The NF for the 2SC5551 transistor is 1dB.?

Everett N4CY

My knowledge of semiconductor theory is rather basic and I was wondering whether amplifiers like the LZ1AQ could be improved by using low noise transistors like the 2N5109, rather than 2N2222 or 2SC5551.

The NF of the 2N2222 is published as 5dB in datasheets but when I searched for the comparitive NF of the 2SC5551 it did not appear to be listed in the ONSemi and other datasheets. There was just an oblique reference to it being 8dB.

Are the noise factors (NF) the same when used in a very low Z amplifier?

Has anyone measured it for the low input Z LZ1AQ amplifier or other similar uses?

Thanks

John

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On Tue, Oct 8, 2024 at 06:32 AM, Fred M wrote:

The ohmic resistance of the loop including the skin effect is rather small with relation to XL, that that it has little influence on the shortcut current.

Agreed.? In the context of a resistance calculation at frequency, skin effect cannot be ignored.? But, as you say, the inductive reactance still defines the overall loop impedance....at least for practical antenna sizes and materials.

Hi Mark,

Yes I have similar but limited results with my MMANA models , creating a field with a dipole source, but? with this setup I cannot compare currents at different frequencies.?? I installed EZNEC today and get the Dan *.ez file, since my main problem is that I still do not learn how to create an antenna model there . But I can run ready models so? now I can test this loop in RX mode with fixed field intensity which is not possible in MMANA. Many thanks to both of you.

Chavdar lz1aq

Mark and All,

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I tried to use Mark's data of 10/4/24 where he ran a single loop model in EZNEC to ascertain whether Chavdar's test simulator is designed correctly:

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f = 15 MHz , Isc = 0.707 mA ( short circuit current ) , Xa = 382.6 Ohms

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If we walk backwards from Norton's model to Thevenin's the predicted voltage of open circuit shall be:?

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? V'oc = Isc x Za, or Isc x Xa? since Xa > Ra

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?V'oc =0.707 mA x 0.3826 KOhms = 0.27 V

Since Chavdar uses two 1 Kohms resistors in his test simulator the V oc across the 50 ohms load ( secondary side of the balun ) shall measure:

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? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? V oc = V' oc x 2000/ 382.6 = 1.414 V?

Meeting this condition we can state the transform from Thevenin to Norton model is done correctly based on equal power requirement.

However, VNA cannot produce across port CH0 the required 1.414 Volts...that will create 0.707 mA of short circuit current.

Do I miss anything ?

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Raphael? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

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The induced voltage in the loop rises proportional with the frequency. The inductive reactance rises also proportional with the frequency. Both compensate each other what, in shortcut mode, results in a constant current versus frequency. The ohmic resistance of the loop including the skin effect is rather small with relation to XL, that that it has little influence on the shortcut current. A loop connected to a real amplifier with Zin = >2 Ohms even more reduces the impact of the loop resistance on the loop current.

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regards

Fred

On Sun, Oct 6, 2024 at 04:55 AM, чавдар левков wrote:

So I will suggest to Mark to ?do another calc.: ?at resonance frequency ?to move the load point at 90 deg on the side of the loop and ?see the value of this current. ?I have made some very rough model of this receiving loop at resonance point? ( creating e.m. field with a small vertical dipole far away from the loop)? and the current at the loop side is almost 1.5 ?times higher than the current at ?point in the bottom of the loop. ?When the loop is "small" the current must be uniform in any point of the loop.

I reran the simulation with the load point located at three different positions: bottom side, front side, and back side.? I obtained the following results:

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At resonance (around 49MHz), the current at the sides is indeed about 1.5 times the current at the bottom.? Dan had a similar result.? At 30MHz, the difference is only 1.3 dB.?

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My opinion:? 1.3 dB of error is acceptable when performing a gain sweep to 30MHz.? Keep in mind the error is much less at lower frequencies and also lower under the more usual circumstance where the load point is at the bottom of the loop.

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- Mark

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Chavdar,

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In order to "complicate" this matter more.

The current theory about usage test simulators for receiving antennae postulates:

"these equivalent circuits ( Thevenin and Norton ) may be applicable to transmit system but not suitable for receive antenna system since the receiving antenna is actually a constant power source.

An accurate model for a receiving antenna should be a power source rather than a voltage or a current source".

Thus, the current NEC or EZNEC tools are for birds unless we are modeling transmit anteenae?

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Raphael

To Dan,

When the loop length is ?close to the wavelength there are standing waves ?- in your last data? from NEC (thanks for that) the ratio Imax/Imin is? 1.4 – the max. ?current ?being at the loop side and min. current at the load. I plotted ?the current vs. segment# and I can ?see a typical ?standing wave pattern. The current variation is not very large so with fixed current source measurement setup we can expect even at these frequencies to get a glimpse of the frequency response of the antenna factor . ??

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About loop as transmission line

Yes, I think that the loop can be considered as a ?transmission line. ?But since the loop is so short compared to the wavelength ?we cannot observe standing wave patterns ?at low frequencies when loop is shorted or opened. ?For me closed and open loop are both standing wave cases. ?By the way if we terminate one side of the small loop with? characteristic impedance of this transmission line ??we will get something? like traveling wave antenna?? - ??Flag antenna ?is a typical example. Flag becomes unidirectional when load R = Zc of this ?tr. line. ?But I will stop here – my knowledge in antenna theory is very very limited, I am just a standard NEC user.

? Chavdar lz1aq

?Hi Raphael,

I am sorry , ignore previous message #20287 it is wrong.

?The ?Current source resistances must be much higher than the anticipated LOAD resistance , ?so 2 KOm which are used are 20 times higher than the maximal expected input impedance of the amplifier (e.g. 100 ohms with input filter at highest frequency) .? NEC data show that the loop deliver ?constant current in very wide frequency range. ??So we will get frequency response of the loop amplifier if we assure constant current at its input . This is very simple? - even we can get almost correct frequency response without inductance at all – just we must feed the input with constant current . ?2 KOm will give current 5% less at 100 Ohm load ?compared to 6 Ohms load.? To increase the accuracy we can use e.g. 20 K? and the error will be 0.5% .? The problem is that 20 K will reduce the signal input substantially? with existing signal generators? and also parasitic capacitances of the resistors cannot be ignored e.g. ?1 pF at 30 MHz? = 5.5 kOm. ?The parallel ?inductance will go into action at low frequencies where its impedance is comparable to R load.

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Again: the internal resistance of the current source must be much higher than the anticipated ?R load (not the? impedance of inductance) . ??

Chavdar

At 7 MHz a 113 inch long .5 inch solid or tubular conductor (either one due to skin effect) the AC resistance (owing to skin effect) is 50 millOhms (0.05 Ohms).

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Verification would be welcome.

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Value from "Round" tool used here:

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de AA5CT Jim

I would agree with Fred.

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Despite what many articles and videos on the internet may suggest, including many professional ones, it is not easy to make practical and repeatable IMD measurements, especially at the IMD levels we are interested in.

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I've struggled with this, and have my own built band pass and notch filters and crystal oscillator based test sources. But even things like dirty connectors, especially things like BNC test leads, can cause results to vary considerably.

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The first step is to check that your IMD test rig is capable of producing results that are at least 10dB better than the device you are testing. Even this isn't really good enough, but it's something to aim for. Then try measuring known devices, such as the Mini-Circuits PGA-103+ or GALI-74+.

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I've always found these perform pretty closely to their published specification, and I tend to use one as a sanity check, before I try measuring something else.

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Regards,

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Martin

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On Mon, Oct 7, 2024 at 04:19 PM, Fred M wrote:

Hi Chin-Leong Lim ( the first time that i got right)

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Besides the TinySA, whose own IP3 is 17 dBm at best (thats your absolute limitation), the Signal generators Output Stages and even the Splitter with its Ferrite nonlinearity are also a source of Intermodulation which limit your measurement dynamics. Again, IP measurement of high IPs is demanding, dont't fall in your own traps.

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regards, Fred