EASA Module Title & No: Electronic Fundamentals - Module 4

Task No: 402

Student number ___________ Student name: ____________________________

Class ______________ Date completed _________________

Practice Assignment

Operating Characteristics of Diode Rectifies Theory;

Diodes commonly permits electron current stream in single direction. Exactly when type N semiconductor material is combined to type P material, a utilization zone is encircled near the crossing point. More voltage ability is required to go current across the consumption area of the diode crossing point.

The more voltage capability required at the depletion district of a diode semiconductor crossing point is the impediment voltage. The hindrance voltage for silicon diodes is about 0.6V (0.5V to 0.7V) and for germanium diodes is about 0.3V.The obstruction voltage is also known as the diode forward voltage drop (VF).

Exactly when a voltage that is negative is associated with the cathode of the diode, electrons in the type N components are compelled nearer to the intersection. Also, the charges that are positive in the P type components of the anode are pulled into the intersection by the extended charge that is negative over the hindrance.

Exactly a time where the associated voltage vanquishes the obstruction voltage, the exhaustion district width is diminished, and electrons move over the intersection to the terminal that is positive of the voltage source. For whatever period of time that the associated voltage outperforms the hindrance voltage, electron stream progresses and the diode is totally forward one-sided.

At the point when a voltage that is positive is associated with a negative voltage the cathode and to the anode, electrons in the type N material are pulled in a long way to the positive source voltage terminal from the junction. Positive charges in the type P component are furthermore pulled in a long way from the intersection toward the voltage terminal of the source that negative. These charge developments lead to augment the width of the consumption area, making the diode to be turn around one-sided (in its off state) with no present stream.

Just in case an AC voltage adequately colossal to conquer the hindrance current is associated with a diode, it leads the diode conducts variations when the voltage of the AC is in the forward one-sided bearing. The diode can't lead in the midst of movements when it is invert one-sided. The ensuing yield voltage is a throbbing DC voltage called swell that streams in a solitary heading or not in the least. The path toward changing over a half-cycle of the AC current to a vibrating DC current is half-wave correction (Figure 1).

53276576200

Figure 1

AC output is turned into a pulsating DC yield through half-wave correction. The circuit comprises of a load resistance and a diode. Contingent upon the manner in which the diode is combined with the circuit, either negative or positive pulsating DC yield can be formed.

Rectification results to the conversion of AC to DC. When transmission is just a single part of each AC cycle it is referred to as half-wave rectification. Because the forward voltage falls of the diode has to be achieved before conduction happens and also current must move in the load, there is reduction in the DC yield when compared to the AC yield.

Half-wave resistance will be seen on an oscilloscope. The oscilloscope estimates the voltage from the wavelength. The observed voltages are converted to RMS using a conversion factor as shown below.

For average voltage Vo(avg) = 0.318 x Vo(pk)

For RMS voltage Vo(rms) = 0.5 x Vo(pk)

The value measured will be more than the calculated value with a multi-meter since the conversion factor used is for full half-cycles.

A half-wave corrector leads alterations in the vibrating DC output also referred to as ripple.A pulsating DC signal is produced through the conversion of both negative and positive changings of an AC signal using a full-wave rectifier, as shown on Figure 2.

596900-22860

Figure 2

Full-wave rectification is carried out using a full-wave bridge rectifier whereby the input to the bridge corrector comes from the secondary transformer coil of a power. The transformer serves to decrease or increase the AC input to the bridge corrector.

The large wave of the DC vibration from a full-wave bridge corrector is reduced to a smoother DC indicator by a capacitor that is electrolytic across the output of the modifier. The capacitor filter transforms fast and discharges slowly to lower the rectifier DC output wave.

A voltage doubler is made up of two pairs of capacitors and diodes, which results to full-wave rectification. The DC output (VO) is the same as double the AC input peak voltage (Vpk). Capacitors are connected in series, the total charge in each capacitor is added giving the output.

Because there are two DC beats for one full cycle of the input AC waveform, the yield beat recurrence of a full-wave corrector is twofold the AC input recurrence..The following relationships apply to full-wave diode bridge rectifiers:

Peak output voltage (Vo(pk)) equals the peak input voltage (Vi(pk)) minus the forward voltage drop (VF) of the two conducting diodes.

Vo(pk) = Vi(pk) 2VF

Output rms voltage (Vo(rms) ) = 0.707 times the peak output voltage.

Vo(rms) = 0.707 x Vo(pk)

Output average (Vo(avg)) voltage = 0.636 times the peak output voltage.

Vo(avg) = 0.636 x Vo(pk)

EXPERIMENT 1

OBJECTIVE

To demonstrate how a diode functions as a half wave rectifier by using a typical half-wave rectifier circuit.

Verification of results with an oscilloscope and a digital multimeter.EQUIPMENT

FACET Base Unit, Semiconductor Devices Circuit Board, Oscilloscope (Dual Trace),

104775316230Generator (Sine Wave).Semiconductor BoardDigital OscilloscopeDMM/Function Generator

Figure 3

LAB PROCEDURE:

The Semiconductor Devices circuit board was inserted into the Base Unit (exercise with due care). The Diodes and Wave Rectification circuit block were located. The circuit was connected as shown on Figure 4 below.

27876524765

Figure 4

The function generator was used to select a sine wave signal with 1000Hz frequency. Then a two-wire probe was used to connect the sine wave signal from the function generator "Output-50" terminals to the oscilloscope Channel 1 and adjusted the Vp-p to 2V.

The rms and average voltage of the input signal were calculated.

157416553340

Vin(rms) (V) Vin (average) (v)

The sine wave signal from the function generator was connected to the GEN (generator) terminals of the Wave Rectification module on the circuit board as shown in Figure 5.

9328152540

Figure 5

The second two-wire probe was connected to Channel 2 of the oscilloscope while the other ends of the probe were connected across R2 (between CR2 and the ground).

The Volt/Div (vertical control) and Time/Div ( horizontal control) were adjusted for both channels to 500mV/Div and 500ms/Div respectively. Then the sine wave and the rectified signal both displayed on the scope were observed.

The waveforms for both Channels were sketched on the following graph :

V

548005-10795

Which channel shows the rectified DC output Ch1

Ch2 1.Why Channel 2 waveform has a gap at the start (OA), and the finish (BC) as shown in Figure 6.

2028190635

Figure 6

Ans:.............................................................................................................................................................................................................

The table below shows related to the voltage of Channel 1 and Channel 2.

144145163195

Channels Vpp(V) Vp(V) Vrms(V) Vmean(V)

CH1 CH2 144145-325120917575-3251202253615-3251203448050-325120144145-1676404353560-167640

2.Explain the difference between the peaks of Channel 1 and Channel 2

Ans:..............................................................................................................................................................................................................

Channel 2 was connected across R1; between CR1 and the ground. The waveforms for Channel 1 and Channel 2 were sketched on the following graph paper (Figure 7).

Note: If the rectified pulse of Channel 2 is negative, reverse the probe connections so that the rectified pulse is positive.

60325039370 V

3. Which pulse of the sine wave does diode CR2 rectify?2630170190526301701530352680970-482602917190-48260

Positive pulse negative pulse

4. Which pulse of the sine wave does diode CR1 rectify?2630170190526301701530352680970-482602917190-48260

Positive pulse negative pulse

The oscilloscope and signal generator were turned off and all circuit board connections removed.

EXPERIMENT 2

OBJECTIVE

To demonstrate full -wave rectification and filtering by using calculated and measured circuit conditions.EQUIPMENT

FACET Base Unit, Semiconductor Devices Circuit Board, Oscilloscope (Dual Trace),

Generator (Sine Wave)

PROCEDURE:

The Semiconductor Devices circuit board were inserted into the Base Unit.

The Full Wave Rectification with Power Supply Filter circuit block was located and connected the circuit shown below in Figure 8.

318770166370

Figure 8

What is the purpose of R1?

Ans:.............................................................................................Select a sine wave from the function generator with a frequency of 100 Hz.

The two-wire probe was connected to Channel 1 of the oscilloscope and adjusted the sine wave Vp-p to 20 Volts. Then the sine wave signal from the function generator was connected to the GEN (generator) terminals of the Full Wave Rectification module. Terminal A, terminal B and the load resistor R1 were jointed as shown in Figure 8 above.

Channel 2 of the oscilloscope was connected to a two-wire probe and displayed the secondary voltage of the transformer. Silver poles were used to hook the probe terminals.

What is the secondary voltage peak to peak Vpp =.......... (V). Is the transformer

2107565571521075651492252156460-425452385060-42545a step up or step down

The probe of channel 2 was connected at the +ve and -ve terminals of the bridge (across the resistor R1). Silver poles or additional link wires were used if needed.

82740516319582740589281082740516205208274052106295The output waveform was sketched on the diagram below.

-571542545

The Full Wave bridge rectifier output voltage were observed.

22967951720852296795285752345055-203202573655-20320Is it a smooth DC output ?Is it a pulsed DC output ?

2289175361952289175-1060452337435-1549402567940-154940Results filled in the table below:

Channel Vpp(V) Vp(V) Vrms(V) Vmean(V)

CH2

The measured DC voltage or Vmean (or VDC) across R1 were compared with the theoretical value :VR1(mean) = 0.636 x VR1(pk) - 2(VD)= .........................

Explain the differences (if there are any).

............................................................................................................

............................................................................................................

1339850494030Only two diodes are forward biased (conduct the AC pulse ) at each half cycle of the AC waveform.

Figure 9

Which two diodes conduct during the positive half cycle?

......... and ...... Which two diodes conduct during the negative half cycle?

......... and ...... What is the approximate forward bias voltage drop across the diode being measured? Vd:

1448435-8890To smooth out the output voltage of the fullwave rectifier, the RC filter was conncetd as shown in Figure 10. Both channels were skecthed on the graph paper shown below.

Note: Keep Channel 1 probe connected to the transformer primary while Channel 2 probe is connected across R1.

209550168275

Figure

Is the DC output voltage across R1 smother with RC filter? Y/N 2473960199517033274001209675375602512096753967480120967541821101209675766445199517014065251995170162052019951702260600199517031140401209675290131519951703114040199517033274001995170375602519951703967480199517076644522580607664452781935766445...