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Power Electronics

12/26/2013 Muhammad Gulraiz Ahmed Roll No : 10-el-15 Power Electronics Labs Submitted to : Sir Ali Abbas
Lab Assignment No 4 Objective: “Obtain maximum undistorted output of voltage AMP, identify cross over distribution and effect of frequency variations on current booster circuit.” Theory: The bi-directional drive of a complementary pair makes it ideally suited for an alternating drive input. This does not need to come from the main supply, or from a high power source, since a voltage amplifier can be used to increase the voltage to the required level, and then the current booster circuit will supply the necessary current to give the power in the load. Apparatus: D-3000 trainer Connecting wires Ammeter Voltmeter Shorting links and connecting leads Circuit Diagram Lab trainer circuit: Procedure : Connect shorting links between the signal generator output & socket 7.1 and between 11.3 and 11.4, and leads between 7.4 and 3.1 and 3.6 and 11.2. Set the voltage Amp. Gain control and the Amp. Offset control to mid-position. Switch On the module Set the signal generator to 1 kHz sine wave. Vary its magnitude noting the effect on lamp Set up the oscilloscope Adjust Voltage Amp. Gain control Adjust Voltage Amp. Offset control Waveforms: Input waveform output of booster Observations: 3v p-p Input = 11.3 v 2.Maximum undistorted Voltage Amplifier output = 7.9 v 3. VRMS = Conclusion: I concluded from this experiment that distortion occur in the middle of current booster output waveform is called crossover distortion and is caused by the non linearity of the transfer characteristics on either side of 0 V.
Lab Experiment No 2 Objective: “To find out the power dissipated in power transistor.” Theory: A transistor is a semiconductor device used to amplify and switch electronic signals and power. It is composed of a semiconductor material with at least three terminals for connection to an external circuit. A voltage or current applied to one pair of the transistor's terminals changes the current flowing through another pair of terminals. Because the controlled (output) power can be much more than the controlling (input) power, a transistor can amplify a signal. Today, some transistors are packaged individually, but many more are found embedded in integrated circuits. Power Dissipation: When a transistor conducts current between collector and emitter, it also drops voltage between those two points. At any given time, the power dissipated by a transistor is equal to the product (multiplication) of collector current and collector-emitter voltage. Just like resistors, transistors are rated for how many watts each can safely dissipate without sustaining damage. High temperature is the mortal enemy of all semiconductor devices, and bipolar transistors tend to be more susceptible to thermal damage than most. Power ratings are always referenced to the temperature of ambient (surrounding) air. When transistors are to be used in hotter environments (>25o, their power ratings must be de rated to avoid a shortened service life. Apparatus: D-3000 trainer Connecting wires Ammeter Voltmeter Procedure : Connect shorting links between sockets 1.1 and 2.1 and 11.1 and 11.2. Connect multineter 1 DC current range to 11.3 and 2.5 to monitor the collector current of TR1. Switch ON the module . Adjust the setting of potentiometer VR1 to give zero collector current Ic ,noting the value of collector voltage Vce and record in table. Repeat the reading of collector voltage Vce for each of the collector currents given in table. Calculate the power dissipated in transistor pt=Vce*ic and resistor lamp Rl=12-Vce/ic. Plot the graphs . Observation And Calculations: Ic (mA) VCE (V) Pt= (Ic*Vce) RL (Ω) 50 11.2 0.56 16 100 10.35 1.035 16.5 150 9.2 1.380 18.6 Conclusion : I concluded from this experiment that when BJT is operated as a digital switch very little power is ever dissipated in the transistor. When off current = 0, power is zero or when on (voltage is very low) power dissipated is very low. This thing can easily be understood by the graph as shown above the graph shows maximum power dissipation at some medium values at peak values or at low values of Ic power dissipation is very low.
Lab Experiment No : 1 Objective: “Determine by calculation or measurement the base-emitter voltage, current gain and input impedance of power transistor.” Theory: The current gain of a bipolar junction transistor (BJT) is defined as the ratio of output current to the input current. β = IC / IB We tend to think of this current gain and the base-emitter voltage (0.6V for silicon ) being constant values for a particular transistor .As soon as you start to look into higher power applications of the bipolar junction transistors you find that this is not quite true . Two multimeters both on DC current range will be used to measure base and collector currents at medium and high values to investigate the effect on β in high power applications . One of the transistor will be then be re-configured to measure the input voltage to allow the calculation of the input impedance ,which should also be found to be surprisingly low. Apparatus: D-3000 trainer Connecting wires Ammeter Voltmeter Procedure : On the trainer connect a shorting link between sockets 11.1 & 11.2 . Connect multimeter 1 on DC current range to sockets 11.3 ( positive ) and 2.5 (common) to monitor the collector current of TR1, and multimeter 2 also on DC current range to sockets 1.2 (positive) and 2.2 (common) to monitor the base current. Switch on the module power supplies. Vary the setting of potentiometer VR1 and note that this controls the brightness of the lamp. Turn VR1 fully clockwise to set the base voltage of TR1 maximum positive .Note the values of collector current (IC) and base current (IB) flowing. Turn VR1 counter-clockwise to find the minimum value of base current to maintain the transistor in saturation (maximum collector current). Observation Table: IB (µA) IC (mA) VBE (V) β Input imp (KΩ) 40.2 4.02 0.81 100 20.149 63.6 6.38 0.82 104.2 12.89 84.6 8.65 0.83 102.12 9.81 103 10.3 0.83 100 8.05 120 12 0.84 100 7 Conclusion: I concluded from this experiment that the collector current and the base emitter voltage of the BJT is depends on the base current hence it is clearly seen in the table that when base current is increase in µA then collector current also increase in mA.
Lab Assignment No 6 Objective: “Calculate voltage gain of Audio Power Amplifier Stages. Perform power and load impedance calculations.” Theory: The simplest way of measuring the power output of an audio amplifier unit is to monitor the voltage across the loud speaker, the impedance of which will be known as an is typically 8 ohm. The output power can then be calculated from the formula Power = (Vrms)2/ R If the voltage across the load is measured using an oscilloscope, the result will be in Vp-p .so it has to be converted into Vrms . Vrms = Vp-p / 2 √2 The lamp will be used as a dummy load, since it is capable of dissipating up to 4W. at the available alternating voltages . To calculate power, we attach an ammeter in series with the load in order to measure the load current IL the power in the load can then be calculated from Power =Vrms x IL. Circuit Diagram: Procedure: I set up the oscilloscope as follows. Time base to 0.2 ms/div Locate the CH1 toward up and CH2 towards down. CH1 Y amplifier gain to 0.5 /div, AC input. CH2 Y amplifier gain to 5V/div, AC input. Set amplifier GAIN CONTROL to maximum (fully clockwise ) and the AMPLIFIER OFFSET control to the mid-point (arrow pointing upward ) Switch on the module power supply. Set the signal generator to 1 KHz sine wave Adjust the OFFSET control to remove distortions If the instability at the peak of the negative half-cycles is a problem ,then connect the 100 nF capacitor C4 across the buffer output as follow Connect the short link between 15.7 and the ground return on the oscilloscope patching panel and the lead between sockets 8.5 & 15. Repeat the Offset adjustment as to reduce the signal generator amplitude control unit and I obtain the max undistorted output waveform. I read the value of output current from ammeter I calculated the value of load impedance =output voltage/ output current. The power in dependent on load impedance Power = V2 / R Observations and Calculations: f=1k V1=3V V2=21V V3=21V V4=10V Amplifier gain=21V/3V=7 Buffer gain=21V/21V=1 Booster gain=10V/21V=0.49 Overall gain=10V/3V=3.33 Vrms =10/2squrt*2=3.53V Io=1.4mA P=VI=4.94w
Lab Assignment No 5 Objective: “Recognize the operation of Audio Amplifier. Recognize wave form in an audio amplifier high output current and output voltage wave form.” Theory: The task of an audio amplifier is to take a small signal and make it bigger without making any other changes in it. This is a demanding task, because a musical sound usually contains several frequencies, all of which must be amplified by the same factor to avoid changing the waveform and hence the quality of the sound. An amplifier which multiplies the amplitudes of all frequencies by the same factor is said to be linear. Departures from linearity lead to various types of distortions. The operational details of amplifiers are buried in the field of electronics, but for audio purposes it is usually safe to say that current commercial audio amplifiers are so good that a normally operating amplifier is seldom the limitation on the fidelity of a sound reproduction system. One must be sure that the amplifier can provide enough power to drive the existing loudspeakers, but otherwise amplifiers are typically one of sound reproduction system. One must be sure that the amplifier can provide enough power to drive the existing loudspeakers, but otherwise amplifiers are typically one of the most trouble-free elements of a sound system. Circuit Diagram: Procedure: Set up the oscilloscope as follows. Time base to 0.2 ms/div Locate the CH1 toward up and CH2 towards down. CH1 Y amplifier gain to 0.5 /div, AC input. CH2 Y amplifier gain to 5V/div, AC input. Connect short links between the signal generator output and socket input 7.1 and between socket 7.4 & 8.1 and 11.3 &11.4 and leads between sockets 8.3 &3.7 ,8.4 and 3.1 and 3.5 & 11.2. Connect CH1 to socket 7.2 to monitor the voltage amplifier input, and CH@ to socket 3.5 to monitor the output to load. Set amplifier GAIN CONTROL to maximum (fully clockwise ) and the AMPLIFIER OFFSET control to the mid-point (arrow pointing upward ) Switch on the module power supply. Set the signal generator to 1 KHz sine wave Adjust the OFFSET control to remove distortions Transfer CH2 to socket 8.2 to monitor voltage amplifier output/buffer input . Observations and Calculations: Srsrxs Sr No. Nno Sr.N0 Collector current IC Collector-emitter voltage VCE Power dissipation PT Load resistance RL 1 10 11.54 0.1204 7.07 2 15 11.56 0.180 0 3 60 12.00 0.689 0 4 65 11.49 0.750 8.5 Waveform:
Lab Assignment No 3 Objective: “Recognize the operation of a complementary Darlington pair Emitter Follower circuit and plot the transfer characteristics, and deduce the base emitter turn on voltage.” Theory: The Darlington Pair configuration has two advantages: Very high current gain Very high input impedance The high current gain is achieved by taking the output current from the first transistor and coupling it directly as the input current of a second transistor to be amplified again. With very high current gain, the input current to TR1 must be very small. This means that the input impedance is very high, since the applied voltage produces such a very small current. When the input voltage is taken positive TR2 conducts and the current flows done through the load to ground. When the input is negative, TR3 conducts and the current flows upward from ground, towards the -12V supply. Apparatus: D-3000 trainer Connecting wires Ammeter Voltmeter Shorting links and connecting leads Lab trainer circuit: Circuit Diagram Procedure : Connect a shorting link between socket 11.3 and 11.4, and leads between socket 1.1 and 3.1 and 3.6 and 11.2. Connect multimeter 1, on dc voltage range to socket 1.2(+ve) and any convenient ground socket (common) to measure V1, the input voltage to the transistors. Connect Multi meter 2, similarly to 3.7 and another convenient ground socket to measure V2, the output voltage feeding the load. Switch ON the module power Supplies. Vary the settings of VR1 and note that the lamp brightness can be controlled through its range as the input voltage is taken either +ve or –ve. Use VR1 to set the input voltage V1 to -10v Record V2 and place values in observation table. Observations & Calculations: INPUT VOLTAGE(V1) OUTPUT VOLTAGE(V2) INPUT VOLTAGE(V1’) OUTPUT VOLTAGE(V2’) -10v -3.3mV 0v -0.2V -8v -2.6mV +1v 0.0019V -6v -1.9mV +2v 0.195V -4v -1.2mV +4v 0.757V -2v -0.5mV +6v 2.14V -1v -0.1mV +8v 4.49V 0v -0.2mv +10v 7V Conclusion: I concluded from this experiment that when the high current gain is achieved by taking the output current from the first transistor and coupling it directly as the input current of a second transistor to be amplified again. When the input voltage is taken positive TR2 conducts and the current flows done through the load to ground. When the input is negative, TR3 conducts and the current flows upward from ground, towards the -12V supply.
Lab Assignment No 7 Objective: “Recognize the operation of pulse width modulator circuit. Determined ON and OFF time for PWM .Determine Duty cycle frequency of PWM.” Theory: Pulse-width modulation, or pulse-duration modulation, is a modulation technique that confirms the width of the pulse, formally the pulse duration, based on modulator signal information. The PWM is actually a square wave modulated. This modulation infects on the frequency (clock cycle) and the duty cycle of the signal. Both of those parameters will be explained in details later but keep in mind that a PWM signal is characterized from the duty clock and the duty cycle. The amplitude of the signal remains stable during time (except of course from the rising and falling ramps). The clock cycle is measured in Hz and the duty cycle is measured in hundred percent (%).These are the basic parameters that characterizes a PWM signal. The first parameter, the clock cycle, is easy to understand. It is the frequency of the signal measured in Hz. The other parameter has to do with the switching time of the signal. Take a look in the following three signals: All three signals shown above are square wave oscillations modulated as per their oscillation width, so called "duty cycle". They have the same frequency (t1), but they differ on the width of the positive state (t2). The duty cycle is the percentage of the positive state compared to the period of the signal. Apparatus: Oscilloscope. Multi-meters. Shorting links and connecting leads. D3000-2.3 power electronic devices-1 module. Signal generator. Circuit Diagram: Procedure: First I connect the circuit as shown in the Figure. Set up the oscilloscope as follows. Time base to 50 ms/div Single trace CH1 operation. CH1 Y amplifier gain to 1v /div, DC input. Then we set the VR1 to the center Position to give the input voltage 0 volt and rate control fully counter clock-wise to set the switching frequency to minimum. Then I connect the multi meter on socket 11.3 and 2.5 to monitor the load current. Now, Switch on the module Power Supply. Rotate the VR1 clock-wise and note the effect of Lamp Brightness. Then I check the reading of current which is giving the ON OFF value. The average current gradually increases as the Duty Cycle is increased. Connect channel of oscilloscope to socket 9.3 to check the output of the PWM. Adjust the VR1 for an approximate duty cycle of 0.5. Slowly increase the setting of rate control to increase the switching frequency, gradually increasing the time based speed. Observe the effect of Lamp response and indicated load current. Observations And Calculations: T = 34 msec F= 1 34 x 10̅ ᵌ =30Hz