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.

step up converter 12V 5V

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 the general microcomputer.

However, in some appliance may have the LCD display inside. For example, 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 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’ »


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


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… Read more »


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



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


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… Read more »


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


Can u Plzz elaborate KILLA CZECH REPUBLIC


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


Wuz Up???????????????


Not Nice Comment Dikshit


@ 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… Read more »


Thanks a lot for helping.
You explained very well.

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