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

AI Thread Summary
The discussion focuses on connecting a photodiode to a differential amplifier and measuring current through an ammeter. The user questions whether the setup creates a closed loop necessary for current flow, noting that the ammeter is connected to ground and a resistor. Clarifications are provided regarding the typical configuration of photodiodes, including reverse biasing and the importance of using an op-amp in the circuit. It is emphasized that the ammeter is part of the circuit and must be properly configured to measure current. The conversation concludes with a reminder about the ammeter's settings and potential fuse issues.
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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.
 
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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.
 
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