question on charge amplication

Discussion in 'General Electronics Chat' started by Nano001, Jan 27, 2010.

  1. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    Hello. I am trying to understand a simple charge amplifier circuit, such as the one attached. I have 2 questions. First, what is the purpose of the feedack resistor. Why can't a normal integrating amplifier be used to charge the feedback capacitor (without resistor). Also, why can't a normal inverting amplifier be used to amplify the current produced by the sensor (such as a photodiode). Why is the integrating-type circuit more attractive for these applications?
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    By using a "charge amplifier", the input node is held at a virtual zero volts because of the virtual ground at the input of the amplifier, therefore none of the charge is lost in "charging" the shielded cable capacitance. This makes it possible to detect tiny Δcharge from the sensor in the presence of large parasitic cable capacitances (long cable run).

    A simple charge amplifier uses just an R in the feedback path. The moving charge (current) from the sensor flows through the R to produce the output voltage from the amplifier (E = IR). The capacitor is sometimes added to act as a low-pass filter to roll-off high frequency noise.
     
  3. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    Thank you Mike. That is what I thought, however every online document I have been reading analyzes the charge amplifier as an integrating amplifier with a capacitor then adds the resistor as optional.

    Also, I am having trouble reading the responsivity of a photodetector I will be using. It is attached. I will be using a red LED, so at 650 nm the responsivity is about .4 A/W. What will the actual current be that is flowing out of the detector when a flash of red light hits it? Does it depend on the area of the photodetector? It is a PIN-10d. Can the background white light of the room be ignored? Thank you.
     
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Your detector will produce a steady-state current due to ambient light. There will be a small pulse riding on it when the flash hits it. Make the charge amp using only a resistor in the feed back path. Set the gain so that the pulse doesn't clip even with the brightest possible room lighting. Use a DC blocking coupling capacitor (differentiator) to extract only the pulse while the DC level due to room lights is blocked...

    Example: 2uA pulse riding on 10uA ambient level.

    btw: there is another way of handling this if you know exactly when the pulse is coming. it is appropriate to integrate the pulse if you can reset the integrator just before the pulse comes. If the pulse is riding on a level due to ambient light, the voltage from the integrator will have to sampled and held, otherwise the ambient level will continue to run up the integral. The integral will contain both the ambient and the pulse. I prefer to do it the way I simulated it, where the ambient is blocked out using the differentiator.
     
    Last edited: Jan 28, 2010
  5. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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    Thank you for that sims Mike. I am currently looking for a high input frequency op amp, once i get this I will implement this circuit and let you know how it goes.
     
  6. Nano001

    Thread Starter Active Member

    Jan 12, 2010
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