Light Bulb Indicator

Introduction: Light Bulb Indicator

About: We design cool circuits.

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

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.

Supplies

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.

Step 1: Design the Circuit

I used 1.5 V light bulbs.

Calculate the maximum current across the R1 resistor:

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

= (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.

Step 2: Simulations

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.

Step 3: Make the Circuit

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 that will definitely reduce the potential voltage across the light bulb.

Step 4: Testing

I connected my circuit to 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 amplifier. A USB Class D amplifier will have poor current output. Thus I will need a mains powered Class D amplifier.

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