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NOISE FIGURE OF A BIPOLAR TRANSISTOR

J

Jeroen Belleman

Jan 1, 1970
0
In theory it's possible to synthesize a 50 ohm resistor using the
Miller effect, and end up with a resistor that has less than 4KTR
noise. Assuming the amplifier you use for the Miller effect is
ultra-low-noise. This could buy you 3db in the limit, assuming you
used this resistor as a termination. Don't know how practical this
really is.

It's quite practical. I've built amplifiers with 50 Ohm matched
inputs and 300pV/rtHz input-referred noise. I use them in the
beam trajectory measurement system of a particle accelerator.

The signal source isn't resistive. It's a capacitive position
pick-up. I'd have preferred to use Hi-Z amplifiers directly on
the pick-ups, but the radiation would kill them.
Back to the OP's topic, generally real products use JFETS. You can
get them with less than 2 nv/root-hz, I think. Some of those parts
have probably gone obsolete.

A BF862 does 0.8nV/rtHz. There are lower noise JFETS, but none
with Yfs/Cg as good as this one. Well, perhaps some RF devices
can beat that, but those lose out on 1/f noise.

Jeroen Belleman
 
R

Robert Macy

Jan 1, 1970
0
Linear Technology's LT1011 [I think it is] has around 1 nV/rtHz input
noise.
Again, from memory Supertex makes some FETs with less than 1nV/rtHz.

The BF862 is around 0.7 to 0.8 nV, with the usual JFET noise corner at 1
kHz or so.  The LT1028 and its ilk are around 0.9 nV, but the original
LT1028A has a nasty noise peak around 300 kHz that they don't tell you
about.  The datasheet noise plot conveniently ends well below that, even
though it's a 100 MHz op amp.  (Marketing again.)

The ADA4898 is a nice well-behaved part with around 0.9 nV noise.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot nethttp://electrooptical.net

THANKS!!!
first 1028, not 1011 Brain dead here.
second, I got bit by that spike! Not having any instrumentation of
value had to 'noodle' out what was wrong and did not understand until
now.

Really sad, because Linear parts are usually 'normal'
 
M

Mr Stonebeach

Jan 1, 1970
0
Two electrically-connected resistors have an equivalent thermal conductivity,
through their Johnson noise, but it's many orders below the thermal conductivity
of any electrical conductor.

The Wiedemann-Franz law, yes. But it doesn't apply if you have
superconductors
available. They conduct electricity but not heat. Or more accurately:
the
electron-mediated heat conductivity freezes out, but phonon-mediated
remains.

But it still is tough...

Regards,
Mikko
 
M

Mr Stonebeach

Jan 1, 1970
0
But you'd still be transferring k(T_hot-T_cold) per hertz just via
I**2*R, even with superconducting wires, right?

Yes, exactly, you would. The cooling effect via the Nyquist
mechanism would be there, and the effect would not get counteracted
by the heat backconduction. As John argued, with non-superconductors
any temperature difference (due to the Nyquist mechanism) would tend
to equilibrate via ordinary heat conduction.

My colleaques one floor up have actually been experimenting with
related effects.

Regards,
Mikko
 
S

Spehro Pefhany

Jan 1, 1970
0
The Wiedemann-Franz law, yes. But it doesn't apply if you have
superconductors
available. They conduct electricity but not heat. Or more accurately:
the
electron-mediated heat conductivity freezes out, but phonon-mediated
remains.

Thermal conductivity of superconducting Nb is not so bad. About 0.25*
RRR at 4.2K. Of course you can make the "wires" extremely thin since
the currents are very low.
 
J

josephkk

Jan 1, 1970
0
Hi all

In an article about low noise PU preamp for magnetic stylus
it was written that the bipolar transistors were chosen for their low noise figure .My question is how exactly a noise figure of a single bip[olar transistor is defined measured .

I assume by the context that by PU you are actually referring to some
audio pick-up preamplifier.

If your intension is to achieve best power match (source impedance =
load impedance) then some grounded base amplifier is the best choice
(a big grounded base 2N3055 or a half doxen grounded base transistors
in parallel).

However, typically magnetic pick-ups are designed for much greater
load impedance to give a flat (after RIAA correction) frequency
response.

So what do you exactlly want to do ?

And you are responding to this scammer WHY?

?-(
 
M

Mr Stonebeach

Jan 1, 1970
0
Thermal conductivity of superconducting Nb is not so bad. About 0.25*
RRR at 4.2K. Of course you can make the "wires" extremely thin since
the currents are very low.

At 4.2K Nb is still quite close to its transition temperature.
Normal
electrons (which do carry heat, as per two-fluid model) freeze away
exponentially
when the temperature is lowered below the transition. At half-Tc the
normal electron fraction still contributes significantly on the heat
conductivity.

Regards,
Mikko
 
G

Glenn

Jan 1, 1970
0
On 12/02/13 21.09, Tim Wescott wrote:
....
I've seen suggestions (in, for example, "Radio Frequency Design" by
Hayward) that you can get simultaneous power match and noise match in an
amplifier using transformer feedback (he called it "advanced feedback
methods"). I'm not sure if this gets you below that magic T/2 noise
temperature, though -- I haven't even done the math on the things, much
less built them and tried them out.

Hi Tim

Do you mean the Norton amplifier?:


Lossless Feedback Amplifiers: Theory and Advanced Techniques:
http://www.ko4bb.com/Manuals/08)_Misc_Ham_Equipment/Trask_-_Lossless_Feedback_Amplifiers.pdf
"...
Perhaps the single most significant development in high dynamic range
amplifiers has been that of the lossless feedback amplifier. Conceived
and patented by David Norton and Allen Podell of Adams-Russell (2), this
to- pology is often referred to as a Norton amplifier, and sometimes as
noiseless feedback.
...."

/Glenn
 
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