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Pulse Train Detection

Discussion in 'General Electronics Discussion' started by zappyton, May 20, 2013.

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  1. zappyton

    zappyton

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    May 20, 2013
    Hi everyone,

    I am building a circuit which takes the output of a photodiode, amplifies it to between 0 and 5 Volts and then outputs the signal, where the goal is to tell when the light source as dropped to a particular value (say half for arguments sake). The problem is the light source flashes for about 150us, at about 3Hz, and I can't think of a way to get this to work. I thought of using a zener maybe, or an op amp for the level detection, but the goal is to output a digital 'good' or 'bad' signal.

    I already have an op-amp in the circuit so adding more of them isn't a problem.
    Any ideas? I'm stumped.:confused:
     
  2. BlinkingLeds

    BlinkingLeds

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    Feb 23, 2013
    Last edited: May 20, 2013
  3. zappyton

    zappyton

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    May 20, 2013
    No, sorry, I don't think I have explained my problem very well. I have to use this photodiode because I specifically am measuring UVC, which is not a problem, and I'm using the amplifier circuit suggested by the manufacturers of the diode, which just ends up being an op-amp output.
    The problem is how do I convert the 150us 3Hz pulses into a DC value so I can have a persistent 'Light is going' signal? I see that the schmitt trigger will give a level, but won;t it just pulse along with the lamp? Or am I misunderstanding something?
     
  4. BlinkingLeds

    BlinkingLeds

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    Feb 23, 2013
    Can you draw you circuit?
    What about a big capacitor to make the 555 on for as long as the signal coming (the consequence is that it will have a short delay after the signal stops before it turns off)
     
  5. zappyton

    zappyton

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    May 20, 2013
    OK, here is the circuit, if you imagine the diode is a photodiode. The op-amp is a 5V single supply rail to rail op-amp.

    The output will be a series of 150us pulses at 3Hz with amplitude around 5V, and when the amplitude of the pulses drops to say, 2.5V then circuit needs to warn the user that they need to send it off to be serviced.

    I have no problem if the signal persists for a little while once the circuit's turned off, but would it be able charge the capacitor in the length of the pulses though? Because presumably the rms voltage is very low, isn't it?
     

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  6. duke37

    duke37

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    Jan 9, 2011
    What do you mean by half light level?
    Are the pulses shorter or less high?

    The pulses are very short so you will need to charge a capacitor at a very high rate using an amplified diode. Show us a sketch of the waveform at full and half level.

    The op-amp circuit you show converts the input current to a voltage and has a low pass filter, this may mean that the pulses are not responded to accurately. I have not calculated the frequency response of that circuit.

    Edit. There is no voltage supply to the photodiode
     
  7. zappyton

    zappyton

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    May 20, 2013
    The pulses will be less high. It hadn't occurred to me about the frequency response of the circuit could be an issue, I'll need to check that.

    The photodiode doesn't need a voltage supply, it generates it's own power from the light (~nA). The op-amp has a five volt supply, which I left out for clarity.

    Unfortunately, I don't yet know the exact pulse shapes, but the width should remain constant, it will just be the amplitude that will vary.

    Also what's an amplified diode? I've never heard of one before.
     
  8. BlinkingLeds

    BlinkingLeds

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    Feb 23, 2013
    Why don't you use a phototransistor it should be better than the diode.
    What sends the signal to the photodiode is the pulse always the same or is it a variable pattern?
     
  9. zappyton

    zappyton

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    May 20, 2013
    Because I have a broadband source, and am only interested in the UVC part of the spectrum, and they are not easy to get hold of. Even getting hold of filters cost several hundred pounds!

    The lamp will be pulsing at a set rate, and it won't vary. It will just pulse until it is switched off.
     
  10. BlinkingLeds

    BlinkingLeds

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    Feb 23, 2013
    That's what i meant.

    What about the signal side? Do you have access to the signal LED if so i guess you could just connect a capacitor across it?.
     

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  11. zappyton

    zappyton

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    May 20, 2013
    So what are you suggesting? That you have a source of UVC sensitive phototransistors? Because if so I am very interested!

    Yes, I have access to the signal side if necessary, so if that would help, that would be fine.

    I'm not sure I understand your circuit. Is it relying on me having a phototransistor instead of a photodiode, or is the photo transistor being switched by a signal from the photodiode?

    Thanks a lot for the help, and I'm sorry if I'm being slow!
     
  12. zappyton

    zappyton

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    May 20, 2013
    Sorry for double posting, but it just occurred to me that 'signal' is ambiguous, so I'll clarify: I can adjust the output from the circuit, but I can't adjust what is fed into the photodiode.

    Thanks!
     
  13. BlinkingLeds

    BlinkingLeds

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    Feb 23, 2013
    OOps sorry i didn't see the uv :)

    What's the photodiode's output voltage? I guess you could use a logic gate fate and then my circuit.

    OR

    Try putting a cap across the uv light source (UV LED?) if the signal has enough power to charge the cap. (where is you signal coming from anyway? is it on the breadboard? if it's not on the breadboard and the signal cannot charge the cap you should wire the p-channel's gate to the signal source with a pull up resistor.

    Oh and what's the uv source? uv led? just a thought here but couldn't you just change the uv led to ir led?

    By the way i'm a newbie in electronics i think i can help you because i had a similar problem before (https://www.electronicspoint.com/simple-phototransistor-switch-circuit-t257481.html)
     

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  14. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    What's the PhotoDiode part number?

    Chris
     
  15. zappyton

    zappyton

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    May 20, 2013
    The UV source is a xenon lamp, and the point of the circuit is to know when the UV has dropped, so switching it with a IR LED isn't really an option. And I don't need to know when it's on, but when the UV has dropped.

    The photodiode I'm using is from SgLux, and it has part number SG01S-C5
     
  16. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    The OpAmp circuit you posted is identical to the Data-Sheet's test circuit, so there's nothing wrong there. Since you want to detect a momentary loss of a pulse train try Googling "555 Missing Pulse Detector".

    Chris
     
  17. BobK

    BobK

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    Jan 5, 2010
    I think the 150nS pulses are going to be a problem. I doubt that a 555 would deal with them. You would need some fast logic to do this. A pulse stretching circuit comes to mind. I think there was a thread on a fast pulse stretching circuit on this forum.

    Bob
     
  18. Laplace

    Laplace

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    Apr 4, 2010
    So the problem is an analog pulse stream of 1/3 Hz with 150 ns 5 V pulses where the 5 V pulses will degrade over time and must be alarmed when the pulse voltage is only 2.5 V.

    Possibilities for a solution are sample-hold-compare and gated-compare. A gated-compare would generate a digital pulse stream from the analog pulses while also comparing the analog voltage to a reference. This would require 2 high-speed comparators so that during the presence of a pulse the analog comparison would be considered valid. But doing analog comparisons within a 150 ns window could be difficult and relatively expensive.

    A simple sample-hold circuit could be nothing more than a capacitor charged by the analog pulse stream through a Schottky diode where the capacitor voltage would rise to the level of the analog pulse voltage (minus the diode drop) with an op amp to compare the capacior voltage to a reference. There would need to be a long time constant discharge resistor in parallel with the capacitor to drift the capacitor voltage back to zero. The problem with this is the large charge/discharge time ratio. With 150 ns charge time and 3 sec discharge time, the charging current would need to be 20 million times the drift discharge current. Suppose the discharge resistor was 22 megohm, then charging current Ic=5V/22Mohm*20 million=4.5A!

    To eliminate the drift discharge resistor use a gated-sample-hold-compare circuit where the hold capacitor is charged through an FET switch turned on by a digital pulse stream and the capacitor voltage is compared to a reference (use an FET input for low leakage current). So this approach would only require two comparators (or op amps used as comparators), one FET switch, and one capacitor, and possibly another op amp as a buffer for charging the capacitor. Can't see any obvious stressing of design limits this way.
     
  19. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    Where did 150nS come from? Did it change from this first post?

    Chris
     
  20. Laplace

    Laplace

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    Apr 4, 2010
    I don't know where it came from. Latent dyslexia?

    With a 150 microsecond pulse length it would be so much easier to charge a capacitor through a diode as the sample & hold mechanism!
     
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