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shot noise source

Hello,

I need to build a white noise source from 1 Hz to 100 kHz (the more the
better)
and came across a photodiode shot noise source which seems fine because
it can
be calculated and I don't have the facilities to calibrate a zener
noise source.

Message from Phil Hobbs
http://groups.google.com/group/[email protected]

Now I need a concrete design idea:


a) use an inverting amplifier with a big feedback resistor (1 G)
Illuminate the diode for 1 nA Photocurrent.
This gives me 1.8e-14 A/sqrt(Hz) current noise,
with 10 KHz bandwidth 1.8 mV after the amplifier.
LF411 has around 10 fA/sqrt(Hz) noise, this is not negligible with
respect to the shot noise.
In addition, johnson noise from the feedback resistor is to large,
since
the voltage is below 50 mV (2kT/e)

b) use 1 mA photocurrent and see how to get rid of the DC before
amplification.
Use the active load circuit from Zetex application note AN3, figure 6.



------|------- +Vee
|1k5 |
..-. |33n
| | ---
| | --- ___
'-' | ||___||
| | | |
| | 330k | |
| | ___ ||\| |
|---|-|___|--------------|-\ | ||
/| | | >------||-
| | GND-|+/ ||
|-------------- |/| out
-
^
|photodiode
|
|
------------ -Vcc
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


Probably I would use a current source to sink the current from the
photodiode
and use negative feedback from the output of the OPA to the light
source to have
a better response at low frequencies

Do you have any comments?

Many thanks for your feedback


Daniel
 
W

Winfield Hill

Jan 1, 1970
0
[email protected] wrote...
Hello,

I need to build a white noise source from 1 Hz to 100 kHz (the
more the better) and came across a photodiode shot noise source
which seems fine because it can be calculated and I don't have
the facilities to calibrate a zener noise source.

Message from Phil Hobbs
http://groups.google.com/group/[email protected]

Now I need a concrete design idea:


a) use an inverting amplifier with a big feedback resistor (1 G)
Illuminate the diode for 1 nA Photocurrent.
This gives me 1.8e-14 A/sqrt(Hz) current noise,
with 10 KHz bandwidth 1.8 mV after the amplifier.
LF411 has around 10 fA/sqrt(Hz) noise, this is not negligible
with respect to the shot noise.
In addition, johnson noise from the feedback resistor is too
large, since the voltage is below 50 mV (2kT/e)

b) use 1 mA photocurrent and see how to get rid of the DC before
amplification.
Use the active load circuit from Zetex application note AN3, figure 6.



------|------- +Vee
|1k5 |
.-. |33n
| | ---
| | --- ___
'-' | ||___||
| | | |
| | 330k | |
|---|-|___|--------------|-\ | ||
/| | | >------||-
| | GND-|+/ ||
|-------------- |/| out
-
^
|photodiode
|
|
------------ -Vcc

Probably I would use a current source to sink the current from the
photodiode and use negative feedback from the output of the OPA
to the light source to have a better response at low frequencies

Do you have any comments?

Many thanks for your feedback


Daniel

Silly man, doing all that work, getting only 1nA. Improve the
optical pathway. Turn up that LED current! Use 1mA, use 20mA,
whatever.
 
1

1

Jan 1, 1970
0
Silly man, doing all that work, getting only 1nA. Improve the
optical pathway. Turn up that LED current! Use 1mA, use 20mA,
whatever.

Do I understand it correctly that increasing the current (and therefore
decreasing Rfeedback) I will never be able to reach shot noise limit
2kT/e, since Uout = sqrt(2 e I) * Rfeedback = sqrt(2 e I) * Ucc/I < 2kt/e
(DC coupling)
So you would go with the current sink?


Thanks


Daniel
 
W

Winfield Hill

Jan 1, 1970
0
1 wrote...
Do I understand it correctly that increasing the current (and therefore
decreasing Rfeedback) I will never be able to reach shot noise limit
2kT/e, since Uout = sqrt(2 e I) * Rfeedback = sqrt(2 e I) * Ucc/I < 2kt/e
(DC coupling)

Think carefully, if you choose a current-sense resistor high
enough so the DC photodiode current drops significantly more
than 50mV dc, such as 1V, etc., the noise developed across
that resistor will be entirely shot noise. You can then ac
couple and amplify to get your desired output noise signal.
 
1

1

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



Think carefully, if you choose a current-sense resistor high
enough so the DC photodiode current drops significantly more
than 50mV dc, such as 1V, etc., the noise developed across
that resistor will be entirely shot noise. You can then ac
couple and amplify to get your desired output noise signal.
ok, my mistake was to assume the _noise_ current had to be larger than
50 mV. With 1 mA photocurrent I have 1.7 µV behind the amplifier (1k
feedback) over 10 kHz bandwidth and this can be AC-coupled with a second
amp.


many thanks


Daniel
 
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