Answer:
f = 1.231 Hertz
t_high = 0.6515 seconds
t_low = 0.326 seconds
Explanation:
The 555 timer is an integrated circuit usually employed in variety of applications relating to timing, pulse generation, and oscillator. The 555 timer are used to provide timing delay, and also as a flip-flop element.
f = 1/(ln(2)*c2*(R1 + 2*R2 )) = 1/(ln(2)*1000uF*(470 + 2*470))= 1.231 Hertz
t_high = ln(2)*(R1 + R2)*C2 = ln(2) * (470 + 470) *1000uF = 0.6515 seconds
t_low = ln(2)*R2*C2 = ln(2) * 470 *1000uF = 0.326 seconds
Given Information:
Resistance = R1 = R2 = 470 Ω
Capacitance = C = 1000 uF
Required Information:
Time period high and low = ?
Frequency = f = ?
Answer:
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%
