A 555 Timer IC Tutorial

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calculated with the simple expression:

intervals of the output pulses. As you

recall from your study of basic electronics,

it takes a finite period of time for a

capacitor (C) to charge or discharge

through a resistor (R). Those times are

clearly defined and can be calculated given

the values of resistance and capacitance.

The basic RC charging circuit is shown in

fig. 4. Assume that the capacitor is initially

discharged. When the switch is closed, the

capacitor begins to charge through the

resistor. The voltage across the capacitor

rises from zero up to the value of the

applied DC voltage. The charge curve for

the circuit is shown in fig. 6. The time that

it takes for the capacitor to charge to

63.7% of the applied voltage is known as

the time constant (t). That time can be

t=RXC

Assume a resistor value of 1 MegaOhm and a capacitor value of 1uF (micro-Farad). The time

constant in that case is:

t = 1,000,000 X 0.000001 = 1 second

Assume further that the applied voltage is 6 volts. That means that it will take one time

constant for the voltage across the capacitor to reach 63.2% of the applied voltage. Therefore, the

capacitor charges to approximately 3.8 volts in one second.

Fig. 4-1, Change in the input pulse frequency

allows completion of the timing cycle. As a general

rule, the monostable 'ON' time is set approximately

1/3 longer than the expected time between

triggering pulses. Such a circuit is also known as a

'Missing Pulse Detector'.

Looking at the curve in fig. 6. you can see that it takes approximately 5 complete time

constants for the capacitor to charge to amost the applied voltage. It would take about 5 seconds

for the voltage on the capacitor to rise to approximately the full 6-volts.

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18/11/00