Math Bridge: RTD Resolution and Self-heating

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Math BridgeSensorsStruggle-friendly runway

Why excitation current helps and heats an RTD

One thread from platinum's resistance slope through ADC count size to the chapter's self-heating limit.

Phoebe, the physics guidePhoebe guides
The one targetConnect current, resistance slope, ADC resolution, and heat.
The chapter casePt100/Pt1000, 1.00 mA, 3.3 V, 12 bits.
What it buys youSee the sensitivity-versus-heating trade-off.

A field team faces an unresolved physical question: Why excitation current helps and heats an RTD They must answer it before changing rtd excitation current in milliamps on the real device. Predict the direction first.

See the relationship before changing it

Set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for rtd excitation current in milliamps is 1.

  2. 2

    Name the relationship. S_V=I(dR/dT); ΔT_count=q/S_V; P=I²R

  3. 3

    Substitute the chapter fixture. Set rtd excitation current in milliamps to 1. The page ledger gives pt100 resolution as 2.093 degrees C.

  4. 4

    Read the result. Keep degrees C beside the value. Use it only inside the technical boundary on this page.

Predict, then change rtd excitation current in milliamps

Try Predict the direction of pt100 resolution. Move one control, calculate, then check your prediction.

1
Chapter baseline
Pt100 resolution

Observe The widget uses the same V=IR, count/sensitivity, and I²R formulas derived above. Reset the control to 1 and compare pt100 resolution.

Explain Only rtd excitation current in milliamps moves here. The other chapter fixtures remain fixed.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only rtd excitation current in milliamps moves. Field effects named in the page's technical boundary stay fixed.

Honesty note. This teaching model changes only rtd excitation current in milliamps. It replays the chapter fixture; it is not a field prediction.

1. Temperature changes resistance

Over a modest span, a platinum resistance temperature detector (RTD) follows R(T)=R₀(1+αT). A Pt100 starts at 100 ohm at 0 C and changes by 100×0.00385 = 0.385 ohm/C; a Pt1000 changes by 3.85 ohm/C.

2. Current turns resistance into voltage

Ohm's law is V=IR. A 1.00 mA current turns 0.385 ohm/C into 0.385 mV/C for Pt100, and 3.85 ohm/C into 3.85 mV/C for Pt1000.

3. Compare that slope with one ADC count

1

Find a converter binq = 3.3 V/4,096 = 0.000806 V = 0.806 mV.

2

Divide by Pt100 sensitivity0.806/0.385 = 2.09 C/count.

3

Divide by Pt1000 sensitivity0.806/3.85 = 0.209 C/count.

4. Try the excitation current

S_V=I(dR/dT); ΔT_count=q/S_V; P=I²R

TryRaise current toward the chapter's under-1 mA limit and watch sensitivity and self-heating move together.

Pt100 sensitivity
Pt100 resolution
Pt1000 sensitivity
Pt1000 resolution
Pt100 power
Pt1000 power

ObserveAt 1.00 mA, Pt100 resolves 2.09 C/count and dissipates 0.100 mW; Pt1000 resolves 0.209 C/count but dissipates 1.00 mW.

ExplainThe widget uses the same V=IR, count/sensitivity, and I²R formulas derived above.

Technical boundaries.

The electrical comparison fixes the converter step and resistance slopes while excitation current changes, treating heating as current squared times resistance. It does not convert that power into temperature error without installed thermal resistance, or include lead resistance, reference error, converter noise, and amplifier gain.

Lead resistance, reference error, ADC noise, and amplifier gain also matter
Needs separate evidence

Use field evidence or a deeper model before release.

5. Work the divider midpoint

A balanced thermistor divider at the chapter's 3.3 V reference produces 3.3/2 = 1.65 V. That check is separate from the RTD slope calculation: the ADC measures voltage, while firmware supplies the sensor model.

6. Name the trade-off

More excitation current makes each degree produce more voltage, but power rises with the square of current. Precision circuits therefore use stable current, amplification, low-noise conversion, and a current low enough not to warm the sensing element.

7. Check yourself

What is one 12-bit count at 3.3 V?
Answer: 3.3/4,096 = 0.000806 V = 0.806 mV.
Why is Pt1000 finer before amplification?
Answer: At the same current its resistance slope produces ten times more millivolts per degree.
Why not keep raising the current?
Answer: Self-heating power grows as I²R and can warm the sensor being measured.
Honesty boundary.

The electrical comparison fixes the converter step and resistance slopes while excitation current changes, treating heating as current squared times resistance. It does not convert that power into temperature error without installed thermal resistance, or include lead resistance, reference error, converter noise, and amplifier gain.

3.3 V reference
Analog-to-digital converter voltage reference
4,096 levels
Twelve-bit converter code count
1.65 V midpoint
Midscale input voltage
Pt100/Pt1000 slopes
Resistance change per degree for compared platinum sensors
1.00 mA limit
Example excitation-current ceiling
2.09
Pt100 temperature change per converter count
0.209 C/count
Pt1000 temperature change per converter count
0.100
Pt100 electrical heating power in milliwatts
1.00 mW reproduce the chapter's ideal examples
Pt1000 electrical heating power

Installed self-heating requires a measured thermal path.

Return to the chapter with the maths unpacked →