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Es is the electromotoric force of the signal.
En is the electromotoric force of the
thermal noise in the source R.
Ea is the equivalent noise electromotoric force of
the amplifier.
Ia is the equivalent noise current of
the amplifier.
The S/N in U, the output voltage from the LNA is determined only by
the the electromotoric forces, currents and impedances on the
source side.
The impedance of the amplifier, whether it is matched or not does not
affect the output S/N.
For simplicity of arguments, assume the impedance is infinitely high
so no current would flow into the amplifier..
Assume also (for now) that all the noise sources are uncorrelated.
If a lossless impedance transformation were applied between the source R and
the LNA that would lift the source impedance by a factor of T,
the signal voltage and the thermal noise of the source would increase by a
factor of the square root of T.
The noise voltage due to Ea would remain unchanged but
the noise voltage due to Ia would increase by a factor of T.
The effect on S/N (power ratio) of a lossles feed impedance transformation
with a factor of T is thus:
S/N = T * Es2 / [ T * En2 +
Ea2 + T2 * 502 * Ia2]
....................... (1)
This simplifies to:
S/N = Es2 / [ En2 +
Ea2/T + T * 2500 * Ia2]
........................ (2)
For the amplifier to have optimum NF for a 50 ohm source impedance,
Ea2/T + T * 2500 * Ia2
must have a minimum for T=1. That means that the derivative of this
expression -Ea2/T2 + 2500 * Ia2
has to be zero for T=1 which leads to:
Ea2 = 2500 * Ia2
......................(3)
Now, we know that for input impedances of 75 and 33 ohms,
T = 1.5 as well as for T = 0.667
the NF degrades by about 0.01 dB for the ATF33143feb amplifier. (SWR=1.5)
In linear power scale that is a factor of 0.9977. From that we get:
0.9977 * Es2 / [ En2 +
Ea2 + 2500 * Ia2] =
Es2 / [ En2 +
Ea2/1.5 + 1.5 * 2500 * Ia2] =
.......................(4)
Es2 / [ En2 +
Ea2/0.667 + 0.667 * 2500 * Ia2]
....................... (5)
Knowing that Ea2 = 2500 * Ia2
we can simplify to:
0.9977 / [ En2 + 2 * Ea2] =
1 / [ En2 + (1.5 + 1 / 1.5) * Ea2] =
....................... (6)
1 / [ En2 + (0.667 + 1 / 0.667 ) * Ea2]
....................... (7)
This is two equations which both give:
0.9977 * (En2 + 2.167 * Ea2) =
En2 + 2 * Ea2
....................... (8)
from which we can compute:
-0.0023 * En2 + 0.162 * Ea2 = 0
....................... (9)
or
Ea2 = 0.0142 * En2
....................... (10)
The noise figure is by definition the actual S/N divided by the S/N
that would have been obtained with an ideal amplifier having
Ea as well as Ia equal to zero.
By use of equation (1) twice with T=1 we get the ratio of the result
without and with the losses included as:
NF = {Es2 / En2} /
{Es2 / [En2 +
Ea2 + 502 * Ia2]}
..........................(11)
By use of equations (3) and (10) we get:
NF = {Es2 / En2} /
{Es2 / [En2 +
2* 0.0142 * En2]}
..........................(12)
This means that NF=1.0284 in linear power scale or 0.12 dB.
This is a lower limit for the NF based on the assumption that
Ea and Ia are uncorrelated.
If we assume that Ea and Ia are 100%
correlated equation 1 would become:
S/N = T * Es2 / [ T * En2 +
( Ea + T * 50 * I)2]
....................... (13)
This leads to:
S/N = Es2 / [ En2 +
Ea2/T + 100 * Ia * Ea +
2500 * T * Ia2]
....................... (14)
Equation (14) is similar to (2).
It has one more term in the denominator but the derivative is unchanged so
equation (3) is still valid but equation (4) becomes different:
0.9977 * Es2 / [ En2 +
Ea2 + 100 * Ia * Ea +
2500 * Ia2] =
Es2 / [ En2 +
Ea2/1.5 + 100 * Ia * Ea +
2500 * 1.5 * Ia2]
.......................(15)
By use of equation (3) this becomes:
0.9977 * Es2 / [ En2 +
4 * Ea2 ] =
Es2 / [ En2 +
( 1.5 + 2 +1/1.5 ) * Ea2]
.......................(16)
From (16) we find:
Ea2 = 0.0146 * En2 ................(17)
Now we can compute the NF as the ratio of (S/N) values computed without
and with the amplifier noise included:
NF = {Es2 / [ En2 } /
{Es2 / [ En2 +
Ea2 + 100 * Ia * Ea +
2500 * Ia2]}
................. (18)
By use of (3) this becomes:
NF = {Es2 / [ En2 } /
{Es2 / [ En2 +
4* Ea2]}
................. (18)
Inserting (17) we find:
NF=1.0584 as power ratio or 0.264 dB.
By a comparison with other amplifiers the NF was determined to 0.18 dB.
That value is traceable to measurements with boiling water and ice that
probably are correct within 0.04 dB.
Now, by a completely different method, we have found that the
NF has to be between 0.12 and 0.26 dB.
(There is a small uncertainty from adopting 0.010 dB NF change for VSWR=1.5
as the differences from table 5. By measuring more accurately and by fitting
a paraboloid surface to the five NF points at the five source impedances
this error can be made very small.)
The NF contour map tells us that the NF of the ATF33143feb amplifier is
0.19 +- 0.07 dB (still assuming that the 0.010 dB change for
VSWR=1.5 is correct)
The interesting aspect of this result is that with access to
a better amplifier with a true NF of a little less than 0.1 dB
one would get the absolute NF to within something like 0.04
dB from measurements at a single temperature, the room temperature.
Maybe a cryogenic amplifier could be used to calibrate the noise head
in a measurement with a conventional NF meter and a circulator
in front of the DUT on higher frequencies where it is difficult
to know the temperature distribution over the different losses
in a hot/cold measurement.
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