
Annex to IOM00052-01 SW_A1x_rev00
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MAXIMUM PERMITTED RF THRESHOLD POWER (Pth)
The RF threshold power (Pth), sometimes called the effective isotropic radiated power (EIRP), as defined in IEC /EN 60079-0, is
the product of the effective output power of the transmitter multiplied by the antenna gain. The maximum threshold powers for
each equipment group as defined by Table 4 in IEC/EN 60079-0 are provided below.
Because most antennas list the gain relative to an isotropic radiator (dBi) instead of the raw power gain, it is often easier to
simply add the antenna gain in dBi to the radio output power in decibel-milliwatts (dBm). Any added cable loss between the RF
output and the antenna may also be factored in.
Pth (dBm) = RF output power (dBm) + Antenna gain (dBi) –Coax cable loss between RF output and Antenna (dB)
The resulting threshold power calculated by the above formula MUST be below the threshold power for the operating area
group rating below.
The above threshold level refer to installation in classified area Ex according to IEC/EN 60079-0 standard.
The use of device differs from one region and/or country to another. The user of the device must take care that the device is
operated only according to local rules and standard or without the permission of the local authorities on frequencies other than
those specifically reserved and intended for use without a specific permit. More detailed information is available at the local
frequency management authority.
MAXIMUM JOULES CALCULATION IN CASE OF COAX CABLE INSTALLATION
In case of use of a coax cable installation for antenna installation the adding cable need to be evaluating to ensure that the
maximum energy stored on cable not exceeded the value allowable per IEC/EN 60079-11:
Max energy (Joules) allowed per
IEC/EN 60079-11
The calculation can be done according to following
equation:
Where:
E = Energy
C1 = Antenna Barrier Capacitance (18 pF)
C2= Coax cable capacitance
R = Impedance (50Ω)
P = RF power output (18 dBm, 63 mW)
1.5 = Safety Factor
Example:
Antenna cable capacitance = 1195 pF
Answer = 0,00858 μJ acceptable for any Group