# Simple 6V to 12V boost converter circuit using transistors

Today we are going to build a 6V to 12V boost converter circuit. Imagine you have a 6V battery, but you need to use it for a 12V load, like LEDs. Looking into your storage, there are only a transistor and simple parts with no IC. What can you do?

Stop thinking of buying a new 12V battery when you could make use of the components you have to solve this problem.

Some people may call it a 6V DC to 12V DC step-up converter circuit. However, the output current is always lower than the input current.

## How it works

Here is a step-by-step breakdown of how it works

The first part of this circuit includes Q1, Q2, R1, R2, R3, R4, R5, D1, C1, and C2. Combined into an astable multivibrator that outputs a square waveform, that is, a positive pulse signal at high frequencies.

Then, the signal flows through R6 to limit a bias current flowing through the base of Q3. It causes Q3 to operate, so high current flows through L1; Q3 now acts like an on/off switch.

This characteristic causes energy to accumulate in L1, and this energy will be released through D1 to rectify the current. Only the positive power is selected and stored in C1, which itself is an output. so causes the output voltage to gradually increase.

But when the output voltage increases beyond VZ of ZD1, which is 12V, current can flow through it and R7 into the base of Q4. This makes Q4 conduct current, like an on switch.

As a result, the current flowing through R6 is pulled to flow through pins C-E of Q4 all the way to the ground. Stopping Q3 from working. The voltage level at the output is reduced to approximately 6V because it is the current flowing through L1 and D1.

When the voltage level is lower like this, there is no current flowing through ZD1. The Q4 stops working. The current flowing through R6 flows through the base of Q3 again. The circuit works in these loops continuously, keeping the output voltage constant at 12V all the time.

With C3 acting as a smooth filter and storing electrical energy for the output’s load.

#### Parts you will need

0.25W Resistors, tolerance: 5%
R1, R4: 1.8K
R2, R3: 4.7K
R5: 1K
R6: 1.5K
R7: 33K
R8: 10K
C1, C2: 0.1µF 50V Ceramic or Mylar
C3: 470µF 25V Electrolytic
D1: 1N4148, 75V 150mA, Silicon Diode
D2: 1N5819, 40V 1A, Schottky Barrier Diode
ZD1: 12V 0.5W Zener Diode
Q1,Q2, Q4: BC549 or BC547, 0.4A 40V Transistors
Q3: BD139 or BD679 or equivalent, 80V 1.5A Transistor,
L1: 100µH to 220µH

#### Experimental creation and testing

My daughter assembled the entire circuit on a breadboard and measured the output voltage. It was 11.6V, or about 12V. When a load was connected to this circuit, the output voltage level dropped.

## Conclusion

In this circuit experiment, we can convert the voltage from 6V to 12V. And it is also great to learn the basics of a boost switching regulator circuit, or, as some call it, a step-up DC converter. It uses only basic electronics components, transistors, no ICs, familiar circuitry such as the astable multivibrator, and the principle of voltage regulation using the Zener diode.

And it is quite certain that it will have low efficiency. Because when used, only 4–5 transistors are used. Unlike ICs, which contain hundreds of transistors assembled to form an integrated circuit that works perfectly according to the manufacturer.

At the next opportunity, we would like to try to improve it or bring transistors to work with ICs as perfectly as possible.

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### 5 thoughts on “Simple 6V to 12V boost converter circuit using transistors”

1. Sir what value is L1?

2. Mir gefällt das

• Hello Dieter Jesse,

Grudene 8e
58644 Iserlohn
Nordrein-Westfalen/ Sauerland
Germany
I speak not so good English

Ist auch ein deutschsprachiger Kollege bei Euch dabei ?