Documentation ref DTUS075 rev A0 06/18/20
-11-
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.
Equipment for
Threshold Power (W)
Threshold Pwr (dBm)
Group I
6
37.7
Group IIA
6
37.7
Group IIB
3.5
35.4
Group IIC
2
33.0
Group III
6
37.7
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
Group I
1500 μJ
Group IIA
950 μJ
Group IIB
250 μJ
Group IIC
50 μJ
The calculation can be done
according to following equation:
Where:
E = Energy
C
1
= Antenna Barrier Capacitance
(18 pF)
C
2
= 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
