Connect with us

555 timer 50% duty cycle

Discussion in 'Electronic Basics' started by panfilero, Dec 27, 2006.

Scroll to continue with content
  1. panfilero

    panfilero Guest

    Hi, does anyone know how I could get a 555 timer to operate in astable
    mode at a 50% duty cycle? All the circuits I've seen are always for
    greater than 50%. thanks.
     
  2. James Beck

    James Beck Guest

    I just googled 555 50% PWM and got :
    http://www.dprg.org/tutorials/2005-11a/index.html

    Look at what is done with the diodes D1 and D2

    Jim
     
  3. Ian Malcolm

    Ian Malcolm Guest

    Also notice that the timing resistor is fed from the Output and the Load
    fed from the Discharge pin (pullup resistor required or directly drive
    a load with low side switching). If you dont need a variable duty
    cycle, leave out the diodes and try a single timing resistor from the
    Output to the junction of Threshold and Trigger. It will probably be
    very close to 50% but might be off by a little. The circuit Jim gave is
    trimmable.
     
  4. John Fields

    John Fields Guest

    ---
    An easy way is to use a 7555 and let the output feed the RC:

    View in Courier

    .. +-------------+
    .. | |
    .. [Rt] +V |
    .. | |8 |
    .. | 6+---+---+3 |
    .. +--|TH OUT|--+-->OUT
    .. | 2|___ _|4
    .. +-O|TR R|O--+V
    .. | +---+---+
    .. [Ct] 1| 7555
    .. | |
    .. GND GND

    Another way is to use the circuit you have and run the output
    through a divide-by-two circuit:


    +-----------+
    | +-----+ |
    +--|D Q|--|--->OUT
    555OUT>-----|> _| |
    | Q|--+
    +-----+
    HC74
    4013
     
  5. Dorian

    Dorian Guest

    mode at a 50% duty cycle? All the circuits I've seen are always for
    The CMOS version of the 555 (e.g. 7555) is far superior to the Bipolar
    version (e.g. NE555) for a variety of reasons. Reason 1 is reduced power
    consumption. Reason 2 is the absence of the quirky short circuit spike (400
    ma) during an output transition. Reason 3 is that output levels of the CMOS
    version approach the supply rails (ground or common and VCC) as they should.
    Also I believe the voltage divider resistors in the bipolar version are 1K
    while they are 100K or more in the CMOS version. This is very useful if
    you're modulating the pin 5 voltage level.

    Hooking the output of the 555 (pin 3) to the RC network will give you a 50%
    duty cycle as John mentioned and this always works best with the CMOS
    version.

    Dorian
     
  6. kell

    kell Guest

    Connect a signal diode like 1N4148 in parallel with the resistor
    between pins 6 and 7, with the cathode (stripe) oriented toward pin 6.
    With this arrangement you can get any duty cycle you want. You can
    even get a fixed frequency, variable duty cycle oscillator if you
    replace the fixed resistors with potentiometer. Connect the ends of
    the pot to pins 6 and 8, the wiper to pin 7, and the diode from pin 7
    to pin 6.
     
  7. If you have the space, feed the output of the 555 into the clock input of an
    edge-triggered J-K FF. Tie J and K high to create a toggle and the output
    will be a nice 50% duty cylce at 1/2 the input frequency.

    Richard
     
  8. jasen

    jasen Guest

    change the voltage on pin 5 or use a different circuit, how much precision
    do you need?

    Bye.
    Jasen
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-