photodiode amplifier and bias current


I have to measure the current of a photodiode.

Input signal is:

  • Maximum current 15uA.
  • Frequency of the pulsed current 4MHz
  • Photodiode capacitance: 1pF

I have done the desing using the AD8067.

I have obteined good results, but i need to improve the SNR. The source of noise is the input voltage noise, so i have been looking for opamps with lower values. The AD8067 has 6.6nV/√Hz, and i have found the ADA4897-1 which it has only 1nV/√Hz. The problem is that the maximum input bias current is -17uA. How does it affect? is there any way of overriding it?

In the analog photodiode wizard ( one of the recommended amplifiers is the AD8048, which it has 3.8nV/√Hz but 3.5uA of input bias current.


Unai Goyarrola

  • Hi Matt,

    I don't care about the DC measurement.

    I am a little bit confused about the bias current input concept. I have always thought about input bias current as a current that it comes/go from the input. What it means, in my case, that the photodiode must supply this current, i am wrong? So my doubt is: if my photodiode output is 15uA, and the input current bias of the ADA4897-1 is 17uA, what will be the current flowing through the gain resistor?

    Thanks for your help,

    Unai Goyarrola

  • 0
    •  Analog Employees 
    on Nov 6, 2013 8:06 AM over 6 years ago

    Hi Unai,

    Is DC measurement important to you?   If not, you may be able to AC couple in the stage following the transimpedance amplifier.   Then offset error created by the bias current of your transimpedance op amp wouldn't matter as much, since you can remove it in the second stage.

    Do you plan on filtering after the transimpedance amplifier?   This would help with noise somewhat, especially if your transimpedance stage ends up having a considerably more bandwidth than the 4 MHz you mentioned.   Anything you can do to restrict your bandwidth will help your noise.

    A couple of other comments: 

    - Given that your photodiode's capacitance is only 1 pF it's likely the input capacitance of the op amp will dominate

    - Note that the AD8067 has a minimum gain of 8, which may affect your stability.   We are in the process of adding more intelligence to Photodiode tool to handle these cases.


  • 0
    •  Analog Employees 
    on Nov 7, 2013 1:53 AM over 6 years ago

    Hi Unai,

    Think of the bias current as an extra current source in parallel with the photodiode current source. 

    Example 1:

    Bias Current = 17 uA

    Photodiode current = 0 uA

    Rf current: 17 uA

    Output voltage (with 39K Rf): -663 mV

    Example 2

    Bias Current = 17 uA

    Photodiode current = 15 uA

    Rf current: 32 uA

    Output voltage (with 39K Rf): -1248 mV

    So the bias current has the effect of shifting your output voltage.   Instead of having an output signal swing from 0V to -585 mV (which you would have with no bias current), your output signal range will be shifted downward by the effect of the bias current.

    Note that in the circuit schematic you attached, you have a resistor attached to the non-inverting input, which is 39k, just like the feedback resistor.  This is presumably there to cancel out input bias errors, so that you only have input offset current errors.  While this will work, you are trading one problem for another.  Unfortunately, that 39K resistor has 25 nV/rt(Hz) noise, which is much larger than the noise of the op amp.   So I wouldn't recommend having it, or would recommend putting a much larger capacitor in parallel with it to limit its noise bandwidth.

    Here is a link to a tutorial on bias current: