04 · Intermediate · instrumentation

The theorem on the laboratory bench

Why a generator can appear to double its amplitude, and why a voltmeter becomes part of the circuit the moment you connect it.

75–105 minIntermediate · instrumentation

In this lesson

Model an instrument output or input as part of the circuit.
Explain 50 Ω generator amplitude conventions with a Thévenin source.
Quantify DMM/scope loading error instead of assuming an ideal measurement.
Recognize when the lumped resistance model must become ZTH(ω) and Zin(ω).

Starting question

Can a voltmeter, oscilloscope or signal generator change the quantity it is supposed to measure?

Yes. Measurement is an electrical interaction. A real source has a finite output impedance, and a real instrument has a finite input impedance. Connecting the two creates a new circuit. Thévenin turns this from an annoying laboratory surprise into a predictable calculation.

predict before connectinginstrument = electrical load

04.0Prerequisite recap — the three relations you need

This lesson is designed to stand on its own, but it assumes that a source can be represented locally by a Thévenin voltage and an output resistance. If that idea is unfamiliar, chapters 01–03 give the full derivation.

V_meas = V_TH · R_in/(R_TH + R_in)

I = V_TH/(R_TH + R_in)

relative loading error = 1 − V_meas/V_TH

04.1Measurement is not observation from outside the circuit

Suppose an unknown device presents a terminal voltage V_TH and output resistance R_TH. Before the instrument is connected, the open-circuit terminal voltage is V_TH. After connection, the instrument input resistance R_in draws current. The terminal voltage therefore changes.

V_meas/V_TH = R_in/(R_TH + R_in)

The phrase “high-impedance input” is therefore incomplete. High compared with what? A 10 MΩ meter is almost an open circuit for a 10 Ω source, but it creates noticeable error for a source resistance of several megohms.

04.2Worked example — why a generator amplitude can appear to double

Many laboratory generators use an output resistance close to 50 Ω and let the user select the load convention used for the displayed amplitude. Consider a generator configured so that a matched 50 Ω load should receive an amplitude of 1.00 V. The following divider model requires an open-circuit source amplitude of 2.00 V.

Figure 4.1 — 50 Ω source driving a matched 50 Ω input

The source model uses a 2.00 V open-circuit amplitude. Equal source and load resistances divide that amplitude by two.

source/load topology verified

open-circuit amplitude = 2.00 Vload amplitude = 1.00 V

V_L = 2.00 · 50/(50 + 50) = 1.00 V

Now replace the 50 Ω load by a 1 MΩ oscilloscope input while leaving the physical source resistance unchanged.

V_scope = 2.00 · 1,000,000/(1,000,000 + 50)

V_scope = 1.9999 V ≈ 2.00 V

Nothing “amplified” the signal. The generator was already producing almost 2 V at open circuit; the matched 50 Ω load reduced that by a factor of two. A high-impedance input removes most of that loading.

04.3Worked example — a DMM can under-read a high-resistance sensor

Assume a sensor is accurately represented at DC by V_TH = 2.50 V and R_TH = 1 MΩ. A voltmeter with an assumed input resistance of 10 MΩ is connected.

Figure 4.2 — A voltmeter is part of the circuit once connected

A 2.50 V sensor with 1 MΩ source resistance is measured by an assumed 10 MΩ voltmeter input. The meter becomes the load.

measurement-load topology verified

ideal open-circuit value = 2.50 Vmeasured value = 2.273 V

V_meas = 2.50 · 10/(1 + 10) = 2.273 V

fractional error = (2.50 − 2.273)/2.50 = 0.0909

loading error = 9.09%

The meter is not “wrong”. It is measuring the voltage of the circuit that exists after the meter has been connected. The error comes from using that loaded voltage as an estimate of the original open-circuit voltage.

04.4Turn an accuracy requirement into an input-impedance specification

A useful engineer does not only calculate the error after choosing an instrument. The calculation can be inverted: start with the maximum acceptable loading error and derive the minimum required input resistance.

let ε = 1 − V_meas/V_TH

ε = R_TH/(R_TH + R_in)

therefore R_in ≥ R_TH · (1 − ε)/ε

Example: a sensor has R_TH = 47 kΩ and the loading contribution to voltage error must stay below 0.25%, so ε = 0.0025.

R_in,min = 47 kΩ · 0.9975/0.0025

R_in,min = 18.753 MΩ

A nominal 10 MΩ input would therefore fail this particular error budget. This is a more useful conclusion than calling 10 MΩ “high impedance”.

Try before opening the answer

A source has RTH = 100 kΩ. What minimum input resistance keeps DC loading error below 1%?

Answer. Rin ≥ 100 kΩ × 0.99/0.01 = 9.9 MΩ.

04.5At higher frequency, input resistance is no longer enough

A real oscilloscope or probe input is not purely resistive. A useful first model is an input resistance in parallel with an input capacitance. For a hypothetical instrument with R_in = 1 MΩ and C_in = 20 pF,

Z_in(ω) = [1/R_in + jωC_in]⁻¹

V_meas(ω) = V_TH(ω) · Z_in(ω)/[Z_TH(ω) + Z_in(ω)]

Take a source whose output impedance is approximately resistive with R_TH = 100 kΩ. At DC, the 1 MΩ input gives a voltage ratio of 0.909. At 100 kHz, the 20 pF capacitance can no longer be ignored.

|X_C| = 1/(2πfC_in) = 79.6 kΩ

Z_in ≈ 6.29 kΩ − j79.08 kΩ

|V_meas/V_TH| ≈ 0.599

phase ≈ −48.8°

The same input that seemed “high impedance” at DC now attenuates the signal strongly and introduces phase shift. The correct concept is therefore input impedance as a function of frequency, not one resistance number.

04.6A bench workflow: predict first, connect second

Identify the source model. Estimate or obtain the source output resistance/impedance and the expected open-circuit amplitude.
Identify the instrument input. Check whether the input is 50 Ω, high-Z, 1 MΩ, 10 MΩ, or a frequency-dependent probe specification.
Write the expected loaded value before connecting. Even an order-of-magnitude prediction catches many setup mistakes.
Check amplitude convention and termination settings. A generator’s “50 Ω” or “High-Z” display setting changes how its displayed amplitude is interpreted; it does not magically remove physical output impedance.
Check safe limits. Never enable a 50 Ω termination or connect an unknown source without verifying voltage, power and current limits for the actual instrument.
Record the configuration. Frequency, amplitude convention, termination, probe mode, cable and instrument input belong in reproducible measurement notes.

04.7Troubleshooting: four common measurement mistakes

ObservationLikely reasoning mistakeQuestion to ask

Scope reads almost twice the generator settingGenerator display assumes a matched load while the scope is high-Z.What load convention is the generator using?

DMM reads below a sensor’s expected voltageMeter input resistance is not large compared with source resistance.What is Rin/RTH?

DC measurement is correct but AC amplitude falls with frequencyInput capacitance or source impedance was ignored.What is Zin(ω)?

Changing termination changes the waveform stronglyThe termination is part of the electrical network.What circuit exists after I press the termination button?

04.8Independent practice — from calculation to diagnosis

Problem A — direct calculation

A 5.00 V source has RTH = 200 kΩ. It is measured with a 10 MΩ voltmeter. Find the reading and loading error.

Solution. Vmeas = 5 × 10/(10 + 0.2) = 4.902 V. The loading error is approximately 1.96%.

Problem B — design from an error budget

A sensor has RTH = 220 kΩ. Specify the minimum input resistance required for loading error below 0.5%.

Solution. Rin ≥ 220 kΩ × 0.995/0.005 = 43.78 MΩ.

Problem C — diagnosis before algebra

A generator set for a 1 V matched-load amplitude gives almost 2 V on a high-Z scope, but about 1 V after 50 Ω termination is enabled. Is the generator defective?

Solution path. No defect is required to explain the result. Model the generator as approximately 2 V open circuit in series with 50 Ω: high-Z approaches open circuit, while 50 Ω termination forms a 1:1 divider. Confirm the actual generator’s load-display convention in its manual.

If these problems are difficult because the source model itself is unclear, revisit E-002 before moving to transistor loading in chapter 05.

Review E-002

04.9What you should now be able to explain

Why connecting a measuring instrument changes the circuit being measured.
Why a 50 Ω source can produce almost twice the matched-load amplitude when observed with high impedance.
How to calculate loading error and convert an error budget into a minimum input resistance.
Why “10 MΩ input” is incomplete at higher frequency when capacitance matters.
Why termination, probe mode and cable/instrument configuration must be recorded as part of reproducible evidence.