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OPA836: TMUX1101, charge sensitive amplifier

Adelson
Prodigy 60 points
Part Number: OPA836
Other Parts Discussed in Thread: TMUX1101

Tool/software:

Hi everyone,

I'm working on a design for a charge-sensitive amplifier with the OPA836 due to its quiescent current, slew rate and bandwidth.

Due to the operating principle, the charges generated by the photodiode must be integrated into the capacitor Cf as quickly as possible, so Rf must be high enough, the input impedance of the amplifier high enough (ideal JFET/CMOS) and Rin as low as possible.

Cin is a decoupling capacitor, defined as Cin 100nF. Cdetector (not shown in the figure) is about 1nF, so Cf must be lower than Cdetector to transfer charges from the input of the charge-sensitive amplifier to the output capacitance Cf..

Also i need to reset the Cf capacitor, so I selected TMUX1101 due to low leakage current, and low charge injection and suitable bandwidth, and it will be controlled by a microcontroller.

For RF=100MEG CF=1.5pF, Cin = 100nF and Rin = 10ohm:

  • Fcmin ≅ 1kHz, Fcmax ≅16 MHz
  • Vref =Vcc/2 
  • OPA836: Single supply Vcc=+3V3; Slew rate ≅ 250V/us (Fig 7.5);Ibias = 650nA; Voffset=65uV;
  • Input signal: Exponential with Trise = 50ns, Tfall = 500ns Vamp=100mV, period = 1MHz.

The Datasheet recommends Rf = 1k, and there is a graph showing the performance for different values. So, working with Rf = 100Meg the amplifier can't be stable? This Rf value is also proportional to the amplifier's common-mode impedance.



  • 0
    TI__Mastermind 26785 points

    Hello Adelson,

      Thank you for providing these details. I have two follow-up questions 

    1. The main purpose is you want to AC couple to avoid saturating the amplifier with DC ambient light, is that correct? Where is Rin value from?
    2. What is the purpose of the switch in the feedback. We usually see this type of application if gain switching is needed, but it looks like the switch is there as a short in the feedback.

       The graph is representing a standard voltage feedback gain configuration, but using the amplifier as a trans impedance configuration (TIA) would be different. I can run through the calculations with your requirements. 1nF photodiode capacitance is larger than what we usually see, you might need to split up the gain stages if that works for your design.

    Thank you, 

    Sima

  • 0 in reply to Sima Jalaleddine
    Prodigy 60 points

    1 - The sensor is a silicon photomultiplier. The main purpose of Cin is to decouple the DC signal of the bias voltage (SiPM bias circuit attached) and set a new offset by Vref. Rin is part of a RC filter, which also has Csipm and Cin. Hence, Rin was set at 10ohm to limit the low cut-off frequency of the RC formed by Rin and Csipm (10ohm and 1nF, respectively) and to set an input current proportional to the detector pulse voltage (Vin-Vref)/Rin.

    2 - The switch in the feedback is necessary to discharge quickly the capacitor, since I'm trying to work as an integrator of the input current. Each input pulse must be integrated and acquired by an ADC. The switch is used as a fast reset to make the amplifier ready to integrate the next incoming pulse.

    This large capacitance is related to the silicon photomultiplier, which are basically a set of single-photon avalanche diodes. They usually have a higher capacitance than other photodetectors.

    Basically, we're trying to acquire the integration of the input signal and reset the signal for the next pulse.

    This topology looks like a TIA, but the signal of interest is from the capacitance voltage (Cf).

    Thank you.