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

The figure reads from left to right. The blue card is rtd excitation current in milliamps. The middle card applies this page's relationship. The green card is pt100 resolution. Walk the arrows once: 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.

RTD excitation current in milliamps changes pt100 resolution An input card leads through the page relationship to the pt100 resolution result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. The widget uses the same V=IR, count/sensitivity, and I²R formulas derived above.

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.

Electrical power is not temperature error until multiplied by the installed sensor's thermal resistance.

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.

These are the chapter inputs, worked results, and named teaching assumptions.

3.3 V reference
Voltage or voltage-step value
4,096 levels
Sensor scale, pressure, or digital result
1.65 V midpoint
Voltage or voltage-step value
Pt100/Pt1000 slopes
Chapter input or worked result
1.00 mA limit
Current or responsivity value
2.09
Chapter input or worked result
0.209 C/count
Sensor scale, pressure, or digital result
0.100
Chapter input or worked result
1.00 mW reproduce the chapter's ideal examples
Named teaching assumption

Installed self-heating requires a measured thermal path.

Return to the chapter with the maths unpacked →