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Home » Simple 12V Boost Converter Circuit using IC LM2698

Simple 12V Boost Converter Circuit using IC LM2698

Last updated on 13 August 2025 by Admin-Lavi Leave a Comment

Simple 12V Boost Converter Circuit using IC LM2698 make output voltage higher than input.

LM2698 IC have switch inside and can handle 18V and 1.35A with 0.2Ω.

Input voltage 2.2V to 17V and output is 2.2V to 17V.

Keep voltage steady even if input changes.

Circuit is good for many input voltages and battery devices.

Example 5V input convert to 12V output.

This article shows how circuit work, design calculation and to make 12V boost with LM2698.

Circuit Working:

Simple 12V Boost Converter Circuit Diagram using IC LM2698

Parts List:

Component TypeValueQuantity
Resistors (All resistors are 1/4 watt unless specified)25k1
30k1
3.5k1
CapacitorsCeramic 4.7nF1
Electrolytic 22µF 25V1
Electrolytic 10µF 25V1
SemiconductorsIC LM26981
1N5819 Schottky Diode1
Inductor 10µH1

IC LM2698 is switching regulator and make input higher and steady to12V output.

Pin 6 power input works with 4.5V to 5.5V DC.

Do not connect load more than 400mA.

Chip gives max 400mA at 12V only.

Inductor L1 save energy when LM2698 switch ON.

When switch is OFF then energy goes through diode D1 to charge C3 and keep output steady to 12V.

Feedback R2 and R3 control output by changing switch ON time.

Pin 3 SHDN connect to GND to turn circuit OFF.

Formulas:

Formula for output voltage:

Vout = 1.26V * (1 + R2 / R3)

where,

  • Vout is output voltage
  • 1.26V is inside reference voltage in IC
  • R2 and R3 are resistors for voltage divider

How to Build:

To build a Simple 12V Boost Converter Circuit using IC LM2698 following steps are required to follow the connections:

  • Get all parts like in circuit diagram.
  • Connect pin 1 of IC1 to GND with resistor R1 and capacitor C1.
  • Connect pin 2 of IC1 to 12V output (Vout) through resistor R2.
  • Connect resistor R2 also from pin 2 to GND.
  • Pin 3 is shutdown pin and connect to ground to turn OFF the circuit.
  • Connect pin 4 and pin 7 of IC1 to GND.
  • Connect inductor L1 between pin 5 and pin 6 of IC1.
  • Connect capacitor C2 from pin 6 to GND.
  • Connect diode D1 between pin 5 and other side of resistor R2.
  • Connect input voltage from 4.5V to 5.5V to joint of pin 6, L1 and positive of C2.
  • Connect positive of capacitor C3 to one side of resistor R2 and negative of C3 to GND.

Conclusion:

Simple 12V Boost Converter Circuit using IC LM2698 is good step-up circuit for low input voltage.

With right parts it gives steady 12V output for many projects.

IC LM2698 is efficient and easy to use and is best for battery devices and power management.

References:

LM2698 SIMPLE SWITCHER® 1.35-A Boost Regulator

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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