Light Bulb Indicator

The circuit in this article indicates the direction of the current flow with two light bulbs.

However, later I connected the circuit to an audio amplifier and recorded the videos with the music.

If you do not care about the direction of the current and only want to see the light bulbs blinking with music then D1a and D2a diodes are optional. Obviously, if you replace those two diodes (D1a and D2a) with a short circuit then you should also remove the second light bulb or connect the second light bulb in parallel with the first light bulb.

Such an indicator can be implemented with LEDs as well. Using LEDs or bright LEDs instead of light bulbs will reduce the cost and improve the performance of this particular circuit. However, some people say that light bulbs provide a unique look.

Components: light bulbs - 5 (1.5 V, 6 V, 12 V), 2.2 ohm resistors (high power) - 2, general purpose diodes - 10, matrix board, insulated wires, light bulb holders.

Tools: wire stripper, pliers.

Optional components: solder, cardboard, bipolar capacitor (from 470 uF to 4700 uF) - 2.

Optional tools: USB Oscilloscope, soldering iron.

I used 1.5 V light bulbs.

Calculate the maximum current across the R1 resistor:

Ir1max = Vr1 / R1 = (VinMax - Vd1a - Vd1b - Vd1c) / R1

We assume that the maximum input voltage is 3 V:

Ir1max = (3 V - 0.7 V * 3) / 2.2 ohms = 0.9 V / 2.2 ohms

= 0.40909090909 A = 409.09090909 mA

Calculate the resistor maximum power rating:

Pr1max = Vr1 * Vr1 / R1 = 0.9 V * 0.9 V / 2.2 ohms

= 0.36818181818 Watts

I have used 1 Watt resistor for my circuit.

Simulations show a maximum light bulb current of about 0.3 A.

IbulbMax = (Vd*2) / Rbulb = 0.7 V * 2 / 5 ohms = 0.28 A

A light bulb should not be modeled as a simple resistor. However, a 1.5 V light bulb with a maximum current rating or 0.3 A can be approximated as 1.5 V / 0.3 A = 5-ohm resistor.

I used two Soviet diodes and one 100 uF bipolar capacitor because I did not have a 1000 uF bipolar capacitor.

The white wire gives the user an option of bypassing the capacitor because the capacitor will definitely reduce the potential voltage across the light bulb. The capacitor is used to remove a possible DC voltage from the source.

I connected my circuit to a 3 V DC power supply, bypassing the 100 uF capacitor and only the first light bulb turned ON. I then reversed the direction of the current (swapping to connectors) and only the second light bulb turned ON. The 3 V input is the maximum for this circuit to prevent failure to light bulbs, resistors, and diodes.

My signal generator could not drive even one light bulb. It could not produce sufficient current (0.3 A) and the light bulbs were dim. I am thinking of buying a Class D or Class B power amplifier. A USB Class D amplifier will have poor current output. Thus I will need a mains-powered Class D or Class B power amplifier.