Shielding Effectiveness
SE = 20 × log₁₀(E_inc / E_trans)
Quick answer
This calculator solves SE = 20 × log₁₀(E_incident / E_transmitted) for Shielding Effectiveness. With Incident Field 0.5, Transmitted Field 1.1, the result is -6.848 dB. It solves in 3 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
Shielding effectiveness quantifies how well an enclosure or barrier attenuates electromagnetic fields, expressed in decibels. It is the ratio of the incident field strength to the field that passes through the shield. Higher SE values indicate better shielding. SE consists of three components: reflection loss (impedance mismatch at the shield surface), absorption loss (energy converted to heat as the wave passes through the material), and multiple reflection correction (significant only for thin shields at low frequencies).
Variables
- SE — Shielding effectiveness (dB)
- E_inc — Incident electric field strength (V/m or arbitrary units)
- E_trans — Transmitted electric field strength after shielding (same units)
Practical Notes
Typical SE requirements: consumer electronics 20-30 dB, industrial equipment 30-60 dB, military/medical 60-100 dB. In practice, SE is limited by apertures (ventilation holes, seams, cable penetrations) rather than the shield material itself. A 1 mm gap in an enclosure can reduce SE by 20-40 dB at GHz frequencies.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| SE | Shielding Effectiveness | dB | dB |
| E_inc | Incident Field | — | — |
| E_trans | Transmitted Field | — | — |
Step by step
- 1. What you know: Incident Field 0.5, Transmitted Field 1.1
- 2. The relation: SE = 20 × log₁₀(E_incident / E_transmitted)
- 3. Substitute: E_inc = 0.5, E_trans = 1.1
- 4. Result: Shielding Effectiveness = -6.848 dB
Worked examples
Minimum range
Given: Incident Field 0.025, Transmitted Field 0.055
Result: Shielding Effectiveness = -6.848 dB
Low range
Given: Incident Field 0.25, Transmitted Field 0.55
Result: Shielding Effectiveness = -6.848 dB
Typical
Given: Incident Field 0.5, Transmitted Field 1.1
Result: Shielding Effectiveness = -6.848 dB
High range
Given: Incident Field 0.8, Transmitted Field 1.76
Result: Shielding Effectiveness = -6.848 dB
Maximum range
Given: Incident Field 8, Transmitted Field 17.6
Result: Shielding Effectiveness = -6.848 dB
Solving for each variable
Solve for Shielding Effectiveness
Incident Field 0.025, Transmitted Field 0.055 → -6.848 dB
Solve for Incident Field
Shielding Effectiveness 0.3 dB, Transmitted Field 0.055 → 0.05693
Solve for Transmitted Field
Shielding Effectiveness 0.3 dB, Incident Field 0.025 → 0.02415
How Shielding Effectiveness changes with Incident Field
| Incident Field (input) | Shielding Effectiveness (result) |
|---|---|
| 0.05 | -26.85 dB |
| 0.125 | -18.89 dB |
| 0.25 | -12.87 dB |
| 0.375 | -9.347 dB |
| 0.5 | -6.848 dB |
| 0.75 | -3.327 dB |
| 1 | -0.8279 dB |
| 2 | 5.193 dB |
| 3.5 | 10.05 dB |
| 5 | 13.15 dB |
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<p><a href="https://vbatpower.com/formulas/shielding-effectiveness-field/">Shielding Effectiveness</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Shielding Effectiveness?
Use SE = 20 × log₁₀(E_incident / E_transmitted). With Incident Field 0.5, Transmitted Field 1.1, Shielding Effectiveness works out to -6.848 dB.
What is the formula for Shielding Effectiveness?
The formula is SE = 20 × log₁₀(E_incident / E_transmitted). Shielding effectiveness quantifies how well an enclosure or barrier attenuates electromagnetic fields, expressed in decibels.
How do you find Incident Field when you know the other values?
Rearrange for Incident Field: with Shielding Effectiveness 0.3 dB, Transmitted Field 0.055, Incident Field is 0.05693.
How do you find Transmitted Field when you know the other values?
Rearrange for Transmitted Field: with Shielding Effectiveness 0.3 dB, Incident Field 0.025, Transmitted Field is 0.02415.
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