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Simple Adjustable Power Supply Circuit

Many electronic things need clean DC power to work good.

This post show how to make one easy Simple Adjustable Power Supply Circuit.

It changes dirty DC power to clean power and it uses one chip called LM338.

This circuit take DC from 5 to 34 volt and gives clean DC from 1.25 to 30 volt.

We can turn one knob to get voltage what our device need.

Circuit Working:

Parts List:

TypeComponentQuantity
Resistors120Ω 1/4 W1
Potentiometer 4.7k1
CapacitorsElectrolytic 10µF 25V1
SemiconductorsIC LM3381
Diode 1N40072

Pin 3 (Vin) of LM338 chip is connected to DC power input and this input can be between 5V to 34V which is not stable.

LM338 has smart parts inside with one voltage reference of 1.25V and one error amplifier.

The error amplifier checks output voltage and compares it with 1.25V reference.

Pin 1 Adjust is connected to ground through one moving resistor called RV1.

By turning RV1 user can change the voltage and this changes output voltage too.

Two resistors R1 and R2 make voltage divider.

Using size of R1 and R2 with the 1.25V output voltage is calculated with the formula.

Pin 2 Vout is output pin which gives final smooth voltage.

User can control output voltage from 1.25V to 30V by changing R1 and RV1.

Maximum current from LM338 is 5 Amps.

Capacitor C1 helps soft start and it slowly gives power to load.

Diodes D1 and D2 stop damage if someone connect power in wrong way.

Resistor, capacitor and LM338 together give clean and stable output voltage.

LM338 is good for devices where power changes and it keeps voltage steady even if load changes.

It can give 5 Amp constant current at max.

Formulas:

This is the formula to find output voltage (Vout) for a circuit using adjustable voltage regulator like LM338:

Vout = 1.25V × (1 + R2 / R1) + Iadj × R2

where,

How to find Vout:

Know the values:

First we must know values of R1, R2 and maybe Iadj.

We pick R1 and R2 depending on how much output voltage we need and check the LM338 datasheet to pick good values.

Calculate the middle part:

First calculate (1 + R2 / R1) and this is from the two resistors.

If R2 is bigger than R1 then Vout goes higher.

Use the formula:

After that use the full formula:

Vout = 1.25V × (1 + R2 / R1) + Iadj × R2

Sometimes we can ignore the last part (Iadj × R2) because Iadj is very small.

Important notes:

The formula has Vout on both sides sometimes so it may not always be easy to solve.

But in most simple cases it works fine.

For many uses just by using the first part 1.25 × (1 + R2/R1) is enough.

To be sure Vout is correct use correct resistor values and sometimes Iadj matters but mostly not.

How to Build:

To build a Simple Adjustable Power Supply Circuit following steps are required to follow while connections:

Testing the Circuit:

Conclusion:

By following these above easy steps we can make a simple adjustable power supply circuit which gives clean DC output and supports voltage we choose and current up to 5 Amps.

Always be safe when working with electronics and for more details go through the IC LM338 datasheet.

References:

LM338 5-Amp Adjustable Regulators datasheet

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