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How to Make a Simple Boost Converter for Various Needs

This converter is incredibly easy to manufacture; it contains only 1 transistor. Made entirely from parts from an old computer power supply board. Can be used to power various LED lamps, LED strips, various devices, etc.

Will need


  • Transistor D1047C (Almost any bipolar one will do) -
    Transistor D1047C

  • Inductor.
  • Resistor 1 kOhm.
  • Diode.
  • Capacitor 100 uF 400 V.
    How to Make a Simple Boost Converter for Various Needs


All parts can be found on the computer power supply.
All parts can be found on the computer power supply

Scheme


Simple Boost Converter Circuit

The simplest circuit using a single transistor with self-excitation. At the output, you can install a zener diode as a diode and then the output voltage will be stabilized to the specified value.
The coil is used ready-made. If you are planning to do it yourself, then take a 0.7 mm wire. The number of turns is 45 with a tap from 30 turns.

Making a DC-DC converter with your own hands


We solder the inductor using similar parameters.
Unsolder the inductor from the board

It doesn’t have a branch from the middle, so we’ll make one ourselves. Having counted the approximate ratio of turns, use a utility knife to scrape off the varnish at the soldering site.
use a utility knife to scrape off the varnish from the wire

Solder the wire and the middle pin is ready.
Solder the wire to the middle of the inductance

We install the transistor on the radiator.
We install the transistor on the radiator

Solder the inductance.
Solder the inductance

Solder the resistor.
Solder the resistor

Solder the diode, capacitor, wires.
Solder the diode capacitor wires

At this point the converter is ready for operation.

Testing the converter in action


We apply a voltage of 1.3 V. And at the output we already have 18 V.
Testing the converter in action

We increase the voltage to 2.2 V and the output is about 50 V
We test the converter in action and get 53 Volts at the output.

With a 4.4 V input, this results in more than 150 V.
We test the converter in action and get 157 Volts at the output.

We try it on a small load, in the form of which a neon lamp is used.
Connect a neon lamp to the output of the converter

Now let's test the converter on a more realistic load - an LED lamp.
LED lamp

The converter powers the lamp well at 5 V.
How to Make a Simple Boost Converter for Various Needs

The converter efficiency is about 98%, power consumption is about 5 W.

Watch the video


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3+three=
Comments (5)
  1. Yvlon Musk
    #1 Yvlon Musk Guests 20 September 2021 18:32
    3
    Thanks a lot. But can you still explain the circuit? With detailed path of current flow through the circuit. I can't understand the logic. Does current flow directly from + through the rectifying diode? to the load.The resulting current is pumped into the transformer, opening the transistor. Through the transistor, the current increases and goes to the load. Where does it become variable? Is the capacitor with the transformer resonating or what? Well, as I understand it, there should be zero on the rectifier diode side, and phase on the source side? It is necessary to understand the basic principles of this device
  2. Nikita
    #2 Nikita Guests 8 November 2021 17:33
    3
    Thank you very much, I am a schoolboy and I assembled this circuit, I always dreamed of assembling it.....
  3. Guest Alexey
    #3 Guest Alexey Guests 19 December 2022 23:39
    3
    Since the efficiency is 98%, then why such a radiator on a transistor? :-))
  4. Nikolay
    #4 Nikolay Guests January 22, 2023 12:47
    0
    I assembled the circuit, either I did it manually, or something was wrong. (I suspect the first)
    It gave 203v from 5, but only at idle. WITH A PRE-CONNECTED load (4 x 3V LEDs in series) it does not rise above 9V, current input is 32mA, output is 5mA...
    1. Dmitry Aleksandrovich
      #5 Dmitry Aleksandrovich Guests 3 August 2023 15:14
      0
      If we calculate the power balance, then in the primary circuit 5 * 0.032 = 0.16 W, in the secondary circuit 9 * 0.005 = 0.045 W. then efficiency = 0.045/0.16 = 0.28 or 28%.

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