Below, we present a stabilized power supply unit which is able to provide up to 2.5 amperes of current, and its output voltage can be adjusted from 5 to 26V. It is a low cost linear power supply unit, based on an op-amp, and the circuit is extremely simple.
The electronic circuit
We use a 24 Volt transformer, a diodes-bridge for rectification, some capacitors for filtering, one Zener-type diode for voltage reference, an operational amplifier and some transistors for voltage regulation. The specific power supply unit is based on the op-amp voltage regulator concept.
The mains voltage (from the power-grid) isn’t always stable, so it is normal to the voltage of the transformer’s secondary winding to vary somehow. Thus, after rectification and filtering we’ll take about 30 to 33 volts. For filtering we use C8. For a full-wave circuit, ripple will be kept within reasonable limits if the product of the values of the load resistor and the filter capacitor is about 0.1s. To stabilize to a voltage of 26 volts – at 2 amps, we will have to use an electrolytic for C8 that has a minimum capacity of: 100000/ (26: 2) = 7700 microfarads. Since this value is not a standard one, we will use a larger capacity, that is, 10000 microfarads. A polyester capacitor of 100nF is connected in parallel to C8 (see C4) and probably you may wonder what difference there might be between a capacity of 10000 uF and one of 1000.1 uF.
This 0.1uF polyester capacitor is not needed for ripple rejecting (filtering) but only to provide an adequate ground path for spurious impulses may present in the mains voltage. If not grounded, they may pass through the transformer, may arrive on the collector of the Q3 with voltage peaks so high that may destroy Q3 after a short period of use.
Having a DC voltage of approximately 32 volts, to calculate the resistance value of R5 to be connected in series to the zener diode D1, so that the zener will absorb a current of not less than 7 mA, we use the formula:
ohms R5 = [(Vin - Vz): mA] x 1,000.
Using the above formula, the value of R5 will be: [(32-4.3): 7] x 1,000 = 3,957 ohms. Since this value is not a standard one, we will use one of 3.9K.
Knowing however that all resistors have a tolerance, the 3,900 ohms resistor could result in practice to as much as 4100 ohms. Considering that the mains voltage could also be lower than its standard value for more than 10%, using a standard 3.9K ±5% value may result in a current less than 7mA. Thus why, we should use a resistor value of 3,300 ohms. With this value, the current through the zener diode will be about 8.4mA (mA = [(Vin - Vz): ohm] x 1.000). Therefore, even if the main’s voltage drops significantly, we‘ll have more than 7mA at the zener.
You may wonder about the use of C3, which is connected at Q2’s collector and base. There is a simple explanation for this: Since Q2 is part of a Darlington pair, there is always the possibility of oscillating due to instability. Self-oscillating may take place at ultrasonic (or higher) frequencies, producing undesirable noise at the output terminals. This capacitor will prevent oscillations on Q2-Q3 Darlington pair.
We have also added a short-circuit protection network, formed by R1, R2 resistors, and Q1 transistor. This network cuts-off power on the output terminals, in case the current exceeds 2.5 amperes.
The output voltage can be varied from a minimum of 5 volts, up to a maximum of 26 volts. It can be controlled from R9 potensiometer. By turning R9’s wiper towards the R7-R8 resistors, the output will get at its maximum value. By rotating towards R4, the output voltage drops to about 5 volts.
There are two more capacitors connected in parallel to the output terminals; C1 and C2. These are used also for filtering and for providing a low impedance path to the ground for high frequency transients. R3 is connected in parallel to these two capacitors, for discharging them every time you turn off the power supply, or when switching from a higher voltage at a lower one. To observe the value of the output voltage, we use a typical voltmeter.
Assembling the unit
To build this project, you will need the printed circuit board provided below. You could begin assembling the circuit by soldering a 8-pin IC-socket for the IC1. Then, you may solder the resistors and the non-polarized capacitors. All resistors are of 1/4W type unless otherwise noticed.
Be careful to place D1 correctly, according the polarity shown on the schematic. After soldering these components, you may place the polarized capacitors (electrolytic), considering again their correct polarity. The longer lead terminal in their bodies, is always the positive one.
Then take the two transistors; Q2 and Q1, and place them without shorten their terminals. Place the bridge rectifier, and then place the PCB terminals used for connecting the transformer, the R9 potentiometer, the LED, the voltmeter, and the output sockets.
To complete the assembly, use a metallic enclosure, and place all components inside it. Place R9, and the voltmeter in the front panel. You could also use an on-off switch in the front panel, for connecting or disconnecting for the mains grid.
Place IC1 into its socket, and check that all its pins to fit perfectly in the holes. Use a large heat sink and place Q3 on it. Then, the heat –sink must be placed at the rear panel of the enclosure. There is an important aspect regarding T3: The metal body of this transistor must be insulated from ground. So, it must be isolated from the metallic heat sink and the box. To do this, use a mica isolator. If you forget to use isolation, the positive voltage terminal will be short-circuited to the ground. Leaving the unit on these conditions for several minutes, will burn the diodes-bridge and the transformer.
To connect T3 to the PCB, use three wires, for the collector, the base, and the emitter, respectively. Check once again by using an ohm meter, that the metal body of the transistor is being isolated from the metal of the cooling fin.
You must know that after using this power supply unit at its maximum current of 2 amperes for more than one hour, the heat-sink will get extremely hot (you could not touch it on bare hand). There is nothing to worry about this. It’s quite normal, especially when setting output voltage at its lower limit. As in any linear regulator, heat losses increase when the output voltage decreases. In fact, the power loss due to heating is the current times the voltage dropped across the regulator.
At extreme heat it’s essential to avoid placing the rear panel near a wall, in order to leave some space for air circulation.
2.5A Power supply unit Printed Circuit Board
Components placement guide