Current to Voltage Opamp for phototransistor

Discussion in 'The Projects Forum' started by MikeB, Dec 27, 2006.

  1. MikeB

    Thread Starter New Member

    Dec 27, 2006
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    I've inherited a project where we require precise measurements of the "breaking" of a infrared beam. The first stage of the analog circuit is where I'm having problems. The circuit itself works, but isn't cost effective for manufacturing, so I'm trying to tweak it to prevent us from having to add components, and stop hand matching transistor / emiter pairs.

    Essentitially the circuit is a voltage divider (R4 = 267k, R5 = 178k ) with a midpoint of 1v hooked to the + terminal of LMV774. As well, there is 0u47 cap to ground. The negative feedback Resistor (R6) can range from 10k to 330k (to bring the circuit into balance, which is 4v on the output), with a 470pF filter cap in parallel. Lastly on the input, before the Phototransistor there is 1.5k Resistor (R3). The emitter of the Phototransistor is connected to ground, and the collector is connected to R3.

    The problem that I'm having is all of the formulas for calculating op-amp circuits that I've found all show a "current source" at the inverting imput, and we have a current drain. Any help on how to caculate out the values for this circuit would be appreciated.

    The final wish of the circuit is to be able to select random pairs of emitters / transistors, populate them, and use 100k programmable pot to set the gain so that each pair falls into a usable region.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    Hi,

    The inverting input of an op amp works at virtual ground. The current through the input resistor/s is nulled by an equal and opposing current through the feedback resistor. The E-book series at the bottom of this page has further material in volume 3.
     
  3. Distort10n

    Active Member

    Dec 25, 2006
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    That does not sound correct. The summing junction will not be held at virtual ground because there is a voltage at the non-inverting node. His description does not sound like the normal transimpedance configuration.
    This is really an exercise in KVL/KCL...or use DC Symbolic analysis on Texas Instruments' TINA spice program.:cool:
    Also, I take it that a 'current drain' is another fancy term for 'current sink.' If so, it should not matter. The only thing that would change is the assignment of the magnitude; e.g. + or - , when writing the equation.
    Can you post a schematic? Your description is running together in these late hours. The voltage divider is on the non-inverting input and the phototransistor is connected to R3 which is going into the summing junction (- input)?
    I have not confirmed by hand the transfer function shown.
     
  4. MikeB

    Thread Starter New Member

    Dec 27, 2006
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    Sorry guys, was just reviewing the circuit for posting, & realized that I left out the voltage source on the output side of the circuit. Here is the circuit, (The phototransistors are shown backwards, but it is my understanding that because we are now using op-amps, we can drop the current supply to ~.25ma & have the same result with the phototransistors in correctly)
     
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  5. Distort10n

    Active Member

    Dec 25, 2006
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    Well...

    Things just got more difficult. You have a 'T' network in the feedback. Do you have some kind of software tool that could derive the symbolic transfer function?
     
  6. MikeB

    Thread Starter New Member

    Dec 27, 2006
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    We don't in house. I have managed to talk to one of the original designers of the circuit, and as he is now retired / and was only one of a few who worked on this project hasn't been overly clear on how it works. I get alot of comments to the effect that if I was still at my desk and had x materials.......
    Whereas I am fast learning in trying to understand this circuit (many many hours reading before finally posting) and easily believe this to be true, getting that information probably won't happen any time soon.
     
  7. Distort10n

    Active Member

    Dec 25, 2006
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    So does anyone know what the orginal design was for? I thought it could be for isolation purposes; however, the phototransistor, LED, and the rest of the circuit share the same common.
    Do you know the purpose of the LED's...what they represent as an input and what V1 and V2 are connected to; e.g., load?
     
  8. MikeB

    Thread Starter New Member

    Dec 27, 2006
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    V1 & V2 feed directly into the A/D pins of a Micro. The circuit continues on after the caps (ie C5) to another series of op-amps that tell the micro to suspend what its doing and make the measurements now. The LED's & Transistors are set up in a Cross-hair pattern (See Drawing in mm). The entire circuit is set up to measure objects falling, and be able to tell with a high degree of accuracy the size, & to a lesser degree shape of the object. Both the amount the beam is broken & length of time that it is broken for go into the calculations. A funnel makes sure the objects fall thru the center, or near the center of the beam.

    The current revision of this circuit has alot of baggage from the "Legacy" attempts to get this working (especially as there were several people involved in its development). I do know from experimentation that it is possible to bring any pair of emitters / transistors into balance, but without understanding what I'm doing, I'm scared that I will create a bigger problem than I will solve.
     
  9. Distort10n

    Active Member

    Dec 25, 2006
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    Ok, I will have to think about it. It is time for me to go to bed.
     
  10. MikeB

    Thread Starter New Member

    Dec 27, 2006
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    Thanks,, any help is greatly appreciated.
     
  11. Distort10n

    Active Member

    Dec 25, 2006
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    I have not been able to think about this much. I can see why you would want to use a pot since it would be simple to tweak to 'zero' out something on each board.
    What is the concern with the phototransistor and LED's? Is it offset that needs to be 'calibrated'?
    There are other options...an integrator like the IVC102 or a current input ADC with switched integrators like the DDC112 that could cut down on circuitry. Is accuracy critical? I have no problem understanding the size of the object being determined from the amount of time the beam is broken; however, I cannot get my head around how to determine that from the length of the beam.
     
  12. MikeB

    Thread Starter New Member

    Dec 27, 2006
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    Here is the waveforms that the circuit develops for two different size objects. Because the objects are very small, (the smallest being 1mm in diameter), we are using the phototransistor in its linear range, (unlike the typical applications where it is either "on" or "off"). The problem is that because of the normal applications, the Phototransistors, & LED's are made to meet minimum spec's, with no regard to maximum, and thus it is hard to keep them in the linear range. Previously to do this we hand matched them, (compared them to a reference LED or Transistor & matched weak tx to strong rx ect). In the posted circuit, the Transistors are installed in reverse to cut down on the sensitivity, (a remenant of when they used a Resistor circuit to measure current), but since we are now using op-amps, it is possible to use them in the normal way. What I want to do, is use the transistors in the normal way, set up the op-amp circuit to give me an optimum gain (with respect to noise, resistance to ambient light ect), and split the current supply into two adjustable current supplies & balance them into the linear range on the TX side. The problem is I don't know what the best quiescent voltage to use is (from experimentation I know that I leave the linear range of the phototransistor as I approach .3v), as well as the best gain resistor (we know from experimentation that over 350k we have problems with the amplification of noise), and lead resistance on the inverting input is.
     
  13. beenthere

    Retired Moderator

    Apr 20, 2004
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    282
    Hi,

    Trying to operate phototransistors in their linear region is too much like nailing jello. You might be interested in another device called a light-to-voltage converter.

    I've attached a pdf describing them to this post. They can let you use visible LED's for illumination. Balancing outputs and such should be a snap.
     
  14. MikeB

    Thread Starter New Member

    Dec 27, 2006
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    Interesting devices. Unfortunately, we are married to the T 1 3/4 mechanical package. Because the circuit CURRENTLY DOES WORK, I don't think I could convince my company to change direction on what we are currently using as it would require well over 10k to have all the mechanical molds changed in China. My problem is not one of getting the circuit to work, but getting so that the manufacturing costs drop. If I can understand how the op-amp circuit works, it would simply my life. Is my best bet to get ahold of a spice program, and model it?

    Thanks
     
  15. Distort10n

    Active Member

    Dec 25, 2006
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    The pains of design. The "T" network is probably being used to limit the bandwidth and is also used to provide a low impedance path to ground.

    In terms of your saturation problem you could use zener diodes to clamp the voltage on the output as long as they can clamp fast enough. Although, I am not sure of their noise contributions.

    Another option to 'ignore' the non-linearities of the phototransistors is to use an op-amp that has voltage limiter pins thus eliminating the need for zeners since the output stage of the op-amp itself is clamping the output voltage.

    I would redesign the entire circuit, but it is what it is.
     
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