(Photodiode Model) Equivalent components when using voltage source as input

[frilance]

Hi!

First I want so say that I'm already familiar with the equivalent electrical circuit of a photodiode you can find in the bibliography.

But the thing is I'm having troubles in something I thought it should be straight forward. I want to build a new OpAmp-based transimpedance amplifier for a photodiode, but before I plug in the PD, I want to check the electrical performance using only electrical signals. That means I want to use a Voltage source as input and perform frequency response measurements.

Well, so let's say that my photodiode is "just" a current source in parallel with a huge resistor (MOhms) and a 17 pF (experimentally measured) junction capacitance.

To simulate the current source with a huge resistor in parallel, I thought I could use the voltage source with HighZ as series resistance. Applying Norton-Thevenin theorem what I get is exactly that (Current source in parallel with a big resistor) but then the amplitude of my current source is really tiny (Inorton=Vth/R) and it would not be sufficient to get the current values I would like to use (from 10uA to 1 mA). Of course then I would connect a capacitor between the neg input of the OpAmp and ground to simulate the junction capacitance.

In the file attached you can see the first "design". The photodiode will be disconnected when the "electrical model" is connected.



Does anyone have any suggestion?

Thanks a lot in advance.

Best regards

 

[Audioguru]

Your photo-diode is not a current nor a voltage source because it is reverse biased. Then light on it causes it to leak and become a resistance. Then it conducts some of its bias to the opamp input.
A photo diode without bias is a tiny solar cell that does generate a voltage and a current.

 

[dick_freebird]

First off, decide whether you want to use reverse bias or
use "photovoltaic" mode. The photocollection is basically
the same, but the (accessible) photovolume and how
much is collected by drift vs diffusion changes with bias.
A reverse biased photodiode will be faster as Cjo is less,
and more collection is via drift in the depletion region.
A diode operating with a large proportion of diffusion
collection may have a "tail" you won't like for high speed
digital uses, depending on lifetimes (which in a silicon
PIN type can be high indeed). App notes from something
in-your-application-niche probably hold useful clues and
cues.

 

[frilance]

Thanks a lot for your answers.

First, sorry, that is something I didn't mention but the photodiode will be reverse biased (high frequency response). The Vbias in the schematic will be 5 V.

Audioguru I don't get why you say the photo-diode is not a current source when reverse-biased.

With the photodiode is reverse-biased, there won't be any current if there is no light (or just leak/dark current, to be precise). Then when an optical signal illuminates the PD it will be absorbed, generating electron-hole pairs that will be swept/drifted by the applied reverse voltage generating then the so-called photocurrent, that's the reason I'm talking about equivalent current source. If the optical signal has an AC amplitude modulation on top of a constant DC value, then we will have an equivalent current source with both DC photocurrent and also an AC component. That's my situation.

The equivalent model can be seen in IEEE article for example **broken link removed**

I know the model, but my question is how to simulate it / implement it using a voltage source (instead of having an optical signal and a photodiode). In the PCB I will have the OpAmp-based TIA + whatever represents faithfully the photodiode when a voltage source is used. A capacitor to ground will be present, for sure, but I'm having problems trying to figure out what about the series/parallel resistors at the input.

I hope I explained me correctly this time.

Cheers

 

[Audioguru]

You want to replace the photo-diode with a voltage source for simulation testing. Then simply calculate (or measure) the photo-diode current, calculate its resistance then replace it with a voltage source feeding the same series resistance.