Cascaded NF when impedances not 50 Ohms ?

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bmalp98

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Hi,

you know these formulas and Excel sheets where one can calculate the casdaded noise figure, they usually assume a 50 Ohms system, i.e. matched inputs and outputs.

Does anyone know how to include other impedance levels into the calculation? As extreme cases, one would have a very small output impedance (voltage mode) or very high output impedance (current mode).

Thanks,
bmalp98
 

Calculate the transfer function from each noise source to output node, multiply it with noise voltage then you get the output noise donated by that noise source. Summing every output noise together then you get the overall output noise in voltage. Divide the output noise by voltage transfer function then you get input reffered noise. Use that input reffered noise to derive NF.

The above is only valid if the noise source are "uncorrelated".
 

Usaually the cascaded system simulator ask for 3 parameters, namely, Gain, Noise Figure, and Intercept Point.
The user try useful that only 3 parameters (and not more) are asked.
Of course, one assumption is the perfect impedence matching.

But... if you'll change the Generator reflection coefficient, the noise figure will change ( this is due to the noise modelling of the 2 port network).
In this case the noise parameters needed for calculation will rise from 1 to 4 (NFmin, ReΓopt, ImΓopt, Rn) so the calculator will became un-useful.
 

Hi again,

OK, I understand that NF actually makes only sense with matched inputs?

So it would then make more sense to indicate noise voltages/currents for voltage/current mode interfaces. Anyone written Excel or Matlab code for cascaded analysis?

Best,
bmalp98
 

what you want is discussed in detail in RF Microelectronics by Behzad Razavi
I think it will help you a lot.
 

Use the formula of Friis for calculation of NF of a cascade, and use available gain as gain factor with the appropriate impedances. Refer to Gonzalez - Microwave transistor amplifiers.
 

Hi,

thanks to those who responded.
Razavy is definitely a good read and answering the question, Gonzalez I will have to get first.

bmalp98
 

bmalp98 said:
Hi again,

OK, I understand that NF actually makes only sense with matched inputs?
Click to expand...

This is not correct. The NF of a 2 port network is a function of the input reflection coefficient. The relationship between NF and the reflection coefficient is a function of 4 real parameters (noise parameters).

regards
 

Hi,

OK, I meant NF can only be easily compared with equal impedance level. OK now?

One little glitch I found in Razavis book p. 45:

eq. 2.101 says for all equal Rs=Rin1=Rout1=Rin2 and Av being the unloaded gain
NFtot = NF1 + (NF2-1)/Av1^2

eq. 2.107 states that for the general impedance case, we can replace Av with Ap, the available power gain, given in eq. 2.104:
Ap1 = (Rin1/(Rs+Rin1))^2 * Av1^2 * Rs/Rout1

Now, Ap1 should be equal to Av1 when again Rs=Rin1=Rout1=Rin2, but is different by a factor of 4.

Did I miss anything?

Thanks for reading up to here.

bmalp98
 

I Think the factor of four is from:
Av=Voltage gain
Ap=Power gain

PWR=V*I=V²/R
(Vrms=V/√2)

so some where you had a (PWR)²--> (V²/R)² or ([V/√2]²/R)² -->

(V²/2*R)²=¼(V²/R)²

Some times book leaves out some assumptions. Also this might not
be the correct answer, but when you go between Power and Voltage
this factor of 2 or 4 does come.

Hope this helps.
 

It could be a matter of your reference planes. The input reference plane for AV1 is to the right of Rs in fig 2.32 (the voltage across Rin1). The reference plane for the available power gain is to the left of Rs (Vin). If Rs=Rin1, the voltage across Rin1 is Vin/2 and hence the power would be (Vin/2)^2 or (Vin^2)/4. That's where the 1/4 factor comes from in your matched system example.
 

Hi toonafishy,

I get your point. But does this prove Razavi wrong? To be consistent in cascading, we have to be clear about this. Regarding fig. 2.32, Av1 and Av2 do not appear unloaded voltage gains, as claimed on the preceding page.

From a practical point of view, would it then be correct to include the factor in eq. 2.104:
Ap1 = (2*Rin1/(Rs+Rin1))^2 * Av1^2 * Rs/Rout1 ?

Looks like splitting hears, but otherwise the said simple Excel sheets would do..

Best,
bmalp98
 

Hi bmalp98

I think you loss a parameter α^2 followed by Av1^2 in eq 2.101.
If Rs=Rin α^2=1/4. So It's consistent.
 

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