Boost Converter Circuit which is the step-up voltage regulator take low voltage and make it high.
It is very useful when power source gives less voltage than needed.
LM2577 is famous IC which helps make step-up circuit easy.
This article for Simple Boost Converter Circuit using IC LM2577 shows how the IC works, give formulas, and shows how to build.
Circuit Working:
Parts List:
| Component Type | Specification | Quantity |
|---|---|---|
| Resistors | 2.2k 1/4 watt | 1 |
| 15k+2.4k 1/4 watt | 1 | |
| 2k 1/4 watt | 1 | |
| Capacitors | Ceramic 0.1µF | 1 |
| Ceramic 0.33µF | 1 | |
| Electrolytic 680µF | 1 | |
| Semiconductors | IC LM2577 | 1 |
| Schottky Diode 1N5821 | 1 | |
| Coil Inductor 100µH | 1 |
This is step-up circuit using LM2577 IC.
It takes 5V input and gives 12V output.
IC LM2577 is switching regulator as it boost voltage.
When switch inside IC is closed then current goes through inductor L1 and stores energy.
Input gives power to inductor.
Capacitor C3 keeps output steady.
When switch opens then inductor releases energy.
Magnetic field collapses and makes higher voltage and this add to input voltage.
Diode D1 pass this boosted voltage to load and charge capacitor.
IC switches fast to around 52 kHz to keep output voltage stable.
Formulas with Calculations:
Below are short formulas and example for Boost Converter using LM2577:
Output Voltage:
Vout = 1.23 × (1 + R2 / R3)
Example: Vout = 1.23 × (1 + 17.4kΩ / 2kΩ) = 12V
Inductor (L):
L = (Vin × (Vout − Vin)) / (f × ∆I × Vout)
where,
- Vin is the input voltage
- f is the 52kHz
- ∆I is the ripple current for 20 to 40% of load current
Capacitor (C):
C = Iload / (f × ∆Vout)
where,
- Iload is the load current
- ∆Vout is the allowed voltage ripple
Diode Current Rating:
Idiode ≥ Iload × (1 + Vout / Vin)
Diode must handle this current safely.
How to Build:
To build a Simple Boost Converter Circuit using IC LM2577 follow the below mentioned connections steps:
- Connect 5V positive to pin 5 VIN of LM2577 through C1 0.1µF to reduce noise.
- Connect ground GND to pin 3 GRND.
- Put 100µH inductor L1 between pin 4 and anode of D1 1N5821.
- Connect D1 cathode to output 12V positive.
- Connect C3 680µF between output Vout and ground.
- Connect R2 15k + 2.4k between Vout and pin 2.
- Connect R3 2k between Pin 2 and ground.
- Connect R1 2.2k and C2 0.33µF in parallel between pin 1 and GND.
Conclusion:
Simple Boost Converter Circuit using IC LM2577 is simple and useful.
To make a circuit only few parts are needed which works well for many projects.
We should use the right values and to build good circuit which gives stable and reliable output.