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Design an Ammeter

  1. Aug 10, 2009 #1


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


    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.
  2. jcsd
  3. Aug 10, 2009 #2


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    Staff: Mentor

    The photodiode is basically a current source, not a voltage source. The light that hits it generates a reverse photocurrent, which you amplify with a current-to-voltage converter circuit. The output of that circuit is a voltage, so you measure that with a voltmeter:


  4. Aug 10, 2009 #3
    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)
    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.
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