DC Boost Converter circuit 3-4V to 12V-13.8V

I like the 12V LED. But I have only 3V of batteries. So, No light. With this circuit, I can get the light of the LED. It is a 3V to 12V boost converter circuit. To turn a DCV source of 3.3V, 3.7V, and 4V into 12V-13.8V at a max current of 100mA. They are a type of switching power supply that uses KA34063 or MC34063.

Which is better than an old circuit. Because of input low voltage, high efficiency with the output adjustable voltage. If you cannot imagine. I will show why you should use this circuit. I have LED light Bulb as Figure 1. It is cheap and super bright, best running with 12V.

12V LED light bulb for Car
Figure 1 LED light Bulb for car

Do you know how many it uses the current? If you do not know. Let’s learn with me.

First, We use the ammeter to measure a current of the LED as Figure 2. Which it has circuit diagram. And Figure 3, is real measuring current.

Measuring the current of 12V LED light bulb
Figure 2 Measuring the current of 12V LED light bulb

Measuring the current of LED
Measuring the current of 12V LED

Figure 3 Real measuring a current of 12V LED light bulb

You will see that the ammeter shows a current of 0.024A or 24mA. So, this LED uses low power consumption (P), about 0.2 watts.

From:

P = V x I
12V x 0.024A = 0.228 watts only.

Second,  if we have a NiMH AA battery, the single battery gives of 1.2V. Although, you connect them in series with three batteries. But a total of 1.2V AA battery are 3.6VDC.

See in Figure 4.

So it cannot be used for the 12V LED light bulb.

Three AA batteries in a series give a total of 3.6V.
Three AA batteries in a series give a total of 3.6V.

Figure 4 as 3 AA battery 1.2V is connected in series

How to add voltage up

We need the DC boost converter circuit to convert input 3.6V or 3.7V or 4V to 12V output.

DC Boost converter circuit, 3.7V-5V to 12V-13.8V
DC Boost converter circuit, 3.7V-5V to 12V-13.8V

As Figure 5 the circuit diagram of DC boost converter

This circuit can help him because:

Special feature

  • Used for a power supply, 3.3-5 volts 750 mA up
  • Can apply voltage 12-13.8 volts depends on the power supply input
  • If using the power supply of 5 volts DC current 300mA can apply the output; voltage is 12 volts DC at 100 mA maximum
  • If using the power supply of 3.3 volts DC current 660mA can apply the output; voltage to 12 volts DC at current maximum to 50 mA
  • Maximum current output: 100 mA
  • Normal Switching Frequency about 43 kHz
  • Can adjust the output voltage
  • There is power on the LED display

How it works

The circuit will show in Figure 5, the operation starts with when we apply the power supply to the circuit. The IC1 will act as the step-up voltage converter. The voltage will increase to pin 1 of IC1-KA34063 through diode D1 to OUT point. We can adjust the output voltage by adjusting at VR1.

How to assemble the circuit

He bought the project as a PCB board so easy to use but you can buy all parts as the list below. Then, you may assemble circuits on the perforated board or DIY Glass Fiber Prototyping PCB Universal Board.

Testing and how to use

As Video below We try to experiment with LED using the 3.6V battery again. The LED will without the light.

Then, He connects the wires or the power cable to LED easy to use.
And Then He tests the circuit

On the PCB; He connects the power line 4, the terminal of the circuit.
As Figure 6: Testing This DC-step-up-converter citcuit using KA34063

Testing DC step up converter circuit using KA34063
Testing DC step up converter circuit using KA34063
  • Apply the 3.6V battery—or 3.7V mobile batts to the input terminal.
  • Apply the LED to the output terminal and measure the LED voltage.
  • Adjust the VR-10K until we read voltage is 12V
  • Measure current of LED is 24mA at voltage 12V
  • Measure current of the circuit is 100mA at 3.9V

Thus we have power input is 0.1A x 3.9V = 0.39W
But we can use load (LED) at power 0.025A x 12V = 0.3W
We will see that use low voltage input make we have high voltage output at low power using.

Application

  • Input terminal connects DC power supply voltage of 3.3V, 3.7V, 4.5V, up tup 5V.
  • The output terminal, we use this as DC power supply to any loads. Which this voltage can change by an input voltage level, and adjusting VR1.

Parts you will needs

Resistors 0.25W
R1: 100Ω -brown-black-brown-gold
R6-R8: 1K -brown-black-red-gold
R2-R5: 1Ω -brown-black-gold-gold

Trimmer Potentiometer
VR1: 10K

Ceramic Capacitor
C1: 680pF 50V

Electrolytic Capacitors
C2: 470μF 16V
C3: 220μF 25V

Semiconductor
D1: 1N5819, 1A, 40V Schottky rectifier Diode
IC1: KA34063 or MC34063,  1.5-A Peak Boost/Buck/Inverting Switching Regulators
Buy MC34063

FAQ

I get many questions from my friends. These may be useful for you, too.

Note: I bought this kit and not design this circuit myself. I am just using it before you. But I am happy that it works well. And, I want to see you happy like me.

You may buy components or Kits here.

Do we use 1N4007 instead of  1N4007?
No, You can’t. Because they are different.
1N4007 is a standard rectifier, 1A 1000V. It works slow.
But,
A 1N5819 is Schottky rectifier,1A 40V. It works faster. So, it is good at switching than 1N4007.
You read: 1N5819 rectifier replaces with 1N4007 rectifier??

You may like it, too!

14 thoughts on “DC Boost Converter circuit 3-4V to 12V-13.8V”

  1. Sir,
    I am most impressed with all your circuits especially teaching your son really this is so GOOD.
    I want to build a project of your and need to buy n IC KA34063 I am not familiar with this IC also WHER C|AN I BUY THIS Please.
    Your reply would be sincerely appreciated.

    Thank you for your projects.
    Regards,
    Chris
    New Zealand.

    Reply
  2. hello, I printed the page to test the diagram. but the coil rating/value was not there… could you please mention the value of the coil . thanks in advance.

    Reply
  3. Assuming I have a 12 volt in and I want 12v out – a UPS, but I want a lot of capacity

    What is the downside of connecting everything in parallel and have a BMS manage the 12v in, to charge the 3 – 8 18650 batteries in parallel and then have a buck-boost converter to take the 3-volt battery bank of 3 – 8 batteries and boost it to 12v.

    In other words, to get 17Ah of 18650 @ 3.7V is a hell of a lot cheaper than to get 17Ah @ 11.1V

    I want to understand what the immediate red flags would be to use 3.7V instead of 11.1V for the battery configuration?

    Reply

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