Volume Resistivity of Conductive Conductor Shielding Volume resistivity is a critical property of conductive conductor shielding materials used in cables. It determines the ability of the material to resist electrical current flow, ensuring safety and performance in electrical systems.
1. Key Requirements - The volume resistivity of the finished extruded, tape, or extrusion-over-tape conductor shielding must not exceed 100,000 ohm-centimeters at the rated temperature of the insulation.
- The test should follow the procedures outlined in ASTM D 257 with specific modifications.
Why It Matters: Low volume resistivity in shielding ensures proper functionality of the conductor, minimal energy loss, and adherence to safety standards.
2. Sample Preparation for Testing - Four samples, each 10 inches (or 250 mm) long, must be taken from the completed cable or conductor shielding.
- The shielding is cut and removed from the conductor to create half-cylinder lengths.
- Potential electrodes are applied at least 2 inches (or 50 mm) apart, with current electrodes set appropriately.
Critical Notes: - Electrodes must ensure proper contact for accurate readings.
- Measurements should be repeated to verify consistency.
3. Volume Resistivity Formulas Volume resistivity can be calculated using the formulas provided, depending on whether dimensions are in inches or millimeters.
If dimensions are in inches:
P = (R * (D² - d²)) / L
If dimensions are in millimeters:
P = 0.039 * (R * (D² - d²)) / L
Where: - P: Volume resistivity in ohm-centimeters.
- R: Measured resistance in ohms.
- D: Diameter over the conductor shielding in inches or millimeters.
- d: Diameter over the conductor in inches or millimeters.
- L: Distance between potential electrodes in inches or millimeters.
4. Example Calculation Given: - Resistance (R): 5 ohms
- Outer diameter of shielding (D): 20 mm
- Inner diameter (d): 10 mm
- Electrode distance (L): 50 mm
Using the formula for millimeters:
P = 0.039 * (5 * ((20²) - (10²))) / 50
P = 0.039 * (5 * (400 - 100)) / 50
P = 0.039 * (5 * 300) / 50
P = 0.039 * 1500 / 50
P = 0.039 * 30
P = 1.17 ohm-centimeters
Interpretation: The volume resistivity of this conductor shielding is 1.17 ohm-centimeters, which is well within the acceptable range for many applications.
5. Testing Parameters - Power in the test circuit should not exceed 100 mW.
- Tests should be conducted at the rated temperature of the insulation.
- Both alternating and direct-current voltage methods are acceptable.
Pro Tip: Ensure consistent temperature conditions during testing to avoid measurement errors.
Conclusion Accurately measuring volume resistivity is crucial for maintaining cable performance and compliance with industry standards. By following the prescribed testing methods and using the correct formulas, manufacturers can ensure their products meet safety and functionality requirements.