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

In summary: If I connect a load resistor to the light, like in the schematic, does that cause the circuit to short out and generate photo current? If you short the circuit, you will generate photo current.
  • #1
qwertypo
4
0
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.
 

Attachments

  • photocurrent.jpg
    photocurrent.jpg
    14.8 KB · Views: 64
Engineering news on Phys.org
  • #2
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.
 
  • Like
Likes qwertypo
  • #3
Merlin3189 said:
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 ?
photo_2022-01-04_20-51-43.jpg


Thanks.
 
  • #4
qwertypo said:
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.
 
  • Like
Likes qwertypo and DaveE
  • #5
qwertypo said:
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.
 
  • Like
Likes qwertypo
  • #6
DaveE said:
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.

Photocurrent.jpg
 
  • #7
berkeman said:
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.
 
  • #8
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?
 
  • Like
Likes Tom.G

1. Why is there no photocurrent in this short circuit geometry?

The absence of a photocurrent in a short circuit geometry can be attributed to the fact that there is no potential difference across the circuit. In order for a photocurrent to be generated, there needs to be a potential difference between two points in the circuit, which is not the case in a short circuit.

2. Can a photocurrent be induced in a short circuit geometry?

No, a photocurrent cannot be induced in a short circuit geometry as there is no potential difference to drive the flow of electrons. The circuit is essentially a closed loop, preventing the generation of a photocurrent.

3. What factors contribute to the absence of a photocurrent in a short circuit geometry?

The main factor contributing to the absence of a photocurrent in a short circuit geometry is the lack of a potential difference. Additionally, the materials used in the circuit may not be sensitive to the wavelength of light being used, or the intensity of light may be too low to generate a significant photocurrent.

4. How does the length of the circuit affect the generation of a photocurrent?

The length of the circuit does not have a direct effect on the generation of a photocurrent. However, a longer circuit may have a higher resistance, which could impact the flow of electrons and thus the potential difference across the circuit.

5. Is it possible to modify the short circuit geometry to sense a photocurrent?

Yes, it is possible to modify the short circuit geometry to sense a photocurrent. This can be achieved by introducing a potential difference in the circuit, either by using a different material or by altering the length or shape of the circuit. Additionally, using a different light source or increasing the intensity of light can also help in generating a photocurrent in a short circuit geometry.

Similar threads

Connect parallel batteries with wrong poles
    • Electrical Engineering
Replies
17
Views
1K
Determining inductive reactance from a 3-phase AC circuit
    • Electrical Engineering
Replies
17
Views
2K
Superposition Principle for two Voltage Sources (confusions about....)
    • Electrical Engineering
Replies
20
Views
2K
Why does a transistor in saturation act like a short circuit?
    • Electrical Engineering
Replies
5
Views
2K
Short Circuit Current on 161kv line
    • Electrical Engineering
Replies
11
Views
2K
Circuit analysis via polyhedron
    • Electrical Engineering
Replies
5
Views
1K
Alternator torque when connected to a phase-shifted AC source
    • Electrical Engineering
Replies
3
Views
561
What is the issue with my boost converter design and how can I fix it?
    • Electrical Engineering
2
Replies
55
Views
3K
Design of a short circuit kit for DUTs (switch, opamp, LEDs)
    • Electrical Engineering
Replies
21
Views
2K
Reflections at a short-circuit
    • Electrical Engineering
Replies
19
Views
3K

Hot Threads

Recent Insights

Back
Top