Why in this short circuit geometry can I not sense a photocurrent?

[qwertypo]

Hi, I'm wondering why shorted circuit geometry like figure 2 did not sense photocurrent?
Even if the the circuit composed like 2, I guess that by the Kirchhoff's Law, voltage should apply to the ampere meter and photocurrent should be sensed. But in real experiment, I found that shorted circuit geometry like 2 cannott sense the photocurrent. What am I missing ? Thanks.

 

[Merlin3189]

Maybe you are forgetting the internal resistance of the photodiode (even though you show it in the picture.)
The diode generates a voltage and a current flows through the short circuit. But if you think about the resistances, the voltage source is in series with the internal resistance and the resistance of the short. The short is a very low resistance (ideally zero), so nearly all of the voltage is across the internal resistance and only a tiny voltage (ideally zero) across the short. The ammeter sees only this very tiny voltage and very little current flows in it. Almost all the generated current goes via the short.

 

[qwertypo]

Maybe you are forgetting the internal resistance of the photodiode (even though you show it in the picture.)
The diode generates a voltage and a current flows through the short circuit. But if you think about the resistances, the voltage source is in series with the internal resistance and the resistance of the short. The short is a very low resistance (ideally zero), so nearly all of the voltage is across the internal resistance and only a tiny voltage (ideally zero) across the short. The ammeter sees only this very tiny voltage and very little current flows in it. Almost all the generated current goes via the short.

I'm really thank you for your answer. But I cannot draw the circuits as you requested. Could you give me some hint to understand the phenomenon as circuit V,R(internal resistance),r(shorted resistance), and ampere meter?

I keep drawing the circuit like the below picture. it shows that the ampere meter should sense the current. How can I draw the right circuit geometry ?

Thanks.

 

[berkeman]

Hi, I'm wondering why shorted circuit geometry like figure 2 did not sense photocurrent?
Even if the the circuit composed like 2, I guess that by the Kirchhoff's Law, voltage should apply to the ampere meter and photocurrent should be sensed. But in real experiment, I found that shorted circuit geometry like 2 cannott sense the photocurrent. What am I missing ? Thanks.

What is the element you are trying to measure? Is it a photodiode or a Light Dependent Resistor (LDR)? If it is a photodiode, you should sense the photocurrent with your ammeter. If it is an LDR, you should be able to sense the change in resistance on the resistance measurement setting of your DVM.

Can you give a link to the datasheet for the component that you have in series with the ammeter and are shining a light on? Thanks.

 

[DaveE]

I'm really thank you for your answer. But I cannot draw the circuits as you requested. Could you give me some hint to understand the phenomenon as circuit V,R(internal resistance),r(shorted resistance), and ampere meter?

I keep drawing the circuit like the below picture. it shows that the ampere meter should sense the current. How can I draw the right circuit geometry ? View attachment 295083

Thanks.

Your schematic and equations are correct (the one on the left). The resistor $r_i$ represents leakage paths around the actual PV junction and will have a extremely high value, maybe $10M \Omega$ or so, so it is often left out of the model.

You should see photo current if nothing is wrong. Is your ammeter sensitive enough? You can also measure the voltage across a load resistor to see the current, although it won't be a true short circuit then. You can test your meter with a PS and resistor. Assuming it is a simple photodiode, you can test it with your DMM just like a conventional diode.

@berkeman is right, you'll want to find a datasheet for your diode.

 

[qwertypo]

Your schematic and equations are correct (the one on the left). The resistor $r_i$ represents leakage paths around the actual PV junction and will have a extremely high value, maybe $10M \Omega$ or so, so it is often left out of the model.

You should see photo current if nothing is wrong. Is your ammeter sensitive enough? You can also measure the voltage across a load resistor to see the current, although it won't be a true short circuit then. You can test your meter with a PS and resistor. Assuming it is a simple photodiode, you can test it with your DMM just like a conventional diode.

@berkeman is right, you'll want to find a datasheet for your diode.

I think I should explain why I come up with the problem. Now I am measuring photovoltaic material. I deposite electrode top and bottom of the PV material, and check the resistance with a multimeter. It shows over Giga ohms order (Out-of-range for multimeter).

Running the PV measurement on that condition showed the PV current value on the ampere meter. However, during the measurement, PV current suddenly was gone. In that case, when I measured the resistance of PV material, the two ends showed kilo ohms order resistance. So I thought that when the PV material is shorted during the measurement, I could not measure PV effect. So I wanted to draw the circuit to understand the phenomenon. But it failed. Thanks.

 

[qwertypo]

What is the element you are trying to measure? Is it a photodiode or a Light Dependent Resistor (LDR)? If it is a photodiode, you should sense the photocurrent with your ammeter. If it is an LDR, you should be able to sense the change in resistance on the resistance measurement setting of your DVM.

Can you give a link to the datasheet for the component that you have in series with the ammeter and are shining a light on? Thanks.

Thanks! I don't have the datasheet now. But I posted more detailed explanation of the situation on the above post.

 

[berkeman]

Try your experiment with a simple LED; you don't even need a PV cell to see a photocurrent. Connect your DVM in current mode across the terminals of the LED and hold the LED under a light. You will see the reverse photocurrent. Then you can compare that to your home-brew PV cell.

Perhaps there is an issue with how you constructed your DIY PV cell?