Need help understanding an Optocoupler circuit

Discussion in 'Homework Help' started by Xray90, Jul 17, 2011.

  1. Xray90

    Thread Starter New Member

    Jul 17, 2011
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    Hello all, could really use some help here. With the following schematic, I need to graph Vs vs. Vo. I need to know the function of R1 and R2. Lastly, if R2 value is increased to 440 ohms, what effect would that have? Here is a link to the data sheet for optocoupler ILD256. Any help is greatly appriciated. http://www.alldatasheet.com/datasheet-pdf/pdf/45334/SIEMENS/ILD256.html
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  2. Xray90

    Thread Starter New Member

    Jul 17, 2011
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    From what I can tell, R1 is a current limiting resistor and 4mA would be applied to A/K junction during positive cycle. Really not sure about R2 or what happens in circuit during negative cycle.
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Do you know what CTR is ??
    As for the negative half, hm you have two diodes ILD256, one diode is ON during positive cycle and the second one during negative cycle.
     
  4. Xray90

    Thread Starter New Member

    Jul 17, 2011
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    From the data sheet CTR is 20% at If=10 mA. Not sure what that even means. Do you know what function R2 has?
     
  5. praondevou

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    Jul 9, 2011
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    R2 gives you a definite threshold on the LEDs. See the attached pdf.

    20% CTR means you get 2mA at the output for 10mA LED current.
     
  6. Xray90

    Thread Starter New Member

    Jul 17, 2011
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    Thanks for the response, that helps quite a bit. Wouldn't the lack of a current limiter on the negative cycle make the optocoupler susceptible to damage from current spikes? I think the question of changing the value of R2 and it's effect on the circuit was a decoy.

    If the input is a 120VAC sine wave, and the optocoupler will always be on, what does the output look like at Vo? 5VDC - .6VDC for transistor breakover?
     
  7. praondevou

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    Jul 9, 2011
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    First I thought you'd need R2 to keep the reverse voltage on your diode under the max. reverse voltage rating, but in your case there are two LEDs in parallel, in opposing directions. If I had only one LED then when it's reverse biased I would have the whole input voltage (170V peak) of the neg halfcycle on it because there is no current.
     
  8. Xray90

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    Jul 17, 2011
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    I think I was wrong on the output at Vo. It would be 5VDC in the off state(if there was an off). Being that it's always on, the value at Vo would always be low, correct? Not sure what that value would be though.
     
  9. praondevou

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    Jul 9, 2011
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    What's with the drawing of R3? Is this a potentiometer or why is VDC going to to middle of R3? The way it is drawn I have always 5V at Vo.
     
  10. Xray90

    Thread Starter New Member

    Jul 17, 2011
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    That has disturbed me too. The output should have been drawn with Vo and C coming off below R3, correct? If this were the case, what would be the output at Vo with U1 constantly on?
     
  11. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Is R3 a potentiometer?

    Below is how I think that circuit should be ... I could be wrong ... but I doubt it.

    R1 and R2 are signal conditioning to the opto-isolator. It's a straight voltage divider.

    Why isn't U1 constantly on is my question to you.
     
    Last edited: Jul 18, 2011
  12. praondevou

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    Agree on R3. D1 is not necessary, the optocoupler has 2 double LEDs, works in both directions.
     
  13. Xray90

    Thread Starter New Member

    Jul 17, 2011
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    That's how I believe the output was meant to be drawn. With that scenario, what value would I see at Vo? It would be a constant value since U1 is always on, correct?

    Doubling the value of R2 from 220 ohms to 440 would have a minimal effect on the circuit, right? Am I wrong in thinking R1 is there for current limiting the input to U1?
     
  14. MrChips

    Moderator

    Oct 2, 2009
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    R1 and R2 act as a voltage divider. Let's do the math.

    120VAC is RMS value. Peak voltage = 120 x 1.414 = 170V
    The peak voltage into the opto-coupler is 170 x 220/(220 + 27000) = 1.37V

    Thus the output of the device would be pulses at 120Hz, 0 to 5V amplitude, difficult to determine the duty cycle.

    When you increase R2 to 440 ohms, peak voltage will double to 2.7V, that is the ON-time is increased.

    Since the transistor circuit is an inverter, the duty cycle will decrease, i.e. the low portion of the output will increase in width.

    This is a useful circuit for deriving a 120Hz time-base clock signal from line frequency.

    Hope this helps.

    (Edit)
    The peak current available to the photo diodes is 170/27K = 6mA
    How much light will the diodes emit at 1.37V @ 6mA, I don't know. R2 will steal 1.37/220 = most of that 6mA.
    Maybe there is not enough current to turn on the photo diodes in which case the output would be a constant 5V. Negative going pulses appear when R2 is increased to 440 ohms.
     
    Last edited: Jul 18, 2011
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