0 to 30V 2A Adjustable DC Power Supply Circuit (Part 1/13)

Required components –

Fig. 2: Block diagram of 0 to 30 V 2A DC adjustable power supply

The circuit is assembled in stages and each stage serves a specific purpose. To step down the 230 V AC, an 18 V – 0 – 18 V transformer is used. The secondary coil of the transformer is connected to a full bridge rectifier. The full bridge rectifier is constructed by connecting four SR560 diodes together designated as D1, D2, D3 and D4 in the schematics. The cathode of D1 and the anode of D2 are connected to one of the secondary coils and the cathodes of D4 and the anode of D3 are connected to the other ends of the secondary coil. The cathodes of D2 and D3 are connected, of which one terminal is taken from the output of the rectifier and the anodes of D1 and D4 are connected, of which another terminal is taken from the output of the full-wave rectifier. A 2A fuse is connected in series to the output terminal at cathode junctions D2-D3 for safety.

A 470 uF capacitor (shown as C1 in the schematic) is connected across the output terminals of the full-wave rectifier for smoothing purposes. For voltage regulation, two zener diodes of 12V and 18V are connected in series and parallel to the smoothing capacitor. A variable resistance is connected in series to the zener diodes for voltage adjustment and a 10 uF capacitor (shown as C1 in the schematic) is connected in parallel to compensate transient currents. Two NPN transistors (shown as Q1 and Q2 in the schematics) are connected as Darlington pair amplifier at one of the output terminals in series to obtain the desired current gains. The output of the Darlington pair is further connected to an NPN transistor (shown as Q3 in the schematics) and resistor (shown as R3 in the schematics) for short circuit protection.

Get the schematic diagram drawn or printed on paper and make each connection carefully. Only after checking each connection made correctly, connect the power circuit to an AC source.

The power circuit operates in well-defined stages, each stage serving a specific purpose. The circuit operates in the following steps –

1. AC to AC Conversion

2. AC to DC Conversion – Full Wave Rectification

3. Smoothing

4. Transient Current Compensation

5. Voltage regulation

6. Voltage adjustment

7. Current amplification

8. Short circuit protection

AC to AC Conversion

The voltage of the main sources (electricity fed by the intermediate transformer after reducing the line voltage of the generating station) is approximately 220-230 Vac, which needs to be further reduced to the 30 V level. To reduce 220 Vac to 30 Vac , a step-down transformer is used.

The circuit experiences some drop in output voltage due to resistive loss. Therefore, a transformer with a high voltage rating greater than the required 30V needs to be used. The transformer must provide 2A current at the output. The most suitable step-down transformer that meets the mentioned voltage and current requirements is 18V-0-18V/2A. This transformer reduces the main line voltage to 36 Vac as shown in the image below.

Fig. 3: 18-0-18V Transformer Circuit Diagram

AC to DC Conversion – Full Wave Rectification

The reduced AC voltage needs to be converted to DC voltage through rectification. Rectification is the process of converting AC voltage to DC voltage. There are two ways to convert an AC signal to DC. One is half-wave rectification and the other is full-wave rectification. In this circuit, a full wave bridge rectifier is used to convert 36 V AC to 36 V DC. Full-wave rectification is more efficient than half-wave rectification as it provides full use of both the negative and positive sides of the AC signal. In the full-wave bridge rectifier configuration, four diodes are connected in such a way that current flows through them in only one direction, resulting in a DC signal at the output. During full-wave rectification, two diodes are forward biased and two other diodes are reverse biased.

Fig. 4: Full Wave Rectifier Circuit Diagram

During the positive half cycle of the supply, diodes D2 and D4 conduct in series while diodes D1 and D3 are reverse biased and current flows through the output terminal passing through D2, output terminal and D4. During the negative half cycle of the supply, diodes D1 and D3 conduct in series, but diodes D1 and D2 are reverse biased and current flows through D3, output terminal and D1. The direction of current in both directions through the output terminal in both conditions remains the same.

Fig. 5: Circuit Diagram showing the positive cycle of the Full Wave Rectifier

Fig. 6: Circuit diagram showing negative cycle of full wave rectifier

SR560 diodes are chosen to build the full wave rectifier because they have the maximum (average) forward current rating of 2A and in reverse bias condition, they can sustain peak reverse voltage up to 36V. That is why SR560 diodes are used in this project for full wave rectification.

As the name suggests, it is the process of smoothing or filtering the DC signal using a capacitor. The output of the full wave rectifier is not a constant DC voltage. The rectifier output has twice the frequency of the main sources, but still contains ripples. Therefore, it needs to be smoothed out by connecting a capacitor in parallel to the output of the full-wave rectifier. The capacitor charges and discharges during a cycle, providing a constant DC voltage as output. Thus, a capacitor (shown as C1 in the schematic) of high value is connected to the output of the rectifier circuit. As the DC that has to be rectified by the rectifier circuit has many AC spikes and unwanted ripples, to reduce these spikes a capacitor is used. This capacitor acts as a filtering capacitor that shunts all the AC through it to ground. At the output, the remaining average DC voltage is smoother and ripple-free.

At the output terminals of the power circuit, a capacitor (shown as C2 in the schematic) is also connected in parallel. This capacitor helps in quick response to load transients. Whenever the current of the output loads changes, there is an initial shortage of current, which can be met by this output capacitor.

The output current variation can be calculated by

Output current, Iout = C (dV/dt) where

dV = Maximum allowable voltage deviation

dt = transient response time

Considering dv = 100mV

dt = 100us

A 10 uF capacitor is used in this circuit, so

C = 10uF

Iout = 10u (0.1/100u)

Iout = 10mA

In this way it can be concluded that the output capacitor will respond to a current change of 10mA for a transient response time of 100 us.

The power supply circuit must provide regulated and constant voltage, without any fluctuation or variation. For voltage regulation, a linear regulator is required in the circuit. The purpose of using this regulator is to maintain a constant voltage at a desired level at the output.

Fig. 9: Voltage regulator circuit diagram for 0 to 30 V 2 A adjustable DC power supply

In this circuit, the maximum voltage at the output must be 30V, therefore a voltage of 30V zener diode is perfect for regulating the output voltage. Here two zener diodes of 12V and 18V are connected in series which provide a total of 30V at the output. A 30V zener diode of 1W rating or a different combination of zener diodes can also be used to obtain 30V at the output.

To adjust the output voltage from 0 to 30 V, a variable resistor (shown as RV1 in the schematic) is connected to the output. The variable probe of RV1 is connected to the collector of the BC547 switching transistor (shown as Q3 in the schematics). By varying this resistor, the emitter of the switching transistor will provide a variable voltage between 0 and 30V.

The zener diode can only supply current in amps. Therefore, to obtain high load current at the output, some linear element must be connected in series with the load that can draw the required current. This circuit uses an NPN bipolar junction transistor as a linear element. A BC547 transistor (shown as Q2 in the schematics) is used to supply sufficient base voltage to the NPN 2N3055 bipolar transistor (shown as Q1 in the schematics). The 2N3055 transistor is capable of supplying 2A current at the output. The transistors are connected in a Darlington pair amplifier configuration to generate the desired current gains. In the Darlington pair configuration, the net current gain is a multiplication of the current gains of the two transistors.

Total current gain (total hFE) = transistor 1 current gain (hFE t1) x transistor 2 current gain (hFE t2)

Therefore, the current gain of the BC547 is 800 and that of the 2N3055 is 20 to 70, taking an average of 50. So,

Total current gain (total hFE) = 800 * 50 = 40,000

This is enough to raise currents from a thousand amps to amp levels.

For short circuit protection, a BC547 switching transistor (shown as Q3 in the schematics) and a resistor shown as R2 in the schematics are connected in series before the output of the circuit.

The following precautions must be taken while assembling the circuit –

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

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

• Capacitor C1 at the output of the rectifier is used to deal with mains noise and remove ripples.

• Capacitor C2 at the output terminals of the power circuit helps deal with rapid transient changes and noise in the output load. The value of this capacitor depends on the voltage deviation, current variations and transient response time of the capacitor used.

• Capacitors used in the circuit must have a voltage rating higher than the input voltage. Otherwise, the capacitor will start to leak current due to excess voltage on its plates and will explode.

• The zener diodes used in the circuit must be rated at 1W otherwise they will be damaged due to heating.

• As the current demand increases on the output load, the 2N3055 transistor will begin to heat up. To overcome this problem, a suitable heat sink must be mounted to dissipate excess heat. Otherwise, the transistor may explode.

• As the circuit was designed to consume a maximum current of 2A at the output, a 2A fuse must be connected to the output of the full-wave rectifier. This fuse will prevent the circuit from drawing current greater than 2A. For current consumed above 2A, the fuse will blow first, cutting off the input power to the circuit.

Once the circuit is assembled, it's time to test it. Connect the circuit to the main sources and change the variable resistance. Take voltage and current readings at the output terminal of the power circuit using a multimeter. Then connect fixed resistors as load and check voltage and current readings again.

During the no-load test, the output voltage at the variable resistance setting varied in value between 0.3 V to 30.3 V. Therefore, when calculating the error, the following error percentage is obtained –

% Error = (Experimental value – Expected value)*100 /Expected value

% error = (30.3 – 30)*100/30

% Error = 1%

When a load is connected to the output the maximum voltage is read as 30V. With a 1K resistance load, the output voltage is read as 29.1 V showing a voltage drop of 0.9 V. The output current is measured as 29.1 mA, so the power dissipation in the load is resistance of 1k is as follows –

Pout = Iout*Iout*R

Pout = 0.0291*0.0291*1000

Pout = 0.84W

If the load resistance used is 470 ohms, a voltage of 28.9 V is measured, showing a voltage drop of 1.1 V, and the measured current is 61.4 mA. Therefore, the power dissipation with 470 ohm load is as follows –

Pout = Iout*Iout*R

Pout = 0.0614*0.0614*470

Pout = 1.7W

This circuit can be used as a power adapter to support a wide range of electronic applications such as broadcasting, digital cameras, printers, laptops and other portable electronic devices. It can also be used as an adjustable DC power supply for electronic devices.

In the near future, High Voltage Direct Current (HVDC) may become a more popular means of power transmission as the focus on renewable energy increases. HVDC is commonly employed only for cross-country and underwater power transmissions. This is done to reduce induced inductance and capacitance losses over long distances. The resistance, inductance, and capacitance of a wire cannot be practically changed. For power transmission within the country, AC is currently the preferred method. Alternating current is preferred for intra-country power transmission despite losses due to inductance and capacitance because reducing AC voltage is much cheaper than reducing DC voltage.

An AC voltage can be easily reduced with the help of a transformer. Therefore, the energy supplied to homes currently is AC voltage. Generating stations provide high AC voltages to reduce energy losses. Just as the AC voltage supplied to homes is 230 V 50 HZ, the generating station supplies 2300 V through the transmission wire, which is stepped down to 230 V by an intermediate transformer. Possibly, the day HVDC from renewable sources becomes a common source of household power supply, semiconductor-based power circuits will be used for voltage reduction and regulation.

Circuit Diagram-Adjustable-0-30V-2A-DC-Power Supply