Question About Duty Cycle and Average Value of a Square Wave

[Mohamed Abdul]

Homework Statement

[/B]
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.

 

[Asymptotic]

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.

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?

 

[Mohamed Abdul]

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?

 

[Asymptotic]

] 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.