core24 min studyLesson 954 in Physics

Impedance loading and signal conditioning

Calculate how an instrument changes a source voltage and how resistance, capacitance, and gain set the loaded measurement response.

Before you begin

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The question for today

What you’ll understand

Use a Thevenin source model to quantify voltage loading.
Include an input capacitance in the loaded transfer function.
Distinguish voltage sensing from impedance matching for power or transmission lines.

Connecting an instrument changes the circuit it observes. A finite input impedance draws current from the sensor and can reduce its output voltage, while cable and amplifier capacitance can introduce frequency-dependent loading. A measurement chain must therefore be analyzed as a connected system. The open-circuit sensor response and the amplifier's isolated voltage gain are not always enough. Signal conditioning aims to preserve the desired information, place it within the next stage's range, and control unwanted bandwidth without creating an unaccounted measurement disturbance.

The input impedance participates in the source circuit

Represent a linear sensor output near its operating point by a Thevenin voltage Vs in series with source impedance Zs. Connecting input impedance Zin produces a divider. For a low-frequency resistive model, a large load resistance compared with source resistance makes the loaded voltage approach the open-circuit value. The remaining fractional error can be calculated rather than dismissed merely because the instrument is called a voltmeter.

This voltage-sensing goal differs from maximum-power transfer into a resistive load, which uses equal source and load resistances and loses half the open-circuit voltage. High-frequency transmission-line termination addresses reflections and must include characteristic impedance and propagation delay. Neither practice implies that every low-frequency sensor should be connected to a matched resistance. The appropriate interface depends on whether the measurand is voltage, current, delivered power, or a traveling wave.

V_{\mathrm{in}}=V_s\frac{Z_{\mathrm{in}}}{Z_s+Z_{\mathrm{in}}},\qquad \frac{V_{\mathrm{in}}}{V_s}=\frac{R_L}{R_s+R_L},\qquad \epsilon_{\mathrm{load}}=1-\frac{V_{\mathrm{in}}}{V_s}=\frac{R_s}{R_s+R_L}

The second and third relations use positive real Rs and RL at low frequency, with Vs nonzero. The error is defined relative to open-circuit voltage. General complex impedances also change phase, and their loading cannot be summarized by this real error expression.↔ Scroll wide equations horizontally.

Capacitance makes the divider a dynamic response

An input often resembles a resistance in parallel with capacitance. The capacitance includes connector, cable, probe, and amplifier contributions. Combining that load with source resistance gives a low-pass response whose DC attenuation is still the resistive divider, but whose time constant contains the parallel combination of the two resistances. Treating the input resistance alone as a complete description misses this bandwidth limit.

A buffer can reduce loading on a high-impedance sensor and drive the following cable, but its own bias current, offset, input capacitance, noise, and stability remain relevant. A bias current through source resistance creates a DC error with a sign determined by current direction. A large output capacitor can affect amplifier stability. Conditioning therefore relocates and modifies error mechanisms; it does not make them disappear.

H_{\mathrm{load}}(s)=\frac{V_{\mathrm{in}}(s)}{V_s(s)}=\frac{K}{1+s\tau},\qquad K=\frac{R_L}{R_s+R_L},\qquad \tau=(R_s\parallel R_L)C,\qquad f_c=\frac1{2\pi\tau}

The model has Rs in series with an ideal source and RL parallel to C at the input node. Resistance is in ohms, C in farads, τ in seconds, and fc in hertz. Parasitic inductance, transmission delay, and frequency-dependent source behavior are excluded.↔ Scroll wide equations horizontally.

Gain and range have to be checked at every stage

Amplification can raise a small sensor voltage above downstream converter noise, but it also amplifies input noise and offsets. Both the wanted waveform and interfering voltages must fit inside the amplifier's linear input and output ranges. A later digital correction cannot reconstruct a waveform after the front end has clipped. Likewise, a filter that suppresses a large disturbance only after an overloaded stage cannot undo the distortion generated there.

A useful chain drawing marks impedances, gain signs, physical units, bandwidths, and expected signal and offset excursions. Each transfer should be the loaded response under its actual connection conditions. Multiplying isolated gains is justified only when those loading interactions are negligible or already incorporated. This discipline connects a circuit calculation to what the instrument actually receives, rather than to an unattainable open-circuit signal that exists only before the cable is attached.

Work it through

Account for a loaded voltage before amplification

A toy sensor has open-circuit voltage 100 mV and source resistance 10 kΩ. It drives a 90 kΩ input resistance at a frequency where capacitance is negligible. An ideal following stage has voltage gain twenty. Find the loaded input, source current, and output; compare with ignoring loading.

The divider factor is $90/(10+90)=0.90$ . Thus the amplifier receives 90 mV, not the 100 mV open-circuit value.
The source current is $100\ \mathrm{mV}/100\ \mathrm{k\Omega}=1$ µA. Its voltage drop across the 10 kΩ source resistance is 10 mV, consistent with the divider.
The output is 20×90 mV=1.80 V. Ignoring loading would predict 2.00 V, so the output is lower by 0.20 V, or ten percent of that unloaded prediction.

The result

The input is 90 mV, source current is 1 µA, and output is 1.80 V; loading creates a ten-percent attenuation in this resistive model.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

An interface factor changes a fixed normalized signal

For each fresh comparison, Reset and set BOTH Position x=1 and Coefficient a. Compare a=0.90 and a=1.00, predicting y=0.90 and 1.00. At this fixed x, interpret a only as the loaded/unloaded voltage factor. The instrument still plots y=ax²; it does not solve an impedance network, sweep frequency, or calculate cable capacitance.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA gain calculation must include the voltage that actually reaches the instrument.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

A large input resistance reduces voltage loading. Matching resistance instead serves other goals, such as resistive power transfer, and capacitance still needs separate consideration.

Your turn

Try it yourself

A different source has Rs=1 kΩ and drives RL=9 kΩ in parallel with C=100 nF. Find the DC loading factor, the time constant, and the pole frequency. At that pole, find the magnitude relative to the open-circuit source and its phase, using the exp(iωt) convention.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

First form the parallel resistance, then distinguish the DC loading factor K from the additional frequency attenuation.

Compare with the solution

The DC factor is K=9/(1+9)=0.90. The parallel resistance is 900 Ω, so τ=(900)(100×10⁻⁹)=90 µs and $f_c=1/(2\pi\tau)\simeq1768.39$ Hz. At fc, $|H|=0.90/\sqrt2\simeq0.63640$ and the phase is −45°. Thus the open-circuit source amplitude is reduced by both resistive loading and the pole. Using only a factor 1/√2 would omit the ten-percent DC loading already present.

Keep the idea

Take it with you

Finite input impedance changes the sensor circuit and can create both gain and phase errors.
Voltage sensing, maximum-power transfer, and transmission-line termination have different objectives.
Loaded transfers and linear-range checks are needed throughout the conditioning chain.

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