LED Clap Switch Circuit – B1P17

Published by frenoy on

We’re now going to learn how to activate an LED with sound. We will be using a microphone to detect sound, so let’s talk about that first.

The microphone we will be using is called an electret condenser microphone. A condenser is another name for a capacitor so the microphone is a capacitor at a very basic level. We have the two capacitor plates as usual but the microphone is designed such that sound waves can change the distance between the capacitor plates, which changes the resulting capacitance.

This is the symbol for a MIC and to use it in a circuit, we would need to connect a biasing resistor. In the absence of sound, the output would be a steady DC value. Sound waves will cause an AC signal to appear on top of this DC value, that nature of this AC signal will depend on the sound waves.

Here’s what the circuit for this project looks like and it’s best we analyse it using the simulation.

We’ve replaced the microphone with a voltage supply and switch. Capacitor C1 blocks the DC component from the microphone output and only allows the AC components to pass through. Resistors R2 & R3 form a voltage divider which biases Q1 at about 1.16V, meaning it is switched ON. The base-emitter voltage drop for Q1 is 0.46V which means capacitor C2 is charged to about 0.7V. Resistors R5 & R6 form another voltage divider and the voltage across R6 is about 0.17V which is not enough to switch ON Q3.

If look at the top section, then we can see that C3 can charge through Q1, giving rise to the following polarity. This is the idle state of the circuit.

A clap generates a sudden change in the microphone output which is amplified by Q1. This output voltage increases the voltage across R6 and causes Q3 to switch ON. This switches ON the LED and also connects the positive end of C3 to ground which places a negative voltage at the base of Q2, switching it ON. When Q2, switches ON, it causes a high current to flow through R6, which increases the voltage drop across R6 and this forces Q3 to stay ON. Q2 & Q3 are thus latched ON. Capacitor C3 eventually discharges and then starts charging with the opposite polarity. As the voltage rises it will cause Q2 to switch OFF which will switch OFF Q3 as well. C3 will then discharge through Q1 and it will charge with the opposite polarity again, bringing us back to the idle state. Another clap will cause this cycle to repeat.

Let’s run the simulation and observe these states. Here we have the circuit in the idle state. Toggling the switch will cause Q2 & A3 to switch ON and C3 to reverse its charge. The transistors will eventually switch OFF and C3 will reverse its charge again, bringing us back to the idle state.

Let’s use the breadboard layout to build and test it.

Since a clap generates a very abrupt sound wave, the latching section of this circuit is necessary so as to extend the ON time of the LED. You can vary the bias point of Q1 and Q2 by adjusting the voltage divider resistors to increase or decrease the sensitivity of this circuit.

Let’s move on to the next project.

Categories: BBox 1