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IV converter

Category: Hardware
Product Number: AD8615

Hello i'm Rona.

I have some problems about IV converter

I checked the Photodiode wizard and obtain below result.

I have some qeustions about this result.

1. Is there any limit value about Rf (1.58Mohm) ?

Can I connect 50Mohm to Rf instead of 1.58Mohm and finish it in one stage?

How I decide using one stage or two stage ? 

2. Looking at the photodiode wizard, it says that AD8615 is not recommended instead of LTC6269-10, but I don't understand why.

Could you explain more details ? I already use AD8615 so I need to why it isn't not recommanded

Thanks alot

  • Hi Rona,

    The circuit the photodiode tool suggests is a good first suggestion.  Feel free to tweak it better to meet your needs, either in the tool itself or by exporting to LTspice.

    There are two reasons the photodiode tool doesn't like to use high value resistors:

    1) With high value resistors any parasitic capacitance on the board can dramatically affect your final response.  We don't want to promise results you won't get on your board.  For slower designs, you can often use higher value resistors, but just be aware that parasitic capacitance will have a big impact on your frequency response.  Many real life designs use bigger resistors.

    2) The other time you cannot have high feedback resistance is when you are trying to get high bandwidths and high output voltages out of your circuit.  Sometimes the op amp just isn't fast enough to do it all on it's own, so need an extra stage to provide more gain.  This appears to be the case in this design.  Looks like you want a transimpedance gain of 50e6 with a 1.6 nF input capacitance at 500 kHz.  This is a tough combination.  It will be very hard to do with one stage.  You will need to relax your requirements if you want to only use one stage.  You can do this by:

    - lowering the capacitance of the photodiode you are using (for example by reverse biasing, see photodiode's datasheet for its capacitance vs. reverse bias graph)

    - lower the bandwidth required (i.e. slow down the pulse width you can accept)

    - reduce the peak output voltage required from the circuit

    - increase the amount of peak current the photodiode is sending

    Given the speed and gain requirements of the circuit with a 1.6nF input capacitance, it is unlikely the AD8615 will work for this design.  The photodiode tool has 5 recommendation levels:

    - The default op amp suggestion.  This is chosen to get close to the bandwidth you are aiming for.  They are typically many other op amps that also will work for the design.  Often another choice can be better for your specific application.  

    - Parts with "recommended" in the photodiode tool's op amp selection guide (open by clicking on the part name button in the left hand column).  The tool thinks any of these will work well

    - Parts with "not recommended" in the selection guide.  The tool will give a reason why it doesn't like the part.  If the problem stated doesn't bother you, then go for it.  

    - Parts with "not allowed" in the selection guide.  This typically happens when the part is so slow in the circuit that the tool doesn't know how to simulate it.  Don't use these.

    - Parts not supported by the tool.   These won't show up in the selection guide at all. We don't support some op amps - typically those without enough input capacitance information to do a good simulation.

    So first step is clicking on the op amp button in the left hand column, locating the AD8615 in the selection guide, and see what the tool says about it.

    Matt

  • Hi Matt,

    Thank you so much for your kind reply. 

    Your answer helped me.

    Additionally, I have a few more questions.

    1. Following your advice, I lowered the target pulse width.

    I read the description of target pulse width but I don't understand exactly what that value means.

    Could I think of it as the time during which the current flows through the photodiode?

    What I want to design is that the light illuminated the photodiode for about 5mS and detect the intensity of the light through the voltage applied to the photodiode during that time.

    I changed the target pulse width to 100uS and 200uS

    Why does the value of Rf(5.9Mohm or 23.7Mohm) Vary depending on the value ? 

    - Target Pulse Width : 100uS

    - Target Pulse Width : 200uS

    Would you help me one more time ?

    Thanks,

    Rona

  • Hi Rona,

    Of course.  

    1) Based on what you put into the "Target Pulse Width", the tool will try to create a circuit just fast enough to create a nice looking pulse of that width:

    If you put in a perfectly square pulse of 50nA at the input for the target pulse width, the pulse response tab shows what you should get at the circuit's output.

    2) Your circuit needs 50 Mohm gain.  This your peak voltage (2.5V) divided by your peak current (50 nA).

    With a 1.6nF input capacitance, the AD8615 isn't fast enough to provide 50 Mohm gain with a 100 uS pulse width.  Each photodiode + op amp combination has a frequency response for a given transimpedance gain.   This sets the minimum pulse width it can do. The higher the gain, the slower the circuit.  So for the 100 us pulse width case, the tool thinks the maximum transimpedance gain the AD8615 can do is 5.9 Mohm to get the speed needed.  The rest of the gain (50M/5.9M = 8.5) is provided by the second stage.   The tool won't let you do a single stage in this case - the AD8615 just isn't fast enough.

    For the 200 uS case, the circuit can be slower, so more gain can be used in the first stage (23.7 Mohm) and less in the 2nd stage (2.1).  At this point the speed requirement is low enough that the tool will also let you do a single stage if you want.  The pulse response shape won't be quite as nice, but you can do it.

    If you widen the pulse width requirement further to 300 us, then the tool will default to all of the gain (49.9 Mohm) in the first stage.  In the 2nd stage, the tool will only do filtering, not extra gain.  Widening the target pulse width further than 300 us will not change the gain. 

    ---

    You can also widen the target pulse width to your 5 ms and see what that looks like in the pulse response.  Here is a one stage vs. 2 stage design.  The difference in response is from the filter in the 2nd stage.  It lowers the bandwidth of the circuit, so your step response is not as sharp, but your noise performance is better.

    Matt

  • Your circuit needs 50 Mohm gain.  This your peak voltage (2.5V) divided by your peak current (50 nA).

  • Hi Matt,

    Thank you so much for your kind reply again.

     

    I tested it with your help and the wizard results below, but the final output fluctuates like the picture below.
    How can I make this not fluctuates?
    I think it's about 60Hz.

    Please help me ! 

  • Hi Rona,

    Divide and attack the problem.  For example, I would recommend disconnecting the photodiode and grounding the input.  If the fluctuation is still there, you know it's an issue in the circuit.  If the fluctuation is gone, you know the issue is in the photodiode.  Keep trying to divide the problem until you can locate the source of the issue.

    Matt