Re: Input capacitance
- From: Phil Hobbs <pcdhSpamMeSenseless@xxxxxxxxxxxx>
- Date: Fri, 03 Oct 2008 11:41:36 -0400
ggherold@xxxxxxxxx wrote:
On Oct 2, 4:30 pm, Phil Hobbs <pcdhSpamMeSensel...@xxxxxxxxxxxx>
wrote:
ggher...@xxxxxxxxx wrote:On Sep 30, 10:00 pm, Phil HobbsThis works for voltage noise, as in the two-point correlation method,
<pcdhSpamMeSensel...@xxxxxxxxxxxxxxxxxx> wrote:
ggher...@xxxxxxxxx wrote:"Thanks. A couple of BF862s will do a good job at AC, true. The
problem
is the amount of extra crapola you have to hang on them to get decent
DC
behaviour. (Also one begins to wonder why one bothers with an op amp
at
that point.)"
Yup, it would be getting very busy.
"The basic issue is that 300K resistors are so very noisy compared to
almost anything else, so you have to use really big ones to overcome
their Johnson noise, which in turn leads to all sorts of bandwidth
problems like this."
Hmm, could you use some sort of coorelation scheme to reduce the
noise? If you had two TIA's running off the same PD current you could
generate two signals Vs+Vn1 and Vs+Vn2, where Vs is the common signal
and the Vn's are the uncorrelated noises from each amp. Now if I had
a "common mode" amplifier, (as opposed to a differential amp), I could
amplify the "common" signal and not the noise... Unfortunately the
only thing I can think of is summing the two signals. Which "I
think" gives me a 1.414 increase in the SNR. Is there any way to do
better?
which does something similar--you measure the voltage twice and
cross-correlate the two measurements. It relies on time-averaging, so
you need stationary statistics. Thus you can do a great job measuring
noise, but not so great measuring signal.
An ensemble average, where you use 10 voltmeters, form the 45 unique
pairwise products, and average those, relaxes the requirement for
stationary statistics, but still measures the squared magnitude.
You win linearly with the number of voltmeters, because the improvement
is sqrt(#pairs) and the number of pairs goes as N**2. Note that it only
works for voltage noise--current noise is a real current that comes out
the input terminals of each amplifier, so they all measure the sum of
their noise currents, and this will survive the averaging since it's
indistinguishable from signal.
Making a linear measurement isn't possible that way, unfortunately. You
can get 3 dB reduction in the circuit noise by connecting a TIA to each
end of the PD, as you suggest (which is an interesting idea). You'd
actually want to subtract the results, because the two currents are
equal and opposite. It requires some fancy footwork on the power
supplies, of course, because you lose much more than 3 dB by not
reverse-biasing the diode.
" I have a new design idea based on using all current feedback thatquiet from a voltage gain perspective (T_noise ~ Tj/beta) but look like
more
or less avoids this problem by making the feedback currents very
quiet.
(This is done by dropping lots of voltage across the resistors in
the
current sources.) Still a partly-baked idea at this point, but it
should be possible to do a better job with a simpler circuit that
way."
Ahh do tell, or post a message/link if things pan out. Are you
talking about using CFA's or some other configuration? I've always
wanted an excuse to learn about CFA's.
Nothing magical. It starts from the observation that BJTs are very
150 kelvin resistors from a current gain perspective, like all ideal
diodes. So if you pick a superbeta transistor and connect the PD to its
base (like a phototransistor), you're stuck with 150 kelvin T_N and
significant nonlinearity. On the other hand, if you enclose it in a
current-feedback loop that pulls all but epsilon times the photocurrent
out of the base circuit, and optimize epsilon, you can win overall noise
temperature like sqrt(beta). All the feedback currents have to be way
sub-Poissonian, so you have to drop a lot of voltage across the emitter
resistors of the current sources, but in principle with a beta of 1000,
you can make a TIA whose noise temperature is of the order of 10 kelvin,
with a reasonably favourable tradeoff of noise vs bandwidth.
Vanilla CFB amps were initially puzzling to me back in the day, because
their bandwidth is not a strong function of the feedback gain. It
became a lot clearer when I imagined making a summing amp from a regular
VFB op amp--if you put a small resistor between the summing junction and
ground, the VFB amp's bandwidth becomes gain-independent as well. (CFBs
do this without the horrible noise degradation that the stupid resistor
would cause.)
Cheers,
Phil Hobbs- Hide quoted text -
- Show quoted text -
"Making a linear measurement isn't possible that way, unfortunately.
You
can get 3 dB reduction in the circuit noise by connecting a TIA to
each
end of the PD, as you suggest (which is an interesting idea). You'd
actually want to subtract the results, because the two currents are
equal and opposite. It requires some fancy footwork on the power
supplies, of course, because you lose much more than 3 dB by not
reverse-biasing the diode. "
Yes I was thinking about this last night. As you say you need to
hook a TIA to each end of the PD, and then you will have to subtract
(and not add) the two signals. And it looks like you lose the
ability to bias the PD. But I "think" I've seen the circuit trick to
beat that. (Though I haven't tried it.) So the non-invering input of
one TIA is grounded, and then you make the non-inverting input of the
other TIA the bias voltage. This does limit your bias voltage to be
less than the op-amp supply voltage, but I don't think that is much of
a constraint. It does look like it might limit the dynamic range a
bit....I'd have to draw all this out. It seems like one could "play
games" with where the ground was defined and still retain most of
dynamic range.
It's potentially a nice solution to the +-5V restriction on many of the good FET op amps e.g. the OPA656.(*) You can run the op amps off +15/+5 and -5/-15 (suitably protected against power supply sequencing funnies, loss of ground, and so on). You can do the level shifting like this:
----------*--RFRFRF-------*---R1R1R1--- +15
| | |
| | +15 |
| | |\ | |
| | | \| |
| ---|- \ | /
| | \ |<
| | >-----|
| | / |\
| +10---|+ / | \
| | /| | | -(1+R_F/R_1)*I_photo +I_bias
| |/ +5 | V
| |
| *------
| | |
| R |
| R |
----- R ------
/ \ R | - |
/___\ | | |
| GND | |----0 Out (plus bias)
| | | |
| R | + |
| R ------
| R |
| R |
| | |
| -5 *------
| |\ | |
| | \| | | (1+R_F/R_1)*I_photo - I_bias
| -10---|+ \ | / V
| | \ |/
| | >-----|
| | / |>
| ---|- / | \
| | | /| |
| | |/ -15 |
| | |
----------*--RFRFRF-------*---R1R1R1--- -15
This gets you back some of the dynamic range you lose by using the +-5V op amps. Also, the current multiplication relaxes the noise requirements on the subsequent stages.
Cheers,
Phil Hobbs
Put BJT follower in the feedback path (base to op amp output, emitter driving R_F), which produces (for level shifting by driving the base with the op ampwith a BJT in series with each feedback resistor
(*) TI wants everybody to use the 657, correctly pointing out that almost all TIAs run at some gigantic noise gain on account of the PD capacitance, so that you don't have to worry about using a decompensated op amp even though it looks like a unity gain application. The thing they don't tell you is that the 657 has twice the input capacitance of the 656, so it's worse at low photocurrents.
.
- Follow-Ups:
- Re: Input capacitance
- From: ggherold
- Re: Input capacitance
- From: John Larkin
- Re: Input capacitance
- References:
- Re: Input capacitance
- From: Phil Hobbs
- Re: Input capacitance
- From: ggherold
- Re: Input capacitance
- From: Phil Hobbs
- Re: Input capacitance
- From: ggherold
- Re: Input capacitance
- From: Phil Hobbs
- Re: Input capacitance
- From: ggherold
- Re: Input capacitance
- Prev by Date: Re: More Vista Woes
- Next by Date: Re: More Vista Woes
- Previous by thread: Re: Input capacitance
- Next by thread: Re: Input capacitance
- Index(es):
Relevant Pages
|