1. [10 points] Calculate the expected frequency, t_high, and t_low for the 555 timer as configured with external 470 ohm resistors and 1000 μF capacitor. Show the equations and your math. Does the timing seen in your circuit match the predictions of the equations?

Given Information:

Resistance = R1 = R2 = 470 Ω

Capacitance = C = 1000 uF

Required Information:

Time period high and low = ?

Frequency = f = ?

t_high = 0.6514 seconds

t_low = 0.3257 seconds

frequency = 1.02 Hz

Explanation:

The 555 IC is a multi purpose IC that can be used in monostable, bistable and astable modes. It is used in a wide variety of applications such as speed control of motors via PWM, Oscillators, Timing circuits etc.

It can generate a series of pulses at fixed frequency that is controlled by R1, R2 and C values.

The high time or on time is the amount of time the capacitor is charging through the resistor and is given by

t_high = 0.693*(R1 + R2)*C

t_high = 0.693*(470 + 470)*1000x10⁻⁶

t_high = 0.6514 seconds

The low time or off time is the amount of time the capacitor is discharging through the resistor and is given by

t_low = 0.693*R2*C

t_low = 0.693*470*1000x10⁻⁶

t_low = 0.3257 seconds

The corresponding frequency of these pulses can be found using

f = 1 / t_high + t_low

f = 1 / 0.6514 + 0.3257

f = 1.02 Hz

Bonus:

The Duty cycle is the amount of time in % when the pulses are high

D = t_high/t_total * 100%

D = 0.6514/0.9771 * 100%

D = 66.67%

answer:

truck

explanation:

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