LTC6560 Noise figure

Hi Giora,

You could calculate the LTC6560 Noise Figure if you knew your APD's noise current density. However, from looking at the literature, the APD noise calculation is complicated and involves many operating characteristics such as APD Gain or Multiplication factor (M), excess Noise Factor (F), etc. as outlined in this document link below:

https://www.hamamatsu.com/content/dam/hamamatsu-photonics/sites/documents/99_SALES_LIBRARY/ssd/si-apd_kapd9007e.pdf

The Noise Figure equation is:

NF = 10log [1+Na/Ni]

where Na is the LTC6560 input-referred noise current density, and Ni is the APD noise density.

Na (from graph above) @ 10MHz =~5pA/RtHz

Assuming your APD noise density at 10MHz is 2pA/RtHz (I've just made-up a number to illustrate the computation) -->

NF (LTC6560) = 10*log[1+5pA/RtHz / 2pA/RtHz ] = 5.4dB

Hope this helps answer your question.

Regards,

Hooman

Hooman,

Thanks for your answer.

Actually I have the noise density of the APD.

Giora

Hi Giora again,

That's great that you know your APD noise current density for your operating conditions.

Correction to my computation earlier: Both Na and Ni are Noise current power densities in A^2/Hz.

Here is the corrected computation of Noise Figure assuming the same conditions (Ni = (2pA/RtHz)^2 and Na = (5pA/RtHz)^2 both at 10MHz):

 NF = 10log [1+Na/Ni] = 10*log [1 + (5pA/RtHz)^2 / (2pA/RtHz)^2 ] = 3.5dB (instead of 5.4dB I'd incorrectly shown earlier).

Regards,

Hooman

Hello Hooman'

Some thing do not add to me:

[1 + (5pA/RtHz)^2 / (2pA/RtHz)^2] = 1 + 25/4 = 6.25

10 log[6.25) = 7.958

you got 3.5.

Can you explain.

Sincerely,

Giora

Hi Giora,

I guess we both made mistakes in our arithmetic (you once, me twice)!

 NF = 10log [1+Na/Ni] = 10*log [1 + (5pA/RtHz)^2 / (2pA/RtHz)^2 ] = 10*log(1+25/4) = 10*log(7.25) = 8.6dB

Check my work please.

Regards,

Hooman

Hi Hooman,

Now it is right.

Thanks.

Regards,

Giora