The overall receiver noise figure is included in the receiver
sensitivity, is it not? I'm referring to
http://www.edn.com/article/CA6442439.html
where they state that:

"the receiver sensitivity S=–174 dBm/Hz+NF+10logB+SNRMIN, where –174
dBm/Hz is the thermal noise floor, NF is the overall-receiver-noise
figure in decibels, B is the overall receiver bandwidth, and SNRMIN is
the minimum SNR. If the total path loss between the transmitter and
the intended receiver is greater than the link budget, loss of data
ensues, and communications cannot take place. Therefore, it’s
important for designers developing end systems to accurately
characterize the path loss and compare it with the link budget to
obtain initial estimations of the range."

Do you have any comments on this?

Yes, the manufacturer has told you that the receiver sensitivity is -96dBm.
(no doubt for a particular BER, which aslo equates to a particular SNR).
Without knowing more about the internals of the receiver you cannot work
back to a NF unless you acuually know what SNR equates to the BER that the
manufacturer stated the -96dBm at. Using Shannon most likely give you an
unreliable optimistic answer.


If you know the rx sensitivity is -94dBm at the data rate you require,
why
go through all of the Shannon stuff, it is not revenant. You have been
told
that the Rx sensitivity is -94dBm use that figure.

That requires a long-winded answer about my project. My study involves
optimizing a multiple technology network where each device is
optimizing their transmission. Using the Shannon theorem allows me to
perform this optimization of data rate while considering physical
layer constraints. It's something I have to work into it
unfortunately. It's functioning correctly for WiMax and ultrawideband
technology.

No it dosen't unless you know more about the receivers and the actual SNR
that they require for a specific BER. Shannon will not tell you that, it
will just give you the 'best possible case' for a particular bandwidth,
which may well be a long way from the truth.


-94 dBm = P_required + G(2.4E9, 500)

P_required is then easily calculated as -94 dBm - G(2.4E9, 500) =
0.0254 dBm. This translates to a transmit power of 1 mW. This required
power is LARGER than the maximum permissible power to transmit at 250
kbps, which is the maximum data rate for Zigbee. So how does this make
sense?

Perhaps because most quoted ranges for Zigbee are in the order of 50m not
500m????

The approximate line-of-sight range for Zigbee is 500m. That's why I
tested both 300m and 500m in my calculations.

Yes and you proved just that!! Path loss 96dB, rx sensitivity -96dBm, 0dBm
tx power, equals receiver operating at its sensitivity limit. So what's the
problem??

Testing other distances, say 300m, we get a maximum transmit power Pt
= 5.9181e-007 = 0.59 uW to transmit at 250 kbps, but a P_required =
-4.41 dBm = 3.6211e-004, which is our maximum transmit power Pt.

How is -4.41dBm greater than 0dBm (1mW) ?? It is about 0.362mW.

I meant that, for my example using a distance of 300m, the transmit
power required to transmit at the maximum data rate of 250 kbps was
0.59 uW. The required power to reach the MIRS at the receiver 300m
away was 3.6211E-4. Therefore, the amount of power required to reach
the receiver (a lower bound on power) was greater than the maximum
power allowed to reach the max data rate of 250 kbps (upper bound on
power). Therefore, the transmission isn't possible. It wasn't for 0
dBm.

I a sorry I don't understand what the hell you are on about. All you need to
know is the path loss at 300m, you aready have the rx sensitivity of -96dBm
so it is a simple subtraction to find the Tx power required (which will be
less than 0dBm).

Your maths is quite simple; you have the rx sensitivity, and can work out
the path loss, thats all you need to do, take one from the other and you
have the required tx power.

That's exactly what I'm doing. The NF is included in the receiver
sensitivity and the transmit power required to reach the receiver
sensitivity is greater than the power I'm allowed to transmit at
because of the 250 kbps maximum for the technology.


That is just not correct, you proved that the power required to reach
sensitivity limit at 500m was 0dBm, which is OK because that is you max tx
power.

At 300m you obviously require less tx power, so lets do the maths:

path loss at 300m, 2400MHz is 90dBm

so required tx power is 90 -96 = -6dBm which is less than your 0dBm max tx
power - no problem.
-6dBm equates to 2.5e-4W or 0.25mW.

If you are trying to say that the -6dBm figure is greater than the number
that you worked out using Shannon for a particular bandwidth and data rate
once you take off the 90dB path loss then you have to look at the accuracy
of your Shannon calculation.

You said that you Shannon Calculation produced a "reasonable transmit power
of 1.6 uW", to achieve 250kB in a 5MHz BW. Now 1.6uW = 0.0016mW = -28dBm
this does not seem reasonable when the manufacturers are quoting -96dBm!!!
You must go back and re-consider your Shannon calculations, this time taking
the actual receiver characteristics into account, such as rx NF, the real
noise bandwidth, number of levels in the modulation scheme etc., etc..

Regards
Jeff