Explain how a capacitor charging through a resistor produces a time delay and how the delay depends on resistance and capacitance

What this dot point is asking

SEAB wants you to explain how a capacitor charging through a resistor produces a time delay, and how that delay depends on the resistance and the capacitance. The central insight is that a capacitor does not charge instantly: with a resistor in the path it charges gradually, and the rising voltage can be used to switch a transistor on after a predictable delay. Bigger resistance or bigger capacitance means a longer delay.

The answer

Why charging takes time

When a capacitor charges through a resistor from a supply, the resistor limits the current. At the start the capacitor is empty, so the voltage across it is zero and the current is largest. As charge builds up, the capacitor voltage rises and opposes the supply, so the current falls. The voltage rises quickly at first and then more slowly, levelling off as it approaches the supply voltage. This gradual rise is the basis of the delay.

Making a switch wait

Connect the rising capacitor voltage to the base of a transistor. The transistor stays off while the capacitor voltage is below about $0.7\ \text{V}$. As the capacitor charges, its voltage climbs, and the moment it reaches $0.7\ \text{V}$ the transistor switches on and turns on the load. The time from pressing the start button to switching is the delay.

What sets the delay

The delay depends on both the resistance and the capacitance:

• A larger resistance reduces the charging current, so the capacitor charges more slowly and the delay is longer.
• A larger capacitance needs more charge to reach the same voltage, so it again takes longer and the delay is longer.

The product of resistance and capacitance, $R \times C$, is called the time constant and measures how quickly the circuit charges. A bigger time constant means a longer delay. Using a variable resistor lets you adjust the delay.

Resetting the delay

To use the delay again, the capacitor must be discharged, for example by a reset switch across it. Once emptied, it can charge up and produce the same delay once more.

Examples in context

Example 1. A delayed corridor light. Pressing a button starts a capacitor charging through a resistor. The light comes on at once and, after the capacitor reaches the switching point and the circuit times out, switches the light off again. Choosing the resistor and capacitor sets how long the light stays on, which is why such lights can be tuned for different corridors.

Example 2. A burglar-alarm exit delay. When the alarm is armed, a capacitor begins to charge, giving the owner a fixed number of seconds to leave before the system becomes active. A larger time constant gives a longer exit delay. The same charging circuit that delays a light also delays an alarm, just with different values.

Try this

• Cue. State two ways to make the delay of a charging circuit longer. Increase the resistance, or increase the capacitance; either raises the time constant $R \times C$.

• Cue. A delay uses a $10\ \mu\text{F}$ capacitor and a $1.0\ \text{M}\Omega$ resistor. Find the time constant. Convert: $R \times C = 1.0 \times 10^{6} \times 10 \times 10^{-6} = 10\ \text{s}$.

• Cue. Explain why a reset switch is placed across the capacitor in a delay circuit. It discharges the capacitor so the delay can start again from zero on the next use.