Question About Duty Cycle and Average Value of a Square Wave

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The discussion revolves around calculating the duty cycle and average value of a square wave using a 555 timer in astable mode. For the first problem, the calculated value of resistor R2 to achieve a duty cycle of 0.7 is determined to be 6 kOhms. In the second problem, the average value of a square wave with a maximum of 10 volts and a duty cycle of 73% is initially calculated as 2.7 volts. However, after further discussion, it is concluded that the correct average value should be 7.3 volts, reflecting the higher contribution of the 10 volts during the duty cycle. The calculations highlight the importance of accurately applying formulas for duty cycle and average voltage in circuit analysis.
Mohamed Abdul

Homework Statement


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1. Using the Schematic of a 555 in astable mode from the Wikipedia entry on '555 Timer IC," resistor R1 = 8 kOhms and capacitor C = 3 micro-Farads. Determine the value of R2 so that the output duty cycle is D = 0.7. Note that D is not expressed as a percentage here for simplicity. Express your answer in kOhms.

2. Determine the average value of a square wave whose minimum value is 0 Volts, whose maximum value is V = 10 Volts and whose duty cycle D = 73 %. Express your answer in Volts.

Homework Equations



DC = high time/low time
high time = ln(2)*C*(R1+R2)
low time = ln(2)*C*R2

average value = D*V1 + (1-d)*V2 <= Not sure if this is the right equation

The Attempt at a Solution



For number 1, I set my high time and divided it by the low time. I managed to cancel out the ln(2)*C and got that duty cycle was (R1+R2)/(R1+2R). After setting that equal to .7, I got that R2 was 6kOhms.

For number 2, I plugged in my values for the last equation, using O volts as V1 and 10 volts as V2. This gave me an answer of 2.7 volts.
 
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Mohamed Abdul said:
2. Determine the average value of a square wave whose minimum value is 0 Volts, whose maximum value is V = 10 Volts and whose duty cycle D = 73 %. Express your answer in Volts.
Mohamed Abdul said:
For number 2, I plugged in my values for the last equation, using O volts as V1 and 10 volts as V2. This gave me an answer of 2.7 volts.

73% of the time at 10 volts, and 27% of the time at 0 volts. Would this average 2.7 volts?
 
Asymptotic said:
73% of the time at 10 volts, and 27% of the time at 0 volts. Would this average 2.7 volts?
] Looking at what you said, it'd make sense that the number should be higher, so should I just reverse the voltages in the equation and get 7.3 volts instead?
 
Mohamed Abdul said:
] Looking at what you said, it'd make sense that the number should be higher, so should I just reverse the voltages in the equation and get 7.3 volts instead?
Yes, an answer of 7.3 volts fits with how I'm interpreting question #2.
 

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