 Hello everyone. Myself Aursang Esmi, Assistant Professor, Electronics and Communication Engineering, Watson Institute of Technology, Sulapur. Today we are going to see the under-regulated power supply design steps. Learning outcome. At the end of this session, students will be able to describe steps to design under-regulated DC power supply and design DC power supply as per output requirements. These are the contents of my presentation. Students should pause video here and recall what is significance of filter circuit with rectifier circuit. One of the output of rectifier circuit is in terms of sinusoidal pulses. It contains both AC component and DC component. So output of simple rectifier is not pure DC signal. It contains ripple. So filter circuit is connected at the output of rectifier circuit to filter out unwanted ripple contents so that we get pure DC output from rectifier circuit. Now this is a block diagram for typical DC power supply. It consists of transformer, diode rectifier circuit, filter circuit and load resistance. A transformer may be a simple transformer or transformer with center tap secondly. A transformer is used to provide required AC voltage at the input of rectifier circuit. Output of rectifier circuit is given to filter circuit. Rectifier circuit converts given AC voltage signal to DC voltage signal and filter circuit is used to filter out unwanted AC components from output of rectifier circuit and provides a ripple free DC voltage to the load resistance. Now let us take one of the examples of power supply design. Let us design power supply using flow rectifier and LC filter circuit to supply DC voltage of 20 volt at 200 milliampere with maximum ripple content of 1 percent. The design steps are the first step, decide overall circuit diagram for required DC power supply. Then second step, selection of filter components and load resistor and third step, selection of transformer. Step number four is selection of diodes and step number five, selection of leader resistor RB. Now let us go for the first step. The first step, let us decide overall circuit diagram for required unregulated DC power supply. The circuit diagram consists of a transformer with center tap secondary or two diodes, diode D1 and D2. So, it is a flow rectifier with center tap secondary transformer inductance L that is connected in series with load resistance and the capacitor C it is connected in parallel to load resistor. Register RB is a leader resistor connected to parallel to capacitor and load resistor. So, it is a power supply using flow rectifier with LC filter with the leader resistor RB connected in parallel to load resistor. The transformer provides a required AC voltage at the input of rectifier circuit as per the requirement of output DC voltage and DC current. Now let us go for the second step. Second step is selection of leader components and load resistor. So, let us decide the values for load resistance RL, inductance L and capacitor C. Output VDC is given by IDC into RL. So, load resistance RL is equal to VDC upon IDC. So, output voltage DC voltage required is 20 volt and output DC current is 200 mA. So, the required value of load resistance it comes by calculation 100 ohm. So, select load resistor RL of 100 ohm with 5 Watt power rating. For flow rectifier with LC filter the ripple factor is given by 1 upon 6 root 2 omega square LC where omega stands for 2 pi f where f stands for frequency of input AC voltage signal. For 1 percent ripple factor calculate the product of L and C using the relation RF equal to 1 upon 6 root 2 omega square LC. So, this comes product of inductance and capacitance is equal to 119.40 into 10 raise to minus 6. Now to maintain continuous current through the circuit and the current flowing in the circuit never becomes 0. So, find some critical value of inductance. The inductance to be used in filter circuits it should not be too large or too small. If inductance is too large the size of inductor increases and the cost of inductor also increases but the ripple factor decreases. At the same time if you decrease value of inductance the size of inductor decreases and cost also decreases but ripple contents increases. So, to select the minimum value of inductance in LC filter circuit so that the current flowing in the circuit is continuous and flows only in one direction and it never becomes 0. So, find out a critical value of inductance LC that is the minimum value of inductance that should be connected in LC filter circuit so that circuit current never becomes 0 and it will not flow in reverse direction. So, critical value of inductance LC is given by RL upon 3 omega. So, for RL equal to 100 ohm a calculate value of critical value of inductance by calculation it comes 0.106 Henry. Generally value of L is selected 10 times of LC so select inductor L of 1 Henry. Then assuming L equal to 1 Henry then find out the value of capacitors C. The product of L and C is given by 119.40 into 10 raise to minus 6 so using value of L equal to 1 Henry the required value of capacitor C it comes 119.40 into 10 raise to minus 6 parade. So, the value of capacitor that should be used in LC filter circuit it should not be too large and it should not be too small. If value of C is large the size of capacitor increases the cost is also increases. So, if value of C is large that increases initial surge current flowing through diode and if value of C is small it increases ripple content. So, capacitor select nearest standard value of capacitor C like 100 microparade. Step number 3 selection of transformer. Select a proper transformer with required secondary voltage to obtain required DC output voltage of a 20 volt at 200 milliampere. So calculate secondary voltage transformer secondary voltage so we are using a transformer with center depth secondary so half secondary winding voltage Vs in RMS is given by Vm upon root 2. Consider voltage drop across conducting diode also for silicon diode assume diode voltage drop is equal to 0.7 volts. So, required secondary voltage from transformer is equal to Vm upon root 2 plus diode voltage drop 0.7 volts. So, to find Vm consider DC output voltage equation for full rectifier with LC filter. So, output Vdc is given by 2 Vm by pi minus Idc into Rs plus Rf plus Rc where Rs is transformer resistance Rf is diode forward resistance and Rc is inductor DC resistance. Generally inductor DC resistance Rc is much larger than Rs plus Rf. So, neglecting Rs and Rf assume value of Rc equal to 10 ohm. So, for required output DC voltage of 20 volt value of Vm it comes 34.55 volts therefore required secondary voltage is 25 volts. So, select transformer with center depth secondary whose primary voltage is 230 volt and secondary voltage is 24.024 volt at 200 mA. Step number 4 selection of diodes select diode with PI rating 2 Vm that is 68 volts average current rating of diode IDC by 2 plus 25 percent of IDC by 2 it comes to 125 mA and P current rating I am equal to Vm upon Rl so it is 340 mA select diode 1n 4002 selection of blader resistor. So, to maintain good voltage regulation under output open circuit condition to limit initial cells current of capacitor blader resist is connected in parallel to capacitor. So, critical value of inductance Lc is greater than Rl upon 3 omega if Rl is disconnected output open circuit condition L is equal to Lc then L should be greater than or equal to Rb upon 3 omega. So, Rb should be less than or equal to 942 L if L is equal to 1 henry then Rb is equal to 942 ohm. These are the references thank you.