Obtaining the series and shunt resistance of a photodiode from the datasheet

[louisnach]

TL;DR Summary

obtaining serie and shunt resistance of a photodiode from the datasheet

Hello,

Hello,

For a project , i need to modele a photodiode with a current source in paralelle with a shunt resistance and in serie with a resistance to use it in a bigger circuit. The photodiode we will use is SFH7050, the datashhet is provideed here https://www.osram.com/ecat/BIOFY® S...catalog_103489/global/prd_pim_device_2220012/

The résistances are not provided but the open circuit voltage and short circuit current for a given luminosity is provided at page 7 (i will only use it with a wavelength of 530 nm). So if i understand correctly can get the shunt résistance by dividing open voltage by short circuit current. Am i wrong ? How can i get the serie resistance ?

Thnaks a lot in advance !

 

[Tom.G]

Summary:: obtaining serie and shunt resistance of a photodiode from the datasheet

So if i understand correctly can get the shunt résistance by dividing open voltage by short circuit current. Am i wrong ? How can i get the serie resistance ?

That calculation will give you the Series resistance, (I calculated 6000 Ohms).

from: https://www.thorlabs.com/tutorials.cfm?tabID=787382FF-26EB-4A7E-B021-BF65C5BF164B
"An ideal photodiode will have infinite shunt resistance, but actual values may range from 10Ω to >10⁹Ω and is the resistance at the origin of the transfer curve."

from: http://www.osioptoelectronics.com/application-notes/an-photodiode-parameters-characteristics.pdf
"Practically it is measured by applying 10mV and measuring the current." That's not something done outside a VERY well equipped lab environment!

Looking at the Dark current curve of your device, pg.15 of the datasheet, implies the shunt impedance is >10⁸Ω >10¹⁰Ω. In the real world that would be ignored, but the simulator you use may require it.

(above links found with:
https://www.google.com/search?q=shunt+resistance+of+photodiode)

Cheers,
Tom

 

[Baluncore]

Looking at the Dark current curve of your device, pg.15 of the datasheet, implies the shunt impedance is >10⁸Ω.

I think you must move the decimal point in the other direction.
The shunt resistance is the dV/di slope of the Dark Current graph on page 15.
In the middle of the graph; dV = 5 volt; di = 0.5 nA;
∴ Rp = 5.0 V / 0.5 nA = 10.0 GΩ = 10¹⁰Ω

 

[Tom.G]

I think you must move the decimal point in the other direction.

Yup. Fixed. Thanks!