Mini adjustable power supply 0V-15V
About power supply
As the name suggests, power supplies are suppliers of energy to any circuit. Every electronic circuit needs an adequate power supply at the input to obtain an ideal result at the output. We need to choose the power supply of any device or circuit according to the power requirements of the device. In this experiment, we are making an adjustable power supply, which will provide voltage in the range of 0 – 15V with 1A as maximum current.
Overview
In this experiment, we are making an adjustable regulated power supply. To reduce any fluctuations and ripples in the output, the power supply must be regulated to be able to provide a constant voltage at the output. Our power supply provides regulated and adjustable voltage at the output.
The power supply we are making receives 220V AC as input and generates a variable DC voltage in the range of 0-15V. This power supply can provide a maximum current of 1A at the output.
Required components
| Required components | Specification | Amount |
|---|---|---|
| Transformer Tr1 | Lower 18V-0-18V | 1 |
| Diode D1-D6 | 1N4007 | 6 |
| Variable Resistance RV1 | 10K | 1 |
| Voltage regulator | LM317 | 1 |
| Capacitor C1 | 470uF 50V | 1 |
| Capacitor C2 | 220uF 50V | 1 |
| Resistor R1 | 820OHM | 1 |
| Resistor R2 | 1K,2W | 1 |
| Fuse | 1A | 1 |
Power supply basics
Every DC power supply needs to follow a few steps to get the proper DC voltage at the output. The diagram below shows these basic steps by which we obtain an AC regulated DC power supply.
Working
• Reduce the mains supply by input transformer
The mains voltage (electricity coming from the government) is approximately 220V, but according to the circuit requirements, only 15V voltage is required at the output terminal. To reduce this 220V to 15V, a step-down transformer is used. The step-down transformer we are using is rated 18V-0-18V/2A. This transformer reduces the main line voltage to 18V as shown in the image below. The circuit suffers some drop in the form of resistive loss and by IC LM317. Therefore, a voltage transformer higher than the voltage required for the application (0-15V) is placed in this circuit and can provide 1A current at the output.
• Rectification
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 using full-wave rectifier. In this circuit, we are using a full wave bridge rectifier to convert 18V AC to 18V DC. As full wave rectifier is more efficient than half wave as it can provide full use of negative and positive pulse of AC signal. In the 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 the image below. The 1N4007 diode is used in full-wave rectification, as it can allow current of up to 1A and 18V supply.
• Smoothing
As the name suggests, it is the process of smoothing or filtering the DC signal using a capacitor. A high value capacitor C1 is connected on the input side after the rectifier bridge to provide pure DC at the output. As the DC which is rectified by the rectifier circuit has many AC spikes and ripples, to reduce these spikes we use a capacitor. This capacitor acts as a filtering capacitor that shunts all the AC through it to ground. At the output, the remaining DC is now smoother and ripple-free.
• Output capacitor
At the output, capacitor C2 is also connected to the circuit. This capacitor helps in quick response to load transients. Whenever the output load current 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)
• Voltage regulation by LM317
To provide a regulated voltage at the output, IC LM317 is used. This IC is capable of delivering a current of up to 1.5, therefore suitable for our 1A requirements. In this circuit, the LM317 will provide an adjustable voltage corresponding to its input voltage. This IC has the good feature of load regulation. It will provide regulation and stabilization of the output voltage regardless of variations in input voltage and load current.
About LM317
It is a positive voltage regulator that provides output in the range of 1.25V to 37V with input voltage up to 40V. It can provide a maximum output current of 1.5A as per the datasheet under ideal conditions.
To set the desired voltage at the output, resistive voltage divider circuit is used between the output pin and ground. The voltage divider circuit has a programming resistor R1 (fixed resistor) and another is the variable resistor RV1. By taking a perfect proportion between the feedback resistor (fixed resistor) and the variable resistor, we can obtain the desired value of the output voltage corresponding to the input voltage.
• Protection diode
A diode can be connected to IC LM317, as in the image below. To prevent the external capacitor from discharging through the IC during an input short circuit. When the input is shorted, the cathode of the diode is at ground potential. The anode terminal of the diode is at high voltage as C2 is fully charged. So in this case, the diode is forward biased and all the capacitor discharge current passes through a diode to ground. This will save the LM317 from reverse current. In this experiment, two diodes are already connected in series at the output, which prevents the IC from countercurrent. Therefore, it is not necessary to connect a protection diode to this circuit.
• Output voltage
The output voltage can be varied using the tuning pin of the LM317 IC. The variable resistor RV1 is used to vary the output voltage from 0V to 15V. Since the minimum output of LM317 is 1.25V, two 1N4007 diodes are connected in series with 1K resistor R2 to make the minimum output close to 0V. Each diode suffers a drop of 0.7 V and the remaining drop is taken up by the 1k resistor. Therefore, at the output we obtain a minimum voltage of 0.3V and a maximum voltage of 15.35V.
Practical Observation
Case 1:
No-load output voltage
By varying RV1 we can vary the output voltage in the range
Vout = 0.34V to 15.35V
%Error= (Experimental value – expected value)*100/Expected value % error = (15.35 – 15)* 100/15 % Error = 2.3%Case2:
When the load is connected to the output setting the voltage at 15.35V
With this, it can be analyzed that when the current demand increases at the output, the output voltage will begin to decrease. As the current demand increases, the LM317 IC starts to heat up and the IC will experience more sags which will reduce the output voltage. Hence, a suitable heat sink is required when the current drawn at the output is increased to dissipate excessive heat from the circuit. The LM317 internally can tolerate 2W of power dissipation above this power, requiring a heatsink.
Application
Points to remember
• The current rating of a step-down transformer, bridge diode and output diode must be greater than or equal to the current required at the output. Otherwise, it will not be able to supply the required 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 LM317 experiences a voltage drop of around 2-3 V. Therefore, the input voltage must be 2-3 V higher than the maximum output voltage required.
• Use a high value capacitor at the input, as a high value capacitor can deal with mains noise. Also use a capacitor at the output, this capacitor helps to deal with fast transient changes and noise at the output. The value of the output capacitor depends on the deviation in voltage, output current and transient response time of the circuit.
• Always use a protection diode when using a capacitor after a voltage regulator IC, to prevent the IC from countercurrent during capacitor discharge.
• The capacitor used in the circuit must have a higher rated voltage than the input voltage. Otherwise, the capacitor will start to leak current due to excess voltage on its plates and will explode.
• For high output load activation, a heat sink must be mounted in the regulator holes. This will prevent the IC from being exploded.
• As our circuit can consume a current of 1A at the output. A 1A fuse must be connected to the rectifier output. This fuse will prevent circuit current exceeding 1A. For currents above 1A, the fuse will blow and this will cut off the electrical supply to the circuit.
Project source code