Math Bridge: EIRP-Capped Patch Antenna

← Back to PID Tuning Applications
Math BridgeControl & GatewaysStruggle-friendly runway

How does a patch save radio power under an EIRP cap?

Gain reduces allowed conducted power, but the valve must stay inside the narrower beam.

Phoebe, the physics guidePhoebe guides
The one targetReproduce the remote valve's 6 dBi power-and-beam trade.
The chapter case20 dBm EIRP, a 0 dBi omni, and a 6 dBi patch.
What it buys youA radio-power claim with a retest condition.

A field team faces an unresolved physical question: How does a patch save radio power under an EIRP cap? They must answer it before changing directional patch gain in dbi 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 directional patch gain in dbi. The middle card applies this page's relationship. The green card is allowed milliwatts. 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.

Directional patch gain in dBi changes allowed milliwatts An input card leads through the page relationship to the allowed milliwatts result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. The cap subtraction and dB conversion drive the saving. The directivity formula exposes the angular condition attached to it.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for directional patch gain in dbi is 6.

  2. 2

    Name the relationship. Ptx=EIRPcap-G; mW=10^(Ptx/10); θ=√(41253/10^(G/10))

  3. 3

    Substitute the chapter fixture. Set directional patch gain in dbi to 6. The page ledger gives allowed milliwatts as 25.119 mW.

  4. 4

    Read the result. Keep mW beside the value. Use it only inside the technical boundary on this page.

Predict, then change directional patch gain in dbi

Try Predict the direction of allowed milliwatts. Move one control, calculate, then check your prediction.

6
Chapter baseline
Allowed milliwatts

Observe The cap subtraction and dB conversion drive the saving. The directivity formula exposes the angular condition attached to it. Reset the control to 6 and compare allowed milliwatts.

Explain Only directional patch gain in dbi 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 directional patch gain in dbi moves. Field effects named in the page's technical boundary stay fixed.

Honesty note. This teaching model changes only directional patch gain in dbi. It replays the chapter fixture; it is not a field prediction.

1. Separate the cap into two parts

EIRP joins conducted radio power and antenna gain. If regulation holds their sum fixed, more antenna gain means less conducted power is allowed. The saving comes with directional focus.

Phoebe: The patch saves transmitter power only while the link stays in its beam.

2. Name the algebra moves

1

Rearrange the capPtx=EIRP−G.

2

Undo dBmmW=10^(Ptx/10).

3

Form the savingreduction=100 mW/Ppatch.

4

Undo dBiGlinear=10^(GdBi/10).

5

Solve the symmetric beamθ=√(41253/Glinear).

3. Work the valve link

0 dBi: 20−0=20 dBm=100 mW; 6 dBi: 20−6=14 dBm=25.1 mW

The conducted-power ratio is 100/25.1=3.98. Six dBi is also 3.98× linear gain. The ideal symmetric beam estimate is √(41253/3.98)=102°. A moved or rotated valve can lose the trade.

4. Try one controlled change

Ptx=EIRPcap−G; mW=10^(Ptx/10); θ=√(41253/10^(G/10))

TryMove only patch gain. The 20 dBm ceiling and 100 mW omni reference stay fixed.

Allowed conducted power
Allowed milliwatts
Power reduction
Ideal beam width

ObserveAt 6 dBi, allowed power is 14.0 dBm or 25.119 mW, 3.98× below the omni reference, with an ideal 102° beam.

ExplainThe cap subtraction and dB conversion drive the saving. The directivity formula exposes the angular condition attached to it.

Technical boundaries.

This is an ideal EIRP and symmetric-beam calculation.

Power
Radio DC power does not scale exactly with RF milliwatts
Pattern
Efficiency, sidelobes, polarization, cable, and enclosure loss remain
Control
Latency, loss, jitter, safety state, and PID stability remain separate

Measure installed EIRP, current, pattern, link reliability, and closed-loop behavior.

5. State the retest trigger

The calculation assumes fixed alignment. Moving the valve, gateway, enclosure, cable, or mounting angle changes the installed pattern, so acceptance evidence must be repeated.

6. Carry the evidence

Record legal EIRP, conducted power, RF and DC current, antenna pattern, cable and enclosure loss, orientation, movement, receiver sensitivity, packet loss, latency, safe state, and control response.

7. Check yourself

Why does 6 dBi leave 14 dBm conducted?
Answer: Under a 20 dBm EIRP cap, Ptx=20−6=14 dBm.
How does 14 dBm become 25.1 mW?
Answer: Milliwatts equal 10^(dBm/10), so 10^1.4=25.1.
Does 3.98× less RF power guarantee 3.98× less battery use?
Answer: No. Radio efficiency, base current, airtime, retries, and other loads remain.
Honesty boundary.

The page makes an RF trade auditable; it does not certify the remote control loop.

20 dBm
Catalog-typical EIRP ceiling used by the chapter
25.1 mW
Ideal 6 dBi allowed RF power
102°
Approximate symmetric beam width

Go deeper in the chapter, then retest the real RF path and PID system after any installation change.

Return to PID Tuning Applications with the maths unpacked →