A field team faces an unresolved physical question: What does a metal cabinet cost in dB and battery current? They must answer it before changing enclosure loss 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 enclosure loss. The middle card applies this page's relationship. The green card is effective gain. 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.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline for enclosure loss is 10.
- 2
Name the relationship. Geff=1.5-10.0=-8.5 dBi Pt,needed=14-(-8.5)=22.5 dBm=178 mW 20 dBm maximum gives only 11.5 dBm EIRP: 2.50 dB short external 3.0 dBi antenna: Pt=11.0 dBm=12.6 mW; current≈12.7 mA
- 3
Substitute the chapter fixture. Set enclosure loss to 10. The page ledger gives effective gain as -8.50 dBi.
- 4
Read the result. Keep dBi beside the value. Use it only inside the technical boundary on this page.
Predict, then change enclosure loss
Try Predict the direction of effective gain. Move one control, calculate, then check your prediction.
Observe dB terms add linearly, but the power and battery-current cost grows exponentially. Reset the control to 10 and compare effective gain.
Explain Only enclosure loss moves here. The other chapter fixtures remain fixed.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
1. Read dB as additions and subtractions
Antenna gain in dBi focuses radiated energy; it does not create energy. Enclosure absorption and detuning subtract from that gain. Conducted radio power plus effective gain gives EIRP.
2. Name every algebra move
Subtract enclosure lossGeff=Gantenna−Lenclosure.
Rearrange the EIRP sumPt,needed=EIRPtarget−Geff.
Convert dBm to milliwattsPmW=10^(PdBm/10).
Convert PA power to currentI=P/(ηV).
3. Reproduce the cabinet comparison
Pt,needed=14−(−8.5)=22.5 dBm=178 mW
20 dBm maximum gives only 11.5 dBm EIRP: 2.50 dB short
external 3.0 dBi antenna: Pt=11.0 dBm=12.6 mW; current≈12.7 mA
The capped internal PA draws about 101 mA at 30% efficiency and 3.3 V, roughly 7.94 times the external case.
4. Try the enclosure loss
TryIncrease cabinet loss while keeping the antenna and target fixed.
ObserveEvery extra enclosure dB demands one extra conducted dB until the radio reaches its limit.
ExplaindB terms add linearly, but the power and battery-current cost grows exponentially.
This is one EIRP ledger, not a complete installed link budget.
- Antenna
- Pattern, cable, connector, ground plane, and orientation still matter
- Radio
- PA efficiency and current vary by output setting and temperature
- Law
- Band, region, duty cycle, and certification set the real limits
Measure total radiated performance and receiver margin in the final enclosure.
5. Test the installed antenna
Compare free-air and installed return loss, radiated power, receiver margin, and burst current across orientations, doors, cables, batteries, and representative sites.
6. Record the form-factor state
Store enclosure, material, antenna, cable, connector, band, firmware power setting, measured loss, EIRP, current, site, and retest triggers.
7. Check yourself
Why is 1.5 dBi not extra electrical power?
Why is the cabinet case 2.50 dB short?
Does an external antenna automatically qualify the link?
The cabinet scenario reproduces the chapter's stated catalog-typical teaching assumptions.
- 14 dBm
- Illustrative sub-GHz target, not a universal legal ceiling
- 10 dB
- Severe illustrative enclosure loss
- 30%
- Simplified PA efficiency for current comparison
Correct, not complete: an EIRP ledger does not qualify an installed radio link.
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