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Micpre of Graham

F

Fred Bartoli

Jan 1, 1970
0
Eeyore a écrit :
Yes. They sound similar though.

Graham

Not quite.
Shot noise has a strict flat PSD and is proportional to current, while
excess noise PSD is almost 1/F and is proportional to current^2.
 
K

Kevin Aylward

Jan 1, 1970
0
John said:
But metal film resistors don't have shot noise current.

Oh dear...
So, if a
transistor's bias current is mostly set by the voltage drop across an
emitter resistor, emitter current won't have much shot noise either.

Ahmmm. The voltage being set across a noiseless resister makes no difference
whatsoever to the internal shot noise of a transistor. Shot noise is the
statistical effect of carriers across the junctions.

The shot noise is

icn = sqrt(2.q.Icdc) acrross the emitter juction transfering through the
collecter
ibn = sqrt(2.q.Ibdc) acrross the emitter juction transfering through the
base.

Thats it. Nothing you can do will eliminate this shot noise.

Why dont you try it in spice? Hint, it dosnt model have a shot noise in
resisters!
So Ic certainly won't. It seems to me that base current shouldn't
either, but I'm not 100% sure about that.

Ic and Ib are both inherent. End of story.
 
J

John Larkin

Jan 1, 1970
0
The shot noise is

icn = sqrt(2.q.Icdc) acrross the emitter juction transfering through the
collecter
ibn = sqrt(2.q.Ibdc) acrross the emitter juction transfering through the
base.

Thats it. Nothing you can do will eliminate this shot noise.

Why dont you try it in spice? Hint, it dosnt model have a shot noise in
resisters!

Which is reasonable.
Ic and Ib are both inherent. End of story.

It is *not* the end of story.

If the metal-film resistor jams zero shot noise current into the
emitter, how can full shot noise emerge from the collector?

I'm not sure about the base current in the case where Ie has no shot
noise.

John
 
W

Winfield Hill

Jan 1, 1970
0
John said:
Kevin Aylward

Which is reasonable.



It is *not* the end of story.

If the metal-film resistor jams zero shot noise current into
the emitter, how can full shot noise emerge from the collector?

Right, thanks the gods. Kevin needs to take some measurements.
I'm not sure about the base current in the case where Ie has
no shot noise.

Surely the base-current shot noise is reduced. Worth checking.
 
J

John Larkin

Jan 1, 1970
0
Right, thanks the gods. Kevin needs to take some measurements.


Surely the base-current shot noise is reduced. Worth checking.


I checked the datasheet input current noise of a few "simple" opamps
(without bias current cancellation) as compared to input bias current,
but they all had much more than shot noise, so that was inconclusive.
Besides, I'd guess that internal to a linear IC, it's hard or
impossible to make a sub-shot-noise current source.

I've been interested for some time in the shot noise reduction in
metal resistors. I'd expect things like carbon comps, with a short
mean free electron path, to be bad, but I can't find decent references
and my measurements indicated well below full-shot. Some day when
things are slow, I'll have to do better experiments.

John
 
F

Fred Bartoli

Jan 1, 1970
0
John Larkin a écrit :
Do you mean noise power?

Shot noise current is proportional to the square root of DC current.

Isn't PSD *power* spectral density?
 
F

Fred Bartoli

Jan 1, 1970
0
Winfield Hill a écrit :
Right, thanks the gods. Kevin needs to take some measurements.


Surely the base-current shot noise is reduced. Worth checking.

I'd have expected the opposite (full base shot noise), so I just took
time to check it.

Added 1K at the emitter for obvious stability issues (about 150p+1M load).


15V >---.
|
.-.
Rc | |
| |1K
'-'
|
| ||
Adjust +---||--> SA
| ||
for Ve=5V |
___ |/
10.7V >----|___|--+-----| 2N3904
| |>
Rb --- |
--- | || ___
| +---||-|___|-> SA
SA <---' | || 1K
|
.-.
Re | |
| |1K
'-'
|
|
===
GND


Rb measured to 157kR, which is 31.8uA

Measurements taken at 10kHz at the emitter:
250nV/rtHz which seemed first wrong but when you realize that with
identical voltage drop in Rb and Re, you have Rb = beta.Re that makes a
nice by two current divider for the base noise current.

So, we have inb = 2*250.10^-9/157000 = 3.18pA/rtHz shot noise current,
which translates to 31.65uA Ib.

That's wonderfully close to the measured 31.8uA, so the base current see
full shot noise.

Then, just for sanity check, I measured the base noise voltage (with a
jfet buffer stage): 250nV/rtHz as expected.
 
F

Fred Bartoli

Jan 1, 1970
0
Fred Bartoli a écrit :
John Larkin a écrit :

Actually that's pretty easy to check with an LF SA (1M input)

15V >---.
|
.-.
Rc | |
| |1K
'-'
|
| ||
Adjust +---||--> SA
| ||
for Ve=5V |
___ |/
10.7V >----|___|--+-----|
| |>
Rb --- |
--- | ||
| +---||--> SA
SA <---' | ||
|
.-.
Re | |
| |1K
'-'
|
|
===
GND


For shot noise to dominate you need:
2q.I.R > 4kT or R.I > 52mV
That's easily satisfied.


Now your SA has about 10nv/rtHz noise floor so you want the measurement
to be higher.

Shot noise across R will be en^2=R^2.2q.I = 2q.R.RI
or en= sqrt(R) sqrt(2q.R_drop).

In the above test, we have 5V across each resistor.
Full shot noise should then develop:
en=sqrt(R)x1.27nV/rtHz

which is 40nv/rtHz for a full collector shot noise.
For base shot noise if you take beta=100, then you'll have 400nV/rtHz,
800nV for beta=400.

A small mistake was made there.
Since we have identical voltage drop across Re and Rb, we have Rb=beta.Re
This makes a current divider for the base noise current.

Then we'll see half that predicted noise voltage at the base point, and
half too at the emitter point (and obviously too at the collector).
 
K

Kevin Aylward

Jan 1, 1970
0
John said:
Which is reasonable.


It is *not* the end of story.

It is as far as internal shot noise goes.
If the metal-film resistor jams zero shot noise current into the
emitter,

This statement makes absolutely no sense. Internal shot noise has
absolutely nothing to do with any ability "to jam" noise free current in
from an
external source. The simplified model of shot noise assumes from the outset
that external sources are noise free. So, its simply irrelevant whether this
is via a metal film resister or teletubbies.

There is a shot noise generator associated with an ideal, otherwise noise
free, DC current source feeding the emitter, irrespective of whence the
current originates. This generator is due to the statistical nature of
charge carriers crossing a junction. This noise generator is directly across
the dynamic impedance re. This drops a voltage right at the base-emitter
junction.

Well, actually, rather than stating that there is a collector noise dropped
across re as one simplified convenient model, it is better to use the output
noise equivalent model, that is:

icn_shotout = (icn_in.re).gm = icn_in

That is, the collecter shot noise is directly at the internal ce nodes. This
internal noise current is a constant set by ICDC, irespective of any
external circuitary. Now..if there is an emitter resister, a simple
calculation on the equivelent circuit will show that the noise actually
generated into the external collecter load is:

in_load = icn_shotout/(1+Re/re), because the controlled source swips some
of the current away.

So, in an actual circuit, the output noise current, dropped across the load
resister, is reduced by Re feedback. However, again, this noise is not a
function of the nature of the driving source or resistance, in addition this
reduced output noise does not imply a better S/N ratio, as the signal also
gets reduced by Re. Of course, if Re generates additional noise, then this
will have to be added into the calculation.
how can full shot noise emerge from the collector?

I hope the above explains what Re does, but draw the equivalent circuit and
do the algebra by all means.
 
K

Kevin Aylward

Jan 1, 1970
0
Winfield said:
Right, thanks the gods. Kevin needs to take some measurements.
Why?


Surely the base-current shot noise is reduced. Worth checking.

I can only assume here Win, that you haven't actually read what I have wrote
and that you are making statements not related to inherent transistor,
junction caused, shot noise. To wit, internal base-current shot (and
collector) noise cannot be reduced by external components. This is trivially
obvious. The effect of such noise may well be modified by the external
circuit, as I stated in my original post, and show in my later one to John.
 
J

John Larkin

Jan 1, 1970
0
That is, the collecter shot noise is directly at the internal ce nodes. This
internal noise current is a constant set by ICDC, irespective of any
external circuitary. Now..if there is an emitter resister, a simple
calculation on the equivelent circuit will show that the noise actually
generated into the external collecter load is:

in_load = icn_shotout/(1+Re/re), because the controlled source swips some
of the current away.

So, in an actual circuit, the output noise current, dropped across the load
resister, is reduced by Re feedback. However, again, this noise is not a
function of the nature of the driving source or resistance, in addition this
reduced output noise does not imply a better S/N ratio, as the signal also
gets reduced by Re. Of course, if Re generates additional noise, then this
will have to be added into the calculation.


That was my point. If the emitter current source is noise-free, it
reduces the shot noise in the collector load; it must, since Ie = Ic +
Ib. Even if Ib has full shot noise, Ib is 1/b of Ie, so its
"diversion" noise current is small. In the case of the differential
mic preamp, if the transistors are biased by dc drops across big
resistors, no-signal emitter shot noise is nil. The AC gain isn't
correspondingly reduced, because AC-coupled emitter-emitter resistance
sets AC gain, and that contributes no shot noise current (although it
does add gain to the equivalent emitter noise.)

I suppose Ib has full shot noise because there's insufficient charge
interaction path in the base of a transistor to get the smoothing
effects seen in metallic conductors, and maybe because holes are
clumsier than electrons, and maybe because the carriers are
essentially thermally driven. Or something.

OK, next question, again for the simple emitter follower with a stiff
base supply and Rbias from emitter to ground. Assume a forced Ie that
is noiseless. Suppose we have some DC base current, say 1/100 of Ie.
Now suppose that base current transiently increases by d above the
normal current because of shot noise. That pumps 100*d current into
the emitter. That raises emitter voltage by 100*d*Rbias, reducing b-e
junction voltage, reducing base current. Since Rbias can be
arbitrarily large, the effective gain here can be large. Since this is
a feedback gain that reduces the base current, it sounds like we can
reduce base shot current by something like beta, 100:1.

If not, why not?

John
 
K

Kevin Aylward

Jan 1, 1970
0
John said:
That was my point. If the emitter current source is noise-free, it
reduces the shot noise in the collector load;

This is so obvious that wouldn't, and I didn't, think for a moment that you
are referring to this external to the transistor noise, as shot noise.
Resisters don't have shot noise, so clearly, one is going to ignore them as
shot noise generators.
it must, since Ie = Ic +
Ib. Even if Ib has full shot noise, Ib is 1/b of Ie, so its
"diversion" noise current is small. In the case of the differential
mic preamp, if the transistors are biased by dc drops across big
resistors, no-signal emitter shot noise is nil.

Yeah, we really have a terminology problem here. "Emitter (or Collector)
shot noise", just means the noise internal to the transiser to me.
The AC gain isn't
correspondingly reduced, because AC-coupled emitter-emitter resistance
sets AC gain, and that contributes no shot noise current (although it
does add gain to the equivalent emitter noise.)

I suppose Ib has full shot noise because there's insufficient charge
interaction path in the base of a transistor to get the smoothing
effects seen in metallic conductors, and maybe because holes are
clumsier than electrons, and maybe because the carriers are
essentially thermally driven. Or something.

I just don't understand what you are trying to say here.
OK, next question, again for the simple emitter follower with a stiff
base supply and Rbias from emitter to ground. Assume a forced Ie that
is noiseless. Suppose we have some DC base current, say 1/100 of Ie.
Now suppose that base current transiently increases by d above the
normal current because of shot noise. That pumps 100*d current into
the emitter. That raises emitter voltage by 100*d*Rbias, reducing b-e
junction voltage, reducing base current. Since Rbias can be
arbitrarily large, the effective gain here can be large. Since this is
a feedback gain that reduces the base current, it sounds like we can
reduce base shot current by something like beta, 100:1.

If not, why not?

All this transient forcing is all darkness to me. Look, draw the small
signal equivalent circuit. Put in i_base_noise generated noise across r_pi
(hfe.re), put i_collector across ce. put in the emitter resister. Add in any
externally generated noise, and churn the handle. If you dont want to do
that, do it is spice. You can even use ideal transistors, which still
generate shot noise. You can even get spice to list out contributions from
each component separately. To get thermally noiseless resisters, you can
use a VCCC connected to itself.

Assuming a stiff base supply, i.e. Rs=o, and neglecting, rbb' as well, In is
the base current noise, I get:

i_out_noise = In. Re.hfe/((1+hfe)(Re + re))

So, for the case of Re=0, we get i_out_noise = 0, as expected, to wit, ib is
s/c directly on the input, so no vbe.gm. For the general case, with hfe >>1:

i_out_noise = In.Re/(Re +re)

So, for large Re, i_out_noise ~ In, where In is due to noise from
sqrt(Ic/hfe.q)

Interestingly, Re allows for Ib to be feed directly into the output circuit,
essentially because ib no longer has a s/c across it.
 
M

Michael A. Terrell

Jan 1, 1970
0
Fred said:
Isn't PSD *power* spectral density?


In Graham's case its "Petty, Snide Donkey" ;-)


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
E

Eeyore

Jan 1, 1970
0
Fred said:
No- it is Petty Simpleton's Demise...

I'm waiting for your design that outperforms mine.

You see, you can't do it can you ?

Graham
 
J

John Larkin

Jan 1, 1970
0
I thought you misspelt that for a minute...

Regards
Ian

;-)

My doctor told me I had "shotty lymph nodes". He pronounced it very
carefully.

John
 
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