Boost Converter Design with MC34063 IC

 I have been interested in power supply circuit for a while now because they are needed in very electronics circuits. I have been writing and studying various methods to build power supply. I wrote how the series and shunt voltage regulator circuits are designed with a zener diode, transistor and op-amps. Then I wrote about linear regulator IC like the LM317, LM7805 etc. Then I studied the SMPS(Switch Mode Power Supply) circuits including boost and buck converter circuits. I tried design of buck and boost voltage converters using Arduino, CD4047 and 555 Timer IC. These are simple circuits. I lately wrote about how to design Arduino-Based Pure Sine Wave Inverter (SPWM, Push-Pull Topology). Here I am writing about the MC34063 IC which is a DC-DC converter IC that can be configured as a step-up (boost), step-down (buck), or inverting switching regulator. It integrates all the key components needed for basic switching regulation—including an internal temperature-compensated reference, oscillator, comparator, switch, and current limiting circuitry—making it a versatile and compact solution for power supply design.

I designed a 5V to 12V boost converter circuit with MC34063 IC whose circuit diagram is shown below.

To design a boost converter using the MC34063 IC that converts 5V input to 12V output, you need to select R1 and R2 based on the output voltage setting formula:

$$V_{OUT}=1.25\times (1+\frac{R2}{R1})$$

Step-by-step:

For V_OUT = 12V, solve for $\frac{R2}{R1}$:

$$12=1.25\times (1+\frac{R2}{R1})\Rightarrow \frac{R2}{R1}=\frac{12}{1.25}-1=8.6$$

So,

$$\frac{R2}{R1}=8.6$$

Choose standard resistor values:

• R1 = 1.2 kΩ → R2 ≈ 10.3 kΩ (close to standard 10 kΩ)

• Or R1 = 1 kΩ → R2 = 8.6 kΩ (you can use a 8.2 kΩ or 8.66 kΩ if available)

Practical Choice:

• R1 = 1kΩ

• R2 = 8.66kΩ (standard 1% resistor)

You can also fine-tune with a trimpot in place of R2 for adjustable output voltage.

Video Demonstration