Adjustable Mini Power Supply 0V-15V
About Power 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 – 15V with 1A as maximum current.
In this experiment, we are making an adjustable regulated power supply. For reducing any fluctuation and ripples at the output the supply must be regulated so that it can provide a constant voltage at the output. Our power supply provides regulated as well as the adjustable voltage at the output.
The supply which we are making takes 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.
|Transformer Tr1||Step down 18V-0-18V||1|
|Variable Resistance RV1||10K||1|
|Capacitor C1||470uF 50V||1|
|Capacitor C2||220uF 50V||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.
• Step down the Mains supply by input transformer
The voltage of Mains (Electricity coming at our come from the government) is approximately 220V but as per the circuit requirement, 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 18V-0-18V/2A rating. This transformer step downs the main line voltage to 18V, as shown in the below image. The circuit takes some drop in the form of resistive loss and by the LM317 IC. Therefore a transformer of high voltage rating than the voltage required for the application (0-15V) is taken in this circuit and it can provide 1A current at the output.
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 18V AC to 18V DC. As full wave rectifier is more efficient than a half wave since it can provide complete use of both the negative and positive pulse 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. The 1N4007 diode is used in full wave rectification as it can allow up to 1A current and 18V supply.
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 at the input side after the bridge rectifier to give pure DC at the output. As the DC which is rectified by the rectifier circuit has many AC spikes and 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 ripple free.
• 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 load 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
• Voltage regulation by LM317
For providing a regulated voltage at the output LM317 IC is used. This IC is capable of providing a current of up to 1.5, thus well suited for our requirements of 1A. In this circuit, LM317 will provide an adjustable voltage corresponding to its input voltage. This IC has the good feature of load regulation. It will provide regulated and stabilized the voltage at the output irrespective of the variations in the input voltage and load current.
It is a positive voltage regulator which gives output in the range of 1.25V to 37V with input voltage up to 40V. It can provide a maximum current of 1.5A at the output as per the datasheet under the optimum condition.
For setting a desired voltage at the output resistive voltage divider circuit is used between the output pin and ground. The voltage divider circuit has one programming resistor R1(fixed resistor) and another is variable resistor RV1.By taking a perfect ratio of feedback resistor (fixed resistor) and variable resistor we can obtain desired value of output voltage corresponding to the input voltage.
• Protection diode
A diode can be connected to the LM317 IC, as in below image. So that it can prevent the external capacitor from discharging through the IC during an input short circuit. When the input is shorted then the cathode of the diode is at ground potential. The anode terminal of the diode is at high voltage since C2 is fully charged. Therefore in this case diode is forward bias and all the discharging current from capacitor passes through a diode to ground. This will save the LM317 from the back current. In this experiment, two diodes are already connected in series at output which prevents the IC from the back current. Hence it is not necessary to connect a protection diode in this circuit.
• Output voltage
The output voltage can vary by using the adjust pin of LM317 IC. The variable resistor RV1 is used for varying the voltage at the output from 0V to 15V. As the minimum output of LM317 is 1.25V,so two diodes 1N4007 are connected in series with the resistor R2 of 1K to make the minimum output near to 0V. Each diode takes a drop of 0.7V and remaining drop is taken by the 1k resistor. Hence at the output, we get a minimum voltage of 0.3V and maximum voltage of 15.35V.
Case 1 :
Output voltage without load
By varying RV1 we can vary the output voltage in the range of
Vout = 0.34V to 15.35V
When load is connected at the output by setting the voltage at 15.35V
By this, it can be analyzed that when current demand increases at the output then the output voltage will start reducing. As the current demand increases then LM317 IC start heating up and IC will take more drop across it which will reduce the output voltage. Hence a proper heat sink is required when current drawn at the output is increased to dissipate the excessive heat from the circuit. LM317 internally can be able to tolerate 2W of power dissipation above this wattage a heat sink is required.
Points to Remember
• The current rating of a step-down transformer, bridge diode, and output diode must be greater than or equal to the required current at the output. Otherwise, it will be unable to supply the required current at the 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, the LM317 take voltage drop of around 2- 3 V. Thus input voltage must be 2-3V greater than the maximum output voltage required.
• Use a high value of capacitor at the input since high-value capacitor can handle mains noise. Use a capacitor at output also, this capacitor helps in handling fast transient changes and noise at the output. The value of output capacitor depends on the deviation in the voltage, output current and transient response time of the circuit.
• Always use a protection diode while using a capacitor after a voltage regulator IC, for preventing the IC from back current while discharging 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.
• For driving the high load at the output, heat sink should be mounted at the holes of the regulator. This will prevent the IC from blown off.
• As our circuit can draw a current of 1A at the output. A fuse of 1A is to be connected to the output of the rectifier. This fuse will prevent the circuit for current greater than 1A. For current above 1A, the fuse will blow off and this will cut the mains supply from the circuit.
Project Source Code
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