Description.

Here is the circuit diagram of a 5V buck switching regulator based on the IC LM2678 from National Semiconductors. LM2678 series of regulators are monolithic integrated circuits which provide all necessary functions required for a buck switching regulator and can drive up to 5A loads. The IC has more than 90% efficiency and has excellent load and line regulation. The LM2678 is available in three fixed output voltages (3.3V, 5V, 12V) and an adjustable output version. The IC is also packed with a handful l of features like thermal shutdown, current limiting and ON/OFF control.

The circuit given here is based on the version LM2678-5.0 which gives an output of 5V.The input voltage for the regulator is fed to the pin 2 of the IC. Capacitors C1 to C4 are input bye-pass capacitors. They also provide current to the ICs control switch when it is switched ON first. Capacitor C5 boosts the gate drive of the internal MOSFET and makes it fully ON. This minimises switching loses and helps to attain high efficiency.Pin7 is the ON/OFF pin and the regulator will shut down if this pin is connected to the ground. Current drain during the shutdown mode will be less than 50uA.Schottky diode D1 is used as a freewheeling diode. When the control switch (internal MOSFET) is switched OFF the current from inductor L1 flows through this diode. Capacitors C6 and C7 and output filter capacitors.

Circuit diagram.

5V buck regulator circuit

Notes

  • Assemble the circuit on a good quality PCB.
  • The power supply for the circuit can be anything between 8 to 40V DC.
  • The feedback wiring must be placed as away as possible from the inductor L1.
  • Do not use loads that consume more than 5A.
  • A heat sink is seriously recommended for the IC.
Author

8 Comments

  1. If you’re not going to answer people, why post something so elementary and ubiquitous, and then ignore people’s questions!

  2. May I know what is the power output of this IC? From this circuit, the output power seem like 5V*5A=25watt. But from my experiment. Vin=32V,Vo=22V(theoretically, calculated by Vo=Vfb*(1+R2/R1)), the current output is limited to 1.73A, and the Vo measure is around 5.77V (not 22V!!) and the power output is 12Watt (far away from 25W). Can you shed some light for me? Many Thanks.

  3. Seetharaman

    Hi Lee if you have already 5volt supply why do you want to generate the same 5 volt? your requirement not clear pl.

  4. anyone know if I can get 5V/1A at output if input is 5V using this regulator?

    I know that my input 5V is below the specified input range (8-40V)

    Thanks

  5. A Nonymous

    Nathan, yes.

    Eugene, probably because he had them.

  6. Eugene Sithole

    why do you have to use the three caps on the input and not just one 50uf 50v cap for the bye pass