3.2 System Components and Circuit Design Based on the various reviews conducted on induction motor protection and the above block diagram which was conceived out of those literature reviews conducted, numbers of components are required in developing the protection system. 3.2.1 Power Supply Power supply is the circuit from which we get a desired dc voltage to run the other circuits. The voltage we get from the main line is 230V AC but the other components of our circuit require 5V DC. Hence a step-down transformer is used to get 12V AC which is later converted to 12V DC using a rectifier. The output of rectifier still contains some ripples even though it is a DC signal due to which it is called as Pulsating DC. To remove the ripples and obtain smoothed DC power filter circuits are used. `A 5V regulated supply is taken as followed: 230V 50Hz 5V DC Transfor mer Rectifier Smoothin g Regulator 5v DC AC Figure 3.2: Block diagram of regulated power supply system Transformer Transformer is the electrical device that converts one voltage to another with little loss of power. Most power supplies use a step-down transformer to reduce the dangerously high mains voltage to a safer low voltage. Here a step down transformer is used to get 12V AC from the supply i.e.230V AC. Rectifiers A rectifier is a circuit that converts AC signals to DC. A rectifier circuit is made using diodes. There are two types of rectifier circuits as Half-wave rectifier and Full-wave rectifier depending Upon the DC signal generated. Here Full-wave bridge rectifier is used to generate dc signal. Smoothing Smoothing is performed by a large value electrolytic capacitor connected across the DC supply to act as reservoir, supplying current to the output when the varying DC voltage from the rectifier is decreasing. The diagram shows the unsmoothed varying DC and the smoothed DC. The capacitor charges quickly to the 3V I (average forward current) =ILDC/2 = 35 mA IFRM (forward repetitive current) = IL/t2 (t1+t2) = 500mA Now from datasheet 1N4001 is chosen For 1N4001 IF (surge) = 30A So Rsurge = VP 14 V I surge = 30 A =0.5Ω Choose standard 1 Ω Designers recommend. It can also be used in circuits to get low DC voltage from high DC voltage.6 + 2 (0.5 (1.5 (1. Fixed output is obtained by connecting the voltage regulator at the output of the filtered DC.peak of the varying DC and then discharges as it supplies current to the output. Mathematical modeling for power supply circuit VREV (reverse voltage in diode) = VP – VF VP= VO max + 2VF = 12.6V V min = 12V – 0.7V = 13.2V) = 12.7) = 14V VREV = 14V – 0.1 * 12V V max = 12V + 0. Voltage regulation Fixed voltage regulator78xx.6V) = 11. produce fixed DC output voltage from variable DC (a small amount of AC on it). Here the capacitor of 470uF is used as a smoothing circuit.4V T= 1 50 HZ =50Sec . C2 = C3 = 0.1µF For proper output of IC7805 Rectification circuit Outputs 12V DC for 70mA load Let’s allow 10present ripple for the rectified output Then: .R ripple = 0. 2 V VR Use standard C1 = 470µF Assume silicon diode Vf = 0.−1 V min Ө1 = sin V max = 65o Ө2 = 90o – 65o =25o t2 = charging time = Ө2 360 *T=1. Vdc = 5v Vprimary = 230v and Vr = 10 percent of Vdc = 0.5v So C1 = (IL*t1)/Vr But IL= Vdc/RL = 5v/500 = 10mA and Discharging time (t1) = 0.7V.17msec T (period) = 1/50 So t1 = 15.t2= 8.6m sec Il 70 mA∗8.5T-t2 t2(charging time) = 1.65v For a full wave rectifiers V reverse = Vp = 6.5v = 310nF Diode selection Peak = Vmax + 2Vforward = 5.65v I foIFRM is repetitive surge current .6 m sec *t = =500 µF 1 1.7 = 6.4m sec t1 = discharging time = Then C1= T 2 .25 + 2*0.5msec C1 = (10mA*15.5msec)/0. t2= 8.65/10 = 0.5 mA From data sheet IFSM = 10 A So Rs = Vp/ IFSM = 6.6 m sec *t = =615 µF 1 1.4m sec t1 = discharging time = Then C1= T 2 .667 Ώ = 1 Ώ 12 Volt DC Power Supply V r = 0.2 V VR Use standard C1 = 1000µF Choose C2 = 10µF .5 + 1.1 * V dc V max = 14.17)/1.17 = 142.3 V Let IL=100mA −1 V min Ө1 = sin V max = 65o Ө2 = 90o – 65o =25o t2 = charging time = Ө2 360 *T=1.7 V V min = 13.6m sec Il 100 mA∗8.IFRM = (T*IL/2)t2 = 10(15. a USB connection. Instead. and amount of RAM and ROM on the chip. The Uno differs from all preceding boards in that it does not use the FTDI USB-to-serial driver chip.Figure 3. simply connect it to a computer with a USB cable or power it with an AC-to-DC adapter or battery to get started. . and a reset button. The Arduino Uno is a microcontroller board based on the ATmega328. It contains everything needed to support the microcontroller. which allows advanced users to reprogram it. it features the Atmega16U2 programmed as a USB-to-serial converter.3: Power supply circuit simulation 3. and universal synchronous asynchronous receiver transmitter (USART) functionality.2 Microcontroller The microcontroller IC which we used is Arduino Uno.2. power consumption. Therefore in order to achieve this task the Arduino mega microcontroller based on ATmega328 was chosen because of its suitability for this project such as speed. It has 20 digital input/output pins of which 6 can be used as PWM outputs and 6 can be used as analog inputs. a power jack. in built ADC. a 16 MHz resonator. This auxiliary microcontroller has its own USB boot loader. an in-circuit system programming (ICSP) header. GND (3): Short for ‘Ground’. They usually have black plastic ‘headers’ that allow you to just plug a wire right into the board. Most of the simple components used with the Arduino run happily off of 5 or 3.3 volts. and the 3. It’s not allowed to use a power supply greater than 20 Volts as you will overpower and thereby destroy the Arduino. each of which is labeled on the board and used for different functions. the 5V pin supplies 5 volts of power. . The Arduino UNO can be powered from a USB cable coming from your computer or a wall power supplies. The recommended voltage for most Arduino models is between 6 and 12 Volts. The pins on your Arduino are the places where it will connect wires to construct a circuit probably in conjunction with a breadboard and some wire. 5V (4) & 3.Figure 3.4: Arduino-Uno and pin configurations Power (USB) and pin configuration Every Arduino board needs a way to be connected to a power source.3V pin supplies 3. There are several GND pins on the Arduino. any of which can be used to ground your circuit.3V (5): As we might guess. The Arduino has several different kinds of pins.3 volts of power. It is sometimes used to set an external reference voltage (between 0 and 5 Volts) as the upper limit for the analog input pins. Digital (7): Across from the analog pins are the digital pins (0 through 13 on the UNO). thermistor and Temperature Detectors (RTDs). Temperature of Stator Windings and Speed of Operation. In this project a thermistor type LM35 temperature sensor is used due to the following main advantages: Lower cost Does not require any external calibration . but it can also be used for something called Pulse-Width Modulation (PWM).Analog (6): The area of pins under the ‘Analog In’ label (A0 through A5 on the UNO) is Analog In pins. 6. The most common type of temperature sensors are. thermocouple.3 Sensors and Transducer Unit This unit consists of several sensors and transducers used to detect the predetermined parameters of the induction machine. AREF (Stands for Analog Reference) (9): Most of the time you can leave this pin alone. PWM (8): the digital pins (3. we mainly monitor four parameters of induction machine that are Voltage. 9. 5. These pins act as normal digital pins. In this work. 10.2. Sl No Parameter Sensors Used 1 Phase Current Current Transformer 2 Phase Voltage Voltage Transformer 3 Winding temperature LM35 sensor 4 Rotor speed LM393 speed sensor LM35 temperature sensor Temperature is a common signal to be sensed. Current. These pins can read the signal from an analog sensor (like a temperature sensor) and convert it into a digital value that we can read. These pins can be used for both digital input (like telling if a button is pushed) and digital output (like powering an LED). and 11 on the UNO). 3. Linear output Precise and accurate output Figure 3. Output. For every 10mv. It can measure the temperature from -55° to +150°C range. The LM35 series are type of precision integrated-circuit temperature sensors. Input source.5: LM35 temperature As shown in the figure above the sensor has three terminals i. Ground iii. it ranges from 2. . the temperature value will be increase in 1 degree. The output of this sensor is linearly proportional the Celsius. The measured temperature from the sensor unit was displayed in LCD through the controller circuit.7V to 5V ii. analogue voltage output ranges from 201mV to 20v The temperature of the motor windings is measured using the LM35 temperature Sensor. This type of single phase rectifier uses four individual rectifying diodes connected in a closed loop “bridge” configuration to produce the desired output. thereby protecting it from damage. which periodically reverses direction. the microcontroller will detect under voltage fault and whenever the voltage is varied to 230VAC. and the relay trips the motor from the AC mains. consequently the microcontroller sends a trip signal to the relay. In this stage we use full wave bridge rectifier . to direct current (DC). whenever the voltage is varied to 200VAC. Rectification stage Rectifier is an electrical device that converts alternating current (AC). In this project the voltage can be varied by using the variable resistor and the output of the voltage monitoring circuit is fed to ADC converter. Here we are used 1N4007. The process is known as rectification and the stage is known as rectification stage.Figure 3. the microcontroller detects over voltage fault. . The voltage transformer will pass through rectification process before fed to the ADC.6: temperature sensor Voltage measurement The voltage given to the induction motor is measured using the voltage transformer with the transformation ratio of 220/5V. which flows in only one direction. The over voltage and under voltage protection circuit is capable of measuring and monitoring voltage from 200 to 230VAC. etc. This capacitor charges up when the voltage from the rectifier rises above that of the capacitor and then as the rectifier voltage falls. (4) Where: Rload = the overall resistance of the load for the supply C= Value of capacitor in Farads f= the ripple frequency this will be twice the line frequency a full wave rectifier is used shown in equation 5.. A current sensor is a device that detects and converts current to an easily measured output voltage. there are two types of current sensing: direct and indirect.……………………… (5) f = 2× 50=100Hz By rearrange equation (5) C ≫ 1 f × Rload C ≫ 1/100Hz×10K C ≫ 1uF For perfect smoothing purpose we take the capacitor value is 100uF Current measurement Current measurement is of vital importance in many power and instrumentation systems. Also. with the advancement in technology.placed across the output of the reciter and in parallel with the load. When a current flows through a wire or in a circuit. communications devices . However. voltage drop occurs. current and voltage regulators. programmable current sources. which is proportional to the current through the measured path. f = 2× line frequency…………. Both of these phenomena are made use of in the design of current sensors. current sensing has emerged as a method to monitor and enhance performance. automotive power electronics. the capacitor provides the required current from its stored charge. viz. Battery life indicators and chargers over-current protection and supervising circuits. Thus. ground fault detectors. while indirect sensing is based on Faraday’s and Ampere’s law. Knowing the amount of current being delivered to the load can be useful for wide variety of applications. current sensing was primarily for circuit protection and control. DC/DC converters. Current sensing is used in wide range of electronic systems. . Traditionally. a magnetic field is generated surrounding the current carrying conductor. The voltage transformation unit consists of diode.Filtering stage To smooth the output of the rectifier a reservoir capacitor is used . The output V1 of the transformer is fed to voltage transformation unit which transforms the input voltage into 5 volts range. motor speed controls and overload protection. linear and switch-mode power supplies. Rload×C ≫ 1� …………………………………………. Direct sensing is based on Ohm’s law. and resistive divider network. Generated magnetic field is then used to induce proportional voltage or current which is then transformed to a form suitable for measurement and/or control system. The burden resister is about 35ohm which if the standard resister for lower current value so the output voltage calculated using equation 2. without disconnecting the circuit to which they are attached. the maximum output voltage is 0.875v Therefore. and closed. The toroid current transformer type which have one turn primary winding (N1=1) and secondary winding 400turns (N = 400). The CT is a type of instrument transformer that is designed to produce an alternating current in its secondary winding which is proportional to the current being measured in its primary. Indirect Sensing involves measurement of the magnetic field surrounding a conductor through which current passes.875V and the maximum secondary current is 30mA. The current consumed by the induction motor is measured using the current transformer. Therefore I1=10A the output current is required to about 25mA therefore by using equation 1 the turn ration is calculated. the buzzer is used to indicate the overload condition of the motor. Current transformer with primary current 5Amps and rated secondary current of 1Amps is used.(1) = 10�/25�A=400 turns Some current transformers have a “split core” which allows it to be opened.Direct Sensing involves measuring the voltage drop associated with the current passing through passive electrical components. � I×R…………………………………………………(2) �= �×R �= 25�A/35ohm=0. CT reduce high voltage currents to a much lower value and provide a convenient way of safely monitoring the actual electrical current flowing in an AC transmission line using a standard ammeter. installed. If the voltage and current exceeds the rated value. . n = �1/ �2 …………………………………………. The measured voltage and current are displayed in LCD. We have to consider 60W load for our system so the current passing through the line is about 10A which is 60W 220V ac induction motor.