just use the data (result) in the second file to complete the assignment ^_^'

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Electrical and Electronics Engineering

The Catholic university of America


just use the data (result) in the second file to complete the assignment ^_^'

The ASSIGNMENT is in the file(Laboratory 4 Assignment)

i will attach some pic of the lab that i did please to put in assignment you are making

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EXPERIMENT 5 HALF WAVE RECTIFIERS OBJECTIVE To study the characteristics and operation of half wave rectifiers and filter circuits. PROCEDURE 1) Connect the circuit in Fig. 1, using a 20 volt p-p square wave at 60 Hz as the input. The square wave should have a minimum value of -10 volts, and a maximum value of +10 volts. Trigger the scope on external, at a sweep speed of 2 msec/division, so as to observe slightly more than one cycle. Set both channels at 5 volts/division, dc, and place both probes on the input. Then, leave one probe on the input, and observe the output with the other probe. The circuit in Fig. 1 is the same as the circuit in Fig. 1 of Experiment 3, and your results should be very similar. When the input is at +10 volts, the output will also be at +10 volts, or, more exactly at +9.3 volts, allowing for the 0.7 volt drop across the diode. When the input is at 10 volts, the output will be at 0 volt, since the diode is reverse biased and does not conduct. The circuit is called a "half wave rectifier" since only one half, in this case the positive half, of the input appears at the output. 2) Connect the circuit in Fig. 2, using, first, the 1 µF capacitor. The positive half cycle at the output should look the same as before. However, during the negative half cycle, when the diode is reverse biased, the output doesn't drop immediately to 0 volts. Instead, the capacitor discharges through the resistor causing the output voltage to decay exponentially to 0 volts, with a time constant of 1 msec. Sketch the waveform at the output. - 20 - 3) Replace the 1 µF capacitor in the circuit in Fig. 2 with a 100 µF capacitor. Exactly the same things happen as before, but now the time constant, 100 msec, is so long that the output doesn't have a chance to decay to 0 volts before the next positive half cycle comes and brings the output back to 9.3 volts. How much does the voltage decrease during the negative half cycle? Sketch the waveform at the output. The amount by which the voltage decreases is called the "ripple." Many times it is desirable for the ripple to be as small as possible, for example in a power supply. Obviously, increasing the time constant will decrease the ripple. Based on your observations with this capacitor, how big a capacitor would be required to reduce the ripple to 0.1 volt? 4) Repeat parts 1 through 3 with a 20 volt p-p sine wave instead of a square wave. The ripple in part 3 should be larger than before, since there is more time available for the exponential decay, but otherwise the results should be very similar. Sketch the waveform at the output. Identify the region in which the output in Fig. 2 undergoes exponential decay. How large is the ripple with the 100 µF capacitor? 5) Connect the circuit in Fig. 3. This uses an additional filter stage to reduce the ripple. The 100 Ω resistor is a trade-off; the larger this resistor is, the greater the time constant, and therefore the greater the reduction in the ripple. However, the dc voltage drop across the resistor also becomes greater because of the current drawn by the 1 kΩ load. What is the ripple observed with this filter stage in place? Switch the scope to ac for this measurement only. This allows the gain to be increased enough to measure the ripple easily. How much does the 100 Ω resistor reduce the dc output voltage across the 1 kΩ load (short it out to see)? How much does the ripple increase when the 100 Ω resistor is shorted out? - 21 - Figures In Out 1k Ω Fig 1 In Volt age Ripp le Out 1 µF, 100 µF 1k Ω Tim e Fig 2 In 100 Ω 100 µF Out 100 µF 1 kΩ Fig 3 - 22 - OLI bal ties : FE RE LE FE EE E . : . TE EEEEEEEEEE 0.1 FUNCTION GENERATOR TTL .. ...... .... 6 PB1 오 26 041 NO DEBOUNCED PUSHBUTTONS 041 NO. C J1 13 S4 છI S6 S7 Sg LOGIC SWITCHES igital Storage cilloscope 50 1 2.00V/ 250V/ O.Os 5.000/ Trig'd A $ 1 -87.50 KEYSIGHT TECHNOLOGIES Acquisition Normal 1.00MSa/s DC DC Channels 1.00:1 1.00:1 Measurements: Freq[2): 60.010HZ Pk-Pk(2): 181mV Phase(1-2): 40.39° Pk-Pk(1): 5.1V 22 Waveform Generator Menu Sine DU.UZ 0.00VPP UU Back M KEYSIGHT InfiniiVision DSO-X 2002A Digital Storage Oscilloscope 70 MHz 2 GSa/s MEGA Zoo 50K wfm 1 2.00V/ 2100V/ O.Os 5.000/ Trig'd -75.0 KEYSIGHT TECHNOLOGIES Acquisition Normal 1.00MSa/s E DC DC Channels 1.00:1 1.00:1 Measurements: Freq(2): 60.002Hz Pk-Pk(2): 185mV Phase(1+2): 0.0° Pk-Pk(1): 5.1V 2011 Autoscale Menu Undo Autoscale Fast Debug All Normal S. M SNG 0372 Eeeeeeee EEEE Saas 28833 3383 38333 29013 19
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Explanation & Answer



Experiment 5
Electrical and Electronic engineering
Catholic University of America

Rectifiers are used to convert alternating current (AC) into direct current (DC). In the half wave
rectifier, the diode is forward biased resulting in a positive wave. The negative wave cycle come
in when the diode is reverse biased. For the positive half cycle, the input voltage is the same as
the output voltage. For the negative half cycle, the output is reduced to zero. The Name half
wave rectifier is because the positive part of the wave appears at the output. When a capacitor is
used, the negative half cycle is not decayed directly, an exponentia...

Excellent resource! Really helped me get the gist of things.


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