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DDC112: The current input range of DDC112

Eagle.Miao
Intellectual 910 points
Part Number: DDC112

Please help confirm the current input range of DDC112 charge to voltage?

Suppose 10Mhz modulated optical signal output na-pa level current through PD diode,

has this device been collected?

Thankyou very much.

  • 0
    TI__Expert 4680 points

    Hi,

    Not sure I understand the questions... What do you mean for na-pa?

    The bandwidth of the device is quite limited so the 10MHz carrier information would be lost, but if you are talking of an AM, you may be able to get the envelop as long as this changes slower than the view time (sampling rate). I.e., the average of the signal during that integration period. Honestly, never thought or tried.

    The current range at any given charge range setting can be obtained by dividing the charge setting  by the fastest/shortest integration time. Notice that that assumes a constant current during the whole integration time. As mentioned there are bandwidth limitations of the device (fast currents will make the input node swing as the amplifier bandwidth can not close the loop to set it as virtual ground). Faster/bigger currents than that could eventually damage the device.

    Regards,
    Eduardo

  • 0 in reply to Eduardo Bartolome
    Intellectual 910 points
    I mean what is the range of current input for DDC112? Thanyou！
  • 0 in reply to Eduardo Bartolome
    Prodigy 10 points

    Hi Eduardo ，

    Thanks for your fast response for this question but I'm still confused for the modulation frequency .

    As the figure3 show , the application is to use this device to collect the charge from PD which photocurrent range is 20pA-1nA , without regard to dark current, and the input optical modulation frequency is about 10mhz.

    -  As you mentioned that ,as long ad the signal change period is shorter than sampling time e.g. 500us or  integration time, the charge will be integrated in the feedback capacitance , then will be read out by AD convertor. It seems  a longer integration time will be helpful to a smaller charge capacity. 

    - Why the bandwidth will be limited , and where is it limited?

    Hope to hear from you , thanks. 

  • 0 in reply to Frank LIU
    TI__Expert 4680 points
    Hi,

    The input amplifier has a finite bandwidth (like any amplifier). This was designed to follow slow changing signals but no more (as you can imagine there are design trade-offs like power). The lost of closed loop gain as the frequency increases, increases the input impedance of the device and as such, the response of the device to that input will not be as one would expect (for instance the input node will not be virtual ground). It may work for the envelope but not to recover the 10MHz signal (I assume you mean MHz, not mHz...) That is obvious anyhow as the sampling rate is << 10MHz. But to see about the envelope, probably you would need to test it...

    Regards,
    Edu