Understanding the Role of Resistors in 555 Astable Oscillators

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In a 555 timer IC astable oscillator, the duty cycle and frequency are influenced by the resistors Ra and Rb, with the formulas Rb / (Ra + 2Rb) for duty cycle and 1/(0.7*(Ra + 2Rb)*C) for frequency. It is common for Rb to be much larger than Ra, which can simplify calculations by allowing the frequency to be approximated as 1/(sqrt(2)*Rb*C) and the duty cycle to approach 1:1. This simplification is beneficial for many applications that do not require precise frequency or duty cycle control. However, for applications needing accuracy, the relative values of Ra and Rb should be carefully considered. Understanding these relationships is crucial for effective circuit design.
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Hello.

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.
 
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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.
 
I see... thanks!
 
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