Unit 4 — Calibration Procedure and Calculations
TM-CAL-012 — Open Handout TM Chapters: Chapter 5, Appendix A ELOs: Execute calibration procedure; compute error and determine pass/fail Estimated time: 30 minutes
Step 1: Read the TM
Open TM-CAL-012. Read Chapter 5 and Appendix A completely.
Then come back here.
Chapter 5 — Calibration Procedure
- Connect resonator BNC to NanoVNA PORT 1. Set NanoVNA for |S11| display, centered on expected frequency, 2 MHz span.
- Identify the |S11| dip minimum: this is resonant frequency.
- Compare to expected: error% = (fmeas − fcalc) / fcalc × 100%.
- If frequency is low, capacitance is high or inductance high: reduce C or L. If high, increase C or L. Capacitor substitution is easier than rewinding.
- Final frequency should be within ±0.1% of design target.
- Verify against WWV/GPS reference for absolute accuracy.
Appendix A — Formulas
Required L for target f and chosen C
L = 1 / ((2πf)2 × C)
Example: f = 7.000 MHz, C = 100 pF
L = 1 / ((2π×7×106)2 × 100×10−12) = 5.17 μH
T50-6 toroid inductance formula
L (μH) = N2 × AL / 106, AL = 3.0 nH/turn2 for T50-6
Key Formulas Summary
- L = 1 / ((2πf)2 × C)
- Example: f = 7.000 MHz, C = 100 pF
- L = 1 / ((2π×7×106)2 × 100×10−12) = 5.17 μH
- L (μH) = N2 × AL / 106, AL = 3.0 nH/turn2 for T50-6
The Calibration Procedure
Chapter 5 specifies 6 calibration steps.
Calibration is a comparison: you apply a known reference value to the instrument under test and record what the instrument reads. The difference is the error. You then either: - Adjust the instrument until the error is within the acceptance criterion, or - Record the error as a correction factor to apply to future readings
An error criterion found in Chapter 5: 100%. Confirm the exact criterion in the TM.
Practice Problems
Work these before checking answers.
P4-1. The reference value is 10.000 V. Your instrument reads 10.043 V. (a) What is the error in volts? (b) What is the error as a percentage of the reference value? Show your work.
P4-2. Using the formula for % error: error% = (measured − reference) / reference × 100 Apply it to: reference = 100.0 kHz, measured = 99,985 Hz. (a) Error in Hz. (b) Error in %. (c) Error in ppm.
P4-3. The acceptance criterion in the TM is ±1%. Your measurement gives an error of +0.8%. Does it pass? State your reasoning.
Practice Problem Answers
P4-1. (a) 10.043 − 10.000 = +0.043 V (b) 0.043 / 10.000 × 100 = +0.43%
P4-2. (a) 99,985 − 100,000 = −15 Hz (b) −15 / 100,000 × 100 = −0.015% (c) −15 / 100,000 × 1,000,000 = −150 ppm
P4-3. +0.8% is within ±1%. PASS. State: "error = +0.8%; criterion = ±1%; result = PASS." Always cite the TM section for the criterion.
Self-Check Questions
SC4-1. How many steps does Chapter 5 specify for the calibration procedure?
SC4-2. What reference value(s) does Chapter 5 apply to the instrument under test?
SC4-3. State the calibration acceptance criterion from the TM. Cite the section.
SC4-4. Write the error formula from Appendix A. Include units.
SC4-5. If the instrument reads 2.3% high, is the error positive or negative? What does a positive error indicate?
Answer Key
SC4-1. Count the numbered steps in Chapter 5. (TM Ch. 5)
SC4-2. See Chapter 5 for the reference values applied. These are the known-good inputs used to check the instrument. (TM §5-1)
SC4-3. See TM Chapter 5 or Chapter 7. Copy the criterion exactly with units and section number.
SC4-4. See Appendix A. Write it exactly as shown.
SC4-5. Positive (instrument reads higher than reference). A positive error means the instrument over-reads — it reports a value higher than the true value. (TM App. A)
Checkpoint
Before proceeding, you must be able to: - State the calibration acceptance criterion without looking - Write the error formula from memory - Work a % error and ppm calculation correctly
→ Proceed to Unit 5