Multisim - Diode

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    BO CO TH NGHIM

    IN THC TNG TV NG DNG

    LAB 1

    DIODE CHARACTERISTICS

    NHM 9

    Trn Hng Giang 40900681

    inh Anh Bo 60900131Lm Ngc Qu 21102833

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    PART1: DIODE V-I CHARACTERISTICS

    1.1. Build the circuit shown in Fig. 1-1 using the 1N4149 diode and a 1k resistor. Vary

    V1 from 0 to 10V in appropriate intervals to obtain enough data points to plot the Forward

    Bias V-I Characteristic of the diode.

    1.2. Measure and record the voltage across the diode (VD) and the current (ID) for each

    data point.

    V1 (V) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5V(V) 0 0.486 0.573 0.593 0.604 0.612 0.618 0.623 0.627 0.630 0.633

    I (mA) 0 0.014 0.423 0.901 1.385 1.873 2.363 2.857 3.348 3.840 4.332

    V1 (V) 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10V (V) 0.633 0.636 0.639 0.641 0.643 0.645 0.647 0.649 0.650 0.652 0.653

    I (mA) 4.332 4.826 5.319 5.813 6.307 6.801 7.295 7.789 8.284 8.778 9.273

    1.3. In order to measure the Reverse Bias V-I Characteristic of the diode, swap the direction

    or polarity of the diode and replace R1 with a 1M resistor as shown in Fig. 1-2. Sweep the

    power supply from 0 to 20V in increments of 2V. Record the voltage drops across D1 and

    the current (ID) for each data point.

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    V1(V) 0 2 4 6 8 10 12 14 16 18 20V(V) 0 -2 -3.99 -5.99 -7.99 -9.99 -12 -14 -16 -18 -20I (nA) 0 -2.1 -4.09 -6.09 -8.09 -10.1 -12.1 -14.1 -16.1 -18.1 -20.1

    1.4. Use the measurements in 1.2 and 1.3 to sketch the V-I Characteristic of the diode.

    1.5. Derive the exponential model parameters IS and n for your diode based on two ormore of your data points.

    Sdng cng thc:

    .1

    D

    T

    V

    V n

    D SI I e

    ( 25 )T

    V mV

    Thay sliu tbng 1.2 vo cng thc trn, v gii hphng trnh (Bng Matlab) Tac:

    (VD1,ID1) = (645 , 680.1) (mV, mA)

    (VD2,ID2) = (647 , 729.5) (mV, mA)

    Ta tm c n = 1,532 vS

    I = 1,425.10-7A

    -0.002

    0

    0.002

    0.004

    0.006

    0.008

    0.01

    0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

    V-I

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    1.6. Compare the three diode models including, the constant voltage drop model, the

    piece-wise linear model, and the exponential model and answer the following questions.

    (1) Based on your data, what voltage would you use for a constant voltage drop model

    at a diode current of 5mA?

    M hnh in p ri l hng s, ng c tuyn ca diode c dng:

    Da vo bng d liu 1.2 ti id= 5mA , vd< 0.653 V nn ta chn vd= 0 mV cho mhnh in p ri l hng s.

    (2)For a piece-wise linear model, what values would you use for VD0 and rD, at a

    current of 5mA?

    M hnh piece-wise linear, ng c tuyn ca diode c dng:

    Da vo th 1.4 ta c VD0 = 610 mV.

    Ta c ng thng mu xanh c dc l1/rD

    Ly 2 im t bng 1.2

    (VD1,ID1) = (0,645 , 680.1) (V, mA)

    (VD2,ID2) = (0,647 , 729.5) (mV, mA)

    =,67,65

    729.568.1= 0,04(m)

    (3)How well do these models predict the diodes V-I behavior?

    Ta thy cc m hnh trn don gn chnh xc hnh dng thV-I.

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    (4) Are there any deviations from the model (e.g. at low currents or high currents)? What

    are the trade-offs involved with each model? Accuracy? Ease of use? Etc.

    Nhng m hnh ny c sai lch so vi thc t, mi m hnh c nhng u, nhc imring:

    - Vi m hnh in p ri l hng s, ssai lch rt ln, V cng ln th sai lch cngln. Nhng m hnh n gin, dsdng.

    -

    Vi m hnh tuyn tnh, ssai lch l tng i, V cng ln th sai lch cng b. Mhnh dsdng.

    - Vi m hnh e m, slch nh, nhng kh sdng do cng thc phc tp.

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    PART II: DIODE V-I CHARACTERISTICS (SWEEP METHOD)

    The sweep method can be used to easily display the V-I characteristic of a diode using

    an oscilloscope. Specialized instruments of this type are known as Curve Tracers. If a curve

    tracer is available, you most certainly can use one for this part of the experiment.

    2.1. Set up the circuit

    2.2 Display the V-I characteristic of the diode on the Oscilloscope by selecting the

    appropriate mode. Set Ch-1 to 200 mV/div and Ch-2 to 5 V/div or 10V/div so you will obtain

    an appropriate characteristic curve of the 1N4004 diode. Sketch the curve.

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    2.3 In the circuit of Figure 1-3, remove the diode and replace it with the 1N5231B,

    Zener diode. Connect the cathode to point A and the anode to point B. Repeat the

    above steps in Part 2.2 above, making oscilloscope adjustments if necessary to obtain an

    appropriate Zener characteristic curve. Sketch the waveform.

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    2.4 How are the V-I characteristics for the 1N4004 and 1N5231B similar? How do they

    differ? Explain.

    - Ging nhau: c tuyn ca Diode v Zener ging nhau khi phn cc thu., I = 0 khi0 Vz, I > 0 khi V

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    PART III: DIODE SIGNAL PROCESSING

    Clipping Circuit

    Often, an input signal may contain large voltage spikes that are too large for sensitive

    circuitry to process (e.g.. during a thunderstorm). In these cases, a signal clipping circuit

    can be employed to prevent the input signal voltage from exceeding a particular value. 3.1

    Build the diode circuit shown in Fig. 1-4.

    3.1. Set up the circuit

    Input and Output Waveforms

    3.2 Set up the oscilloscope to view VIN on Channel 1 (A) and VOUT on Channel 2 (B).Set the vertical scale to 2V/div for each channel. Set up the time base of the scope to showat least one full cycle of the sine wave.

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    Nhn xt:Ta thy in p ct tng dn khi ta tng gi trngun DC.

    C sthay i l do trong mch ny, Vct =VDC + 0,7

    3.5 Change the display format of the oscilloscope to XY. This will display channel 1(VIN) on the horizontal axis, and channel 2 (VOUT) on the vertical axis. The result will be

    a plot of the Voltage Transfer Characteristic. Sketch the characteristic curve.

    VDC= 2.5V VDC= 3V

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    3.6 Design a circuit to clip the signal at both +0.7Volts and -0.7Volts. Is a DC power

    supply necessary for this design? Explain.

    Ta thc hin mch nh sau:

    Kt qum phng:

    Vi mch trn ta thp 2 mch Diode mc thun v mc ngc mc song song. Khi ta c 2 gi trct:

    Vct1= VDC+ 0.7 v Vct2= -VDC- 0.7

    Do c mch ct ti 0.7 V v -0.7 V ta cho VDC= 0 tc l khng cn ngun DC.

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    PART IV: DIODE SWITCHING FREQUENCY LIMITATIONS

    Rectifying Circuit

    Diodes are often used in circuits to convert Alternating Current (AC) into Direct Current

    (DC). In these applications, the physical size (cross-sectional area) of the diode is increased

    to handle the higher current levels. Diodes used for these applications are called rectifiers.

    One trade-off for increasing the size of the diode is that maximum switching frequency of

    the diode is reduced. This part of the lab investigates the switching frequency limitation of

    a rectifier diode.

    4.1 Build the diode circuit shown in Fig. 1-5.

    Input and Output Waveforms

    4.2 Set up the oscilloscope to view VIN on Channel 1 (A) and VOUT on Channel 2 (B).

    Set the vertical scale to 2V/div for each channel. Also, set up the time base of the scope to

    show at least one full cycle of the sine wave. Sketch the resulting waveforms. Is the diode

    performing the rectifying function? Explain.

    Gii thch: Khi V1>0diode phn cc thun, chodng i qua R1, do

    VR1>0. Khi V1

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    4.3 Now increase the frequency of the signal from 100Hz to 1kHz, 10kHz, 100kHz, and

    1MHz. Adjust the time base of the scope accordingly to constantly view at least two complete

    cycles of the wave. Sketch the waveforms at each frequency. What happens to the output

    voltage as frequency is increased? At what frequency does the 1N4004 diode cease to

    operate as a rectifier?

    f = 100Hz

    f = 1kHz

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    f = 10kHz

    f = 1MHz

    f = 100kHz

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    Khi tn sf=2.5kHz th thbt u sai lch, diode ngng hnh ng nh mtchnh lu.

    4.4 Using the V-I characteristic of the diode 1N4149 obtained in part 1, determine iDand vD in the circuit of Fig. 1-5 graphically.

    Theo bi 1.6, ta tnh c RD, in p ri V0

    Khi 0.D D Dv i R V

    =>0D

    D

    D

    v Vi

    R

    Mt khc: 1 1D Rv v v

    => 1 1.D Dv v i R

    => 1 1.

    D

    D

    D

    vv v R

    R

    =>

    1

    1

    1 1

    .

    1

    D

    D

    D

    D

    v Rv v

    R R R

    R

    f = 2.5 kHz

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    =>

    1 0

    0 1

    01

    1

    .D

    D D

    D

    D D

    D

    D D

    Rv V

    v V R Ri

    R R

    Vvi

    R R R