Transimpedance Amp - first pole location

Discussion in 'General Electronics Chat' started by jbriaris, Apr 26, 2013.

  1. jbriaris

    Thread Starter New Member

    Apr 11, 2013
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    Hi All

    I'm modeling an example photodiode (transimpedance) amplifier in LTspice using the ADA4505 from Analog Devices. The theory suggests that the location of the first pole (actually a double pole) will be located at frequency f_{\rm fp} where

    f_{\rm fp} = \sqrt{\frac{f_{\rm c}}{2\pi R_{\rm f} C_{\rm i}}}

    where f_{\rm c} is the crossover frequency of the opamp, R_{\rm f} is the feedback resistance of the transimpedance amplifier, and

    C_{\rm i} = C_{\rm j} + C_{\rm id} + C_{\rm icm}

    where C_{\rm j} is the photodiode's junction capacitance, C_{\rm id} is the differential input capacitance of the opamp, and C_{\rm icm} is the common-mode input capacitnce of the opamp.


    • Is the crossover frequency f_{\rm c} of an opamp the same as its gain bandwidth product (GBP)?


    From the ADA4505 datasheet; GBP = 50kHz, C_{\rm id} = 2.5pF, C_{\rm icm} = 4.7pF. From the photodiode datasheet C_{\rm j} = 5pF.

    Using a feedback resistance R_{\rm f} = 100M, and substituting these values into the above equation yields a first pole frequency f_{\rm fp} at 2,554Hz. However, the equivalent simualtion in LTspice appears to show a first pole frequency (peak) at 320Hz (see attached) - an order of magnitude less.

    • Any ideas what I'm missing or doing wrong?
    Thanks! :)
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Try setting the DC on your current source to 25nA. That will set the output at 2.5V. When I did this, The peak moved to about 950 Hz. Still not as calculated, but closer.

    I ran the sim with a single pole, 50kHz GBW op amp, and the peak was at 2.55kHz, as calculated.

    The difference is apparently partially the result of the second pole inside the op amp.
     
  3. crutschow

    Expert

    Mar 14, 2008
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    Not sure if that equation is correct. And the large peak indicates the circuit is unstable. A small feedback capacitor in parallel with the feedback resistor will eliminate that. Here is an article on calculating the proper value.
     
  4. Ron H

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    As I said, a simulation of a single pole op amp gave the same results as the equation. AFAIK, there are no single pole op amps available in the real world.
     
  5. jbriaris

    Thread Starter New Member

    Apr 11, 2013
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    Great, thanks.

    I've been following the theory in Jerald Graeme's book Photodiode Amplifiers, and perhaps, aspointed out, that equation is only valid for an ideal single-pole opamp - the text doesn't seem to make this clear.

    Thanks for the link crutschow. Yes, it is unstable, picking a phase compensation capacitor was my next job, but doing some investigating first.

    The link provides a formula for the phase compensation capacitor which differs slightly to the one offered by Graeme, i.e., 2\sqrt{2} cf 2 in the denominator - perhaps a slightly different phase margin in the compensation.

    Since the premise of the transimpedance amplifier is to create a virtual ground at the inverting input, would you recommend using a split-supply opamp? :confused:

    Also, how did you model a single-pole opamp in spice? Do you use something like this - http://www.ecircuitcenter.com/Circuits/opmodel1/opmodel1.htm

    Thanks again.
     
  6. jbriaris

    Thread Starter New Member

    Apr 11, 2013
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    Agreed. Using LTspice's default opamp with GBW = 50K. :)
     
  7. crutschow

    Expert

    Mar 14, 2008
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    Obviously you can't go below ground at either input or output with a single-supply op amp. But a photo-diode only outputs signal in one-direction, thus if connect the diode so it puts a negative current into the (-) summing-junction, giving a positive output, then you should be OK with a single supply.
     
  8. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Yep. You got it.
     
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