Optical-tester project

Discussion in 'The Projects Forum' started by Al.ro92, Sep 7, 2016.

  1. Al.ro92

    Thread Starter Member

    Sep 7, 2016
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    Hey guys,

    My colleagues and I are currently busy on a small project we've been wanting to do for a while now.

    At our job we make specific machine parts and assemble some of them.

    One of the things we make is the shield for an optical sensor for a microbalance, which we also assemble.

    After we assemble the sensor, which consisits of 2 phototransistors and an LED, we also have to test it with an optical tester we got from the company for which we build these.

    The tester looks old and outdated, and has some pretty old components. My collegues and I (we are no EE but do have some knowledge about the subject out of interest) think we can build a new one using some newer components, that's our challenge.

    image.png

    The first thing we did was open it up to see what's inside and try to understand what goes on in there.

    image.jpeg image.jpeg

    image.jpeg

    Next thing we did was try to identify all the components. It gets a power supply of 230v (Europe), the step down transformer, lowers this to 24v ( we think, still have to measure) and is then rectified with the diodes on the board. This we want to replace with an adapter. The sensor only needs 4,5 v.

    We still don't know what type of resistors and what component is next to it, the silver one.
    We think it's a capacitor, tried looking the serial number up for the data sheet but couldn't find anything.

    image.jpeg image.jpeg
    We still need to figure that out to completely understand what the tester really does.

    Briefly what it does is check if the voltage produced by the phototransistors is according to the specs we got from the instructions, if it doesn't then we usually have to adjust the light shield a little more.

    Right now we are thinking what step to make next, any tips or ideas are welcome.

    Thanks!
     
  2. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    The silver thing is an electrolytic capacitor unless I am very much mistaken.
    Why on earth do you replace it as long as it is working?

    You would have to trace out the full circuit and reproduce the existing circuit exactly - unless you can fully understand its function from the circuit then you might be able to redesign it to perform exactly the same function. But, again, why on earth would you?
     
  3. Sensacell

    Well-Known Member

    Jun 19, 2012
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    Looks like it was built in the 80's- judging from the parts.

    You could probably do it better-faster-cheaper now, assuming you know exactly what it does.
    That's the tricky part, focus on defining exactly what it needs to do, and the precision it requires.
     
  4. Al.ro92

    Thread Starter Member

    Sep 7, 2016
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    Thanks for the reply! We don't really want to replace the old one, just try to make a new one to see if it works and if it does then we would have 2 testers which we could use when there's a lot of work. We are going to try to trace out the full circuit like you said, and reproduce it using newer parts, maybe also make it for 230v instead of 110v.

    Any tips on how to trace out the full circuit without having to disassamble it even more?
     
  5. Al.ro92

    Thread Starter Member

    Sep 7, 2016
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    Yes that's what we also think, right now we are trying to understand what the function of every part is. There are still a few things unclear, like those two blue resistors, do they have a special function for wich they have a different color? today after work we are going to try to draw the schematic and simulate it on a program we can easily get our hands on like Pspice or something.
     
  6. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    The blue resistors are 1% tolerance which were pretty rare in those times so that tolerance is important to the design. They are much bigger than the standard resistors on that board. This may be because they are rated for higher voltage or higher power than the standard resistors. I think a higher voltage is unlikely as a reason for choosing them. They may be higher power rating so they have a lower temperature rise and hence keep within that 1% tolerance. It may be that 1% resistors were only available in that size (could they be wirewound resistors?).
    To decide, and hence choose good replacement parts, you need to understand the circuit.

    As to dismantling you will need to be able to see the other side of the circuit board as there may be links underneath. I can't see from the photographs but there may also be components hidden underneath it. Before undoing the screws on this board make a drawing of where all the wires connect to the board as, while you are moving the board around to see what is going on, one or more wires may break off and you need to be sure where to reconnect them. This comes from long experience of doing similar things :oops:
     
  7. Al.ro92

    Thread Starter Member

    Sep 7, 2016
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    Ok, so we have two options; 1. Trace the full circuit and rebuild one just like the old one but with some newer parts.
    2. Fully understand how the circuit works, and then build a new one that does the same.

    We think the first one might be somewhat easier, although we know what the tester does, there are some parts we still have not figured out what they do in the circuit. Either way we are going to have to dissmantle it. We tried making a schematic but were not able to trace where all the wires go, we will need to dismantle it first.

    Here are some pictures from the sides where you can see a little of what's underneath.
    IMG_0332.JPG IMG_0333.JPG IMG_0334.JPG IMG_0335.JPG

    Judging from your experience, which one is the most reasonable and doable option?
     
  8. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    That depends on how certain you can be, from your derived schematic, exactly what it does and how it does it, taking into account how accurately it does it. Those 1% resistors suggest that it does have critical sections.

    There will likely be some parts which are easy to understand and replace - for instance a regulated power supply (that TO220 device dangling by its wires perhaps). There may also be parts where it is not clear exactly why a particular part of the circuit is as it is and that part would be better replaced part for part (new components with the same specification).
     
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  9. hp1729

    Well-Known Member

    Nov 23, 2015
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  10. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    Actually, I have an IDEA of what it's doing.

    First the photo-transistor creates a current. To measure a voltage, you have to drop the current across a resistor.

    The 4.5 V may be the drop across a resistor, to fix the LED current. Reducing this voltage may check the linearity or dark current.

    Next, your looking at the current from each leg and adjusting mechanically, so they are close.

    The third part, might actually be measuring the difference of the sensors to insure that one is slightly higher than the other.

    So the actual test circuit would look something like:

    PS(power supply) LED resistor ground (the voltage across the resistor sets the current)

    PS photo-transisistor #1 - resistor#1 - ground

    PS photo-transisistor #2 - resistor#2 - ground

    The voltages across R1 and R2 are adjusted mechanically which effectively matches the current with one slightly higher. This is where the 1% resistors come into play.

    The last step would be the difference between the voltages across R1 and R2 as a verification.

    The silver cylindrical thing that has a 150-25 is a 125 uf, 25 V electrolytic capacitor (polarity is important)

    The small 3-leaded thing is a 2N2405 transistor.



    What is the part number of the sensor?
     
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  11. Al.ro92

    Thread Starter Member

    Sep 7, 2016
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    Hey thanks for the info, I still have to read it a couple of times and do some more research to understand everything but I think I'm getting it. I will try drawing a schematic too:)

    The 4.5 V may be the drop across a resistor, to fix the LED current. Reducing this voltage may check the linearity or dark current.
    Next, your looking at the current from each leg and adjusting mechanically, so they are close.

    Do you mean with each exactly? Is the each leg of the LED or the phototransistors?

    What is the part number of the sensor?[/QUOTE]

    I think it's this one, ill check tomorrow at work again.

    http://www.osram-os.com/Graphics/XPic3/00101768_0.pdf
     
  12. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    I know little about the set-up, but I think your mechanically adjusting the sensors, so that they see ALMOST the amount amount of light. The ZERO, so to speak is slightly biased toward one sensor.

    You start out with the max amount of light and then the minimum amount (sensitivity) and make sure you get a response.

    The big part of your drawing needs to include the sensor and what the sensor is connected too.
     
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  13. Al.ro92

    Thread Starter Member

    Sep 7, 2016
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    Ok, so if I get you right what we are doing is adjusting the sensor by checking if the values (by a max. and min. amount of light) from the phototransistor are within the given range in the specs, when those values are ok we can finish adjusting the lightshield in the back. (see pic)

    Image-1.jpg

    The drawing consists, like you said, of the sensor and what it's connected too, in this case the tester.

    We will try to draw the schematic of the sensor first, and then the tester which will be more complicated we think.
    So far we got this for the sensor: IMG_0347.JPG

    Please let me know if we are on the right track.

    Thanks for the help so far!
     
  14. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Can you post a copy of those instructions, or a link to them?
    Do you choose the values/types of the resistors/transistors/LED, or are they already decided/provided by the client?
     
  15. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    The sensor schematic looks like what one would expect. What are those three resistances?
    I'll bet at least two out of the three are precision resistors. Give the elements arbitrary labels like R1, R2, Q1, Q2, LED1 etc.

    I'm hoping that that and the datasheet can help make sense of the numbers.

    What you have makes sense.

    In a guess, and you can do this yourself. You can make the required adjustments and independently check where those numbers are coming from.

    I'll look at all of the info a little later.

    This is what I THINK your doing.
    1. varying Vcc from 4.5 to 70 mV; Yea the 70 mV seems ODD
    2. You look at the voltages across each resistor attached to photo-transistor
    3. You look at the difference in those voltages.

    So, when your instructions say adjust to 4.5 V; See if you can independently measure 4.5 V using probes.
    I think that one of the values your measuring is across on of the resistors of the photo transistor and the other is the voltage across the other one. Finally the difference between these two readings or Vout1-Vout2. Again have different labels. So, I can say the voltage across R1, the voltage across R2 and Vout1-Vout2, Vcc-Vout1 and Vcc-Vout2. Those are what I think your measuring or something similar.

    The switchbox makes it a little easier and the instructions were written before auto-range meters were common and meters were expensive.

    So, get the values of the resistors, and label the schematic a bit better. Effectively calibrate one and we'll measure some voltages independently for Vcc = 4.5 and Vcc = 70 mV (again, an odd value, so it may not be Vcc)

    In the US, we can buy a DVM for about $5.00 USD. (not auto-range though). Let' just say you had 3 DVM's set up. Now, there would be very little switching. You might also be able to get away with DPM's or digital panel meters. The cheap DVM's could run off of wall-warts.

    More sophisticated techniques might use bar graphs.
     
  16. Al.ro92

    Thread Starter Member

    Sep 7, 2016
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    I drew the schematic of the sensor again, apparently I had to flip the resistors to get a high output when light is detected.
    . IMG_0349.JPG

    Do you mean that I have to measure those 4.5v where I drew the voltmeter on the schematic?

    The actual values of the resistors are not known, I guess we will have to use some formulas or a lot of testing to get those values.
    Although when we opened the tester:
    1.jpeg

    I measured these 4 resistors:
    groep.PNG

    kleine2.PNG The one above is 1KΩ, the other is 10.5KΩ

    groot2.PNG The one above is 83Ω, the other is 322Ω

    Then you have the precision resistors and a potentiometer.

    Do you mean the voltages across R1 and R2, the voltage difference between Vout1 and Vout2, the voltage difference between Vcc and Vout1,
    and the voltage difference between Vcc and Vout2?

    If the above schematic is ok, I will build the circuit (after I get my hands on some resistors) on a breadboard and measure the ouputs for Vcc=4.5v and Vcc=70 mV and see what kind of values I get.

    Thanks for the all the info so far, we really appreciate it and are learning a lot from this project.
     
  17. Al.ro92

    Thread Starter Member

    Sep 7, 2016
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    I made another schematic, with the voltmeters: schema.PNG


    Are we getting what you are saying right?
     
  18. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    The bands on the resistors color code the values: A brown, black, orange, gold is 10, 3 zeros, 5%
    So, 10,000 ohms 5% and probably 1/2 Watt carbon composition. Wattage is determined by size.

    or-black-brn-gold

    is 30, 1 zero, 5% OR 300 ohms.

    Vio, green, black gold is 85 ohms 5%, probably 2 Watt,

    brown, black, red, silver is 10K 10%

    i thought the resistors may need to be flipped.

    You show 4 resistors and there are three in your circuit.

    I was expecting the 1K and 10K to be the same size.
    The 330 makes sense for the LED (5-2)/330 is about 10 mA. (the numbers make sense)

    You don't really have to breadboard the circuit.

    Just do the calibration on one and measure the following with another meter.

    Vcc at 70 mV and 4.5 V

    measure the voltage across
    R1, T1 = 4.5 total or Vcc; e.g. The voltage across R1 + the voltage across T1 = Vcc
    R2, T2 = 4.5 total or Vcc
    RLed, VLed 4.5 total
    Gnd to Vcc = Vcc
    Vout2-Vout1 = (Watch the sign); might be reversed of what your supposed to measure.

    Now compare with the values you are supposed to measure with the box.
     
  19. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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    That;s what I think the basic idea is. Put a 4th meter between Vout1 and Vout2 so you don't have to do the subtraction.

    The 70 mV value still bothers me,
     
  20. Al.ro92

    Thread Starter Member

    Sep 7, 2016
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    We also think it's pretty strange, that's why we want to build the circuit on a breadboard (it's also fun), or on a simulation software to see what kind of values we get with that input.
     
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