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555 Astable Circuit question

  1. Feb 7, 2007 #1

    I have a question about an astable oscillator made with a 555 timer IC. According to a National Instruments website, the off-duty cycle for an astable oscillator is Rb / (Ra + 2Rb) and the frequency is 1/(0.7*(Ra + 2Rb)*C), where Ra is the resistor in between pin 7 and the power source, and Rb is the resistor in between pins 7 and 2/6, and C is the capacitor in between pins 2/6 and ground. I believe this is a standard setup for an astable oscillator. However, I've noticed that oftentimes Rb >> Ra, so my question is, is there a reason for this? Because if Rb >> Ra and you calculate the duty cycle and frequency of your circuit, the influence of Ra in these cases is often negligible.
  2. jcsd
  3. Feb 7, 2007 #2
    Most applications want a 1:1 dutycycle. If you set Rb >> Ra then you can approximate frequency to 1/(sqrt(2)*Rb*C) and ignore the duty cycle as it will approximate to 1:1...
    Unless your application needs really accurate frequency and 1:1 duty cycle making Rb>>Ra is a good working simplification.
  4. Feb 7, 2007 #3
    I see... thanks!
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