Switching Regulator Circuit using IC uA78S40 is called a buck converter.
It change high input voltage to lower output voltage with smartly and easily.
IC uA78S40 chip is good for power supply.
It has built-in parts like voltage reference that change with temperature, oscillator for duty cycle, strong switch, separate op-amp and its own diode.
Input can be 25V DC and output 5V at 500mA.
This article will show how step-down converter work with uA78S40 IC.
Circuit Working:
Parts List:
| Component Type | Value/Part Number | Quantity |
|---|---|---|
| Resistors | 1.2k 1/4 watt | 1 |
| 3.6k 1/4 watt | 1 | |
| 0.33Ω 1/4 watt | 1 | |
| Capacitors | Ceramic 470pF | 1 |
| Electrolytic 470µF 25V | 1 | |
| Electrolytic 100µF 25V | 1 | |
| Semiconductors | IC uA78S40 | 1 |
| Schottky Diode 1N5822 | 1 | |
| Inductor 220µH | 1 |
This circuit take 25V DC input and change to steady 5V DC output with 500mA.
Main part is uA78S40 IC which is special regulator that can step down, step up or invert voltage.
Input 25V smooth by capacitor C3 to stop sudden voltage change.
IC uA78S40 keep voltage steady by switching ON and OFF fast which is controlled by capacitor C1.
Resistor R3 watch current to keep output right.
Inductor L1 store energy when switch is ON and gives energy when switch is OFF to keep power flow.
Schottky diode D1 change current direction and capacitor C2 smooth output to 5V.
Resistors R1 and R2 set output voltage by feedback to IC.
Formulas with Calculations:
Output Voltage Formula:
Vout = Vref × (1 + R2/R1)
For uA78S40
where,
- Vref is 1.25
- VR1 is 1.2kΩ
- R2 is 3.6kΩ
- Vout = 1.25 × (1 + 3.6/1.2) = 1.25 × 4 = 5V
Inductor Calculation:
L = ((Vin – Vout) × D) / (Iout × f)
where,
- Vin is 25V
- Vout is 5V
- Iout is 0.5A
- f is 40kHz
Duty cycle D = Vout / Vin = 5 / 25 = 0.2
L = (20 × 0.2) / (0.5 × 40000) = 4 / 20000 = 200µH
Use 220µH in circuit or standard value
Current Sense Resistor R3:
Rsc = Vsense / Ilimit
where,
- Vsense is 0.165V
- Ilimit is 0.5A
- Rsc is 0.165 / 0.5 = 0.33Ω
Use 0.33Ω resistor for R3.
How to Build:
To build a Switching Regulator Circuit using IC uA78S40 follow the below steps for connections:
- Get all parts from above circuit.
- Pin 1 of IC1 uA78S40 connect to cathode of diode D1 and anode of D1 connect to GND.
- Pin 2 of IC1 connect to GND.
- Pin 3 of IC1 connect one end of inductor L1 and other end of L1 connect to 5V output.
- Pin 8 and Pin 9 of IC1 connect together.
- Pin 10 of IC1 connect to junction of R1 and R2 and also connect to GND.
- Pin 11 of IC1 connect to GND.
- Pin 12 of IC1 connect to one end of capacitor C1 and other end of C1 to GND.
- Pin 13 of IC1 connect to 25V input.
- Resistor R3 connect between pin 13 and pins 14, 15, 16 of IC1.
- Positive of capacitor C3 connect to +25V and negative to GND.
- Positive of capacitor C2 connect to 5V output and negative to GND.
Conclusion:
This Switching Regulator Circuit using IC uA78S40 smartly changes big DC voltage to steady small 5V DC.
Circuit is good for small power needs with reliable 5V supply.
Switching regulator with loss less power work better than normal linear regulator.