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Adjustable 0-30V 2A power supply

By Diksha Nama September 7, 2022

Adjustable 0-30V 2A Power Supply

Basics of Powers Supply

As the name suggests power supply are power providers for any circuit. Every electronic circuitry needs a proper power supply at the input for its optimum result at the output. We need to choose the power supply of any device or circuit as per the power requirements of the device. In this experiment, we are making an adjustable power supply, which will give voltage in a range of 0 – 30V with 2A as maximum current.

Overview

This article describes an adjustable power supply of 0V to 30V. The supply provides a regulated DC with the voltage in the range of 0 to 30V at the output. The output voltage can vary by a variable resistance and able to provide a maximum current of 2A.

The power supply is an adjustable linear regulated type in which the output voltage is constant and can vary mechanically. In this type of supply, a linear regulator element in series with the load is connected to the output. Linear element like BJT or FET is used for providing the correct voltage at the output.

In this supply the BJT (bipolar junction transistor) (2N3055) works in its linear mode, in the linear mode, it acts as a variable resistance. This variable resistance helps in providing the appropriate voltage at the output for any current under operating range. For drawing high loads at the output high current transistor is required. This 2N3055 transistor increases the output current of the power supply.

Components Required

Components Required Specifications Quantity
Transformer TR1 Step up of 18V-0-18V/2A 1
Diode D1-D4 SR560 4
Zenner Diode D5 12V, 1W 1
Zenner Diode D6 18V, 1W 1
Transistor Q1, Q3 BC547 2
Transistor Q2 2N3055 1
Capacitor C1 470uF, 50V 1
Capacitor C2 10uF, 50V 1
Resistor R1 2k, 1W 1
Resistor R2 0.3 Ohm, 5W 1
Potentionmeter RV1 10k 1
Fuse 2A 1

Basics of power supply

Every DC power supply needs some steps to follow for getting proper DC voltage at the output. Below diagram shows these basic steps by which we get a regulated DC power supply by AC.

Basics of power supply

BLOCK DIAGRAM

Working

• AC to AC conversion

The voltage of Mains (Electricity coming at our home from the government) is approximately 220V but as per the circuit requirement, only 30V is required at the output terminal. To reduce this 220V to 30V, a step-down transformer is used.

The circuit takes some drop in the form of resistive loss. Therefore a transformer of high voltage rating greater than the voltage required for the application (30V) is taken and which can provide 2A current at the output. The best suitable step-down transformer that meets our voltage and current requirement is 18V-0-18V/2A. This transformer step downs the main line voltage to 36V, as shown in the below image.

AC to DC conversion

• AC to DC conversion – Rectification

The rectification is the process of converting AC to DC. There are two ways to convert an AC signal to DC. One is through half wave rectifier and another is by using a full wave rectifier. In this circuit, we are using a full wave bridge rectifier for converting the 36V AC to 36V DC. As full wave rectifier is more efficient than a half wave since it can provide complete use of both the negative and positive side or part of AC signal. In full wave bridge rectifier configuration, four diodes are connected in such a way that it generates a DC signal at the output, as shown in below image. During full wave rectification, at a time two diodes become forward biased and another two diodes are reversed biased. We have chosen the SR560 diode because they can allow 2A current through them when forward biased and in reverse biased condition, they can sustain reverse 36V supply. Due to this, we are using SR560 diodes in rectification purpose.

AC to DC conversion- Rectification

• Smoothing

As its name suggest it is the process of smoothing or filtering the DC signal by using a capacitor. A capacitor C1 of high value is connected to the output of rectifier circuit. As the DC which is to be rectified by the rectifier circuit has many AC spikes and unwanted ripples, so to reduce these spikes we use a capacitor. This capacitor acts as a filtering capacitor which bypasses all the AC through it to ground. At the output, the DC which is left is now smoother and ripples free.

Smoothing

• Output capacitor

At the output, capacitor C2 is also connected to the circuit. This capacitor helps in fast response to load transients. Whenever the output loads current changes then there is an initial shortage of current, which can be fulfilled by this output capacitor.

The output current variation can be calculated by – :

Output current, Iout = C (dV/dt)
dV = Maximum allowable voltage deviation
dt = Transient response time
Consider dv = 100mV
dt = 100us
In this circuit C = 10uF
Iout = 10u (0.1/100u)
Iout = 10mA

By this we can say output capacitor will respond for 10mA current change for a transisent response time of 100us

Output Capacitor

• Voltage regulation

As it is necessary for every circuit that at the output, it should provide a voltage which is regulated and constant without any fluctuation or variation. For voltage regulation a linear regulator is needed in the circuit, the work of this regulator is to maintain a constant voltage of required level at the output.

In this circuit, the maximum voltage at the output is 30V so a 30V zener diode is perfect for voltage regulation at the output. Here two zener diode of 12V and 18V are used in series which will provide a total of 30V at the output.

(Note: Can also use 30V zener diode or a different combination of zener diode for getting 30V at output)

• Voltage adjustment

For adjusting output voltage from 0 to 30V a variable resistor (RV1) is connected to the output. By varying this resistor we can get voltage at the output as per the requirement between 0 to 30V.

• Output current

The zener diode can provide current in milliamperes only.Therefore for deriving high load current at output some linear element must be connected in series with the load. This circuit uses an NPN bipolar junction transistor as a linear element. The transistor BC547 is used for providing the sufficient base voltage to the 2N3055 BJT. 2N3055 is capable of providing 2A current at the output. For short circuit protection Q3 and R3 is used in the circuit.

Practical results

Without load when output voltage varies by RV1
Vout = 0.3V to 30.3V
Error in the Vout
% Error = (Experimental value – Expected value)*100 /Expected value
% Error = (30.3 – 30)*100 / 30
% Error = 1%
When load is connected at the output when output is at maximum voltage i.e. 30V
With load of 1K
Load R = 1k
Then output voltage
Vout = 29.1V (voltage drop 0.9V, 30-29.1)
Then output current
I out = 29.1mA
Power dissipation at load of 1k
Pout = Iout*Iout*R
Pout = 0.0291*0.0291*1000
Pout = 0.84W
If load R = 470 ohm
Then Vout = 28.9V (voltage drop 1.1V, 30-28.9)
Output current = 61.4mA
Power dissipation at load of 470 ohm
Pout = Iout*Iout*R
Pout = 0.0614*0.0614*470
Pout = 1.7W

Points to Remember

• The current rating of the transformer, bridge rectifier and transistor must be greater than or equal to the output current requirement. Then the only circuit can provide sufficient current at output.

• The voltage rating of a step-down transformer should be greater than the maximum required output voltage. This is due to the fact that, circuit takes voltage drop due to some resistive loss. Thus input voltage from transformer must be 2-3V greater than the maximum output voltage.

• Use a capacitor C1 at the output of rectifier since this capacitor can handle mains noise.

• Use a capacitor (C2 in this experiment) at the output of the regulator, this capacitor helps in handling fast transient changes and noise at the output. The value of this capacitor depends on the deviation in the voltage, current variations and transient response time of the capacitor.

• The capacitor used in the circuit must be of higher voltage rating than the input voltage. Otherwise, the capacitor starts leaking the current due to the excess voltage at its plates and will burst out.

• The zener must be of 1W otherwise, it will start up heating and will damage.

• As the current demand increasing at the output, the transistor 2N3055 will start heat up. To overcome this problem a proper heat sink must be mounted across it to dissipate the excess heat. Otherwise, the transistor will blow off.

• As our circuit can draw a current of 2A at the output. A fuse of 2A is to be connected to the output of the rectifier. This fuse will prevent the circuit for current greater than 2A. For current above 2A, the fuse will blow off and this will cut the input supply from the circuit.

Project Source Code

 

Project Source Code

###

//Program to 

###

 



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