I've been trying to wrap my head about this problem for days and can't
seem to figure it out.

Zigbee technology allows you to transmit at 250 kbps. I'm comparing
the required power to transmit at that rate to the required minimum
power based on the path loss equations to reach the minimum receiver
sensitivity of the 802.15.4 physical layer for Zigbee of roughly -94
dBm. Simple enough right?

To analyze the required power to transmit at 250 kbps for Zigbee, I've
just re-arranged the Shannon capacity formula of Rate = w * log_2 (1 +
(Pt * G)/(w*N + I)). Here, w is the bandwidth, Pt is the transmit
power, Pt * G is the received signal power, N is the white Gaussian
noise and I is the sum of interference powers.

For Zigbee, these values are as follows:
w = 5 MHz
N (in dB) = 10*log10(k) + 10*log10(T) where k is boltzmann's constant
of 1.23E-23 and T is the temperature in Kelvin of 300 degrees. In
Watts, N is approximately 3.6900e-021.

I've assumed no interference at this stage, so I = 0. Simple enough so
far right?

So to calculate the required transmit power to reach a rate of 250
kbps, I've just re-arranged this equation to yield:
Pt = [2^(250 kbps/w) - 1]*(w * N + I)/G and plugging in the above
values gives me a reasonable transmit power of 1.6 uW.


You seem to be missing the noise figure of the receiver in the calculation,
and making assumptions about the receiver noise bandwidth that may not be
correct.


The reason I haven't talked about G yet is that I'm addressing it
here. G is the channel quality (path loss) and is defined as follows:

G(freq,D) = GrdBi+ GtdBi- 20*log10(4*pi*freq*d0/c) -
10*pathLossExponent*log10(D/d0) (in dB)

I'm using unity gains so GrdBi = GtdBi = 0 dBi. Other parameters a
freq = 2.4 GHz
d0 = 1m as the reference distance
c = 3.0 x 10^8
pathLossExponent = 2, and
D = 500 metres

Hence, G (in Watts) * Pt is the received signal power. G(2.4E9, 500)
is the path loss transmitting at 2.4 GHz a distance of 500m away. G
(2.4E9, 500) = -94.0254 dB


-96db Seems about right for *free space* path loss.


Now the first thing that I mentioned was the receiver sensitivity for
Zigbee of -94 dBm, Analyzing G (in dB). The required transmit power
considering this path loss is easily calculated as:

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.


-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????

So you have an rx sensitivity of -94dBm and a path loss of 94dB, so with a
tx power of 0dBm you have a receiver operating at its sensitivity limit. So
you will have a link with a Bit Error Rate of whatever the manufacturer
quoted the -94dBm sensitivity figure at (10% BER?????).


However, 1mW happens to be the maximum transmit power of Zigbee of
roughly 0 dBm,

Exactly 0dBm actually

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.

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.

Of course your formula only gives the free space path loss, reality is
likely to be considerably worse!!!

73
Jeff