Answer Calculations from PE commissioned by AT&T: the GQ2410-06621 antenna does have 10 total input ports, however, our analysis of the AT&T DAS node includes only the frequencies and power levels at which AT&T proposes to operate.  For this site, three Ericsson radio units are proposed to be installed - two Model 2203’s, which will operate in the PCS (1950 MHz) band, and one Model 2205, which operates in the unlicensed 5 GHz band.  Each unit is connected to the antenna by two ports, and the four remaining CBRS band ports are not proposed to be used. As the report states, the maximum effective radiated power in any direction would be 85.4 watts.  This ERP value includes the antenna gain, so it would be the output value of the calculator you linked to, rather than the input.  We calculated the ERP as follows:

Each Ericsson Model 2203 has two ports, each with a maximum Transmitter Power Output (TPO) of 5 watts.  The maximum TPO in the PCS band is therefore 20 watts (13 dBW).

The connections between the Ericsson radios and the Galtronics antenna result in power losses that we estimate to be 0.55 dB.

The maximum gain of the GQ2410-06621 antenna is 8.9 dBi (6.75 dBd) in the PCS band.

The ERP in the PCS band can therefore be calculated as 13 dBW - 0.55 dB + 6.75 dBd = 19.2 dBW = 83 watts.

Unlicensed operation in the 5 GHz band is limited to an ERP of 2.4 watts in order to comply with FCC Title 47 §15.247.

Original question: For their 5G rollout, AT&T plans to erect a small cell site in my neighborhood. Based upon information provided by AT&T, the Galtronics Model GQ2410-06621 will be installed.

Looking at the specifications sheet for the Galtronics Model GQ2410-06621, it has 4 ports for AWS/PCS/WCS Band at 1695-2360 MHz, where the maximum power per port is 100 Watts. This Galtronics model also has 2 ports for 5 GHz WiFi (2x ports for U-NII Band 5150-5925 MHz), where the maximum power per port is 1 Watt.

My question is, how does one calculate the maximum effective radiated power in any direction for this Galtronics small cell site? Professional engineer consultants calculated the maximum effective radiated power in any direction would be 85.4 Watts, representing simultaneous operation of 2.4 Watts for 5 GHz WiFi and 83 Watts for PCS service.

Here's a quote from the Site and Facility Description: AT&T is proposing to install one Galtronics Model GQ2410-06621, 2-foot tall, cylindrical antenna, on a cross-arm to be added to the side of a new utility pole. The antenna would employ no downtilt and would be mounted at an effective height of about 24 feet above ground. The maximum effective radiated power in any direction would be 85.4 Watts, representing simultaneous operation of 2.4 watts for 5 GHz WiFi and 83 Watts for PCS service.

For the Galtronics Model GQ2410-06621 (https://galtronics.com/wp-content/uploads/2019/10/Datasheet-GQ2410-06621-Rev_1.0.pdf), I have attached the specifications.enter image description here

  • \$\begingroup\$ Depends on the antenna pattern (omni, bidirectional, ect), one thing to keep in mind is the power falls with the distance squared, so there is a rapid drop of in power the further you get away from the antenna. \$\endgroup\$ – Voltage Spike Oct 7 '19 at 17:03

Ok, so considering the PCS stuff first, they are basically saying that on axis in the horizontal plane of the antenna they the EIRP is ~83W, so +19dBW, allow for the ~9dB of antenna gain and the power they are feeding the thing is right around 10W = +10dBW.

In the 5Ghz band they are running 2.5W = +4dBW EIRP, with an antenna gain of ~4.7dB, so they are running a little less then 1W into the thing.

To calculate power in any given direction you take the gain values from a polar plot in the appropriate axis (The more interesting one is generally the vertical plot) add the value from the polar plot to the input power in dB (Some polar plots are drawn in terms of gain below maximum which you would need to fiddle with a bit), which gives you the effective EIRP in that direction...

Now you take that number, convert from dBW to watts =(10^(dB/10)) and calculate the power flux density by dividing the power in watts by the area of a sphere at whatever distance you are interested in. Results will be in Watts per square meter (SMALL Number).

My take, it is a flea powered little radio, your personal cell phone produces far more power density where you are then that thing ever will.

Just for shits and giggles, lets pretend that it was on its side and so aiming that EIRP at the ground 24 feet below, (It isn't these things are designed to be anaemic at high vertical angles), but lets pretend... So 85W EIRP, at 24 feet, area of a sphere of radius 24 feet is 4PiR^2 = ~7200 square feet, 85W/7200ft^2 = 0.012W/Ft^2 = 12mW per square foot on axis, reality will be much lower unless you are directly in line with it 24 feet up.

Pony little toy radio set, get them to put a proper one in that actually means it.

  • \$\begingroup\$ You've got your signs incorrect on the antenna gain, that's a positive gain. I calculate 83W=49.2dBm + 9dBi antenna gain = 58.2dBm = 660W EIRP, so your calculations seem to be a little off, I get 91mW/ft^2, so it's a fair amount of power at 24ft. Considering LTE requires 6 to 9 dB of PA backoff, that's a respectable amount of fully modulated power. \$\endgroup\$ – rfdave Oct 9 '19 at 0:09
  • \$\begingroup\$ @rfdave The OP says "Professional engineer consultants calculated the maximum effective radiated power in any direction would be 85.4 Watts", looks to me like the number we have is EIRP not transmitter output, so getting to transmitter output by subtracting the antenna gain is surely correct (If not particularly relevant). If the 85W had been antenna input power then yea, but the question as stated is the other way around... \$\endgroup\$ – Dan Mills Oct 9 '19 at 0:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.