Interpreting Noise Figure Data

In this document:

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there is in the equation below Figure 6 on page 4 a variable that is a Beta symbol with the subscript "re". I know what Beta without the subscript is, but this is the first time I have seen that subscript with it as it is in that equation's numerator. This has resistance units. What transistor parameter value is this? I suspect it is one I already know by a different symbol. Could this be the incremental emitter resistance I know as the symbol r with the geek letter pi subscript?

In that equation there is the symbol Rs which I know to be the source resistance the noise figure was measured with. I need to figure out what this value is from data available for this transistor:

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which has numeric data available for the Smith Chart data shown in the spec sheet PDF as a download.

Is calculating Rs matter extracting it from reflection coefficient data by solving for the real part of Zs in the second equation at:

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?

Or is Rs simply the typical 50 ohm source impedance which is also the typical characteristic impedance of cable?

In the tuning networks for making noise measurements is the load in that reflection coefficient equation always only a reactive element without a real part? And is the source always only a real element with no reactive part?

Reply to
Artist
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You're probably supposed to read the equation below Figure 6 as the product of Beta times re:

B * re

Rs needs to come off the data sheet of the signal source (not the DUT). 50 ohms is a nominal value that's good for rough approximations.

Thank you,

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Don Kuenz, KB7RPU 
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Reply to
Don Kuenz

Why isn't it 25 ohms? In any practical measurement, that's the parallel impedance of a 50 ohm source with a 50 ohm termination. Without the termination, you could have all kinds of reflections which would cause indeterminate results.

Reply to
Steve Wilson

You need to impedance match each stage to minimize reflection and maximize power transfer. 50 ohms is used as a compromise between the best power handling (30 ohms) of air dielectric coax and the lowest loss (77 ohms).

Thank you,

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Don Kuenz, KB7RPU 
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Reply to
Don Kuenz

In a matched 50 ohm system, the Thevenin equivalent is 25 ohms (50||50 = 25).

Why isn't Rs 25 ohms instead of 50?

Reply to
Steve Wilson

If you connect a source with Rs of 25 ohms to 50 ohm coax you create an impedance mismatch.

Thank you,

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Don Kuenz, KB7RPU 
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Reply to
Don Kuenz

Because you've already taken Rs into account? Rs || Rl = 50 || 50 = 25

Reply to
krw

I'm not talking about a generator source of 25 ohms. I specified 50 ohms.

Reply to
Steve Wilson

What are you asking? It's not clear at all.

Reply to
krw

OK. Settle down. Why do you want to Thevenize the node where a 50 ohm source connects to 50 ohm load? Do you intend to insert parallel load at that node? And, what in the heck does Rs (Zs) mean to you?

Thank you,

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Don Kuenz, KB7RPU 
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Reply to
Don Kuenz

Consider a 50 ohm cable coming into a scope, via a tee, with a 50 ohm terminator attached.

What impedance does the scope "see", looking out of its input? Assuming the

50 ohm cable is connected to a 50 ohm source (say, your average function generator).

This, by the way, is not a unique characterization. There is the Thevenin equivalent used for high impedance loads (like a scope), and there is the Norton equivalent used for low impedance loads (say, a grounded-base amplifier). In the latter case, you series terminate the input, and the amplifier input sees 2*Zo or 100 ohms.

For amplifiers having input impedances between zero and infinity ohms, the zero-reflection condition lands somewhere inbetween. Obviously, a 50 ohm amplifier needs no additional termination whatsoever, and sees 50 ohms at its input. (However, if you were to probe the input node, you would measure its Thevenin impedance as 50 || 50 = 25 ohms, as you would at any other point along the cable.)

Incidentally, probing a transmission line can be done by current or voltage (or both). Suppose you break the signal path and insert a current transformer; it sees 50 + 50 ohms.

Suppose you add an RF power tap; when matched, it sees 50 ohms, because it does not use just voltage or current, but a combination of both, to sense the signal.

So, it generalizes, and one must always be specific in what conditions they are talking about.

In this case, I think Steve is talking about adding a terminator in parallel, to swamp the reflection from the amplifier, which most likely has a relatively high input impedance, and therefore would have a lower SWR with the terminator.

Tim

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Reply to
Tim Williams

No. I thought I was clear at the beginning.

I'm talking about a 50 ohm source, coupled through a 50 ohm coax, to a 50 ohm matched load such as a minicircuits broadband microwave amplifier.

The 50 ohm source is then in parallel with the 50 ohm load, giving an effective source impedance of 25 ohms.

This would change the noise figure, since thermal noise is proportional to the square root of resistance. For example, the noise generated in a 50 ohm resistor in 10 GHz bandwidth is 90.7 microvolts. However, the noise generated in a 25 ohm resistor in 10 GHz bandwidth is only 64.15 microvolts.

If the noise generated at the input of the amplifier is constant, then the noise figure has to change. This is illustrated in the Siliconix AN-106 paper at

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As a side issue, you can measure the input noise of a amplifier by connecting a signal generator to the input and a rms voltmeter to the output. Turn the signal generator off and measure the rms output voltage.

Turn the signal generator on and increase the output until the rms voltmeter increases by 3 dB. The signal generator output power is then equal to the amplifier input noise, and the signal generator displays the input noise directly.

However, if the amplier input noise is extremely low, you have to include the noise generated by the source and load impedance, and the bandwidth. In this case, the source and load impedance in parallel is 25 ohms.

Reply to
Steve Wilson

That's correct. That's precisely what you want to do. Set gamma-s equal to gamma-opt and crank out the number. (A thousand pardons for botching my first reply to your question. Your second question was skimmed too fast by me during my first take.)

Thank you,

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Don Kuenz, KB7RPU 
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Reply to
Don Kuenz

I routinely measure noise in amplifiers with noise figures well below 3dB. That means that the noise of the amplifier proper is well below the noise of the (room temperature) 50 Ohm source. I do that by connecting either a terminator cooled to 77K, or one at room temperature. An ideal noiseless amplifier would then show a 10log(293/77) = 5.8dB change in output level. A real amplifier won't show that much of a change, because its own noise washes out the change in input noise. With a few lines of algebra, you can then work out the amplifier's own contribution.

Looking into the amplifier input, one sees 50 Ohms with a noise temperature well below room temperature, in my case in the 35K ballpark, despite the fact that the amplifier is at room temperature.

If you were to measure the noise level on the connection between the amplifier and the room temperature source, you would see an EMF of sqrt(k(293+77)*50) V/rtHz at a source impedance of

25 Ohms. A real 25 Ohm room temperature resistor would show sqrt(4 k 293*25), which is more.

Jeroen Belleman

Reply to
Jeroen Belleman

Ah sorry, sqrt(k(293+35)*50) V/rtHz, of course.

Jeroen Belleman

Reply to
Jeroen Belleman

On 3/11/2018 12:35 AM, Don Kuenz wrote: > Artist wrote: > > > >> >> Is calculating Rs matter extracting it from reflection coefficient data >> by solving for the real part of Zs in the second equation at: >> >>

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>> >> ? >> > > That's correct. That's precisely what you want to do. Set gamma-s equal > to gamma-opt and crank out the number. > (A thousand pardons for botching my first reply to your question. > Your second question was skimmed too fast by me during my first take.) > > Thank you, > > -- > Don Kuenz, KB7RPU > "To invent, you need a good imagination and a pile of junk" > - Thomas Alva Edison >

Thank you for you reply. Much appreciated. It leads me to more questions:

The reflection coefficient provides two knowns, which are the magnitude, and the phase angle. The noise figure provides a third known.

Assuming the tuning network can be modeled as an unknown series resistance between source and DUT input, and an unknown reactive element connected from the DUT input to ground (can it?), I would then have three unknowns, with rbb being the third unknown.

The remaining DUT input hybrid pi elements that affect the reflection coefficient rbe (aka re in that equation, or aka r subscript pi ?) Ce (aka C subscript pi), Cu, can be calculated for the collector current the NF was measured at using the spec sheet parameters.

Is solving the above for the three unknowns the right track?

The next question regards the collector's load resistance. I assume the NF is a ratio of noise voltages. The transistor's output is a current source. So there must be a load resistor on the collector to convert the collector current to a voltage. But wouldn't that contribute thermal noise? So why would the load resistance not be included in that equation?

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Reply to
Artist

Noise Figure data for the BFU520A does not go below 400MHz. My interest is in calculating noise from 0 to 40 MHz. How valid would rbb calculated from 400 MHz data by the means discussed in this thread be from 0 to 40 MHz?

I seek a transistor noise model for Q1 in the topology shown Figure 9, page 4, of

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.

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Reply to
Artist

Am 13.03.2018 um 00:47 schrieb Artist:

What about Re? It is in series to Rb and at small currents it will dominate. Now we know how it scales with f(Ie), so not everything is lost if we have s-parameters at several currents. But it starts to get clumsy.

Noise data are generally of lesser quality than the s-parameters themselves. Their generation involves differences of similar numbers, so small errors have great consequences. The gamma of the noise head plays a major role here, esp. the difference between ON and OFF state.

15 dB noise sources are usually not good enough unless they are followed by a 10 dB metrology grade attenuator. At least not for noise factors below 1 dB.

As long as the transistor has decent gain, that usually can be ignored. Its thermal noise could probably be computed away when the s-parameters are all given since we then know the effective gain of the stage.

Rb is part of the spice parameters, but the fact that there is a number does not mean that someone has really measured that. The p-spice PARTS program to convert easily imprecise data sheets to even worse spice parameters was well known to generate the same arbitrary RB value for just about every transistor. And easy means easily published.

No doubt JT will pop up and cry that's not true.

regards, Gerhard, DK4XP

Reply to
Gerhard Hoffmann

Page 61 of this document talks about how to extract rbb:

idea2ic.com/PlayWithSpice/pdf/G U M M E L - P O O N.pdf

Thank you,

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Don Kuenz, KB7RPU 
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Reply to
Don Kuenz

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