Question

Using a 555 timer as an oscillator, answer the following. When solving, rather than leaving the control voltage (pin 5) open, connect another power supply voltage at +2V to the control voltage pin. Also, assume that the power supply voltage is +5 volts, C T =1nF, R 1 =50.5KΩ, R 2 =93.8KΩ, and the diode configuration for the 555 timer is used. Solve for the duty cycle in % of the output waveform, and the frequency of the output waveform.

Answer 1

The duty cycle of the oscillator is approximately 64.97% and the frequency is approximately 333.33Hz.

Step-by-step explanation:

The duty cycle of an oscillator can be determined using the formula:

Duty Cycle = (R2 / (R1 + R2)) * 100%

For this given circuit, R1 is given as 50.5KΩ and R2 is given as 93.8KΩ. Plugging these values into the formula:

Duty cycle = (93.8KΩ / (50.5KΩ + 93.8KΩ)) * 100% = 64.97%

The frequency of the oscillator can be determined using the formula:

Frequency = 1.44 / ((R1 + 2R2) * C)

Plugging in the given values of R1 (50.5KΩ), R2 (93.8KΩ), and C (1nF):

Frequency = 1.44 / ((50.5KΩ + 2 * 93.8KΩ) * 1nF)

After solving this equation, the frequency comes out to be approximately 333.33Hz.