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Dual +ve edge

Discussion in 'Electronic Basics' started by [email protected], Jun 18, 2005.

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  1. Guest

    I need a low power very low offset voltage comparator opamp which will
    take 0-5v as supply voltage. It should take -5v to 5v as its input.
    Will the -5v input wth 0-5v supply burn the opamp??? PLease let me know
    about ny suitable opamp. I will use the opamp 4 the purpose described
    below. Any critical thoughts abut my plan is most welcome...

    I need to produce +ve edges from each of the +ve and -ve edge of an
    square wave input. The simplest idea was to diffenciate the square wave
    and full wave rectify the output. I cannot do this b cos I'm restricted
    to a single power supply. Instead I designed the followin circuit: A
    capacitor and a resistor is connected in series.The other end of the
    cap is connected to square wave input. The other end of the resistor is
    grounded. The +ve input of one opamp is connected to the -ve input of
    another opamp and they r grounded. The other +ve and ive inputs of the
    opamps r also joined and they both r connected betn the cap and
    resistor. Each of the output of the opamps r connected togethe via
    rectifiers. THis is the output. PLEASE DRAW THIS SIMPLE SCHEMATIC AND

    Please help me ASAP.
  2. Chris

    Chris Guest

    Good morning, AlienOnEarth. Sorry about the delay. I can detect the
    note of panic in your post.

    First off, I'm hearing you say you want to interface a -5V/5V square
    wave with a +5V only system and output a pulse both for the positive
    transition of the square wave and the negative transition. That
    shouldn't require comparators at all -- you simply use a diode and a
    resistor to rectify the input to a +4.3V/0V square wave.
    ` Vin Vout
    ` o->|-o----o
    ` |
    ` .-.
    ` | |
    ` | |
    ` '-'
    ` |
    ` ===
    ` GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    You can then use standard digital logic to get your pulses (I'm
    assuming this is a digital logic interface).

    But, if you want to use a dual comparator to do this, it becomes pretty
    easy with DC level shifting.

    ` VCC
    ` +
    ` |30K* VCC
    ` VCC .-. +
    ` + | | |
    ` 30K*| | | .-.
    ` .-. '-' VCC | |
    ` | | | + | |
    ` | | | |\| '-'
    ` Vin '-' o-----|-\ | Vout
    ` || ___ | | | >-----o-----o
    ` o---||--|___|-o-----|--o--|+/ |
    ` || 120K | | | |/ |
    ` 100pF .-. | | |
    ` 20K*| | | | |
    ` | | .-. | |
    ` '-' | |3.3K |
    ` | | | | |
    ` === '-' | |
    ` GND | | |\ |
    ` | '--|-\ |
    ` | | >-----'
    ` o-----|+/
    ` | |/|
    ` | ===
    ` .-. GND
    ` 20K*| |
    ` | |
    ` '-'
    ` |
    ` ===
    ` GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    You can see by the diagram that you've established a DC level of 2V for
    the input signal. The input resistor and cap then cause approx. 1V
    excursions above and below that DC level with the +5V/-5V excursions of
    your input square wave. You then use two comparators as a window
    comparator to get pulses if the input signal goes above the higher
    divider voltage, or below the lower one. (Resistors with an asterisk
    are 1% or better).

    I'm not sure why you'd need a low offset comparator for a square wave
    that slams 10V either way. With this setup, you don't have to worry
    about the differentiator GND. Low power is relative, and up to you.
    But the voltage dividers in the above take less than 0.5 mA, and any
    single supply dual comparator should do the job. An LM393 will only
    take another 1 mA or so with a 5V supply.

    I hope this has been of help.

    By the way, I'd like to add a few style notes. First, many respondents
    don't even bother answering posts that are written in cell phone text
    messaging style or all caps, so if you're actually posting from a cell
    phone, the emergency must be truly dire (all caps usually indicate
    either a hostage situation, a fire, or shouting). Please note this in
    the body of your text, but only after you call the police or the fire
    department. If not, plz uz std wrds -- r 2 hrd 2 rd.

    Also, you might want to download Andy's ASCII circuits -- it's much
    easier to get a point across with a picture, even within the
    limitations of ASCII art. The program is incredibly easy to use, and
    is practically intuitive if you've ever done anything in CAD.

    Good luck
  3. John Fields

    John Fields Guest


    | Y--+------A
    +--B | EXOR Y-->OUT
    | [R] +--B
    | | |
    [1K] +---+
    | |
    | [C]
    | |

    Any gate will work at the junction of the diode and resistor since
    its purpose is to provide a totem-pole output to source and sink
    current to charge and discharge the cap.
  4. Guest

    thanx a lot chris, Use net is a wonderful service. It's my 1st time in
    the groups so I was ignorant about its precepts.

    Any way I will use "+5v/0v not +5v/-5v square wave".As u guessed, it
    will be a digital logic interface. You said I can use standard digital
    logic to get pulses from a rectified +5v/-5v squarewave. So it implies
    I can also get the desired pulses from +5v/0v input using digital
    logic. How can I do that??
  5. Ban

    Ban Guest

    You invert the digital pulse train and then with a small capacitor you can
    differentiate both outputs, so the pulse length is short compared to the
    original frequency. this signal you can feed into two comparators and
    or-them by two diodes or another gate.
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