Power Supply System Design Using a Voltage Divider - Lab Report Example

Power Supply System Design Using a Voltage Divider Power Supply System Design Using a Voltage Divider Introduction 1.1 The voltage divider The voltage divider is also refereed to as potential divider. The electronic circuit that makes up the voltage divider produces an output voltage as a fraction of the input voltage. A simple voltage divider is made up of two resistors that are connected in series. 1.2 Purposes of using the voltage divider. 1) The voltage divider is mainly used to reduce the voltage supplied to electrical appliances. Different electrical appliances require varying voltages but the power supplied by voltage sources are constant. It is therefore necessary to reduce the voltage to suit the given application. 2) The voltage divider is also used to create a reference voltage. 3) The voltage divider can also be used as a signal attenuator at low frequency. Background information 2.1 Different types of voltage dividers There are different types of voltage dividers that are used to divide the voltage into different ratios. Some of the most common voltage dividers include; Resistive voltage divider Low pass resistor-capacitor filter Inductive voltage divider Capacitive voltage divider 2.2 The calculations and derivation using the voltage divider The diagram below shows a simple restive voltage divider. The circuit comprises of two resistors (a) and (b), connected in series. The series resistor circuit is then connected to a constant source of power. The voltage is the taken from the middle of the two resistors. Fig 1 showing the principle for the resistive voltage divider. For two resistors connected in series as shown in the diagram below, the total resistor is given by; Fig 2 showing 2 resistors in series Total voltage in the circuit = sum of the voltage drop across the two resistors V= voltage drop across R1 + the voltage across R2 According to ohms law, the current through any conductor is directly proportional to the potential difference between the two points and is inversely proportional to the resistance of the circuit. Form ohms law V=IR Where (V) is the voltage (I) is the current (R) Is the resistance The resistance is given by R = V/I(from ohms law ) Thus the total voltage across the two resistors R1 and R2 is given by V (total) = IR1 +IR2 V (total) = I (R1 + R2) The ratio hence The current in a series circuit is given by The corresponding voltage drop across resistor (say R1) is given by For a voltage divider two types of voltage results, the voltage when there is no load and the voltage under load. In no load condition the voltage is given by; When a load is applied, a load resistance is introduced to the circuit, the equation changes to; Where V1 is the input voltage and Rl is the resistance of the load circuit. From the equations outlined above it follows that if, (R1=R2) then Aim and objectives The main aim of conducting this experiment is to study and practically develop a potential divider using resistors. The specific objectives include To outline a detailed study of the voltage divider using resistors To design the voltage divider To carry out a detailed engineering design for the voltage divider considered herein To evaluate the designed voltage divider Theoretical design The design specifications provided are; Input voltage Vin= 15.0 V Voltage under no load Vout= 6 Load resistance = 1000 The maximum voltage drop = 10% of the output The resulting circuit for the voltage divider is as shown in the figure below Fig 3: showing the outline of the theoretical design From the parameters given the resistance across R2 and Rload can be calculated from The resistance ratios should then be given by The ratio of the resistance R2 to the total resistance is given by 2:5 Let R1 be 300Ω Then, on no load conditions we have the resistance R2 should equal to; This is the resistance if there was no load. However on the introduction of a load, there will be a voltage drop which depends upon the load resistance. From the specification, the maximum allowable voltage drop is 10% 10% of 6V = 0.6 V 6V-0.6V = 5.4 V Voltage drop due to the load Let’s assume that a resistance (Rc) is the total resistance of the load and the voltage divider resistor R2 that is Rc = Rl + R2 Then Where V1 is the input voltage and Vout is the output voltage and R1 and R2 are the resistors of the voltage divider. Solving the equation, we have; Thus, the combined resistance of R2 and RL(load resistance) is equal to 168.75. The load resistance is however equal to 1000Ω, and the two resistors are in parallel, then Solving for R2,we have Taking the nearest common resistance we get = 220Ω Power dissipation Power dissipated is given by P= VI And V= IR (ohms law) Combining the two equations we get P = (IR) ×I P= I2 R The power dissipated by the load The current flowing in the load is; V= IR V = 6V maximum = 5.4 V minimum Current at maximum voltage Power dissipated at operating voltage = I2R = 0.00542×1000= 0.029160 watts ii) Power dissipated by the 300Ω resistor Voltage across the 300Ω resistor is Or V= 9.6V P= I2 R I = = 0.032 P= I2 R = 0.0322 ×300= 0.3072 watts. (iii)Power dissipated by the R2 resistor Voltage across the resistor = 5.4 V P= I2 R = 0.0242 × 220Ω = 0.12 watts Engineering design The range of E12 resistors available include; 100Ω, 120 Ω and so on The E12 series has a tolerance of 10% which means that the resistor values increases as shown below. The resistor can take values between 90 Ω and 110 Ω. So after 100 Ω resistor the next resistor is 120 Ω. Design. 1) a constant source of power is used to supply the circuit with power 2) the resistorR1 = 300 Ω can be obtained from the E12 series by Using the 330 Ω resistor which has a tolerance of 10% this means that 330-33= 297 so the 300 Ω resistors fall in the 330 Ω category. 3) the resistor 220 Ω resistor is available in the E12 family Power dissipation The resistors available for design have 0.25 watts rating. Checking if they are suitable for R1 and R2 we have; Resistor R1 =300 the calculated power dissipation is 0.3072 watts, which is much bigger than the 0.25 watts resistors available for the design. From the calculations, it can be seen that the increase in current causes the increase in power dissipated. By joining two resistors of value 560 Ω. We can get the 300 Ω. Two resistors are joined in parallel as shown in the figure below Fig 4: showing the two resistors in parallel The current will flow through two paths; hence the power dissipation in each of the resistors will reduce by half. Considering that the two resistors are 600 Ω resistors. We get P=I2R = 0.0162 × 600= 0.15 watts which is less than 0.25 watts (current considered is half since two parallel resistors are used.) For the resistor R2, the power dissipated P= I2 R = 0.0242 × 220Ω = 0.12 watts Which is lower than 0.25 watts, this means that the 220 Ω can be used directly. Design drawing The experimental design drawing is as shown in the figure below Fig 5 showing the design drawing without load The drawing showing the design plus load is shown in figure below Fig 6: showing the voltage divider connected to the load Design evaluation Experimentation To determine the voltage flowing between the different points and the current and hence the evaluation of the power rating the following experimental procedure was followed. Materials used Bread board 6 multi meter (100MΩ for voltage measurement, 1.000µΩ for measuring voltage) 3 resistors (560 Ω) the resistor colors (green, blue, brown, gold) 2 resistors (220Ω) the resistor colors (red, red, brown, gold) Procedure Each of the resistors was measured and its resistance determine. To determine this, the resistances were measured using a multi-meter. The reading of these resistances were recoded. The different components were connected as shown in the diagram below Fig 7 showing the connection of different components To determine the currents flowing when the load was connected, a 1000 resistor was connected in parallel to the 220Ω resistor. This is shown in the diagram below; Fig 8 showing the connection of different components with load included Results When connected, the results obtained were as shown in the diagram below Fig 9: Simulated results showing the voltages and current obtained when no load was connected The measurements taken are as shown in the table below Location of multi meter Type of measurement Results Current flowing through the 560 ohms resistor Current 15.0 milliamps Voltage across 560 Ω resistor Voltage 8.4 V Voltage across the load resistor Voltage 6.6 V Current flowing to the load Current 30 milliamps After connecting the load the following measurements were taken Fig 10: Simulated results showing the voltages and current obtained when load was connected As it can be seen from these simulated results the voltage across the load resistor was 5.876. This corresponds with the theoretical design. A summary of each multi-meter reading is given in the table shown below Location of multi meter Type of measurement Results Current flowing through the 560 ohms resistor Current 16.29 milliamps Voltage across 560 Ω resistor Voltage 9.124 V Voltage across the load resistor Voltage 5.876 V Current flowing to the load Current 5.876 milliamps Power dissipated The power dissipated across the resistor R2 is given by Current: 26.714 mA Power dissipated = I2 R = 26.7142 × 220Ω = 0.14 watts Power dissipated by R1 is given by Current: 16.29 mA Power dissipated= 16.292 × 560 = 0.148 watts References The E12 resistors series. [Online]. Available at http://oldradios.co.nz/downloads/The%20E12%20Resistor%20Series.pdf . Accessed 30 august 2009 Logwell. 2009. Standard EIA Decade Resistor Values Table. [Online]. Available at http://www.logwell.com/tech/components/resistor_values.html . Accessed 30 august 2009 Read More