Here is a 1.5V to 5V boost converter circuit. Why it is interesting. Now, small electrical appliances have a microcomputer that requires the 5V power supply and low current less than 10mA.

The 1.5V AA battery is cheap and popular. How we can use it as a power supply of that tiny microcomputer.

Sure, we have many ways to step-up voltage. Using ICs may be a good idea. But If we are beginners or who love experiment basic circuits.

In this circuit, we use only small transistors.

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.

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

Learn how this works

If you got it, Read next

Single output version

See in the circuit below. It is the single output (5V 10mA) version.

The important is Q1, L1, and D1 like the previous circuit. But additional the Zener diode ZD1 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.

Small LCD display Supply(Dual Voltage)

However, some appliances may have an LCD display inside. For example, game players, MP3 players, etc. 

They require the -5V power supply.

We can do it easily. It becomes 5V Dual power supply, +5V at 10mA, and -5V at 0.5mA (approximately).

Look at the circuit below.

We add C4, D2, and D3. They are the simple charge pump converter, to make a negative voltage, -5V. The current of this section is about 0.5 mA.

How the circuit works

First of all, apply the current(1.5V battery) to the circuit. The current cannot flow through L1. Because Q1 does not get the bias current.

Then, Q2 starts working with the current from 1.5V. And, the current flows from the emitter to the base of Q2 through R5 to be the full circuit.

The Q2 provides the current out of the collector to be a biased current to Q1. The Q1 is conducting.

You may like to see these too.

• DC to AC Converter circuit
• DC Boost Converter circuit 3-5V to 12V-13.8V
• 12V to 24V step-up converter using TDA2004

Now, the L1 connects to the ground like a normal boost converter. But the working of Q1 makes the voltage across collector-emitter is lower. They work looks like a switch.

So, It will add quality conducting of Q2, with R3 and C2. Make this system is high performance. Also, Q1 is good working.

And, the L1 passes 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.

Parts you will need

0.25W Resistors, tolerance: 5%
R1, R4: 470K
R2, R3: 100Ω
R4: 2.2K

Ceramic Capacitors
C2: 10nF (0.01µF) 50V
C4: 0.1µF 50V

Electrolytic Capacitors
C1,C5: 10µF 25V
C3: 1µF 25V

Semiconductors and others:
Q1: BC548 or BC550, 45V 0.1A, NPN TO-92 Transistor
Q2: BC558 or BC560, 45V 0.1A, PNP TO-92 Transistor
D1,D2,D3: 30V 200mA, Schottky Diode
ZD1: 5.1V 0.5W Zener Diode
L1: 270uH coil, 200mA

This concept optimizes the performance of the power supply circuit.

Keep reading:

• 3V to 5V boost DC converter
• USB 5V to 1.5V Step Down Converter
• High power LED flashlight with 1.5V battery

Credit: LCD display photo

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