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Photodiode wich is fast enough to detect +50Mhz analog (sinus) signal??

Y

Yannick

Jan 1, 1970
0
Careful, using a sufficiently-wideband opamp can insure the summing-
junction impedance will be low compared to the total shunt capacitance.
Resistors have 0.05pF to 0.1pF of self capacitance, this should be your
total feedback capacitance. With 3k resistor you'd have a -3dB rolloff
at 530MHz. You want high R for low noise, so we'll try 100k, yielding
a 16MHz rolloff. Then we can apply the standard R-C-R trick (this is
more than 30 years old) to get a flat frequency response to 75MHz, or
whatever you decide your bandwidth should be.

R-C-R trick to get a flat frequency responce, i never heard of this
before , can you explain how it works? is it butterworth filter
specification that you mean?

thanks,

Yannick
 
W

Winfield Hill

Jan 1, 1970
0
Yannick wrote...
Winfield wrote...

R-C-R trick to get a flat frequency responce, i never heard of this
before, can you explain how it works?

I posted the circuit previously in this thread (20 July) and described
in detail how it works. Here's the circuit again:

| Rf R2 adjustable
| ,---/\/\---+---/\/\--/\/\----,
| | '--||--' | C2 R3 | nA-sensitivity wideband
| | Cf '--||--/\/\-- gnd | transresistance amplifier
| | |
| | __ ,-||--/\/\--+ correction network details
| input O--+---|+ \ | __ | R2 C2 = Rf Cf
| | >-+-/\/\-+-|- \ | R3 C2 sets bandwidth
| ,-|-_/ | | >-----+---
| | | gnd --|+_/
| gnd --/\/\--+-/\/\--' composite amplifier

The undesired Rf capacitance Cf is canceled by the R2 C2 network.
R3 is used to limit the upper frequency of this cancellation.

R2 C2 and R3 constitute the standard R-C-R trick. I thought of this
about 18 years ago, and have used it with great success since then.
Later I learned that it had been described in an old Keithley manual,
and probably in many other places years before that.

BTW, the circuit above will outperform (sensitivity, bandwidth, SNR,
phase accuracy) any of the resonant schemes you've been contemplating
here. It's not true that feedback makes things more noisy. In this
circuit feedback (and a high-performance composite amplifier) insures
that all of the signal current is used by the amplifier, rather than
becoming uselessly drained away by the input-node capacitance. Thus
feedback actually improves the SNR. Using a resonant input doesn't
solve the capacitance problem because if a high enough Q is used for
a solution, it simply creates insurmountable phase-error problems.
 
F

Fred Bloggs

Jan 1, 1970
0
Winfield said:
Yannick wrote...



I posted the circuit previously in this thread (20 July) and described
in detail how it works. Here's the circuit again:

| Rf R2 adjustable
| ,---/\/\---+---/\/\--/\/\----,
| | '--||--' | C2 R3 | nA-sensitivity wideband
| | Cf '--||--/\/\-- gnd | transresistance amplifier
| | |
| | __ ,-||--/\/\--+ correction network details
| input O--+---|+ \ | __ | R2 C2 = Rf Cf
| | >-+-/\/\-+-|- \ | R3 C2 sets bandwidth
| ,-|-_/ | | >-----+---
| | | gnd --|+_/
| gnd --/\/\--+-/\/\--' composite amplifier

The undesired Rf capacitance Cf is canceled by the R2 C2 network.
R3 is used to limit the upper frequency of this cancellation.

R2 C2 and R3 constitute the standard R-C-R trick. I thought of this
about 18 years ago, and have used it with great success since then.
Later I learned that it had been described in an old Keithley manual,
and probably in many other places years before that.

What R-C-R trick? It is the standard RC compensator turned around, and
as anyone with any observational ability knows, the transfer ratio of
currents reverse through the network is identical to the transfer ratio
of voltages forward through the network.
 
W

Winfield Hill

Jan 1, 1970
0
Fred Bloggs wrote...
What R-C-R trick?

Right, a trick is no longer a trick once you think of it
or someone tells you about it.
 
C

colin

Jan 1, 1970
0
Winfield Hill said:
Yannick wrote...

I posted the circuit previously in this thread (20 July) and described
in detail how it works. Here's the circuit again:

| Rf R2 adjustable
| ,---/\/\---+---/\/\--/\/\----,
| | '--||--' | C2 R3 | nA-sensitivity wideband
| | Cf '--||--/\/\-- gnd | transresistance amplifier
| | |
| | __ ,-||--/\/\--+ correction network details
| input O--+---|+ \ | __ | R2 C2 = Rf Cf
| | >-+-/\/\-+-|- \ | R3 C2 sets bandwidth
| ,-|-_/ | | >-----+---
| | | gnd --|+_/
| gnd --/\/\--+-/\/\--' composite amplifier

The undesired Rf capacitance Cf is canceled by the R2 C2 network.
R3 is used to limit the upper frequency of this cancellation.

R2 C2 and R3 constitute the standard R-C-R trick. I thought of this
about 18 years ago, and have used it with great success since then.
Later I learned that it had been described in an old Keithley manual,
and probably in many other places years before that.

BTW, the circuit above will outperform (sensitivity, bandwidth, SNR,
phase accuracy) any of the resonant schemes you've been contemplating
here. It's not true that feedback makes things more noisy. In this
circuit feedback (and a high-performance composite amplifier) insures
that all of the signal current is used by the amplifier, rather than
becoming uselessly drained away by the input-node capacitance. Thus
feedback actually improves the SNR. Using a resonant input doesn't
solve the capacitance problem because if a high enough Q is used for
a solution, it simply creates insurmountable phase-error problems.



--
Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)

i like your idea and admit that phase variance is a big issue for his
method, and i now see the significance of the R-C-R, but i cant agree with
you on this one issue of SNR is beter even than a tuned ciruit.

the capacitance will rob most of the signal before it gets to the amplifier,
the amplifier will add its own input noise
to this small signal. whatever feedback topology u consider the noise has
already been added by the time the feedback gets back to the inputs.

the feedback cant improve the SNR. it cant just selectivly amplify the
signal and not the noise. it merley recovers the signal strength lost into
the capacitor by extra amplification of the signal and noise.

the photodiode is like a curent source and the capacitor is an impedance. if
you convert this to the thevenin equivelent at the frequency of interest you
end up with a voltage source and src impedance and a normal inverting
amplifier with gain of Rfeedback/Rsrc.
the strenght of this thevenin voltage against amplifier input noise is what
sets the SNR. as the frequency rises the voltage will fall but so will the
src impedance hence the gain wil rise in proportion. the response remains
flat therefore but the SNR deterioates with rising frequency.

as you pointed out the dominant noise is set by the amplifier input
noise/ZCin*rfeedback.
ie this is the extra amplification thats going on as the frequency rises
wich is amplifing the input noise along with the reduced signal strenght.

tuning negates the efect of this capacitance without ading noise, hence
making the signal stronger at the input to the amplifier therefore the SNR
is better for the tuned frequency.

Colin =^.^=
 
P

Phil Hobbs

Jan 1, 1970
0
Winfield said:
BTW, the circuit above will outperform (sensitivity, bandwidth, SNR,
phase accuracy) any of the resonant schemes you've been contemplating
here. It's not true that feedback makes things more noisy. In this
circuit feedback (and a high-performance composite amplifier) insures
that all of the signal current is used by the amplifier, rather than
becoming uselessly drained away by the input-node capacitance. Thus
feedback actually improves the SNR. Using a resonant input doesn't
solve the capacitance problem because if a high enough Q is used for
a solution, it simply creates insurmountable phase-error problems.

Win,

I don't think this is true. Circuit elements are linear, which means that
for analysis purposes, you can replace the output of the amplifier with a
voltage source for frequency response purposes, and with ground for noise
purposes. The elements act independently. This is the origin of Thevenin's
theorem, for example.

The currents coming out of the amplifier output may cancel the signal nicely,
but as with any other currents coming from independent sources, the noise
powers always add. The compensation current is generated by performing a
measurement of the input current and controlling the amplifier output to
cancel it out. The compensation current is then no more accurate than the
measurement of the input current with the feedback disconnected.

The net is that feedback can improve the frequency response, but can't do
anything for the SNR. I've gone through the analysis for transimpedance amps
and bootstraps, and this principle certainly holds there. The complexity of
your circuit makes analytic treatment messy, but I don't see any different
physics operating.

Cheers,

Phil Hobbs
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Phil Hobbs <pcdhSpamMeSenseless@us
..ibm.com> wrote (in <[email protected]>) about
'Photodiode wich is fast enough to detect +50Mhz analog (sinus)
signal??', on Wed, 11 Aug 2004:
The net is that feedback can improve the frequency response, but can't
do anything for the SNR. I've gone through the analysis for
transimpedance amps and bootstraps, and this principle certainly holds
there. The complexity of your circuit makes analytic treatment messy,
but I don't see any different physics operating.

The trouble with determining what effect feedback has or doesn't have on
noise is that there are umpteen ways of considering the subject.

For example, the amplifier with feedback has less gain and a wider
bandwidth than the amplifier without feedback. To make a fair comparison
by my judgement, you must compare the noise output of the amplifier with
feedback to the noise output of another circuit incorporating the
amplifier which has the same lower gain and extended bandwidth (which
may not be possible).

But your judgement may favour a different comparison. Who is 'right'?

{I am, of course. (;-)}
 
W

Winfield Hill

Jan 1, 1970
0
John Woodgate wrote...
The trouble with determining what effect feedback has or doesn't have
on noise is that there are umpteen ways of considering the subject.

For example, the amplifier with feedback has less gain and a wider
bandwidth than the amplifier without feedback. To make a fair comparison
by my judgement, you must compare the noise output of the amplifier
with feedback to the noise output of another circuit incorporating the
amplifier which has the same lower gain and extended bandwidth (which
may not be possible).

This issue is nicely solved by referring all noise measurements
to the input where they can be compared to alternate approaches
and to the expected signal. Differing Bandwidth issues are dealt
with by using noise spectral densities.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Winfield Hill
about 'Photodiode wich is fast enough to detect +50Mhz analog (sinus)
signal??', on Wed, 11 Aug 2004:
John Woodgate wrote...

This issue is nicely solved

For sufficiently optimistic values of 'nicely', yes.
by referring all noise measurements
to the input where they can be compared to alternate approaches
and to the expected signal.

I agree that it is very useful to do that.
Differing Bandwidth issues are dealt
with by using noise spectral densities.

But if I have two configurations that have the same noise spectral
density but different bandwidths, the one with greater bandwidth
produces more noise.
 
W

Winfield Hill

Jan 1, 1970
0
John Woodgate wrote...
But if I have two configurations that have the same noise spectral
density but different bandwidths, the one with greater bandwidth
produces more noise.

Perhaps if measured with an RMS meter at its output, or if presented
to a threshold comparator, etc. Then you have a painful valid point.
But in many serious applications the signal will be post processed,
and the effects of any bandwidth differences obliterated. Assuming
an excess bandwidth is available from the input stage.
 
Y

Yannick

Jan 1, 1970
0
| Rf R2 adjustable
| ,---/\/\---+---/\/\--/\/\----,
| | '--||--' | C2 R3 | nA-sensitivity wideband
| | Cf '--||--/\/\-- gnd | transresistance amplifier
| | |
| | __ ,-||--/\/\--+ correction network details
| input O--+---|+ \ | __ | R2 C2 = Rf Cf
| | >-+-/\/\-+-|- \ | R3 C2 sets bandwidth
| ,-|-_/ | | >-----+---
| | | gnd --|+_/
| gnd --/\/\--+-/\/\--' composite amplifier

The undesired Rf capacitance Cf is canceled by the R2 C2 network.
R3 is used to limit the upper frequency of this cancellation.

I see, the feedback impedance is normally (with only Rf) decreased as
frequency increases due the resistors capacitance (+-0.1Pf). Now you
will increase the impedance after Rf due ground coupling with C2 and
R3 thus this will compensate for the loss in impedance at the front
andso will give a total feedback impedance wich stays the same as
frequency increases (for a given bandwidth ofcourse).

Now why using 2 amplifiers ? i see the second amplifier is an
integrator configuration, increases in frequency will lower the
feedback impedance andso lower the gain, is this for decreasing higher
harmonics of the signal ?

What opamps should i use? I am planning to work with the opa655

As for the noise, every amplifier will raise noise due his own noise +
increases the noise at the input with his gain. Can you explain how
this feedback gives a better SNR , i don't understand this !

thanks for the help soo far, i learned a lot

Yannick
 
W

Winfield Hill

Jan 1, 1970
0
Yannick wrote...
Winfield wrote...

I see, the feedback impedance is normally (with only Rf) decreased as
frequency increases due the resistors capacitance (+-0.1Pf). Now you
will increase the impedance after Rf due ground coupling with C2 and
R3 thus this will compensate for the loss in impedance at the front
andso will give a total feedback impedance wich stays the same as
frequency increases (for a given bandwidth ofcourse).

Now why using 2 amplifiers ? i see the second amplifier is an
integrator configuration, increases in frequency will lower the
feedback impedance and so lower the gain, is this for decreasing
higher harmonics of the signal ?

A composite opamp is needed here to get the extrapolated f_T up
into the GHz region so there's enough gain to drive the summing
junction to an impedance well below that of the paralleled input
capacitance, at 50MHz, when using high-value feedback resistors.

f_T >> 2pi Rf Cin fc^2 = 8GHz for 50MHz, 100k and Cin = 5pF.

Can you go back and read my post in this thread on July 20th, or
should I repost the relevant parts?

BTW, a 50MHz resonating tank needs a parallel inductance of 1uH
for say 10pF (the inductor and the preamp FET have capacitance).
If this has a Q of 75 at 50MHz, that's equivalent to a parallel
24k resistor, which makes 2x more current noise than a 100k
feedback resistor. BUT, the tank nicely steps up the voltage
by making it appear the photocurrent is dropped across the 16k,
where a noisy JFET can be used to further amplify it...
 
C

colin

Jan 1, 1970
0
BTW, a 50MHz resonating tank needs a parallel inductance of 1uH
for say 10pF (the inductor and the preamp FET have capacitance).
If this has a Q of 75 at 50MHz, that's equivalent to a parallel
24k resistor, which makes 2x more current noise than a 100k
feedback resistor. BUT, the tank nicely steps up the voltage
by making it appear the photocurrent is dropped across the 16k,
where a noisy JFET can be used to further amplify it...
Thanks,
- Win


ah now here is a point i hadnt bothered to consider before, the johnson
noise introduced by the resistance in the tuned circuit,
the inductor im using is a plastic tube cored toriod, i cant imagine it has
much loss. therefore this means most of the loss comes from the capacitance
wich is there and generating noise even without the tuned circuit. but i
would gues some of the loss comes from the resistance in series with both
the photodiode junction capacitance and mosfet input capacitance wich i
asume is already included in the devices noise figures.

incidently, is it worth considering putting several lower value resisters in
series to reduce the .3pf parasitic capcitance?
it makes it harder to analyse the efective resistance to use to calculate
the johnson noise especialy as the extra resistors arnt given, but it seems
likely the r-c-r will lower the feedback restance as seen by the input
therefore higher noise current, although the noise from the feedback
resistor isnt the most significant.

Colin =^.^=
 
W

Winfield Hill

Jan 1, 1970
0
colin wrote...
ah now here is a point i hadnt bothered to consider before, the
johnson noise introduced by the resistance in the tuned circuit

It's most useful to reduce all noise sources to their equivalent
input current noises, so they can be directly compared to your
photocurrent signal. The Johnson current noise density from a
resistor is i_n = sqrt (4kT/R) = sqrt (1.6E-20 / R). Since R is
in the denominator, this means you need high parallel resistance
to get low noise. Please see AoE page 431, etc, for more detail.
the inductor im using is a plastic tube cored toriod, i cant
imagine it has much loss. ...

It has a great deal of copper-wire resistance loss, which is
greatly increased at 50MHz over the DC value, from skin depth
and proximity loss. A 1uH inductor has 314 ohms of reactance.
If the effective ac resistance of the coil is 4.2 ohms, which
is pretty low, it'll have a Q of 75, which is pretty good at
50MHz. This can be conveniently modeled as a 75 * 314 = 23.5k
parallel loss resistance in your tank circuit. Which creates
about 2x more noise current than a 100k resistor.
 
Y

Yannick

Jan 1, 1970
0
I have done some noise calculations for my transimpedance amplifier to
detect signals up to 50Mhz , i have still some questions , this is
what i have found already :

for example i want to detect 2na current through my photodiode (C =
5pf at 20v reverse bias). Soo the noise in the feedback resistance
should be lower then 2nA. In² = 4KTdf/R = > df = 50Mhz, In = 5nA ,K =
1.38*10^-23 and T = 290°K
soo i get R >= 200Kohm to have less then 2na current feedback noise.

But then you also have equivalent voltage noise at the input of the
amplifier wich is 6nv/sqrt Hz for the OPA655 => this noise and the
Total input capacitance will give noise integration : vn² = KT/Ctot
with Ctot = Cd+Cf+Cin = 5pf+0.2pf+1pf = 6.2pf.

veq² = 645pV => veq = 25.4µv then this noise is increased due opamp
voltage gain R2/R1 with R2 = 250K and R1 = opamp input impedance
parrallelel with photodiode impedance wich will be about 500ohm for
50Mhz and 6.2pf total capacitance wich will give R2/R1 = 250K/500 =
500 voltage gain, soo Vnoise = 12.7mv at the output for this noise
contributor . is this correct of am i doing it all wrong?


i have also found something to reject daylight with succes, i put a
coil with a resistor from 20Volt power supply to my photodiode to
reverse bias it, then i use a 1nf capacitance between the photodiode
and inverting input of my OPA655. is this a good way of rejecting
daylight? i understand this resistor has to be high because of current
noise wich will be added to the total noise like the feedback resistor
does.

Further i was thinking of just using discrete components like a simple
HF transistor in place of the OPA655, then using a current mirror to
drive the basis current to get a low Rpi (small signal resistance)
(rpi = dDvbe/Dib)
so that the Rload of the photodiode is small, i didn't see at the
noise figure of the Transistor yet , anyone tried this ?

thanks for the help soo far,

Yannick
 
C

colin

Jan 1, 1970
0
Yannick said:
I have done some noise calculations for my transimpedance amplifier to
detect signals up to 50Mhz , i have still some questions , this is
what i have found already :

for example i want to detect 2na current through my photodiode (C =
5pf at 20v reverse bias). Soo the noise in the feedback resistance
should be lower then 2nA. In² = 4KTdf/R = > df = 50Mhz, In = 5nA ,K =
1.38*10^-23 and T = 290°K
soo i get R >= 200Kohm to have less then 2na current feedback noise.

But then you also have equivalent voltage noise at the input of the
amplifier wich is 6nv/sqrt Hz for the OPA655 => this noise and the
Total input capacitance will give noise integration : vn² = KT/Ctot
with Ctot = Cd+Cf+Cin = 5pf+0.2pf+1pf = 6.2pf.

veq² = 645pV => veq = 25.4µv then this noise is increased due opamp
voltage gain R2/R1 with R2 = 250K and R1 = opamp input impedance
parrallelel with photodiode impedance wich will be about 500ohm for
50Mhz and 6.2pf total capacitance wich will give R2/R1 = 250K/500 =
500 voltage gain, soo Vnoise = 12.7mv at the output for this noise
contributor . is this correct of am i doing it all wrong?


i have also found something to reject daylight with succes, i put a
coil with a resistor from 20Volt power supply to my photodiode to
reverse bias it, then i use a 1nf capacitance between the photodiode
and inverting input of my OPA655. is this a good way of rejecting
daylight? i understand this resistor has to be high because of current
noise wich will be added to the total noise like the feedback resistor
does.

Further i was thinking of just using discrete components like a simple
HF transistor in place of the OPA655, then using a current mirror to
drive the basis current to get a low Rpi (small signal resistance)
(rpi = dDvbe/Dib)
so that the Rload of the photodiode is small, i didn't see at the
noise figure of the Transistor yet , anyone tried this ?

thanks for the help soo far,

Yannick


It seems amplifier input noise voltage combined with detector capacitance is
the limiting factor, wich is why i went for very low noise discrete stage.
its a shame you cant go for tuned input stage, however the key to getting
rid of the noise is to use a narrow pas filter, or if you converting it to a
dc voltage you can just use a low pass filter with low cut off frequency
wich should reduce your noise greatly.

low noise discrete mosfets have about 1db noise wich i think equates to
about 1nv/hz-1/2. however i found one of the largest problems to be EMI
pickup.

I roughly calulated that at 3 meters with 5mw of laser power reaching the
target and 10% being reflected back in a 90' cone and a 1mm receiving area ,
that 50pw would be received by the photodiode wich at about 1A/W = 50pa.
with a 15 mm lense that increases to 12.5 nw, and an APD shld get at least
1.25 ua or much more, but my optical system is a bit of a lash up (lenses
glued to the pcb etc) and so i think i am seeing a lot less than this as its
very critical to focus it precisly, and it doesnt stay focused. i need to
make a machined housing to mount the lenses, detector and act as heatsink
for the laser etc.. but this probably isnt going to hapen.

AC coupling the signal from the photodiode is one way to reduce daylight
signal, or using an inductor in the feedback path, or as you sugest to bias
it, however dont forget as Win reminded me that inductors have resistance at
ac depending on their Q wich will add to noise, as well of course as well as
any paralel capacitance.

you could look at this circuit for ideas ..
http://www.imagineeringezine.com/PDF-FILES/fetamp0.pdf i think your idea of
using a curent mirror is a good idea, and would be easy in this circuit, you
might need to use a negative bias voltage for the mirror or bias the input
transistor a bit higher. you could easily arange the curent mirror so at the
frequencies you want to amplify it only fed back a smal fraction, thus
avoiding the need for the inductor (wich by the way are horible at picking
up EMI). you can get narow band light filters to filter out most of the
daylight and just alow your laser light through, but i dont think this wld
be necessary.

however stil the best way would be to mix it opto-electronicaly so it is
converted to a lower frequency before it is converted to an elctrical signal
wich is what im working on trying to do now and its looking quite good.

i gave up trying to get proper 455 khz ceramic filters or using resonators
as filters, and i found that the 20khz filter i made out of 4 inductors was
horible becuase its phase varied with signal strenght, i used toroids to
avoid interfernce but these are ungaped and so a bit variable and also means
the hysteresis of the feritte cuases a non linear efect wich i didnt
consider. Also ive gone for two IF now, wich makes it more complicated but
was necesary. it means i am not so limited to what high frequency i could
use. also the 2nd stage is now driven at constant amplitude so linearity
isnt an isue, obviously if i was to redisgn i would use an inductorless
filter circuit.


Colin =^.^=
 
Y

Yannick

Jan 1, 1970
0
It seems amplifier input noise voltage combined with detector capacitance is
the limiting factor, wich is why i went for very low noise discrete stage.
its a shame you cant go for tuned input stage, however the key to getting
rid of the noise is to use a narrow pas filter, or if you converting it to a
dc voltage you can just use a low pass filter with low cut off frequency
wich should reduce your noise greatly.

yes thats why i want my cutoff frequency at 50Mhz or a little bit more
soo jhonsson noise at higher frequencies doesn't gain much andso rms
noise will be much lower.
I roughly calulated that at 3 meters with 5mw of laser power reaching the
target and 10% being reflected back in a 90' cone and a 1mm receiving area ,
that 50pw would be received by the photodiode wich at about 1A/W = 50pa.
with a 15 mm lense that increases to 12.5 nw, and an APD shld get at least
1.25 ua or much more, but my optical system is a bit of a lash up (lenses
glued to the pcb etc) and so i think i am seeing a lot less than this as its
very critical to focus it precisly, and it doesnt stay focused. i need to
make a machined housing to mount the lenses, detector and act as heatsink
for the laser etc.. but this probably isnt going to hapen.

hmmm if i calculate it i get different values : lets see :

5mw , 10% reflected = 500µw

then this power will be dissipated on a half sphere surface = 2*PI*R²
, this will give us 2*PI*300² in cm².

soo 500µw/(2*PI*300²)= 884.19 pw/cm²

with a 15mm lens we get a detection surface of 1.5cm²*PI = 7cm² soo we
get 884.19*7pw = 6.24nw , this is 1/2 what you get, hmmm who is wrong
???
AC coupling the signal from the photodiode is one way to reduce daylight
signal, or using an inductor in the feedback path, or as you sugest to bias
it, however dont forget as Win reminded me that inductors have resistance at
ac depending on their Q wich will add to noise, as well of course as well as
any paralel capacitance.

i know they have capacitance and the larger the coil the more but the
resistance is always the same, isnt it? What do you mean with
resistance for ac? (i know the impedance is increasing with frequency
but the resistance???)and if i place a large resistor in serie with
the coil then the noise isn't a problem or is it?

you could look at this circuit for ideas ..
http://www.imagineeringezine.com/PDF-FILES/fetamp0.pdf i think your idea of
using a curent mirror is a good idea, and would be easy in this circuit, you
might need to use a negative bias voltage for the mirror or bias the input
transistor a bit higher. you could easily arange the curent mirror so at the
frequencies you want to amplify it only fed back a smal fraction, thus
avoiding the need for the inductor (wich by the way are horible at picking
up EMI). you can get narow band light filters to filter out most of the
daylight and just alow your laser light through, but i dont think this wld
be necessary.
hmmm very very interesting, it seems he is using a JFET for large
input impedance and low noise, then coupled this to a common basis
(cascoding) for low Rload and voltage gain whereafter he is buffering
it with a common collector. but why is the 4.7K at the end ? i am
gonne try this without the tuned stage ofcourse:)

however stil the best way would be to mix it opto-electronicaly so it is
converted to a lower frequency before it is converted to an elctrical signal
wich is what im working on trying to do now and its looking quite good.

hmmmm mix it optoelectronicaly , don't get it :)
i gave up trying to get proper 455 khz ceramic filters or using resonators
as filters, and i found that the 20khz filter i made out of 4 inductors was
horible becuase its phase varied with signal strenght, i used toroids to
avoid interfernce but these are ungaped and so a bit variable and also means
the hysteresis of the feritte cuases a non linear efect wich i didnt
consider. Also ive gone for two IF now, wich makes it more complicated but
was necesary. it means i am not so limited to what high frequency i could
use. also the 2nd stage is now driven at constant amplitude so linearity
isnt an isue, obviously if i was to redisgn i would use an inductorless
filter circuit.

u guys arent normal i hope i will ever be as good but this will take
years, i see that as a student in his last year of master course
doesnt know much about electronic design wich is quit frustrating:)


Yannick
 
C

colin

Jan 1, 1970
0
Yannick said:
hmmm if i calculate it i get different values : lets see :

5mw , 10% reflected = 500µw

then this power will be dissipated on a half sphere surface = 2*PI*R²
, this will give us 2*PI*300² in cm².

soo 500µw/(2*PI*300²)= 884.19 pw/cm²

with a 15mm lens we get a detection surface of 1.5cm²*PI = 7cm² soo we
get 884.19*7pw = 6.24nw , this is 1/2 what you get, hmmm who is wrong
???

you asumed a half sphere , i asumed a cone with a 90'angle (makes the maths
pretty easy as its roughly a circle of diameter 3m compared to a circle of
15mm) depends entirly on how reflective the surface is as tio wich is right
but its just to get a genral idea to at least be the right order of
magnitude. i tend to skimp as much maths as i can get away with.
i know they have capacitance and the larger the coil the more but the
resistance is always the same, isnt it? What do you mean with
resistance for ac? (i know the impedance is increasing with frequency
but the resistance???)and if i place a large resistor in serie with
the coil then the noise isn't a problem or is it?

wel the resistance of the inductor at dc is detemined by the measurable dc
resistance of the copper, but as the frequency increases the losses go up
due to skin efect and dielectirc losses, so the efective resistance
increases. this is the real part of the impedance and it generates johnson
noise, but as its in series with the inductor it is to some extent blocked,
or you can convert it to a parallel resistance.
hmmm very very interesting, it seems he is using a JFET for large
input impedance and low noise, then coupled this to a common basis
(cascoding) for low Rload and voltage gain whereafter he is buffering
it with a common collector. but why is the 4.7K at the end ? i am
gonne try this without the tuned stage ofcourse:)

yes the jfet and comon base are very much like a cascode pair with the top
transistor turned upside down, so that it acts as a level shifter, the 4.7k
load determines the open loop gain as the drain signal curent is flowing
through it. i think mosfets have lower available noise, compared to jfets.
hmmmm mix it optoelectronicaly , don't get it :)

i asume u are using a mixer to do some detecting ? well if u do this before
it is converted to a electrical signal using optical devices then u avoid
the capacitance, a mixer can be just a simple variable gain stage (as long
as you can vary the gain fast enough).
u guys arent normal i hope i will ever be as good but this will take
years, i see that as a student in his last year of master course
doesnt know much about electronic design wich is quit frustrating:)


Yannick

I'm sure you will be, it was my hobby since i was about 10, then shortly
after university i decided it didnt pay enough so i went into software but
now i prefer to do electronics.

Colin =^.^=
 
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