40 LED AC 230V Bright Light Circuit Diagram | Transformer-less

The transformerless 40 LED AC 230V bright light circuit diagram presented here is a cheap solution for achieving bright illumination using LEDs without the need for a transformer. This circuit design harnesses the power of alternating current (AC) directly to drive the LEDs efficiently, resulting in an eco-friendly and cost-effective lighting solution. By eliminating the transformer, the circuit reduces complexity and saves space while delivering a powerful output. The LED light circuit incorporates 40 high-intensity LEDs, ensuring a brilliant and uniform illumination. With its simple yet robust design, this circuit diagram represents a breakthrough in LED lighting technology, offering a reliable and energy-efficient solution for various applications.

#### Components Required:

Resistors:R1 - 100E, 2W

R2 - 390K, 1/2W

Capacitors:

C1 - 0.27uF or 0.22uF, 400V or 630V

C2 - 10uF, 400V

BR - Bridge Rectifier ( DB107 or W04)

MOV - 275/20

LED - Ultra bright white LED(40 numbers)

General purpose PCB

The forward current of the LED that I have used in this project is 20mA.

We can connect 1W zener diode parallel to the output to get a stabilized output. Number of Leds in the output can be reduced or increased accordingly.

__:Always ask the forward voltage and current of the LED you are going to use before purchasing.__

**Note**If you are going to use LED's of another current rating, please use the following calculation.

$X_{c}= \frac{1}{2\pi fC} Ohms ---- (1)$

$X_{c}= \frac{V_{RMS}}{I} Ohms ----(2)$

Where,

f - Frequency in Hertz (Hz)

C - Capacitance in Farads (F)

$V_{RMS} \text{- Mains Voltage Volt(V)}$

I - Current in ampere (A)

Known values,

f - 50Hz

V - 230V

C - 0.27uF (This is the value of C1)

π - 3.14

Comparing (1) and (2),

Known values,

f - 50Hz

V - 230V

C - 0.27uF (This is the value of C1)

π - 3.14

Comparing (1) and (2),

$$\frac{1}{2\pi fC} = \frac{V_{RMS}}{I}$$

$$\Rightarrow I = V_{RMS} \times 2\pi fC$$

$$\Rightarrow I = 230 \times 2 \times 3.14 \times 50 \times 0.27 \times 10^{-6} [\because \mu = 10^{-6}]$$

$$\Rightarrow I = 19.4994 \times 10^{-3} A$$

Rounding off, we get:

$$\Rightarrow I = 19 \text{ } mA$$

$$\Rightarrow I = V_{RMS} \times 2\pi fC$$

$$\Rightarrow I = 230 \times 2 \times 3.14 \times 50 \times 0.27 \times 10^{-6} [\because \mu = 10^{-6}]$$

$$\Rightarrow I = 19.4994 \times 10^{-3} A$$

Rounding off, we get:

$$\Rightarrow I = 19 \text{ } mA$$