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

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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

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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

Reply to
rubbishemail
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(I posted this two hours ago without a result, so please don't be offendet if you see two messages)

Hello,

I need to build a white noise source from 1 Hz to 10 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

formatting link
snipped-for-privacy@SpamMeSenseless.us.ibm.com

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

formatting link

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

Reply to
rubbishemail

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

--
 Thanks,
    - Win
Reply to
Winfield Hill

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.

--
 Thanks,
    - Win
Reply to
Winfield Hill

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

Reply to
1

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

Reply to
1

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