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Home » Switching Regulator Circuit using IC uA78S40

Switching Regulator Circuit using IC uA78S40

Last updated on 21 June 2026 by Admin-Lavi Leave a Comment

The Switching Regulator Circuit using IC uA78S40 works as a buck converter that smartly and efficiently reduces a high input voltage to a lower output voltage.

Also, IC uA78S40 chip is good for power supply and 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.

Furthermore, input can be 25V DC and output 5V at 500mA; also this article will show how step-down converter work with uA78S40 IC.

Circuit Working:

Switching Regulator Circuit Diagram using IC uA78S40

Parts List:

ComponentsValuesQuantity
Resistors1.2k 1/4 watt1
3.6k 1/4 watt1
0.33Ω 1/4 watt1
CapacitorsCeramic 470pF1
Electrolytic 470µF 25V1
Electrolytic 100µF 25V1
SemiconductorsIC uA78S401
Schottky Diode 1N58221
Inductor 220µH1

To begin with, this circuit take 25V DC input and change to steady 5V DC output with 500mA; further, its main part is uA78S40 IC which is special regulator that can step down, step up or invert voltage.

Here, capacitor C3 smooths the 25V input to prevent sudden voltage changes and the IC uA78S40 keeps the voltage steady by switching on and off rapidly under the control of capacitor C1.

Then resistor R3 watch current to keep output right.

After that, inductor L1 store energy when switch is ON and gives energy when switch is OFF to keep power flow.

Now Schottky diode D1 change current direction and capacitor C2 smooth output to 5V and then 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

So 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

Hence, use 0.33Ω resistor for R3.

How to Build:

To build a Switching Regulator Circuit using IC uA78S40 follow the below steps for connections:

  • First, get all parts from above circuit.
  • Next, pin 1 of IC1 uA78S40 connect to cathode of diode D1 and anode of D1 connect to GND.
  • Also, pin 2 of IC1 connect to GND.
  • Then pin 3 of IC1 connect one end of inductor L1 and other end of L1 connect to 5V output.
  • Now pin 8 and pin 9 of IC1 join together.
  • After that, pin 10 of IC1 connect to junction of R1 and R2 and also connect to GND.
  • Further, pin 11 of IC1 connect to GND and also pin 12 of IC1 connect to one end of capacitor C1 and other end of C1 to GND.
  • Then pin 13 of IC1 connect to 25V input and resistor R3 connect between pin 13 and pins 14, 15, 16 of IC1.
  • Finally, positive of capacitor C3 connect to +25V and negative to GND and then positive of capacitor C2 connect to 5V output and negative to GND.

Conclusion:

To conclude, this Switching Regulator Circuit using IC uA78S40 smartly changes big DC voltage to steady small 5V DC.

In addition , circuit is good for small power needs with reliable 5V supply and also a switching regulator with loss less power work better than normal linear regulator.

References:

Datasheet IC uA78S40

Filed Under: Mini Projects, Power Supply Circuits, Voltage Regulator

About Admin-Lavi

Lavi is a B.Tech electronics engineer with a passion for designing new electronic circuits. Do you have questions regarding the circuit diagrams presented on this blog? Feel free to comment and solve your queries with quick replies

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