Simple LED Boost Driver Circuit with IC MP3302 chip is good boost converter for LED lights in many projects.
It can light 27 LEDs in 9 groups and each group have 3 SMD 2835 LEDs.
Battery are used in Lithium-ion.
MP3302 IC is special because it gives steady current and control many LEDs together.
IC is made for LED driving and can give 150mA output current.
It works with 3 to 6 volts input and is good for battery power.
This circuit is good for LED backlight, display lights and normal LED lighting.
Circuit Working:
Parts List:
| Component Type | Value / Part Number | Quantity |
|---|---|---|
| Resistors | 1Ω 1/4 watt | 1 |
| 0.56Ω 1/4 watts | 2 | |
| Capacitors | Ceramic 0.47µF | 1 |
| Electrolytic 10µF 25V | 1 | |
| Semiconductors | IC MP3302 | 1 |
| Schottky diode MBR0520 | 1 | |
| SMD 2835 LEDs | 27 | |
| Inductor 4.7µH | 1 |
MP3302 LED driver works like boost converter as it raise input voltage to power many LED strings.
Circuit uses input voltage 3V to 6V.
Capacitor C1 filter power to keep smooth.
Inductor L1 save energy when internal switch SW pin is ON.
When switch is OFF then L1 send energy through diode D1.
Diode stop reverse current back to circuit.
Circuit keep steady current in LED strings with feedback.
Feedback FB pin check voltage on resistor R1 to keep current same.
Enable EN pin for PWM dimming.
We can change LED brightness by changing control signal duty cycle.
Formulas with Calculations:
Inductor Selection:
Use formula:
L = (V_IN × (V_OUT – V_IN)) / (I_LED × f_SW × V_OUT)
where,
- V_IN is the input voltage
- V_OUT is the output voltage which depends on LED forward voltage
- I_LED is the LED current
- f_SW is the switching frequency
LED Current Calculation:
To find R1 for 150mA LED current:
I_LED = 0.2V / R1
where,
- I_LED =is 150mA = 0.150A
- Reference voltage is 0.2V
Calculate R1:
R1 = 0.2V / 0.150A = 1.33Ω
Output Voltage Calculation:
V_OUT = N × V_LED
where,
- N is the number of LEDs in series
- V_LED is the forward voltage of one LED
Resistor Selection for R2 and R3:
Use ohms law:
R = (Vsupply – Vf) / I_LED
where,
- Vsupply is 12V
- Vf is 3V
- I_LED is 0.150A
Calculate R:
R = (12V – 3V) / 0.150A = 9V / 0.150A = 60Ω
Power Dissipation in Resistor:
P = I² × R
P = (0.150A)² × 60Ω = 0.0225 × 60 = 1.35W
How to Build:
To build a Simple LED Boost Driver Circuit with IC MP3302 follow the below mentioned steps for connections and assembling:
- Gather all the components as shown in circuit diagram
- Connect L1 between SW pin and IN pin of IC1 MP3302.
- Connect C1 from positive supply 3V to 6V to GND.
- Connect GND pin of IC1 to circuit GND.
- Connect FB pin to junction of R1 and LED resistors.
- Connect diode anode to junction of L1 and SW pin of IC1.
- Connect diode cathode to LED anodes.
- Connect C2 across LED anodes and GND.
- EN pin used to turn circuit ON/OFF and control dimming.
Conclusion:
Simple LED Boost Driver Circuit with IC MP3302 is good for powering many LED strings with steady current.
Design is easy and brightness can change by PWM.
It is efficient boost converter which is perfect for battery powered lights and other LED uses.
References:
Simple open-circuit protection for boost converters in LED driver applications