Electronics Lab Experimentals

Kurdistan Iraqi Region Ministry of Higher Education Sulaimani University College of Science Physics Department

Electronic Lab. Experimentals Prepared by Dr. Dlear Rafic Saber Dr. Omed Ghareb Abdullah

2007 – 2008

Exp. (1) Characteristics of Ordinary Crystal Diode Theory: The term of diode refer to an electronic device having two external electrodes to which electrical connections may be made. Such an arrangement is called a P-N junction diode. It includes the boundary or P-N junction and some of N-type and of P-type semiconductor material on each side of the boundary. If a voltage is applied to the diode as shown in figure (1), holes in Pregion move across the junction into N-region and electron in N-region move across the junction into P-region. External voltage that applied to the diode with the polarity positive side to side and, negative to N-side is called forward biasing. Forward biasing reduces the strength of barrier potential. The result is a current (flow of both electrons and holes). That is mach greater than that the resistance of the germanium (pure semiconductor) alone would allow. When the positive terminal of applied voltage is connected to N-side and negative terminal to the P-side the diode is said to be reverse biased. The barrier region contains no free carriers and it has higher resistance than remainder of crystal on either side of the barrier. In fact, the P- and Nregions have such comparatively low resistance that they act much like metal contacts between the external connecting leads and barrier at the common boundary. The external voltage source attracts electrons and holes away from the junction. This result increases the potential charge carriers must overcome in traversing the P-N junction. -

+ ++++++

---------------------

P ++++++ - - - - - - N ++++++ ++++++ ++++++

R

+

Fig.(1): Forward biased P-N junction.

-

Vf Barrier

+ Region + + + + + +P + + + + + ++ + + + + ++ + + + + ++ + + + + +Negative ions

+++++-

--------- N ------------Positive ions

-

+

Vr

Fig.(2): Reverse biased P-N junction.

Procedure: Forward: 1- Connect the circuit as shown in Fig. (3). 2- Fix the voltage V f = 0 volt, using the potentiometer ( P1 ). 3- Increase the voltage in step (0.1 in each step) and note the current ( I f ) and record it for each step. 4- Draw a graph of ( I f ) against V f . 5- Determine threshold voltage Vt at which current starts through diode. 6- Calculate dynamic resistance r = ∆V ∆I at two different voltage values.

2 KΩ

mA

220 Ω

+15 V

IN4007

P1

Vf

Fig.(3) Reverse: 1- Connect the circuit as shown in Fig. (4). 2- Fix the voltage Vr = 0 volt, using the potentiometer ( P1 ). 3- Increase reverse voltage Vr in step (0.2 in each step) and record current ( I r ) for each step. 4- Draw a graph of ( I r ) against Vr . IN4007

0 150V

µA

Vr

Fig.(4) Discussion: 1- Why the forward current is very small in the beginning, and increases so fast when the voltage reaches to a certain value? 2- In forward state the voltage reaches to a fixed value explain why? Does this value of voltage is same for all type of semiconductor diode? 3- Does the current flow in reveres state? Why? Explain the advantage and disadvantage of this phenomenon?

Exp. (2) Characteristics of Zener Diode Theory: The P-N junction diode normally dose not conduct when a revers biased is applied. If the revered biased voltage is increased up to particular voltage it starts to conduct heavily. This voltage is called breakdown voltage. High current through the diode can permanently damage it. To avoid high current, we connect a resister in series with the diode in the circuit. This resistance is extremely important and its value is quite critical. Reverse voltage breakdown must be taken into account in practical application of diode. One type of breakdown process that can supply carriers to support conduction when the voltage across the junction is large, called the Zener effect. It arises when the barrier potential becomes so large that electrons may be ripped from covalent in the barrier region. Thus, electron-hole pairs are created in the ion layers at the junction. The hole moves quickly into the P-region under the influence of the electric field in the barrier region and electron is swept into the N-region. These moving charges constitute an increase in reverse current in the junction. When the electric field is strong enough to break a single covalent bond, it is sufficiently strong to break many, once the Zener effect has started. The Zener effect occurs when the diode is heavily doped and the breakdown occurs at a low voltage about 10 volts or so. When the breakdown occurs in lightly doped crystals or at high voltages, the mechanism is something other than the Zener effect. Operation in the breakdown region does no damage to the diode, provided the current is limited to the value set by the manufacturer. Procedure: 1- Connect the circuits as shown in Fig. (1), fix the potentiometer ( P1 ) at maximum right side. 100 Ω

+15 V

2 KΩ

P1

Fig.(1)

mA

ZF4.7

V

2- Increase slowly the supply voltage using ( P1 ), and record the voltage and current, (Note that these values are negative). Once breakdown occurs V remain constant even though I increases. 3- Tabulate your data in table. 4- Replace the Zener diode ZF4.7 with ZF6.2, and then repeat steps (1-3). 5- Draw graphs for both cases between ( I ) and ( V ) as I = f (V ) , on the same paper. 6- Calculate the Dynamic resistance in the following current ranges, ⎯→ 20 mA ) and, ( 20 ⎯⎯ ⎯ ⎯→ 25 mA ). ( 5 ⎯⎯ ⎯ The students can use any ranges of current for determining the Dynamic resistance. Discussion: 1- What do you understand from the graphs? 2- What is we meaning by the Dynamic resistance? 3- As we know that the Zener diode is used for voltage rectification, what is the basic factor which controls the rectification process from your results? Table (1): V (volt) I (mA)

Exp. (3) Half Wave Rectification and Filtration

Volts

Volts

Theory: The characteristic of an ideal diode shows that it carries forward current, its voltage drop is zero regardless of how large the current may become. Thus an ideal diode behaves as a short circuit to current in the forward direction. However, it behaves as an open circuit when the voltage drop across the diode is negative. No reverse current passes through the diode regardless of the magnitude of the reverse voltage. When an alternating voltage Fig.(1A) is applied to a crystal diode connected in series with a resistor, the diode acts as a half-wave rectifier by current flow only the positive half of each alternating voltage cycle. During the negative half of each cycle the diode act in reverse region and the current can not pass through the diode. Thus only the positive half wave of the input wave is rectified as shown in Fig.(1B), so these circuits are used to converts an AC voltage in to DC voltage.

Time

Time 1 Cycle

1 Cycle

Fig(1B): Output (Rectification)

Volts

Fig.(1A): Input Voltage

Time 1 Cycle

Fig(1C): Output (Rectification & Filtering)

Procedure: 1- Connect the circuits as shown in Fig.(2). In the beginning replace the resistor RL and capacitor C from circuits.

IN4007 +

R

C

Vrms=10 V

VDC

CRO

Fig.(2) 2- Use the first channel (CH1) of the oscilloscope to obtain the input voltage, and draw it. 3- Draw the output voltage as a function of time, on graph papers, and record V pp , and V DC . (Note: to measure the peak to peak voltage V pp of input voltage, put the channel of the oscilloscope on AC, and select the suitable scale for measuring, while in the case of output voltage put the channel of the oscilloscope on DC). 4- Connect RL and C with the following values, and repeat step (3). R L (Ω )

C ( µF )

100

10

1000

10

100

100

1000

100

VDC (volt )

V pp (volt )

5- Calculate ripple r factor for all previous cases. 1 1 r= 2 3 fCR L Discussion: 1- Discuss briefly the role of diode, RL , and C in the circuit. 2- Find the efficient of the circuit. 3- How can you estimate the value of V DC from figure (without using voltmeters)?

Exp. (4) Full Wave Rectification and Filtration

Volts

Volts

Theory: The circuit in Fig.(1) is called full wave rectification circuits because it utilizes the full wave of input voltages to supply to the load. Each diode is forward biased on an alternate half cycle of the input voltage, and each conducts current through the load in the same direction. Thus the full wave of input voltage is rectified and produces two half sine waves of load voltages and current for each cycle of input sine wave.

Time

Time 1 Cycle

1 Cycle

Fig(1B): Output (Rectification)

Volts

Fig.(1A): Input Voltage

Time 1 Cycle

Fig(1C): Output (Rectification & Filtering)

Procedure: 1- Connect the circuits as shown in Fig.(2). In the beginning replace the resistor RL and capacitor C from circuits.

IN4007 12 V Vin (1) + C

6V

R

VDC

CRO

Vin (2) IN4007 0V

Fig.(2) 2- Use the first channel (CH1) of the oscilloscope to obtain the input voltage, and draw it. 3- Draw the output voltage as a function of time, on graph papers, and record V pp , and V DC . (Note: use oscilloscope to measure the peak to peak voltage V pp , and voltmeter to measure V DC ). 4- Connect RL and C with the following values, and repeat step (3). R L (Ω )

C ( µF )

100

10

1000

10

100

100

1000

100

VDC (volt )

V pp (volt )

Discussion: 1- Discuss briefly the role of two diodes, RL , and C in the circuit. 2- Compare the results of this experiment with previous experiment (Half Wave Rectification and Filtration), in terms V pp , V DC , and the frequency of output value. 3- Find the ripple factor. 4- Calculate the efficiency of the circuit.

Exp. (5) Full Wave Rectification and Filtration by using Bridge Rectification Theory: Four diodes may be connected in bridge circuit as shown in Fig.(1). This type is widely used because it does not require center tap, although it produce full wave rectification.

Procedure: 1- Connect the circuits as shown in Fig.(1). In the beginning remove the resistor RL and capacitor C from circuits.

6 Vrms

D4

D1

D2

D3 + C

R

VDC

CRO

Fig.(1) 2- Use the first channel (CH1) of the oscilloscope to obtain the input voltage, and draw it. 3- Draw the output voltage as a function of time, on graph papers, and record V pp , and V DC . (Note: use oscilloscope to measure the peak to peak voltage V pp , and voltmeter to measure V DC ). 4- Connect RL and C with the following values, and repeat step (3).

R L (Ω )

C ( µF )

100

10

1000

10

100

100

1000

100

VDC (volt )

V pp (volt )

Discussion: 1- Discuss briefly the role of two diodes D1 and D2 during the first half of the wave, and role of two diodes D3 and D4 during the second half of the wave. 2- Compare between the most important differences of this experiment and the previous experiment (Full Wave Rectification and Filtration), in terms V pp , V DC , and the frequency of output value. (Note that the Vrms is the same in both experiments). 3- Explain, the bridge rectifier is better than the full wave rectification using two diodes.

Exp. (6) Diode as a Voltage Limiter (Clipper) Theory: The diode is a useful circuit element in applications requiring voltage levels to remain with prescribed amplitude levels. A circuit which uses a diode as a voltage limiter is illustrated in Fig.(1). This circuit is designed to prevent the output voltage from exceeding the value of the DC source voltage marked E . The circuit operation is easy to understand, when the diode is reverse biased by external source no current can pass through the diode and the output voltage will follow any variation of the external source. However, when this value reaches the amplitude of E1 + V f , the diode will be forward biased. Current can then pass through the diode, and the output voltage will be equal to E1 + V f during periods of forward biasing. If the internal voltage drop of the diode is very small compared to the DC source, then the output voltage will be practically limited to the value of E1 as required. It is sometime necessary to limit voltage variations in both positive and negative ranges about some zero reference voltage. The additional circuitry and wave form of the output voltage are shown in Figs.(2-3). The diode limiter might be used to alter the wave form of a sinusoidal as the input voltage. If E1 and E2 are chosen equal magnitude and much smaller than the input amplitude, then the output will be very nearly a square wave of voltage. Procedure: PART 1: Clipping the positive part of the sine wave: 1- Connect the circuits as shown in Fig.(1). 2- Let Vin = 10 V pp and plot it on a graph paper. Use the same signal for all other parts. 3- Take E = 2 volt , then plot the output voltage. 220 Ω IN4007 CRO

10 Vpp E=2V

Fig.(1)

PART 2: Clipping the negative part of the input wave: 1- Connect the circuits as shown in Fig.(2). 2- Repeat the step (3) in the first part. 220 Ω IN4007 Fig.(2)

CRO

10 Vpp E=2V

PART 3: Clipping the negative and positive parts of the input wave: 1- Connect the circuits as shown in Fig.(3). 2- Plot output voltage for; E1 = E2 = 3volt , and E1 = E 2 = 0 220 Ω D1

D2 CRO

10 Vpp E=2V

Fig.(3)

E=2V

Discussion: 1- Discuss and analyze all circuits operation in Figs.(4-8). 2- Estimate the output voltage in first and second cases, when E = 0 . 3- What is the signification of the clipping circuits?

Exp. (7) Diode as a Voltage Clamper Theory: In many electronics applications, it is necessary or desirable to eliminate either all positive or all negative amplitudes of input voltage wave form. The circuit which can perform this alteration of input wave form is called a clamper circuit. The clamping circuit can be define as that circuit which clamps the out voltage at a fixed level or to restore DC component, which is not present in the input signal. The output voltage in clamping circuit is similar in shape and magnitude with the input voltage, but there is a difference in the positive peak or the negative peak level. The diode voltage limiter can be extended to this application, simply by making E1 equal to zero; the positive output voltage will be limited to internal voltage drop of the diode. The negative voltage will be clamped by D1 and E1 at the output terminal. If the other voltage in the circuit are very large compared to the internal diode drop, which is usually than 1 volt , then it can be said that the output voltage is clamped at zero volts. The diode limiter might also be said to be a clamper, where its output is clamped to positive E1 and negative E2 . For these applications sometimes employ the terms voltage limitation and voltage clamper synonymously. Procedure: PART 1: 1- Connect the circuits as shown in Fig.(1). 2- Chose input voltage Vin = 7 sin(wt ) ; use the same signal for all other parts. 3- Draw the input and output voltage. +

2µF

6 Vrms

IN4007

CRO

Fig.(1)

PART 2: 1- Connect the circuits as shown in Fig.(2). 2- Draw the output voltage. +

2µF

6 Vrms

IN4007

CRO

Fig.(2)

PART 3: 1- Connect the circuits as shown in Fig.(3). 3- Put E = 3 volt . 2- Draw the output voltage. +

2µF IN4007

6 Vrms

CRO

Fig.(3)

E=2V

PART 4: 1- Connect the circuits as shown in Fig.(4). 3- Put E = 3 volt . 2- Draw the output voltage. +

2µF IN4007

6 Vrms

CRO

Fig.(4)

E=3V

Discussion: 1- Discuss the work of the circuit in part one, then compare it with part two. 2- Compare the state in part one, with part three. 3- What is the advantage of clamping circuits?

Exp. (8) Half Wave Voltage Doublers Theory: For some application we need the DC voltage more than the maximum input voltage; In the previous cases we get the output DC voltage less or equal to the maximum input voltage. For the cases when the load current very less, we can use the doubling voltage circuit to get DC voltage whose value is double of the input peak voltage. The circuit for the half wave voltage doubling is shown in Fig.(1). On the negative half cycles C 2 charges through D2 . On positive half cycles the input current is flows through D1 because D2 will be in reverse bias. The capacitances are usually equal C1 = C2 . The DC voltage across RL nearly double the peak value of input voltage for large value of RL . The resistance RL should be at least ten times the sum of the source resistance and internal resistance of diode. +

C2

D1

10µF 6 Vrms

Vin

Voltage Clamper

D2

C1

+

10µF

RL Vout

CRO

Voltage Rectifier & Filterer

Fig.(1) Procedure: 1- Connect the circuits as shown in Fig.(1). 2- Draw the input AC voltage, and measure its V pp . 3- Put RL equals the following values ( 100Ω , 220Ω , 470Ω , 680Ω , 1KΩ , 4.7 KΩ , 10KΩ , and 100KΩ ) for each value of RL record and draw the output DC voltage. 4- Draw a graph between ( VDC output ) as a function of ( RL ). 5- Replace C 2 by 100µF , then repeat the steps (3-4).

Discussion: 1- What is the role of C1 , C 2 , and RL in the circuit. 2- Discuss the effect of RL in the circuit.

Exp. (9) Full Wave Voltage Doublers Theory: The maximum output voltage of full wave rectifier is 0.636 times the maximum instantaneous of the AC input voltage. It is possible to double the output of full wave rectifier by employing capacitors in what is called a voltage doublers circuits. D1 10Vrms

D2

VDC3

VDC1

VDC2 +

RL

+

10µF

10µF

Fig.(1) Procedure: 1- Connect the circuits as shown in Fig.(1). 2- Remove the resistance in this step. 3- Measure VDC 1 , VDC 2 , and VDC 3 without the load. 4- Put RL equals the following values ( 100Ω , 220Ω , 470Ω , 1KΩ , 2.2 KΩ , and 10KΩ ), and measure VDC 3 in each steps. 4- Draw a graph between ( VDC 3 ) as a function of ( RL ). Discussion: 1- What is the role of D1 , and D2 in the circuit. 2- Discuss the effect of RL in the circuit. 3- What is the relation between VDC 1 and VDC 2 , with VDC 3 . 4- Estimate the out put VDC in the following circuit: +

C1

+

2µF 10 Vrms

C3

+

2µF D1

D2

2µF D3

C2

+

C5

2µF

D4

D5

C4

+

2µF

VDC