Using AD8561

Discussion in 'General Electronics Chat' started by OSOO, May 15, 2014.

  1. OSOO

    OSOO Thread Starter Member

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    Hi,
    I have AD8561 IC ,and I don't know how to use it in my comparator circuit ,why it has 2 output pins ? and what is the function of the latch pin ? do I have to use it in my circuit ?

    for example this circuit:

    [​IMG]
  2. crutschow

    crutschow AAC Fanatic!

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    The circuit has differential outputs which are used where a high speed differential signal is needed. You can use just one output, if desired.

    The latch is also an optional input, used to latch the output in its last state, if desired.

    But I wouldn't think you need such a high speed comparator for that optical application. Note that you need a pull-down resistor for the output of detector to ground to rapidly pull the voltage to 0V when the detector turns off.
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  3. OSOO

    OSOO Thread Starter Member

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    So why my circuit is not working , I connected pin 1 to 6v power supply and pin4 and pin6 to ground ,and V+ the reference voltage from the voltage divider which ~1v ,(I used a 46k resistor rather than 100k in the circuit schematic above ) ,the V- is connected to the output of the phototransistor though a potentiometer which is 3v ,when the light is interrupted this voltage become ~0v ,but the output voltage still high even if the reference voltage is greater the V- voltage ,why !?
  4. crutschow

    crutschow AAC Fanatic!

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    Do you have a resistor from the photo-transistor output to ground as I mentioned in my previous post. Otherwise the leakage current at that node may maintain the signal voltage high or, at best, it will very slowing leak to zero. If you measure that voltage, the input resistance of the voltmeter will pull the node low and you will measure 0V.

    Edit: I don't see a pot in your schematic. Please post an updated schematic.
  5. OSOO

    OSOO Thread Starter Member

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  6. ronv

    ronv Senior Member

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    You may need to ground the latch pin. The chip also has 2 grounds - in your case V- and ground. Add some resistance in series with the LED or it may burn your IC or the LED.
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  7. OSOO

    OSOO Thread Starter Member

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    Thank you rony
    the problem is solved
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