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.
Note:Always ask the forward voltage and current of the LED you are going to use before purchasing.
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:
$$\Rightarrow I = 19 \text{ } mA$$
