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Voltage to current converter

  1. Sep 7, 2012 #1
    Can someone advice how a V to I converter (with IC741 or any other easily available Op Amps) can be made for converting few mV to pA. The converter is to be used to convert the output from a photo-diode preamplifier to a lock-in amplifier with current input. Thank you for the help, in advance.
     
  2. jcsd
  3. Sep 7, 2012 #2

    berkeman

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    Welcome to the PF.

    The photodiode's output is a photocurrent. So you already have a current to input to your lock-in. You can look at the photodiode's datasheet to see what kind of photocurrent you will get for your illumination levels...

    What kind of frequency response are you hoping to get, BTW? If you need much bandwidth, you may need to implement an I-->V-->I converter, so that you can put the photodiode in reverse voltage bias to improve the bandwidth...
     
  4. Sep 7, 2012 #3
    The 'photodiode' has a built in transimpedance amplifier which does the I->V conversion.

    Bandwidth should be less than 300 Hz.
     
  5. Sep 7, 2012 #4

    berkeman

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    Ah, got it. Are you sure that you want just pA output? And if your voltage signal is just a few mV, it will probably be a pretty noisy signal, right? Is that why you need to use the lock-in?
     
  6. Sep 7, 2012 #5
    Thanks!

    few tens of pA also would be okay. Anything larger might saturate the lock-in's input.

    Exactly. It is noisy. I hope to improve S/N by the lock-in.

    I have already tried passive V->I conversion (with a resistor) and it seems to work better (with a test optical signal at a larger power level than the signal I want to measure) than directly using the lock-in's voltage input. At the moment I think, an active V->I conversion might improve the situation. But I have no idea how to get this done.
     
  7. Sep 7, 2012 #6

    berkeman

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    So the current input to the lock-in amp takes a positive input current that is ground referenced? (If not, the circuit will change)

    What is the input impedance of the current input to the amp? If the impedance is known, then you can just use a voltage amplifier to drive the input... If the impedance is not known or constant, then you can still use opamps to make a V-->I converter circuit (it's just a bit more complicated).
     
  8. Sep 7, 2012 #7
    The lock-in accepts both AC as well as DC inputs. In this case I guess, the DC input is preferred as the signal to be measured does not go below zero.

    I have not understood the second question. Are you referring to the lock-in's current input? If yes, it is less than 250 k (@ 100 Hz).
     
  9. Sep 7, 2012 #8

    berkeman

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    So the input impedance is < 250 kOhms, but is not specified and constant? Okay, then you can't use the input impedance as the V-->I converter.

    So basically what you need to do is have the opamp circuit drop the input voltage across a known resistance, to convert that voltage into a current. And it has to do that non-ground-referenced, so that the current can be made before going into the current input of the amp.

    The figure below shows one way to do it using two opamps (ignore the numbers on those opamps in the figure for now):

    http://www.maxim-ic.com/images/appnotes/4394/4394Fig03.gif
    4394Fig03.gif

    The feedback opamp on top is sensing the differential voltage across the current sensing resistor, and the input opamp at the lower left is ensuring that the voltage drop across the sense resistor matches (to within some gain factor) the input voltage. That gives you a V-->I conversion, with gain set by the current sensing resistor, and the resistors in the differential gain stage on top.

    Are you able to take a first cut at the resistor values on your own? BTW, you should not use a bipolar opamp like the LM741, since the input bias currents and input offset currents are going to be way higher than the pA signal that you want to produce. You should use something like a low-noise FET opamp, like the TL071 or better:

    http://www.ti.com/product/tl071

    .
     
  10. Sep 7, 2012 #9
    >>Are you able to take a first cut at the resistor values on your own?

    Not yet. Any help would be appreciated. What are the selection criteria?

    Thank you very much for the enlightening inputs. I see that the solution is not that simple as I thought it to be.
     
  11. Sep 7, 2012 #10

    berkeman

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    I'll take a cut at the resistor values when I get a chance. Something like converting +10mV to +10pA ?

    BTW, it seems like a much simpler (and better?) option would be to use a simple photodiode all by itself (without an internal transimpedance amp), and hook it straight to the current input on the lock-in amp. Connect the [STRIKE]cathode[/STRIKE] anode of the photodiode to the input of the amp, and the [STRIKE]anode [/STRIKE] cathode of the photodiode to the ground of the lock-in amp (assuming that the input is single-ended with respect to ground).

    There are lots of good candidate photodiodes to choose from. Can you say what the light source is?

    EDIT -- fixed the polarity of the photodiode -- the photocurrent flows from cathode to anode.
     
  12. Sep 7, 2012 #11

    berkeman

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    Also, are you able to say anything about the application? What are you trying to measure?
     
  13. Sep 8, 2012 #12
    I want to measure optical power levels at tens of fW power level.

    What you suggested is the best way to do this. However, I do not have the instrumentation (monochromator with the required spectral resolution) to do this readily available.

    What I have is an optical spectrum analyser which has a good spectral resolution and an analog amplified (pw - 100 mW scaled as mV to 5V) output from its internal photodiode. The signal corresponding to few pW optical power is already noisy and with the lock-in I could still detect it with a reasonable S/N. The question is - can we go two orders of magnitude below (tens of fW or better) and still measure with a good S/N?
     
  14. Sep 10, 2012 #13

    f95toli

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    Which lock-in are you using?
    I've never come across a lock-in with only a current input.

    Also, which frequency are you modulating at? Note that this can have a significant effect on the S/N; you should if possible try to go above the 1/f knee of the eletronics. If you can do that , you will "only" have to worry about shot noise (and perhaps Johnson noise in the transimpedance amp).
     
  15. Sep 13, 2012 #14
    The lock-in has both current and voltage inputs. The signal to be measured could not be detected by using the voltage input, but could be detected with a passive VI converter (a resistor) by using the current input. I hope, by a suitably designed VI converter the S/N can be improved further.

    ~230 Hz
     
  16. Sep 13, 2012 #15

    f95toli

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    I have never come across a situation where the current input was better than the voltage input. I am pretty sure you are going in the wrong direction by putting a V-to-I in front of th lock-in, especially since there will be a I-to-V inside the lock-in; you will be doing two conversions which can never be a good thing, it will just add noise. There must be something else going on.

    What happend when you increased the sensitivity of the lock-in? Increased the time-constant?

    Could it be that the output impedance of voltage output of the diode is a bit strange?
    There are all sort of of trick you could try, depending on the impedances involved.
    there are e.g. "passive amplifers" which in reality are just transformers. You can also experiment with some basic filtering before the lock-in to see if that helps (too much noise can saturate a lock-in, so it is usuallt best to get rid of as much noise as possible).


    Also, which lock-in are you using (brand/model)?

    230 Hz is a bit low but should be OK, it is usually better to operate at a few kHz; especially if you are using an analogue lock-in.
     
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