Sound activated LED

Question

I have some questions about sound activated LED ciurcuits.

Q1) How do LEDs turn on and of in this circuit?

My answer: "The condenser microphone consists of a capacitor and a resistor. When sound is produced, the capacitance changes as the diaphragm vibrates, and the voltage applied to the resistance inside the microphone changes. This changes the base voltage of transistor Q1, and as a result, when the base voltage increases, a current flows between the emitter and the collector. This in turn increases the base voltage between Q2 and Q4, and the LED turns on as a current flows through the transistor emitter and collector."

Q2)

a) How to calculate the internal resistance of the microphone, where Vm = 2.5V?

My answer: "From left voltage regulator and microphone parts, Ri can be calculated by voltage divider rule. Vm = 5 * Ri/(Ri+Rm) -> Ri = VmRm/(5-Vm) = (2.5 * 10k)/*(5-2.5) = 10kohm"

b) How to calculate the Rm(resistance of 10kohm), if the microphone is connected directly to 9V voltage without LM78L05.

My answer: "By using voltage divider rule, Vm = 9 * Ri/(Ri+Rm) -> Rm = (9-Vm)*Ri/Vm = 26kohm

Q3) What is the role of the capacitor 100nF between the microphone and the transistor 2N2222(Q1) in the circuit?

My answer: ?

I don't understand why 100nF is necessary

Please let me know if there's something wrong....

Answer 1

There are two 100 nF caps in the schematic. This is why reference designators are so important.

Probably, the one in the question is the coupling capacitor (sometimes called a blocking capacitor, but that is not as descriptive a term in this application). This is not a bypass capacitor. The microphone element output is a small audio signal sitting on about 2.5 V of DC. This DC level would cause the first transistor (reference designators - !) to be saturated on continuously, so the LEDs never would light.

Ignoring the audio for the moment, when the circuit is powered up there is approx 2.5 V on the cap's left side. The transistor is biased on through the 1 M resistor, so the transistor base is sitting at approx. 0.6 V. Adding the audio, any audio negative peaks greater than about 0.4 V will turn off the transistor. Now the 5K resistor can pull up the four output transistor bases and there shall be light.

Note that there is no reason for the 5 V regulator in this circuit. If you double the value of the microphone bias resistor, it can be connected directly to the 9 V source.

Also, you can cut the circuit's current draw in half by having only two output transistors, and putting two LEDs in series for each transistor, and reducing the value of the LED current-limiting resistors. Overall, not an optimized circuit design.

BTW, your questions sound a lot like schoolwork. Next time, say that in the question so we know how to respond.

Answer 2

The 100nF is called a bypass capacitor. It's role is to block the DC bias when there is no signal. Without it, Q1 stays conducting no matter what the microphone is doing. It essentially passes the changing AC signal from the mic and stops the DC current that would flow that's powering (biasing) the previous mic stage.