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Home » Simple 10V Switching Power Supply Circuit with IC LM5007

Simple 10V Switching Power Supply Circuit with IC LM5007

Last updated on 24 March 2025 by Admin-Lavi Leave a Comment

A switching regulator is a great circuit for changing power, skillfully controlling voltage levels by quickly turning a power transistor on and off.

The IC LM5007 is a special high voltage buck regulator made for situations where you need to lower voltage efficiently from a wide range of input voltages.

This built-in step-down switching regulator includes all the important parts to make a reliable and budget friendly switching regulator circuit.

The simple 10V Switching Power Supply Circuit using the LM5007 provides a steady 10V DC output from an input voltage range of 12V to 72V DC making it perfect for both industrial and automotive uses.

Also the LM5007 IC has a maximum output current limit of 700mA.

Circuit Working:

Simple 10V Switching Power Supply Circuit Diagram with IC LM5007

Parts List:

ComponentValueQuantity
Resistors (1/4W)
R1200k1
R2100k1
R33k1
R41k1
R51Ω1
Capacitors
C1PPC 1µF 100V1
C2, C3Ceramic 0.1µF2
C4Ceramic 0.01µF1
C5PPC 15µF 100V1
Semiconductors
ICIC1 LM50071
D1Diode 1N40071
Inductor
L1Inductor Coil 100µH1

The IC LM5007 acts like a buck converter, which means it takes a higher direct current voltage and lowers it to a steady 10V output.

Before the voltage gets to the integrated circuit, it passes through capacitors C1 and C2 to make sure everything is stable.

The LM5007 IC works by quickly turning its internal MOSFET on and off at a high speed, which helps send energy to the output through inductor L1.

Resistor R1 helps set up the conditions needed for the circuit to start, while resistor R2 is important for controlling how long the circuit stays off.

Capacitor C3 is used as a decoupling capacitor to improve how well the circuit works.

The fast switching of the internal MOSFET helps move energy efficiently through inductor L1.

Diode D1 is to protect the circuit from any sudden voltage spikes that might cause harm.

The output voltage is kept steady by a resistor divider made up of R3 and R4 and capacitor C5 helps smooth out the final output signal.

Additionally the pairing of resistor R5 and capacitor C5 forms a filter network that boosts the overall performance.

Formulas with Calculations:

Below are the formulas with calculations in Simple 10V Switching Power Supply Circuit with IC LM5007:

Output Voltage Calculation:

The output voltage Vout​ is determined by resistors R3 and R4:

Vout = 1.225V × (1 + R3 / R4)

where,

  • R3 is 3k
  • R4 is 1k

Substituting values:

Vout = 1.225V × (1 + 3000 / 1000)

Vout = 1.225V × (1 + 3)

Vout = 1.225V × 4

Vout = 10V

Inductor Selection Formula:

L = (Vin – Vout) × D / (fs × Iout)

where,

  • Vin is the input voltage
  • Vout is the output voltage
  • D is the duty cycle
  • fs is the switching frequency
  • Iout is the output current

For practical purposes a 100µH inductor is chosen

Output Capacitor Selection Formula:

C = Iout / (8 × fs × Vripple)

where,

  • Iout is the output current
  • fs is the switching frequency
  • Vripple is the allowable output voltage ripple

For a C5 15µF capacitor ripple is kept low.

How to Build

To build a Simple 10V Switching Power Supply Circuit with IC LM5007 follow the below mentioned steps:

  • Assemble all the components as mentioned in the above circuit diagram
  • Connect pin 1 of IC1 to Vout 10V DC
  • Connect a cathode of diode D1 from pin 1 of IC1 and anode of D1 diode connect to GND
  • Connect inductor coil L1 from pin 1 of IC1 through the junction of capacitor C4 and resistor R3
  • Connect one terminal of capacitor C4 from pin 2 of IC1 and other terminal connect to the junction of diode D1 and coil L1
  • Connect resistor R3, R4 and resistor R5 and capacitor C5 in parallel through the junction of L1 coil and Vout 10V DC
  • Connect pin 3 of IC1 to pin 4 of IC1 and GND through resistor R2
  • Connect pin 5 of IC1 between resistor R3 and R4
  • Connect pin 6 of IC1 to pin 8 of IC1 through resistor R1
  • Connect pin 7 of IC1 to GND through capacitor C3
  • Connect pin 8 of IC1 to Vin of 12V to 72V DC
  • Connect capacitor C1 and C2 in parallel to GND from the junction of resistor R1 and Vin of 12V to 72V DC

Conclusion:

This Simple 10V Switching Power Supply Circuit with IC LM5007, which uses the LM5007 effectively reduces voltage from a broad input range of 12V to 72V.

The circuit ensures steady voltage regulation with very little power loss making it perfect for industrial power supplies, automotive uses and embedded systems that need a 10V power source.

References:

LM5007 75-V, 0.5-A DC/DC Buck Converter With 80-V Integrated Power MOSFET

Filed Under: Power Supply Circuits

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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