regulatory domain and tx power and turbo mode

RouterOS Wireless Networking
1

Hi,
can someone tell me how is calculated the tx power when I use “regulatory domain” ?

I mean in my country (like many others) I can use 1W in the ragne 5.47-5.725.

Then I select:
frequency-mode=regulatory-domain
country=switzerland
antenna-gain=23
tx-power-mode=default

with band=5ghz the TX power will me set at 4dbm

with band=5ghz-turbo the TX power will me set at 7dbm

can some one explain me the meaning of this?
Is correct that in turbo mode is allowed more power ?
or I misinterpretate the values?

Thanks for your help
regards
Stefano

2

Atheros Turbo mode uses two channels at the same time. Therefore the 3dB difference between the calculated power levels. Although it seems back to front to me!

Ron.

3

Hi Ron,
ok this means the tx power in “turbo mode” is calculated as the sum of the two channels.

But wich is the real eirp ?
2 W ?
Is this legal?

regards
Stefano

4

Working backwards as you know the maximum EiRP permitted is 1Watt…

1000mW = 30dBm
Antenna gain = 23dBi
Coax loss = x dB
RF Connectors loss = y dB
RF Pigtails Loss = z dB

30-23 = 7dBm

Now ADD the cable and connectors losses, makes it 7+x+y+z dBm is the setting you can use for maximum permitted EiRP

E.g. (just examples! You need to work these out for your own setup)
Coax Loss (10m of LMR400 at 5.8GHz) = 3.6dB loss
RF Connectors, lightning arrestors etc = 1dB loss
RF Pigtail = 0.15dB Loss

7+3.6+1.0+0.15 = 11.75 dBm

So for this example, setting a power level of 11dBm will ensure you do not exceed 1W EiRP when using the above equipment losses and gains.

Ron.

5

hi nest,
this is correct.
But assuming this, why mikrotik is setting tx power 7dbm (in my example) for the “turbo mode” and 3 for the “normal” mode?

for me this is unclear.

In your example will means that in tha case of “turbo mode” I can power to 2x11.75 ?

is realy unclear what mikrotik is calculating when you use “regulatory domain” with and without “turbo mode” .

regards
Stefano

6

I agree that the 4 and 7 appear (to me) to be reversed. Ask support!

7

If you transmit at X dB and 20 MHz then you will receive at Y dB.
If you transmit at X dB and 40 MHz then you will receive at Y-3 dB.

Remember, doubling the channel width halfs the power (and Z-3 dB is, of course, half the power of z db).
So, by doubling both the channel width and the transmit power, the power at the receiver is the same.