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How to Illuminate 18 LEDs from a 3.7V Battery

Powering 18 LEDs from a 3.7V battery is like having a party with lots of tiny light bulbs!

To make sure all the bulbs light up nicely, we need a special plan.

This plan involves putting the LEDs in smaller groups and connecting them in a smart way.

By connecting them in lines series and rows parallel, we can adjust the voltage and current so all the LEDs get exactly what they need to shine bright without burning out.

Circuit Working:

Parts List:

CategoryDescriptionQuantity
Resistors1k, 10Ω (1/4 watt each)1 each
10k (1/4 watt)2
CapacitorsCeramic 10nF1
Electrolytic 100μF 25V1
SemiconductorsTransistors BC547, 2N22221 each
Inductor1mH1
LEDsWhite straw hat 5mm, 20mA18
Other ComponentsON/OFF switch1
Power SourceLi-ion Battery 3.7V1

This circuit operate 18 white LEDs using a 3.7V Li-Ion cell and incorporates components from an old Compact Fluorescent Lamp.

An intriguing feature of this circuit is its use of two transistors to maintain constant brightness with the BC547 providing feedback to sustain oscillation.

Despite the seemingly high 10k base resistor the circuit was tested for 12 hours on a 1200mA Hr cell, and the brightness remained consistent.

Brightness is controlled by the drive strength of the 2N2222 transistor.

The transistor is activated by the 10k resistor delivering minimal current.

However, due to the transistors gain of 100 to 300 the collector current can reach up to about 100mA.

Essentially, the circuit drives the LEDs beyond their capacity and the BC547 regulates the current to the necessary brightness level.

The BC547 acts to switch off the 2N2222 at a specific point in each cycle reducing its overall on time.

This switching is timed by the 10k and 1k resistors which form a voltage divider that produces a base voltage.

When the 2N2222 turns on a voltage develop across the 10 ohm resistor that when added to the base voltage slightly delays the transistors activation.

The 1nF capacitor determine the circuits oscillation frequency.

Through experimentation with these four components the desired brightness can be achieved and maintained throughout the cells lifespan.

All LEDs are connected in series in each string and the brightness depends on matching each string.

By swapping some LEDs you can adjust the brightness to ensure they all emit equally.

Formulas:

Using a boost converter, you may raise the voltage from 3.7V to a higher voltage that can be used to drive 18 LEDs in series or parallel arrangement from a 3.7V battery.

The boost converters important formulae is provided below:

Output Voltage Vout:

A boost converters output voltage is determined by:

Vout ​= Vin​​ / 1−D

where,

Inductor Current IL:

IL ​= Vout ​* D * T / ​ L

where,

T is the switching cycle period, while L is the inductance.

Duty Cycle D:

The duty cycle, which is determined by dictates how long the switch is closed as opposed to open.

D = Vout​−Vin​​ / Vout​

Choose Inductor L:

Based on the intended output current and the converters operating frequency, choose an inductor.

The following formula may be used to estimate inductor value:

L ≥ ​(Vout​−Vin​) * Vout​​ / f * ΔIL

where,

This formulas offers a foundation for building an efficient boost converter circuit to light up to eighteen LEDs using a 3.7V battery.

The precise LED forward voltage requirements and intended operation circumstances will determine the component selections and detailed calculations.

How to Build:

How to Illuminate 18 LEDs from a 3.7V Battery find out the below mentioned steps for connection of components.

Note:

Conclusion:

Illuminating 18 LEDs from a 3.7V battery requires a well thought circuit design that consider the voltage and current requirements of the LEDs.

Overall, the goal is to create a circuit that efficiently utilize the battery power to provide adequate illumination from the LEDs.

References:

Is 3.7V LiPo suitable to power arduino+led strip?

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