Controlling AC

Discussion in 'The Projects Forum' started by SudokuCkt, Dec 4, 2006.

  1. SudokuCkt

    Thread Starter New Member

    Dec 4, 2006
    2
    0
    Okay so I really have no clue why this is not working. I have the following ckt and could really use someone's expertise.

    I have an independent 5 V running the rest of the ckt versus the RC control you see up top.

    I have a varying 50 - 110 AC voltage as my input. My question into Pin 2 of IC1 what is the voltage? And is it necessary to have the 100 pF Cap tied to back to the input from output pin 1? Tieing a 10k pot to pin 6 causes what to happen exactly.
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Does this mean there is no 3.3v on the top of R2?
     
  3. SudokuCkt

    Thread Starter New Member

    Dec 4, 2006
    2
    0
    I connected it to 5.0 volts.
     
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    I don't know about you thingmaker3 but this circuit has several things about it that don't look quite right.

    One of the more obvious irregularities is the hookup of IC2. At first, I thought maybe it was an opamp being used as a comparator but that seems unlikely since what would serve as a reference would be the grounded terminal 2. That makes no sense since the positive input of IC2 can not go below ground as far as I can see.

    The most likely explanation is that IC2 is mislabeled and the positive and negative connections should be swapped. That would turn it into an inverting amplifier. There would still be a problem since the negative and positive terminals would never be able to reach equi-potential as is the nature of an opamp. It seems like that once IC2 was recofigured as I have suggested, the positive terminal would need to be connected to a positive voltage rather than ground.

    IC1 seems to be in reasonably good shape since the input is probably swinging symmetrically above and below ground.

    What are you thought on the circuit?

    hgmjr
     
  5. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    I could easily be wrong, but I think proper operation would have IC1 pin2 experiencing roughly 3vp-p on a 2 volt offset - resulting in pulses on pin 1. But with 5 volts on top of R2, there would be a 3v offset on pin 2 - pin 1 output would be steady.
     
  6. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    Just to put in a guess, IC1 & 3 are probably comparators. IC1 is going to put out a square wave. The base frequency will be set by current through R2, which will cause the output to go negative until the LED on pins 1 & 2 of the IC2 coupler conducts. The light will cause the photodiode between pins 3 & 4 to conduct, pulling the voltage off IC1 pin 2, and causing the output to fall to zero.

    The current though the 470K resistor coming from the 100-to-1 divider will make the circuit a voltage to frequency converter. IC3 will follow merrily along at that frequency, simply being isolated by IC2 from the high voltage side. Vr1 & 2 adjust its operating point and duty cycle.

    The 100 pF cap on IC1 is too small to do integration, so it's probably a feed-foreward noise eliminator. I couldn't guess about the 10K pot on pin 6 until I see the arrangement.
     
  7. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    This "10K pot on Pin 6" is confusing to me. Do we mean the inverting input of the second OpAmp? (IC3 inverting i/p?) If so, where would the the other two connections of the 10K pot hook up?
     
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