Many electronics circuits and projects for learning in simple ways.

1.5V to 5V boost converter circuit for micro computer

Now, small electrical appliances have a microcomputer that requires the 5V power supply. Why it can use only one 1.5V battery? There is 1.5V to 5V boost converter circuit inside them. Let’s start to learn the simple boost converter.

How it works

Also, we can use the nickel cadmium/nickel metal hydride battery of 1.2V, alkaline battery to these appliances.

The microcomputer requires a low current power supply. This 1.5V to 5V boost converter circuit is small. So, it can be in the game players, MP3 portable, DVD players.

1.5v to 5v boost converter circuit for micro computer

This 1.5V to 5V boost converter circuit as above is applied from this boost converter into a new circuit.

The important is T1, L1, and D1 like the previous circuit. But additional the Zener diode D4 is regulation to makes the stable output.

This circuit converts the 1.2V or 1.5V input voltage up to the 5V.  While it can apply the current of 10 mA. This current level is sufficient for to the general microcomputer.

However, in some appliance may have the LCD display inside. For example, the game players, MP3 players, etc.  They require the -5V power supply.

Using C4, D2, and D3 is the simple charge pump converter, make -5V. The current of this section is about 0.5 mA.

To begin with, apply the current to the circuit.  The current cannot flow through L1.  Because  T1  does not get the bias current.

Then, T2 starts working with the current from 1.2V.  It flows the emitter to base of T2 through R5 to full circuit.

The T2 provides current out of the collector to be a biased current to T1. The T1 is conducting.

Now the L1 connects to the ground like a normal boost converter. But the working of T1 makes the voltage across collector-emitter is lower. When the results of this work as a switch.

Then, T2 has the quality conducting.  R3 and C2 make this system is high performance. Also, T1 is good working.

The L1 pass the high current until the maximum in linear form.  After that, the polarity of the voltage across L1 changes.

The performance of this circuit is about 60%.  The advantage of this circuit.  The voltage drop across it is low. While the transistor switch on the voltage across between collector-emitter is only 0.2V. But it can supply the high current. This concept optimizes the performance of the power supply circuit.

‘Keep reading:1.5V to 5V boost DC converter’ »

This Post Has 12 Comments

  1. pls admin can I use dis cct as a charger for my nokia phone?

  2. Hallo,

    I dont quite understand this circuit.

    What is the job of L1?

    What is meant with:
    “When start apply the current do not flow through L1, because not have bias for T1
    But the current start at T2 by current from voltage source of 1.2 volts. Flow into emitter pin out to base pin of T2 through R5 go to full circuit.”

    I managed to translate it a little bit, but i dont quite get how the current goes from emitter to base?

    Tried translation:
    “At the start the current does not flow through L1 because T1 does not have its base.

    But the current that is applied to T2 by the source Voltage (1.2V)
    flows into the emitter from T2 to its base (?)through R5 (1k Ohm) to 0V to complete the circuit.

    T2 has a base current (?) >THE current by ARROWHEADS<(whats meant by this???)

    So T1 is getting a current onto its collector???

    —"

    Im willing to translate this for you… but i need more infos, insights as to how this circuit works.

    Also would you like to change your circuit images to not have such a solid text over it. And instead use a little bit more transperancy? so that its not as much blocking the circuit?

  3. Hi there,
    If I need only the +5V, should I remove C3, C4, D2 and D3?

    Thanks.
    Nelio.

  4. Can i replace bc560 with bc547
    & bc 550 with bc548?

  5. Terrible english

    1. Hello, Dikshit Thakuria
      I am sorry for my English. I hope my English is improved. I will try to rewrite it.

  6. tested and working, authors english is didaster.
    to EDWIN: you can replace the transistors with similar ones.. bc560 cannot be replaced with bc 547, cause its PNP!!! i used bc557c (the “c” is not important, versions differ in h21e) for T2 PNP and bc548b as T1 NPN.. the circuit works, it stops to oscillate as you rise the input voltage, it has to be low as described. the critical part is the inductor, i used 330uH / 0,5A, Rdc (max) = 1 Ohm (https://www.gme.cz/tlumivka-radialni-09p-331k-fastron), it didnt work with 2.2mH with around 30 Ohms Rdc.. also didnt work for higher voltage (switched zener for 10 volt one), it was still around 6V out..

    to NELIO: yes, i did remove those too..

  7. can u plzzz elaborate ????? @KiLLA CZECH REPUBLIC

  8. Can u Plzz elaborate KILLA CZECH REPUBLIC

  9. Wuz Up???????????????

  10. @ Kocy

    T2 is a PNP transistor and you need to make the base voltage lower than the emitter by around 0.65V. This is being done by putting a pull down resistor R5 while the emitter is connected to 1.2V. This is the base current path that the author was trying to explain in order to turn it on and provide base current to T1 to turn it on also. With T1 starts to conduct, it provides additional base current path to T2 from 0V through R3 and C2 to make its collector further conduct. T1 continues to conduct until it could no longer support the current rise of L1 due to its hfe so it will suddenly turn off…. this process is repeated over and over which is called an oscillation and thus the circuit is an oscillator.

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