en300328

dear friends,

I need some help to understand the EN300 328 which states ,

2400 -2483.5 MHz Min 15 hop Channel 100 mW EIRP if FHSS

or

2400 -2483.5 MHz. PSD 10 mW in 1 MHz BW EIRP if DSSS

can anybody explain this if considering i have 2dBi antenna , hom much conducting power i can fed to the antenna.

I think information is insufficient, it guess should say that these are upper limits.
Kindly confirm to move ahead and suggest you.

Actually the answer is in the definition of the spec:

4.3.2.1 Definition
The maximum spectral power density is defined as the highest level of power in Watts per Hertz generated by the transmitter within the power envelope.
4.3.2.2 Limit
For equipment using FHSS modulation, the maximum spectral power density shall be limited to -10 dBW (100 mW) per 100 kHz e.i.r.p. For equipment using other types of modulation, the maximum spectral power density shall be limited to -20 dBW (10 mW) per MHz e.i.r.p.

The EIRP power includes the gain of the antenna. So, whatever gain you have on the antenna the EIRP shall meet the spec.

EIRP[W] = 10^(Ant_Gain[dBi] / 10) * Output_Power[W]

Yes yo are right my friend

But it says that

EIRP(dBm)=P(dBm)+G(dBi)+10log(1/x)

here P is the average conductive power fed to the antenna , G is the antenna gain and x is the duty cycle in case of FHSS . Does this mean that by reducing duty cycle (x) , actually i may fed higher power at the antenna ? Also, is there any limit on how lower i may go . How do we calculate duty cycle ?

Now in case of DSSS,

they say 10mW/1MHz , is this mean that in a bandwidth of 5MHz i may transmit 50mW or so on.

Please bear with me and help me out.

thanks in advance.

I guess duty cycle is not criteria for determining the limits.
I mean let say the average power is less than 10mW with a duty of 0.001% then the peak power will definately a voilation.
So it derives EIRP(dBm)=P(dBm)+G(dBi)

so if say Average power is 10mW ,

peak power can be ~ (Average power/dutycycle) isn't ?

Ahhh i am confused

yes, you are right

Limit=20dBm (100mW)
Antenna Gain,A=2dBi
duty cycle,x=0.001%=0.00001

So max average power allowed ,Pav=20-A-10log(1/x)
=20-2-50
=-32dBm=6.31e-7W
AND PEAK power, Ppk=Pav/x=63mW

so, 63mW will be maximum PEAK power. Does this calculation make sense .

By the way the above calculation done on the basis only if duty cyce is considered as defiend in the standard EN300 328 V1.7.1 (2006-10) defined in 5.7.2.2

Please comment.

Also see my doubt on power in case of modulation other than FHSS

Your calculation is ok. The only question is how did you arrive to a 0.001% duty-cycle !?
What is your doubt regarding other modulations as DSSS and OFDM? Actually measurements for these modulations are easier to make compared to FHSS.

Hello Vfone ,

the duty cycle=0.001% is just an example as given by "kspalla" above my reply.

May i conclued in FHSS duty cycle is important factor to consider(if we do the conductive measurment)

other doubt regarding DSSS or OFDM (other than FHSS) is that about "limit is 10mW/1MHz does this mean that in a bandwidth of 5MHz i may transmit 50mW or so on. "

Because the conducted spec is looking for ?the average output power of the transmitter shall be determined? you have to take in consideration the duty cycle for calculation. See the spec 5.7.2.2 for procedure.

Actually 10mW/1MHz means 2mW/5MHz (7dB lower)

For FHSS , now it's clear to me .Thanks a lot

For other modulation how does 10mW/MHz means 2mW/(5MHz) .

I have attached one paper to understand the isuue . I hope it will help all.