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how to test the AD8605 as TIA (I to V) design using current source?

manikandan.devarasu@hcl.com
on Oct 2, 2019

Hello,

I am trying to test the Trans-impedance amplifier design using AD8605 for the photo-diode application circuit.

currently i am using an NI 9265 module to generate current source up to 20mA in my setup 

below is my circuit:

 

when i try to set the current to 500 micro Amps from the sources and feed to inverting terminal of AD8605; the output voltage of I-V amp is in few millivolts (~14mV).

But actual result is expected to be saturated at high level 3.3V.

is this correct or i am doing something wrong ?

please provide me the way to test the Trans-impedance circuit output with the current sources(micro amps range).

thank you,

mani

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  • +1
    •  Analog Employees 
    on Oct 2, 2019 11:44 PM over 1 year ago

    Hi Mani,

    With a single supply setup as yours, make sure the current source input direction is such that it'd move the output higher (positive direction). Otherwise, you're forcing the output further towards ground, which it won't move.

    So, input current source I1 has to be positive in the direction of the arrow on your schematic, for the correct output shift direction.

    Regards,

    Hooman

  • 0 on Oct 3, 2019 3:44 PM over 1 year ago in reply to Hooman

    Hello Hooman,

    Thanks for the response.

    I tried the current sourcing direction in other way,in that case the TIA output seems to be at high when input is fed with very little current in microamps(40uA).

    Do you have any protocol / procedure to generate and feed the TIA circuit to validate the design.

    Thanks for your time.

    Regards,

    Mani

  • +1 on Oct 3, 2019 4:40 PM over 1 year ago in reply to manikandan.devarasu@hcl.com

    Hi Mani,

    Solving of this problem depends from tolerance and value of current which you need. As start, you can try high-ohm resistor and external voltage source in series. 

    Transimpedance amplifier has low input impedance, which equal Feedback resistor value divided by (1+Aol), here Aol is open-loop gain of op-amp. Therefore, input current will be equal External voltage divided by (External resistor plus inpur resistance of TIA). Keep the right current direction - from TIA input.

    I hope that this help you.

    Regards,

    Kirill 

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  • +1 on Oct 3, 2019 4:40 PM over 1 year ago in reply to manikandan.devarasu@hcl.com

    Hi Mani,

    Solving of this problem depends from tolerance and value of current which you need. As start, you can try high-ohm resistor and external voltage source in series. 

    Transimpedance amplifier has low input impedance, which equal Feedback resistor value divided by (1+Aol), here Aol is open-loop gain of op-amp. Therefore, input current will be equal External voltage divided by (External resistor plus inpur resistance of TIA). Keep the right current direction - from TIA input.

    I hope that this help you.

    Regards,

    Kirill 

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