Photodiode, I don't know how it works here.

Thread Starter

Xenon02

Joined Feb 24, 2021
355
Hello !

I've got a problem understanding one thing.
I've got this graph
1659018867743.png
I know that the more light the photodiode takes the more current it creates. Just like the photodiode is like a current source but what happens if I cross this current U_B/R_L ?

1659018976794.png
Something like this ?
Because when I give more light it looks more like it not photodiode anymore it;s more like photovoltaic.
 

DickCappels

Joined Aug 21, 2008
9,297
That is exactly what’s happening -the junction becomes forward biased so the most voltage you can get is one diode drop per junction.

I have a friend who makes solar cells. One day at lunch, in answer to a question he remarked “We are in the diode business. Really big diodes.”
 

Irving

Joined Jan 30, 2016
3,127
Photodiodes can be used either in photoconductive or photovoltaic modes. If the diode is reversed biassed it is in photoconductive mode. Initially, when dark, the diode leakage is minimal, but as the light intensity increases the leakage current increases. This is the fastest response mode and is used for beam interruptors, IR receivers, etc. Normally its arranged so the junction never becomes forward biassed even at the highest light level.

https://en.wikipedia.org/wiki/Photodiode
 
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Thread Starter

Xenon02

Joined Feb 24, 2021
355
Photodiodes can be used either in photoconductive or photovoltaic modes. If the diode is reversed biassed it is in photoconductive mode. Initially, when dark, the diode leakage is minimal, but as the light intensity increases the leakage current increases. This is the fastest response mode and is used for beam interruptors, IR receivers, etc. Normally its arranged so the junction never becomes forward biassed even at the highest light level.
But If I make the diode make more current like in my image 6mA then it is photovoltaic ?
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
Or am I wrong ?
I mean if I cross this current that is Ub/RL then I change the photodiode from photodetector to photovoltaic ? Like there in my picture Ub/RL was max 5mA so when I will have 6mA because I gave enough light to make it so I change the mode ?
 

Irving

Joined Jan 30, 2016
3,127
But If I make the diode make more current like in my image 6mA then it is photovoltaic ?
A typical photodiode wouldn't generate that much current. The open-circuit voltage of a photodiode is about 430mV and the short-circuit current at 0v maybe <200uA.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
A typical photodiode wouldn't generate that much current. The open-circuit voltage of a photodiode is about 430mV and the short-circuit current at 0v maybe <200uA.
Okey but I mean the example I gave. I want to know theoretically if I changed the mode from photodetector to photovoltaic ?
I mean here What happens if I cross the Ub/RL. Is it photovoltaic ?

1659022825158.png
1659022981537.png
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
Its not how a photodiode works; there is no external current unless its zero-biassed. Maybe this series of articles will help:
https://www.allaboutcircuits.com/te...odiodes-the-nature-of-light-and-pn-junctions/
I still don't get it.
But tell me in the most simple way you can. What happens If I give enough light to get Ub/RL, and give a little bit more light than for the Ub/RL ? What happens ?

Will this happen ? Or what will happen ?
1659025746455.png
First image there is Ub/RL. So what happens if I give a little bit more light ? Like in the second one. Does it work like that ?
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
I feel a little bit dumb because I feel that this is very easy but my mind can't proccess it.
Because I know how the photodetector works to the moment of Ub/RL. But I don't know what happens if (theoretically) I will cross this value like in this picture Ub/RL = 5mA but now I give more light, and what now?
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
Academic.
I want to understand it. Just want to know what happens when I cross this Ub/RL that's all. The moment I have Ub/RL and when I give more light what will happen then.
 

Danko

Joined Nov 22, 2017
1,441
Just want to know what happens when I cross this Ub/RL that's all. The moment I have Ub/RL and when I give more light what will happen then.
When you cross Ub/RL, current through photodiode is maximal.
When you give more light, current still equal Ub/RL. Photodiode is saturated.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
When you cross Ub/RL, current through photodiode is maximal.
When you give more light, current still equal Ub/RL. Photodiode is saturated.

Aaaaa good to know.
Because I was thinking that maybe the diode will produce even more current because this photodiode as a photodetectore mode is working like current source.
 

Danko

Joined Nov 22, 2017
1,441
Because I was thinking that maybe the diode will produce even more current because this photodiode as a photodetectore mode is working like current source.
In this circuit battery produces voltage, RL defines maximal current, but photodiode only limits current at level, defined by light intensity.
 
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Irving

Joined Jan 30, 2016
3,127
As has been said, there is a limit on the actual current that will be output. For a photodiode that's intended to work in reverse-biassed or zero-bias configurations that limit will be at around 1mW/cm2 or 10W/m2 which is why data sheets generally only show values up to that point. 10W/m2 is typically medium to bright indoor lighting (1000W/m2 is generally taken as the maximum irradiation from the sun overhead on a cloudless day, and is used for solar panels and other true photovoltaic cells).

You can think of a photodiode as a perfect diode in parallel with a current source and a parallel load resistor Rp, plus a series resistor Rs to the external connection. The current source is linear with respect to the light power up to a certain limit but can't go above that (there only being so many electrons in the depletion layer that can be freed). The current source biasses the diode in the forward direction. Once the diode is starts conducting the current mainly flows through it giving an output voltage of typically 0.4-0.6v; when the output is shorted the photocurrent flows externally and the voltage is near to zero. These two states are illustrated in the simulation below:

1659093508507.png

The following simulations show what happens when the diode is is nominally reverse-biassed, but with different load resistances, RL < RLmax, RL = RLmax and RL > RLmax. In the middle case the diode just remains reverse or zero-biassed at max illumination, while last shows the diode going into forward-bias at high illumination.

1659096840430.png
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
As has been said, there is a limit on the actual current that will be output. For a photodiode that's intended to work in reverse-biassed or zero-bias configurations that limit will be at around 1mW/cm2 or 10W/m2 which is why data sheets generally only show values up to that point. 10W/m2 is typically medium to bright indoor lighting (1000W/m2 is generally taken as the maximum irradiation from the sun overhead on a cloudless day, and is used for solar panels and other true photovoltaic cells).

You can think of a photodiode as a perfect diode in parallel with a current source and a parallel load resistor Rp, plus a series resistor Rs to the external connection. The current source is linear with respect to the light power up to a certain limit but can't go above that (there only being so many electrons in the depletion layer that can be freed). The current source biasses the diode in the forward direction. Once the diode is starts conducting the current mainly flows through it giving an output voltage of typically 0.4-0.6v; when the output is shorted the photocurrent flows externally and the voltage is near to zero. These two states are illustrated in the simulation below:

View attachment 272487

The following simulations show what happens when the diode is is nominally reverse-biassed, but with different load resistances, RL < RLmax, RL = RLmax and RL > RLmax. In the middle case the diode just remains reverse or zero-biassed at max illumination, while last shows the diode going into forward-bias at high illumination.

View attachment 272488
So the last one even if it's reverse biased it is going into forward and it is becoming fotowoltaic ?
Even though it looks like that :

1659099511839.png
It is going from reverse bias to forward bias what does it mean ?
In this circuit battery produces voltage, RL defines maximal current, but photodiode only limits current at level, defined by light intensity.
I can understand that more light won't produce more current. The RL give the limit.
 

Irving

Joined Jan 30, 2016
3,127
Once the diode starts conducting, because the load resistance doesn't allow enough current from the battery to maintain reverse bias, the photocurrent is absorbed by the diode pn junction and voltage across the diode is clamped at -0.5v approx. leaving the external current fixed at (5 - -0.5)/RL or about 110uA.

You were earlier considering the diode in photovoltaic mode behaving like a battery or voltage source, but its not, its more akin to a poor current source with a voltage clamp across it. Definitely non-linear.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
Once the diode starts conducting, because the load resistance doesn't allow enough current from the battery to maintain reverse bias, the photocurrent is absorbed by the diode pn junction and voltage across the diode is clamped at -0.5v approx. leaving the external current fixed at (5 - -0.5)/RL or about 110uA.

You were earlier considering the diode in photovoltaic mode behaving like a battery or voltage source, but its not, its more akin to a poor current source with a voltage clamp across it. Definitely non-linear.
As I understand it.
Even though in your simulation the reverse bias is going into forward bias it is still photodetector and not a photovoltaic yes ?
 
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