Photodiode's legs to both the input for a differential amplifier

Click For Summary
SUMMARY

The discussion focuses on connecting a photodiode to a differential amplifier for current measurement. The user describes amplifying the voltage difference from the photodiode and converting it to current using a resistor before connecting it to an ammeter. The conversation clarifies that a complete circuit is formed when the ammeter is properly integrated, emphasizing the importance of correct configuration and settings on the multimeter. Additionally, it highlights the operational principles of both solid-state and vacuum photodiodes in various configurations.

PREREQUISITES
  • Understanding of operational amplifiers (op-amps) and their configurations
  • Knowledge of photodiode operation in photoconductive mode
  • Familiarity with current measurement techniques using ammeters
  • Basic electrical circuit theory, including closed loop principles
NEXT STEPS
  • Explore the design and application of differential amplifiers in signal processing
  • Study the characteristics and configurations of solid-state and vacuum photodiodes
  • Learn about resistor selection for optimal voltage drop in photodiode circuits
  • Investigate the setup and troubleshooting of multimeters in ammeter mode
USEFUL FOR

Electronics engineers, hobbyists working with photodiodes, and anyone involved in designing or troubleshooting circuits that utilize differential amplifiers for current measurement.

cks
Messages
164
Reaction score
0
Hihi,

I put a photodiode's legs to both the input for a differential amplifier

http://en.wikipedia.org/wiki/File:Op..._Amplifier.svg

Then, the voltage difference of the photodiode is amplified and its output is then connected to a resistor which is not shown in the above picture. The purpose I put a resistor is to convert the voltage output to current. Then, I connect in series of the output from the resistor to an ammeter to obtain the current. After which, the other end of ammeter is connected to ground.

May I ask, If I put one end of the ammeter to the ground and the other end to the resistor, will the system works, because from my high school knowledge of electric circuit, current flows only when a closed loop is formed, but I didn't see any closed loop here.

Thank you.
 
Last edited by a moderator:
Physics news on Phys.org


Often, the photodiode is put in series with a battery (photoconductive mode) and the current measured with an opamp in the current-measuring mode. For a solid state photodiode, the reverse bias is a few volts. For example, the negative end of a battery is connected to ground, and the positive end connected to the cathode of the photodiode 9reverse biased). The anode is connected to the op-amp inverting input, and the positive input of opamp tied to ground (inverting configuration). See second illustration in (inverting amplifier)
http://en.wikipedia.org/wiki/Operat...ations#Negative_impedance_converter_.28NIC.29
Photodiode replaces Rin resistor. Choose Rf resistor so that the expected photodiode current produces about 1 volt drop across it.
If you are using a vacuum photodiode (e.g., 935), the bias is usually 50 to 100 volts (or more), and the anode is tied to the + end of battery, and the cathode (light sensitive electrode) tied to the op amp inverting input.
 


If I understand the OP correctly, you do have a complete loop. Remember, when using a multimeter as an ammeter, the ammeter itself is part of your circuit. However, you must be physically pluged into the ammeter part of you multimeter (usually yellow) and your dial must be set to be an ammeter. Remember, you ammeter has a fuse in it that might already be blown.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
4K
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 23 ·
Replies
23
Views
5K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 28 ·
Replies
28
Views
6K