by two orders of magnitude when the allowed loading error changes from 1% to 0.01%.

02.5The same meter can be excellent on one source and poor on another

Assume a meter input resistance of 10 MΩ only as a numerical example—not as a universal DMM property. For a 5 V source:

Source resistance RTHReadingLoading error
100 Ω4.99995 V≈0.0010%
100 kΩ4.9505 V≈0.990%
1 MΩ4.5455 V9.09%
10 MΩ2.5000 V50%

There is no context-free statement that “10 MΩ is high.” What matters is RIN/RTH together with the required accuracy.

Limiting-case check

As RTH→0, Vmeas→VTH: an ideal zero-resistance voltage source is not loaded in voltage by any finite positive RIN. As RIN→∞, Vmeas→VTH. As RIN→0, Vmeas→0 in the model.

02.6Do not assume input resistance is fixed across all meters, functions or ranges

Instrument input behavior is a specification, not a universal constant. A particular meter may provide a nominal resistance on some DC voltage ranges, a much higher input resistance on other ranges or modes, or a deliberate low-impedance voltage mode intended to suppress ghost voltages. The exact implementation is model dependent.

Therefore a rigorous workflow is:

  1. Identify the exact instrument model.
  2. Identify the selected function and range.
  3. Read the input-resistance/input-impedance specification for that condition.
  4. Use the specification in the loading model.
  5. Record the function/range when the result depends materially on it.

The term input resistance is appropriate for the purely DC resistive model used here. Later, with time-varying signals, the more general input impedance can depend on frequency.

02.7You can correct known loading—but correction does not erase uncertainty

If RTH and RIN are known, invert the divider relation:

VTH=Vmeas(1+RTH/RIN)

Example: a measured value of 2.273 V with RTH=1.00 MΩ and RIN=10.0 MΩ gives

VTH≈2.273·1.1≈2.500 V

But the correction factor itself depends on the values used. If the source resistance is poorly known, or the meter input differs from its assumed value, the corrected result inherits that uncertainty. A large correction is therefore not equivalent to a good measurement.

A better engineering strategy is often to choose a measurement method that perturbs the source much less in the first place.

02.8When the simple RIN model stops being enough

The finite-resistance model is powerful, but it is not universal. It can become incomplete when:

  • the source resistance is so high that input leakage currents, connector contamination or board-surface leakage produce comparable voltage errors;
  • the measured quantity is changing fast enough that the instrument input can no longer be represented by one DC resistance;
  • the meter uses a special input mode whose behavior is not equivalent to the assumed simple resistance;
  • the source itself is nonlinear, time dependent or changes operating point when loaded.

At that point the model should become more detailed, not the algebra more confident. The rule is: use the simplest model that has been justified by the circuit and the instrument specification.

02.9Mastery exercises

  1. A 3.300 V source has RTH=2 kΩ and is measured by RIN=10 MΩ. Find Vmeas and loading error.
  2. A 2.50 V sensor has RTH=1 MΩ. Compare readings from 10 MΩ and 100 MΩ inputs.
  3. For RTH=47 kΩ, find the minimum RIN for loading errors below 1%, 0.1% and 0.01%.
  4. A meter reads 1.800 V from a source whose RTH=500 kΩ. If RIN=10 MΩ, estimate VTH by correcting loading.
  5. Two nominally accurate meters disagree on a high-resistance node. List three checks to make before declaring either meter defective.
  6. Why is an input-resistance value from one voltage range not automatically valid for every other range?
  7. A node has RTH=20 MΩ and the available meter has RIN=10 MΩ. Is numerical correction alone a satisfactory strategy? Explain.
  8. Derive the approximation εL≈RTH/RIN from the exact expression and state its required condition.
Solutions

1. Vmeas=3.300·10 MΩ/(10 MΩ+2 kΩ)=3.29934 V; εL≈0.019996%.

2. With 10 MΩ: 2.273 V and 9.09% error. With 100 MΩ: 2.475 V and 0.990% error.

3. 1%: 4.653 MΩ; 0.1%: 46.953 MΩ; 0.01%: 469.953 MΩ.

4. VTH=1.800·(1+0.5/10)=1.890 V.

5. Check each meter’s actual RIN in the selected range, confirm the same two nodes/reference and connection, and estimate loading from the source resistance before comparing accuracy specifications.

6. Input circuitry can change with function/range/mode; only the manufacturer’s specification for the selected condition is authoritative.

7. RIN/RTH=0.5, so the meter strongly disturbs the node: Vmeas/VTH=1/3. A large correction would depend strongly on imperfectly known source and input parameters. Use a less invasive method if possible.

8. Divide numerator and denominator by RIN: εL=(RTH/RIN)/(1+RTH/RIN). If RTH/RIN≪1, the denominator approaches one.

Chapter 03 will move to current measurement and treat shunts, burden voltage, range choice, fuse/protection consequences and measurement-induced current error at the same depth.