5V low dropout regulator circuit using transistor and LED

This is a fixed DC regulator, 5v low dropout regulator. It uses a transistor and LED as main parts. It needs at least 6V for the input. And there will be at least 1V inside the circuit. And the output voltage will always be 5V at 0.5A.

5V Low Dropout Regulator circuit

Suppose that we have a 6V battery. But it cannot use to a digital circuit that needs the 5V.

I think many ways:

  • First, I will use a resistor to low down voltage. But it is low current and not constant.
  • Second, I will connect two 1N5402 diodes in series to reduce the voltage. But it doesn’t work. Because of low current output.
  • Third, We need to use a high current circuit. The 5V low dropout regulator is best now.

This simple circuit uses normal part, transistors, and LED.

We can change an output voltage by VR1. At the same time, an output current is 500mA.

5V low dropout regulator circuit using transistor and LED

How it works
In general regulator circuit, need an input voltage is higher than an output of about 3V.

But this is a special circuit, 6V battery input is higher than the 5V output about only 1V only.

As circuit below. The load connects the collector of Q4, transistor output. This makes transistor works in saturation. So, a voltage drop across the collector and emitter is very low.

And this current depends on a type of transistors, setting about 0.5A, a voltage is 0.2V. But for all circuit need plus with the voltage across R6. (Current limiting resistor)

At voltage across R6 is about 0.5 volts. The Q3 is activated and the output current is limited there.

LED-D1 is second function is to indicate working and the reference voltage. Which is set at 1.5 volts to 1.6 volts at the emitter of Q1, for the base current to drive it works.

The divided voltage circuit consists of Q4, VR1, and R5. They are the difference between the reference voltage and the output voltage to Q1.

Then Q1 conducts more or less. And this current makes the Q4 provide the current under such conditions to the Q2 base pin of T4.

The capacitor C1 filters the output current to smooth up.

If we cannot find a transistor BD438. We may replace it with BD136, BD138, and BD140. However, these transistor has saturation voltage is slightly higher than BD438.

For LED-D1 should be red to match the reference voltage. If different color values may be wrong.

The BD438 transistor Q1 is quite hot. It needs to mount a heatsink. We may use a TIP3055 or 2N3055 instead (easy to buy all the time).

This is the first circuit, use transistor as the key, And second, we use a popular IC in the main. So easy… Read below!

What is more? See More Power supply circuits

Keep reading: ‘LM317 Low dropout regulator 5V 2A’ »

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I always try to make Electronics Learning Easy.

1 Comment

  1. joynal


    I need circuit low volt cut off. 12v battery when 10v than cut off.

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