**Para Versão Original em Português, Clique Aqui!**This is an adjustable power supply circuit from

**1.2V**to**37V**and**6 amps**of current, with short circuit protection, equipped with adjustable positive voltage stabilization circuits of three terminals**LM317**, plus a booster circuit, using the**TIP36C**, which is an inexpensive power transistor.What makes this power supply special is the implementation of a short-circuit protection circuit, for which a

**BD140**PNP transistor is used.### You may be interested in:

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## How the circuit works

Resistor

**R1**, which is a load sensing resistor, receives a small current flowing through it. As long as the current in the output circuit does not reach a certain current calculated through**R1**, the circuit behaves like a normal voltage regulator, because at small "*calculated*" currents there is no voltage drop in the load sensing resistor, so the**Boosters TIP36C**transistor does not trip.As the current in the circuit increases, the voltage across resistor

**R1**increases. When this voltage reaches about**0.6 V**, the "*transistor cut-off voltage*", the power transistors turn on and current flows through them, with the threshold determined by the maximum current supported by the power transistors.However, we have implemented a current protection circuit that consists of a circuit equipped with a

**BD140**transistor with a resistor that acts as a current sensing resistor that serves to polarize the transistor and, depending on the value detected, limit the output current of the entire circuit according to a simple*formula that serves to set this threshold current.***Ohm's Law**## Formula 1st Ohm's Law

The

**states that the potential difference between two points of a resistor is proportional to the electric current flowing in it, and that the ratio of electric potential to electric current is always constant for ohmic resistors. The formula is as follows:***1st Ohm's Law***V = R * I****V**- Voltage or electric potential**R**- Electrical resistance**I**- Electrical current

Knowing

**Ohm's Law**, we can now calculate the values of the load sense resistors that activate the power stage and the bias resistors of the protection transistors that form the short circuit protection circuit.## Calculating the load resistors

First, we need to know the current of the

**LM317**voltage regulator, which is**1.5 amps**according to the datasheet.**LM317**=

**1.5A**

Let us calculate

**R1**. We know that using**Ohm's law**, we get the following expression:**V = R * I****V**= The cut-off voltage of transistors**Q2**and**Q3****TIP36C**is**0.6V**. This is the cut-off range of the transistor. Let us call**Q2**and**Q3**of**Qeq**

**I**= This is the current of the regulator

**IC1**. Let us set the operating current of

**IC1**to

**600mA**, which is

**0.6A**. This current is enough for the

**IC**to work unhindered.

### Then:

**R1**=**Vbe_Qeq / I_CI1****R1**= 0.6V / 0.6A**R1**= 1 Ohm

## Calculation of the protection circuit resistance

Similarly, we need to know the total current of the selected power supply so that there is an interruption in this range. Our power supply for

**6 amps**.Power supply =

**6A**Let us calculate

**R2**. We know that**Ohm's law**gives us the following expression:

**V = R * I****V**= The cut-off voltage of the transistor**Q1**is**0.6 V**. "This is the cut-off range of the transistor".**I**= The total current of the power supply, which is**6A**.

### Then:

**R1**= Vbe_Q1 / I_ps**R1**= 0.6V / 6A**R1**=**0.1 Ohm**

## Current of the power transistors

**Q2**+

**Q3**=

**25A**+

**25A**=

**50A**

However, the total power of the

**TIP36C**transistor is**125W**, which means it operates at a current of**25A**to**5V**. Remember the above formula,**P = V * I**;**P**=**5V*****25A**=**125W**.

For this circuit with a maximum voltage of

**37V**and transistors with a maximum power of**125W**, we look as follows:**Pmax**= V * I:**Imax**= P / V = > Imax = 125W / 37V = > Imax =**3.37A**- How are two transistors together Imax =
**6,74A**

Therefore, our circuit works with two

**TIP36C**transistors to get**6 amps**at the output.**Figure 2**shows the schematic of the adjustable power supply circuit with short circuit protection. Those who follow us already know this circuit very well, the difference is exactly in the implementation of the protection circuit, as we can see below.

Fig. 2 - Schematic diagram Adjustable power supply circuit with short circuit protection |

## Components List

- CI1 ................ Voltage Regulator LM317
- Q1 ................. PNP Transistor BD140
- Q2, Q3 .......... Power Transistor PNP TIP36C
- D1 ................. Bridge Rectifier 50A - KBPC5010
- D2, D3 .......... Rectifier Diode 1N4007
- R1 ................. Resistor 2W / 1Ω
- R2, R4, R5 ... Resistor 5W / 0.1Ω
- R3 ................ Resistor 1/4W / 220Ω
- C1 ................ Electrolytic Capacitor 5600uF - 50V
- C2, C3 .......... Polyester/Ceramic Capacitor 0.1uF or 100nF
- RV1 .............. Potentiometer 5KΩ
- P1, P2 ........... Screw Terminal Type 5mm 2-Pin Connector
- Others .......... Wires, solders, printed circuit board, etc.

**Source: fvml.com.br**## Download

We provide the files with the **PCB**, the schematic, the PDF, GERBER and JPG, PNG and provide a direct link for free download and a direct link, "* MEGA*".

**Click on the direct link to download the files: ****Layout PCB, PDF, GERBER, JPG**

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