Fig. 1 - Symmetrical Adjustable Power Supply 1.25V to 47V 10 Amps with Short Circuit Protection |

This is a

**Symmetrical Adjustable Power Supply**can vary its output voltage from**1.25V to 47V**, based on the**LM317HV**Linear Voltage Regulator Integrated Circuit for positive voltage and the**LM337HV**for negative voltage, which together with the NPN transistors**TIP 35C**and the PNP transistor**TIP36C**provide a current of**10 amps**.This schematic was taken from our partner fvml.com.br, what we did was a small change in the supported current capacitance, you can check the original circuit here

**Click Here!**## High Voltage Adjustable Regulator

The

**LM317HV**and**LM337HV**voltage regulators are adjustable 3-terminal voltage regulators capable of delivering currents of 1.5A or more over an output voltage range of**1.25V**to**50V**.The

**LM317HV**and**LM337HV**offer overload protection such as current limiting, thermal overload protection, and safe-area protection, that make the device breakdown-proof. The overload protection circuit remains fully functional even if the setting terminal is disconnected.Remember, we limit the maximum output voltage of the power supply to

**47V**, because the**LM337HV**negative voltage regulator supports a maximum of**50V**, unlike the**LM317HV**which supports up to**60V**.## How the Circuit Works

After rectification and filtering, which are the first basic operations of the circuit, the total voltage coming from the transformer and rectified enters the first output block, the voltage regulator, which is controlled by the

**LM**s Integrated Circuit and mirror image "*Same function, just in a negative way*".**R1**and

**R2**are

**10 ohm**resistors that have the function of

**Load Sensor**, they receive the current flowing through the circuit, and while this current does not reach the current calculated across the resistors

**R1**and

**R2**, the circuit behaves like a normal voltage regulator, because for small "

*calculated*" currents there is no voltage drop across the

**Load Sensing resistor**, so the Boosters Transistors

**TIP36C**and

**TIP35C**are not activated.

When the current in the circuit increases, the voltage across resistor

**R1**increases, when this voltage reaches about**0.6V**"*transistor turn-off voltage*", the power stage is activated and current flows through it.## The Protection Circuit

The output short circuit protection circuit is formed by transistors; Q1

**BD140**PNP and Q2**BD139**NPN, each for an output bias voltage.They regulate the maximum current "

*Calculated*", which is fixed at**10 amps**, and work together with**resistors***0.06ohm***R3**and**R4**as a current sensing resistor, which is used to polarize transistors**Q1**and**Q2**, so that, depending on the determined value, they limit the output current of the entire circuit according to a simple formula from Ohms Law, which is used to set this limiting current.## Formula 1st Ohm's Law

The

**First Ohm's Law**states that the potential difference between two points of a resistor is proportional to the electrical current established in it, and the ratio of electrical potential to electrical current is always constant for ohmic resistors. The formula is given by:**V = R * I****V**- Voltage or Electrical Potential

**R**- Electrical Resistance

**I**- Electrical Current

Armed with the knowledge of the

**ohms law**, we can now calculate the values of the**Load Sense resistors**, which activates the power step, and the bias resistors of the protection transistors, which is the**Short Circuit protection circuit**.### Load Resistor Calculation

First, we have to know the current of the

**LM317hv**Voltage Regulator, which according to the datasheet is**1.5 amps**.**LM317HV & LM337HV = 1.5A**

Let's calculate

**R1**, knowing that the same calculation is done for**R2**. We know that**Ohm's Law**gives us the following expression:

**V = R * I**

**V**= The cutoff voltage of transistors

**Q3, Q4**&

**Q5**, which follows the same principle for set

**Q6, Q7 & Q8**, is

**0.6V**"

*Which is the Transistor cutoff region*". Let's call

**Q3**,

**Q4**&

**Q5**as a

**Qeq**.

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

**CI1**, let's put the working current of the

**CI1**at 300mA, which is equal to 0.3A, with this current we won't need to put a heatsink on it.

Then:

**R1**= Vbe_Qeq / I_CI1

**R1**= 0.6V / 0.3A

**R1**=

**2 ohms**

### Protection Circuit Resistor Calculation

Likewise, we have to know the total current of the chosen source so that there is a cut in this region. Our source is for

**10 Amps**.**Power Supply = 10A**

Let's calculate

**R3**, knowing that the same calculation is done for**R4**. We know that Ohm's Law gives us the following expression:**V = R * I**

**V**= The cutoff voltage of transistor

**Q1**, which follows the same principle as for transistor

**Q2**, is 0.6V "

*Which is the Transistor cutoff region*".

**I**= It is the total current of PS, which is

**10A**.

Then:

**R1**= Vbe_Q1 / I_ps

**R1**= 0.6V / 10A

**R1**=

**0.06 ohms**

### Power Transistors Current

Q3 + Q4 + Q5 = 25A + 25A + 25 = 75A

**NOTE:**Remembering that the power of

**TIP36C**transistors is

**125W**, this means that it works with current from

**25A to 5V**, remember the formula above,

**P=V*I**;

**P = 5V * 25A = 125W**.

**For this circuit**with a maximum voltage of

**47V**, and transistors with a maximum power of

**125W**, we look like this:

**Pmax = V * I:**

**Imax**= P / V =>

**Imax**= 125W / 47V =>

**Imax**=

**2.66A**

How are three transistors together Imax =

**7.98A**And that's why our circuit uses three

**TIP36C**transistors to achieve**10 amps**at the output.In

**we have the schematic diagram of the adjustable power supply circuit with short circuit protection, so those who accompany us already know this circuit very well, the difference is exactly the implantation of the symmetry of the circuit and the protection circuit, as we can see below.***Figure 2*## The Power Transformer

The transformer must be symmetrical, i.e.: "

*3 wires*". The transformer must be able to supply at least**10A**at the output. The primary, "input voltage", must match the voltage in your area;**110V or 220Vac**. The secondary, "output voltage" should be**36 - 0V - 36Vac**.## Material list

U1 ......................... Voltage Regulator LM317HV

U2 ......................... Voltage Regulator LM337HV

Q1 .......................... PNP BD140 Transistor

Q2 .......................... NPN BD139 Transistor

Q3, Q4 ................... PNP TIP36C Power Transistor

Q5, Q6 ................... NPN TIP35C Power Transistor

D1 ......................... 50A Rectifier Bridge - KBPC5010

D2, D3 .................. 1N4007 Diode Rectifier

R1, R2 ................... 2W / 2Ω Resistor

R3, R4 ................... 5W / 0.06Ω Resistor

R5, R6 ................... 1/8W / 5KΩ Resistor

R7, R8 ................... 1/8W / 120Ω Resistor

R9, R10, R11 ......... 5W / 0.1Ω Resistor

R12, R13, R14 ....... 5W / 0.1Ω Resistor

C1, C2 ................... 10uF - 63V Electrolytic capacitor

C3, C4 ................... 1000uF - 63V Electrolytic Capacitor

C5, C6 ................... 5600uF - 63V Electrolytic capacitor

RV1 ....................... 5KΩ Potentiometer

P1, P2 .................... Connector 3 screw terminal 5mm 3 Pins

Others .................... Wires, Solders, pcb, etc.

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