Can a Photodiode Operate Below 0.65V Without a Load Resistor?

[wirefree]

Homework Statement

Obtain open-circuit voltage for the shown circuit

Relevant Equations

1. Diode equation with ratio of diode to thermal voltages in the exponent
2. Ohm's Law

I greatly appreciate this opportunity to submit a query, and apologise for posting two simultaneous, but unrelated, queries on the forum.

I simply wonder if a lower than prescribed diode voltage (0.65V) for a given I_naught be considered for the case when there is no load resistor.

wirefree

 

[berkeman]

So by "open circuit" you mean if there is no load resistor connected?

Also, photodiodes are not normally operated in forward bias. Does the problem really specify a photodiode? If so, which one? Can you link to the datasheet?

For a good answer to this question, it's best to check out representative datasheets to see what they list for the v-i curves and values of Is, IMO.

 

[berkeman]

As an example, here is the first datasheet on my Google search for photodiode datasheet:

https://www.vishay.com/docs/81503/bpv10nf.pdf
All the specs I see in that datasheet are for reverse-bias or zero-bias.

But for a simple standard diode like the 1N4148:

https://www.vishay.com/docs/81857/1n4148.pdf
You will see specifications for both forward and reverse-bias:

 

[DaveE]

I'm not clear about what Io is. As berkemen said, it's unusual to see in a photodiode circuit. However, without the load resistor, the diode current will be larger and so will it's forward voltage, not less. The diode voltage can be determined from the diode equation in part 1, plus (in practice) maybe some series resistance.

 

[wirefree]

Many thanks for the responses. I am encouraged to enquire more.

My query arises from the given question, where the current source in parallel with a diode is simply a "model" for the photodiode.

I wondered what current would flow through the diode when there's no load attahced. Would it simply be Io, in which case Threshold Voltage breaches the given threshold of 0.65V by a few mV on the lower side.

I believe I've addressed berkeman as well as DaveE.wirefree

 

[berkeman]

The model has a problem, IMO. The current source should be in series with the photodiode, not in parallel. When there is no load attached to an illuminated photodiode, there is no current flow (aside from a reverse leakage current balanced by a slight net forward photocurrent). When you attach a load to the photodiode terminals, you get a photocurrent that generates a voltage across the load. That voltage is limited by the characteristics of the photodiode (or photovoltaic cell or LED, etc.).

Here is a picture that I took for another thread a couple months ago. It shows that when you shine a light on a common LED (a desk lamp is shining down from above in the picture), it generates a voltage across the DMM's input resistance.

 

[DaveE]

when you shine a light on a common LED (a desk lamp is shining down from above in the picture), it generates a voltage across the DMM's input resistance.

Yes, all diodes are photodiodes to some extent. I recall a troubleshooting problem I had once that was due to light on a diode in a clear package (the old axial 1N4148, I think). Switched to a diode in black plastic and solved it, but not before I was very confused for a while.

 

[willem2]

The circuit from the first post is fine. If you remove the load resistance, all of the current from the current source will go through the diode. You can calculate the voltage with the diode equation. Ohms law won't be needed because there are no resistors left in the model.

Photodiodes are normally used in reversed bias if accurate measurements are needed. If you connect the photodiode in series with a resistor in reverse bias, the voltage across the resistor will be proportional to the photocurrent.
If you do not use a reverse bias, the voltage will be set by the diode equation, and you will find that it will vary only little even with a large variation of the photocurrent.

This circuit without reverse bias is used for solar cells. If you used reverse bias the cell would consume electric energy instead of producing it.

 

[DaveE]

Photodiodes are normally used in reversed bias if accurate measurements are needed.

If you mean linearity of the response, then the most accurate configuration is to force zero volt bias across the diode. However, you are correct that they are usually used in the reverse bias mode to speed up the response. The zero volt bias has large junction capacitance and thus slow response. The worst choice for linearity is forward bias (voltage mode), like the way a PV generates power.

 

[berkeman]

Yes, all diodes are photodiodes to some extent. I recall a troubleshooting problem I had once that was due to light on a diode in a clear package (the old axial 1N4148, I think). Switched to a diode in black plastic and solved it, but not before I was very confused for a while.

Yep, I've been there too. One of the worst ones when we were bringing up a new ASIC in prototype ceramic packages that had open tops that you could glue metal lids on. We had the lids off trying to figure out a strange behavior when we realized that it was the light falling on the open IC that was causing the really weird behavior of the IC. Duh!

 

[wirefree]

Many thanks to all senior members and mentors who have posted here. I am proceeding with the information you have provided.

I am required to find two quantities next: (1) voltage across the diode with no load, and (2) voltage across a load that is at a resistance that affords maximum power transfer.

(1) I have obtained per willem2's suggestion. It is merely the diode threshold voltage, as dictated by I_naught, which, as I mentioned, came out to be a few mV lower than the prescribed 0.65V.

I need some clarifications pertaining to (2). For maximum power, of course, Maximum Power Theorem holds. My concern is: which resistance do I match the load against?

All thoughts appreciated.
wirefree

 

[willem2]

I need some clarifications pertaining to (2). For maximum power, of course, Maximum Power Theorem holds. My concern is: which resistance do I match the load against?

The Maximum Power Theorem does not hold for here. That only works if you have a voltage source in series with a resistor.
You'll have to find a equation for the voltage V across the diode. If you know that you can compute the current through the load, so you also know what the current through the diode is, and you can use the diode equation to derive an equation for V. Once you have that find the load resistance R^L that maximizes V²/R^L

 

[wirefree]

Thank you, willem2, for taking this forward.

It seems I didn’t comprehend your solution from your earlier post.

Voltage across the diode I already have, or don’t I? I believe we agreed that the voltage is back-calculated for the current Io, and it comes out to be 0.626 mV.

Next, I am tackling, with your guidance, the resistance issue. And I found this to get me started...

It’s explained in the document from which the above diagram is culled that Rf can be assumed to a mere ~50 Ohms.

To address your concern about the Max Power Thm holding true for voltage source in series with a resistance, clearly Thevenin’s gives me an equivalent model for the circuit given. To go ahead from that is where I seek advice.

Thank you.