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RC Time Delays and the 555 Monostable

Timing in Electronic Systems

Many electronic systems need a controlled time delay: a car alarm that waits two seconds before triggering, a washing machine that holds a motor on for thirty minutes, a burglar-alarm sounder that turns off after ten minutes. The simplest way to generate a time delay in electronics is to let a capacitor charge through a resistor. This RC network forms the foundation of both simple delay circuits and the 555 timer.

RC Networks and the Time Constant

When a capacitor is connected in series with a resistor and a voltage is applied, the capacitor does not charge instantly. It charges gradually — fast at first, then slower as the voltage builds up. The rate of charging is controlled by the time constant of the circuit, given the symbol τ (tau):

Key Fact

$$\tau = R \times C$$

where \(\tau\) is in seconds, \(R\) is in ohms, and \(C\) is in farads. After one time constant, the capacitor has charged to approximately 63% of the supply voltage. After five time constants, it is considered fully charged (over 99%).

The charging graph is an exponential curve — it starts steep and then levels off as the capacitor approaches the supply voltage. Discharging through a resistor follows the same shape in reverse: the voltage falls quickly at first and then decays slowly towards zero. You do not need to evaluate the exponential equation numerically, but you must be able to recognise the curve shape and read values from a charging or discharging graph.

For example, with \(R = 100\) kΩ and \(C = 10\) μF:

$$\tau = 100{,}000 \times 10 \times 10^{-6} = 1\text{ s}$$

After 1 s the capacitor has reached 63% of the supply. After 5 s it is essentially fully charged. This 1-second time constant is useful for human-timescale delays such as door-sensor delays or alarm hold times.

Simple RC Delay Circuits

A simple RC delay circuit connects a resistor in series with a capacitor. The voltage across the capacitor rises exponentially. If a comparator or transistor switch is connected across the capacitor, the circuit will trigger once the capacitor voltage exceeds a threshold — creating a time delay from the moment the supply is applied.

The weakness of a plain RC delay circuit is that its threshold can be uncertain and its output is not a clean logic level. For precise, stable timing, the 555 timer is used instead. The 555 contains two internal comparators, a bistable latch, a discharge transistor, and a voltage divider — all inside a single 8-pin IC.

The 555 Timer in Monostable Mode

In monostable mode, the 555 produces a single output pulse of fixed duration each time it receives a trigger. The pulse duration depends on an external resistor \(R\) and capacitor \(C\). The trigger is normally a brief negative-going pulse applied to pin 2.

Key Fact

555 monostable pulse duration:

$$T = 1.1 \times R \times C$$

where \(T\) is in seconds, \(R\) is in ohms, and \(C\) is in farads. The 1.1 factor reflects the internal threshold level of the 555 (the capacitor charges to two-thirds of the supply voltage before the output resets).

What happens during a monostable cycle, step by step:

  1. Initially, the output (pin 3) is LOW and the internal discharge transistor holds the capacitor discharged (pin 7 to ground).
  2. A trigger pulse takes pin 2 below one-third of the supply voltage. This flips the internal bistable and turns the output HIGH.
  3. The discharge transistor turns off. The external capacitor \(C\) begins charging through \(R\) towards the supply voltage.
  4. When the capacitor voltage reaches two-thirds of the supply voltage, the internal comparator resets the bistable. The output goes LOW and the discharge transistor turns on, rapidly discharging \(C\).
  5. The circuit is back in its stable state, ready for the next trigger. The output pulse lasted for time \(T = 1.1RC\).

A 555 monostable must produce a 5-second pulse. The designer has a 100 μF capacitor. What value of R is needed?

Step 1 — Write the formula: \(T = 1.1 RC\).

Step 2 — Rearrange for \(R\): \(R = \frac{T}{1.1 \times C}\).

Step 3 — Substitute: \(R = \frac{5}{1.1 \times 100 \times 10^{-6}} = \frac{5}{1.1 \times 10^{-4}} \approx 45{,}455\) Ω.

The nearest E24 standard value above this is 47 kΩ. With \(R = 47\) kΩ the actual delay would be \(T = 1.1 \times 47{,}000 \times 100 \times 10^{-6} = 5.17\) s.

If the question uses a variable resistor with a fixed series resistor, you must subtract the fixed resistor value before calculating the variable-resistor setting. For example, if the circuit has a fixed 10 kΩ in series with a variable resistor and the total \(R\) needed is 47 kΩ, the variable resistor must be set to \(47{,}000 - 10{,}000 = 37\) kΩ. Forgetting this subtraction is one of the most common exam errors.

A plain RC network charges a capacitor, but the voltage across it rises gradually and never produces a sharp logic-level transition on its own. The delay is approximate and sensitive to supply voltage.

The 555 monostable uses an RC network internally, but the 555’s comparators and bistable latch convert the gradual capacitor curve into a clean digital output that switches sharply between LOW and HIGH at precisely \(T = 1.1RC\). It is also buffered — the output can source or sink up to 200 mA to drive LEDs, relays or other loads directly without additional transistor stages.

The next lesson extends the 555 into astable mode, where it oscillates continuously without any trigger to produce a repeating pulse train.

Practice Questions

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