5V low dropout regulator circuit using transistor and LED

This is 5V low dropout regulator circuit using transistor and LED only so easy,lowest voltage input is 6V so across it is 1V only, make output is 5V 0.5A

I have a 6V battery, but it needs to be applied to digital circuits that used to 5 volt regulated.
In first, I will reduce this voltage with a single resistor, but the load current is not constant.
The other way to use is a second, I would use 2 x 1N5402 diodes to connected in series to reduce the voltage to the load. But it don’t work because low current output.

I think it is two circuits. first circuit, use transistor is the key, And second we use a popular IC undertake the main. Read below!

I selected this simple circuit is the 5V Low Drop Out Regulator Low Volt. It used normal electronics part are transistors and resistors. While we can adjustable the P1 for control output voltage at current 500mA.

When the BD438 transistor Q1 is very hot, so the cooling pad. And we use numbers instead TIP3055 number 2N3055 (easy to buy all the time).

How it works
The general regulator circuit, need input voltage is higher than the output voltage. So cannot apply to the input voltage is higher just little than the need voltage. But this circuit you can have the five volt output from input voltage only six volt of the normal battery.

5V Low Drop Out Regulator Low Volt

As circuit picture below. The load is connected with collector pin of the transistor output. Connection this way makes transistor works in a saturation. As a result the voltage drop across the collector pin and emitter pin very low (because to be the saturation voltage).
-And this current depends on a type of transistors, Which is set about 0.5 amperes, has a voltage drop is 0.2 volts. But for all circuit need plus with voltage across R6. (Current limiting resistor)

At voltage across R6 is about 0.5 volts. T3 is activated and the output current is limited there. LED-D1 is the second function is indicate working and the reference voltage. Which is set 1.5 volts to 1.6 volts at emitter pin of T1, for base current to drive it works.

From divided voltage circuit, which consists of R4, P1 and R5 based on the difference of the reference voltage and the output voltage to T1 conduct more or less and this current makes T4 provide the current under such conditions to T2 base pin of T4. The capacitor C1 is used to filter the output current.

If find transistor BD438 does not, to replace with a BD136, BD138 and BD140. However, a number of transistor in the saturation voltage is slightly higher than BD438. For LED-D1 should be red to match the reference voltage. If the a different color value may be wrong.

One Comment

  1. joynal November 27, 2012

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