Shankersinh Vaghela Bapu Institute of Technology Diploma EE Semester III 3330905: ELECTRONIC COMPONENTS AND CIRCUITS INDEX Sr. No. Title Page Date Sign Grade 1 Obtain I-V characteristic of Diode. 2 To measure ripple factor at the output of a. Half wave rectifier with and without filter capacitor. b. Full Wave rectifier with and without filter capacitor. c. Bridge rectifier with and without filter capacitor. 3 To verify performance of various Clipper circuits. 4 To verify performance of various Clamper circuits. 5 Obtain I-V characteristic of Zener Diode. 6 7 To obtain input and output characteristics and calculate gain of CE amplifier circuit. To obtain input and output characteristics and calculate gain of CB amplifier circuit. 8 To obtain characteristic of transistor as a switch circuit. 9 To obtain the transfer characteristics of FET. 10 To obtain frequency response of single stage transistor amplifier.
PRACTICAL - 1 AIM: Obtain I-V characteristic of a Diode. EQUIPMENTS: [1]. Experiment board [2]. Ammeter - 2 (0-30 ma) and (0-100 A) [3]. Voltmeter - 2 (0-1 V) and (0-10 V) [4]. Power supply - 1 (0-10 V) THEORY: A p-n junction diode is formed by joining a p-type and n-type semiconductor material. It is a two terminal, uni-junction, uni-directional device. If the positive terminal of the supply is connected to anode (p-region) and negative terminal is connected to cathode (n-region), the diode is said to be forward biased. If the connections are reversed, the diode is said to be reverse biased. Unbiased diode: When the diode is unbiased, diffusion of carriers take place because the concentration of hole is more in p-region and less in n-region and concentration of electron is more in n-region and less in p-region. Hence, holes from p-region diffuse to n-region and electrons from n-region diffuse to p-region. As soon As these charge carriers cross the junction, they recombine with opposite polarity charges. These recombined carriers neutral in charge and they oppose the further movement of charge carriers from one region to another. The region near the junction, occupied by the recombined charges, is known as depletion region or space charge region or transition region. The difference of potential across this region i.e. barrier potential is 0.6 V to 0.7 V for Silicon diode and 0.2 V to 0.3 V for Germanium diode. Forward biased diode: When the diode is forward biased and the applied bias is less than barrier potential, no current flows. As the applied potential increases, the charge carriers gain sufficient energy to cross barrier potential and enter the other region. The holes enter the n- region and the electrons enter the p-region, crossing the p-n junction. This crossing of charge carriers results in a flow of current. This is called drift current. Reverse biased diode: When the diode is reverse biased the majority charge carriers are attracted towards the terminals of applied potential away from the p-n junction. It results in widening of the depletion region. Under this situation, no current flows across the p-n junction. But, there will be a very small current across the p-n junction due to the thermally generated minority carriers, called reverse saturation current. It is independent of applied potential and increases with temperature. When the applied reverse voltage increased beyond limit, it results in break down. During break down, the diode current increases tremendously for a particular voltage. When a p-n junction diode is heavily doped, it is known as Zener diode. When the Zener diode is forward bias, it exhibits the same characteristics as p-n junction diode. In reverse biased condition, when reverse voltage increases beyond breakdown voltage, current through diode rises sharply. Beyond breakdown very small rise in voltage results in large current through the diode. Due to applied reverse potential, an electric field exists near the junction. This field exerts a strong force on the bound charges, which breaks the covalent bonds and
releases free charge carrier. These newly generated electron-hole pairs result in sudden increase of reverse current. As the level of doping increases, the breakdown voltage reduces. Such diode exhibits negative coefficient of temperature. Zener diode, connected in reverse biased condition, is used as a voltage regulator. Once the reverse voltage crosses breakdown voltage, voltage across the zener diode remains constant even if current through the diode increases sharply. PROCEDURE: [1] Connect the circuit as per the circuit diagram (1). [2] Increase the power supply voltage such that the voltage across the diode changes by 0.1V till the power supply meter shows 20V. [3] Note down the corresponding ammeter readings. [4] Plot the graph: I against V. [5] Find the dynamic resistance from the slope of the graph: r = V / I at I = 10mA. [6] Connect the circuit as per the circuit diagram (2). [7] Increase the power supply voltage such that the voltage across the diode changes by 1V till the power supply meter shows 20V. [8] Note down the corresponding ammeter readings. [9] Plot the graph: I against V. [10]Find the dynamic resistance from the slope of the graph: r = V / I at V = 10V. [11]Repeat steps 1 to 10 for Zener diode. OBSERVATION TABLE: JUNCTION DIODE Forward bias Sr.No. Diode voltage, V Diode Current, I Reverse bias Sr.No. Diode voltage, V Diode Current, I RESULTS: Dynamic resistance R= V/ I Forward bias Reverse bias Junction Diode
CIRCUIT DIAGRAM: 1K + 0-30 ma A + 0-30 V 0-1 V + V OR FIGURE :1 10K 0-100uA + A + 0-30 V 0-10 V + V OR FIGURE :2 CONCLUSION:
PRACTICAL - 2 AIM: [1] To measure ripple factor at the output of a (i). Half wave rectifier with & without filter capacitor (ii). Full wave center taped rectifier with & without filter capacitor (iii).bridge rectifier with & without filter capacitor EQUIPMENTS: [1]. Experiment board [2]. C.R.O. [3]. Multimeter THEORY: The process of converting an alternating voltage/current to a unidirectional voltage/current is called rectification. A diode offers very low resistance when forward biased and very high resistance when reverse biased. Hence, it can be used as rectifier. The rectified output is a pulsating unidirectional voltage/current. A filter is necessary after rectifier to convert pulsating waveform to dc. Half wave rectifier: During positive half cycle, the diode is forward biased. Hence, current flows through load resistor. During negative half cycle, the diode is reverse biased and is equivalent to open circuit. Therefore current through load is zero. Thus, the diode conducts only for one half cycles and results in half wave rectified output. Half wave rectifier is simple and low cost circuit but it has very high ripple, lower efficiency. Full wave rectifier: The full wave rectifier consists of a center-tap transformer, which results in equal voltages above and below the center-tap. During the positive half cycle, positive voltage appears at the anode of D1 while negative voltage at the anode of D2. So diode D1 is forward biased and it results in current through load. During negative half cycle, the positive voltage appears at the anode of D2 and hence it is forward biased resulting in current through load. At the same instant a negative voltage appears at the anode of D1, thus reverse biasing it and hence D1 does not conduct. The current through the load during both half cycles is in the same direction hence it is sum of individual currents. The individual currents and voltages are combined in the load and therefore their average values are double that obtained in a half wave rectifier. In full wave rectifier, ripple is reduced, efficiency is improved. And as equal current flows through secondary during both the half cycle, core does not saturate. The demerits of full wave rectifier are : output voltage is half the secondary voltage and diodes with high PIV ratings are required. Bridge rectifier: This circuit does not required center-tap transformer. During the positive half cycle, diodes D1 and D2 are forward biased and D3 and D4 are reverse biased. Thus, current flows in the circuit due to D1 and D2. During the positive half cycle, diodes D1 and D2 are forward biased and D3 and D4 are reverse biased. Thus, current flows in the circuit due to D1 and D2. During the negative half cycle, diodes D3 and D4 are forward biased and D1 and D2 are reverse biased which results in a current in the same direction. Thus the current flows for the whole cycle across the load in one direction resulting in full wave rectification. Since bridge rectifier does not required center-tapped transformer, its cost, weight and size are lesser compared to a full wave rectifier. Diode with lesser PIV can be used in bridge rectifier compared to full wave rectifier.
FORMULA: For half wave rectifier Vdc = Vm / Vrms = Vm / 2 Vrms' = (Vrms 2 - Vdc 2 ) (without filter) Vrms' = Vm' / 2 (with filter) For full wave rectifier (center tapped & bridge ) Vdc = 2Vm / Vrms = Vm / 2 Vrms' = (Vrms 2 - Vdc 2 ) (without filter) Vrms' = Vm' / 2 (with filter) where, Vm - peak value of the pulsating waveform Vrms - root mean square value of the total output Vrms' - root mean square value of ripple Vm' - peak value of the ripple PROCEDURE: [1] Complete the circuit as per the fig.(1) for half wave rectifier. [2] Connect the CRO across the load. [3] Switch on the supply. [4] Keep the CRO in ground mode and adjust the horizontal line on X-axis. [5] Switch the CRO to DC mode and draw the waveform. Note down its amplitude Vm. [6] Calculate Vdc, Vrms and Vrms' from the formula given in above table. [7] From Vrms' and Vdc, find out ripple factor. [8] Now, connect the capacitor filter at the output of rectifier, draw the resulting waveform and measure the peak value, Vp. [9] By keeping the CRO in AC mode, observe the ripple output and measure the peak value Vm'. [10]Calculate Vdc = Vp- Vm'. [11]Calculate Vrms' from the formula given in above table and from Vrms' and Vdc, Calculate the ripple factor. [12]Complete the circuit as shown in fig.(2) for full wave centre tapped rectifier and repeat the steps 5 to 11. [13]Complete the circuit as shown in fig.(3) for bridge rectifier and repeat the steps 5 to 11. OBSERVATION TABLE: WITHOUT FILTER : Rectifier Type Half wave rectifier Full wave (CT) Bridge rectifier Vm measured (Volts) (dc mode) Vdc Calculated Volts Vrms calculated (Volts) Vrms' calculated (Volts) Ripple factor = Vrms'/Vdc
WITH FILTER : Rectifier Type Half wave rectifier Full wave (CT) Bridge rectifier Vp measured (Volts) (dc mode) V'm Measured (Volts) (ac mode) Vdc calculated (Volts) V'rms calculated (Volts) Ripple factor = V'rms/Vdc RESULT: Half wave rectifier Full wave rectifier Bridge rectifier C=0.1uF Ripple factor with out filter Ripple factor with filter Theoretical Practical Theoretical Practical =1/(23fCR L) =1/(43fCRL) =1/(43fCRL) & RL = 10K CONCLUSION:
CIRCUIT DIAGRAM: HALF WAVE RECTIFIER: Fig. (1) 230 V ac, 50Hz 0V o/p without filter O/p with Filter FULL WAVE RECTIFIER: Fig. (2) 230 V ac, 50Hz 0V O/p without Filter O/p with Filter BRIDGE RECTIFIER: Fig. (3) 230 V ac, 50Hz o/p without Filter o/p with Filter
PRACTICAL - 3 AIM: To verify performance of various Clipper circuits EQUIPMENTS: [1] Circuit board [2] Regulated power supply [3] Audio signal generator [4] C.R.O. THEORY: The circuit used to limit or clip the extremities of the ac signal is called the limiter or clipper. Limiters can transform a sine wave into a rectangular wave. It can limit either the positive or negative amplitude of an ac voltage. It can also perform other useful wave shaping functions. The unidirectional characteristics of semiconductor diodes permit to serve as limiter. In Figure 1, a sine wave signal of amplitude V in is applied to the input. During the positive half cycle of the input signal, diode D is reversed biased and cannot conduct. Hence the output obtained at the output terminals is zero. During the negative half cycle of the input signal, diode D is forward biased and current can flow in the circuit. Hence at the output terminal, the output voltage V o will be developed which will be having full negative half cycles and clipped positive half cycles. This circuit is known as positive series clipper because the diode is in series with the output terminal and the output is clipped on the positive side. By reversing the polarity of diode negative series clipper can be realized which can give the output voltage V o with full positive half cycles and clipped negative half cycles. In Figure 2, a sine wave signal of amplitude V in is applied to the input. During the positive half cycle of the input signal, diode D is reversed biased and cannot conduct. The diode will offer very high resistance R r to the current. The resistors R and R r will form the voltage divider circuit and most of the input voltage will be available at the output terminal because the resistor R r is very large as compared to R. During the negative half cycle of the input signal, diode D is forward biased and current can flow in the circuit. Hence at the output terminal, the output voltage V o will be almost zero. Thus at the output terminal, the output voltage V o will be developed which will be having full positive half cycles and clipped negative half cycles. This circuit is known as negative parallel clipper because the diode is connected in parallel with the output terminal. By reversing the polarity of diode positive parallel clipper can be realized which can give the output voltage V o with full negative half cycles and clipped positive half cycles. The circuit of Figure 3 will provide partial limiting of the negative half cycle of the input signal and is known as biased parallel clipper, while the circuit of Figure 4 will provide partial limiting of both positive and negative half cycles of the input signal and is known as double biased parallel clipper. PROCEDURE: POSITIVE AND NEGATIVE CLIPPER [1] Connect the circuit of Figure and apply a sine wave voltage of 100 Hz (V in max.9v) at the input terminal using audio signal generator. [2] Observe the output voltage Vo and draw the waveforms. [3] Repeat steps (1) & (2) for figure 2.
BIAS CLIPPER [1] Connect the circuit as shown in figure. [2] Apply a sine wave of 100 Hz and maximum 10V input keeping Vr = 0 to 3V dc. [3] Observe the output by changing Vr in step of 1 V. [4] Draw the output waveform of each. DOUBLE BIAS CLIPPER (1) Connect the circuit as shown in figure. (2) Apply a sine wave of 100 Hz and maximum 10V input keeping V r1 = 1V & V r2 = 3Vdc. (3) Change input from -5V to 5V dc in steps of 1V and measure the output voltage. (4) Tabulate the same. (5) Plot output voltage Vo against input voltage Vin.. OBSERVATION: Sr. No. V in (Volt) V o (Volt) CONCLUSION:
CIRCUIT DIAGRAM:-
PRACTICAL 4 AIM: To verify performance of various Clamper circuits. EQUIPMENTS: [1] Circuit board having diode, resistor and capacitor [2] Audio signal generator [3] C.R.O THEORY: The circuit that does not change the shape of the input waveform but only adds a D.C. level to the input waveform is known as clamper, d.c. Restorer or dc inserter. In the circuit of Figure 1, during the negative half cycle of the input sine wave signal of 10 V (peak-to-peak), the diode D will conduct and the capacitor C will charge to the 5Vpeak value of the negative half cycle through the low resistance path of the forward biased diode. During the positive half cycle of the input sine wave signal, the diode D can not conduct and the capacitor C will try to discharge through R. If the discharge time constant RC is large as compared to the time period of the input sine wave, the capacitor will lose very little of its charge and will hold the charged value of 5 V across it. Now during the next negative half cycle of the input sine wave signal, the negative input voltage will be cancelled by the positive voltage across the capacitor and diode cannot conduct. Hence, effectively 5 V is added by this positive clamper to the input sine wave. The output wave will have d.c. Level of 5 V and it will vary from 0 V to 10 V. Negative clamper works on the similar principle as that of the positive clamper and can add a negative D.C. Level to the input wave. In the circuit shown in Figure 2, the polarity of the diode D has been reversed. The capacitor C will charge during the positive half cycle of the input sine wave signal and will hold the peak value of the half cycle across it. The circuit effectively will add - 5 V to the input sine wave and the output will vary from 0 V to - 10 V with d.c. Value equal to - 5 V. In the circuit of Figure 3, a 3V battery is connected in series to reverse bias the diode. The diode will conduct only after 3 V of the input during the positive half cycle. Thus the capacitor will charge only to 2 V (5-3V). The circuit will effectively add - 2 V to the input and the output will vary from - 7 V to 3 V with D.C. Value equal to - 2 V. This circuit is known as biased clamper. PROCEDURE: [1] Connect the circuit of Fig. 1 and apply sinusoidal input signal of 10 Vpp value to the circuit. [2] Observe the output signal on C.R.O. keeping it on D.C. Position. [3] Repeat the step (1) for the circuit of Fig. 2. [4] Now connect the circuit of Fig. 3 and keep the biasing battery at 3 V. [5] Apply sinusoidal input signal of 10 Vpp value to the circuit of Fig. 3 and observe the output on C.R.O. [6] Draw observed waveforms & actual dc level on graph paper. [7] Write down conclusion based on your observation.
CIRCUIT DIAGRAM: CONCLUSION:
EXPERIMENT - 5 AIM: Obtain I-V characteristic of the Zener Diode. EQUIPMENTS: 1. Bread board 2. Diode, Resistor and Capacitor 3. DC power supply 4. Function generator 5. C.R.O THEORY: The Zener diode is designed to operate in reverse breakdown region. Zener diode is used for voltage regulation purpose. Zener diodes are designed for specific reverse breakdown voltage called Zener breakdown voltage (VZ). The value of VZ depends on amount of doping. Breakdown current is limited by power dissipation capacity of the Zener diode. If power capacity of the Zener is 1 W and Zener voltage is 10V, highest reverse current is 0.1A or 100 ma. If current increases more than this limit, diode will be damaged. Forward characteristics of the Zener diode is similar to normal PN junction diode. CIRCUIT DIAGRAM: Reverse Bias Forward Bias
PROCEDURE: 1. Connect the power supply, voltmeter, current meter with the diode as shown in the figure for reverse bias. You can use two multimeter (one to measure current through diode and other to measure voltage across diode) 2. Increase voltage from the power supply from 0V to 20V in step as shown in the table. 3. Measure voltage across diode and current through diode. Note down readings. 4. Reverse the direction of diode in the circuit and repeat the procedure. 5. Draw observed waveforms & actual dc level on graph paper. CONCLUSION:
PRACTICAL 6 AIM: To obtain input and output characteristics and calculate gain of CE amplifier circuit. EQUIPMENTS: [1]. Ammeter - 2 (0-10 ma) and (0-200 ua) [2]. Voltmeter - 2 (0-10 V) and (0-2 V) [3]. Power supply - 2 (0-10 V) THEORY: Bipolar junction transistor is a three terminal two junction semiconductor device. In transistor, the conduction is due to both polarity charge carriers. Hence it is called bipolar device. BJT can be either pnp or npn. An npn transistor consists of one p-layer sandwiched between two n-layers. Similarly, a pnp transistor consists of one n-layer sandwiched between two p-layers. Three terminals of transistor are known as Emitter, Base and Collector among which emitter is heavily doped and base is lightly doped. Depending on which terminal is made common to input and output, the transistor is classified into three configurations: Common base, common emitter, Common collector. Common emitter configuration: In Common Emitter configuration, the emitter is common to both input and output. The input characteristics relate base current IB and base to emitter voltage VBE for constant VCE. When VBE is less than cut-in voltage, IB remains almost zero and after cut-in voltage, IB increases with VBE just like in forward biased diode. If VCE is increased, IB decreases for a given value of VBE. The output characteristics relates collector current IC and collector to emitter voltage VCE for constant IB. For given value of IB, IC increases with VCE and then it becomes almost constant. The factor = IC / IB is known as current gain. PROCEDURE: Input Characteristics: [1] Connect the circuit as per the circuit diagram (1). [2] Set VCE = 2V(say), vary VBE insteps of 0.1 V and note down corresponding IB. [3] Repeat the above procedure for VCE = 5V, 10V. [4] Plot the graph : VBE against IB for constant VCE. Output Characteristics: [1] Connect the circuit as per the circuit diagram (2). [2] Set IB = 40 A(say), vary VCE insteps of 1 V and note down corresponding IC. [3] Repeat the above procedure for IB = 80 A, 120 A. [4] Plot the graph : VCE against IC for constant IB.
OBSERVATION TABLE: [A] INPUT CHARACTERISTICS: Sr. No. VCE = 2 V VCE = 5 V VCE = 10 V VBE (Volts) IB (A) VBE (Volts) IB (A) VBE (Volts) IB (A) [B] OUTPUT CHARACTERISTICS: Sr. No. IB = 40 A IB = 80 A IB = 120 A VCE (Volts) IC (ma) VCE (Volts) IC (ma) VCE (Volts) IC (ma)
CIRCUIT DIAGRAM: 0-10mA A + 470 10K + 0-1 ma A BC 107 + + 0-10 V + 0-10 V + V 0-1 V V 0-10 V CONCLUSION:
EXPERIMENT - 7 AIM: To obtain input and output characteristics and calculate gain of CB amplifier circuit. EQUIPMENTS: 1. Bread board 2. Diode, Resistor and Capacitor 3. DC power supply 4. Function generator 5. C.R.O THEORY: In a common base configuration, base terminal is common between input and output. The output is taken from collector and the input voltage is applied between emitter and base. The base is grounded because it is common. To obtain output characteristics, we will measure collector current for different value of collector to base voltage (VCB). Input current is emitter current Ie and input voltage is Veb. To plot input characteristics we will plot Veb versus Ie. Current gain for CB configuration is less than unity. CB configuration is used in common base amplifier to obtain voltage gain. Output impedance of common base configuration is very high. CB amplifier is used in multi - stage amplifier where impedance matching is required between different stages. CIRCUIT DIAGEAM: Input Characteristic: Output Characteristic:
PROCEDURE: For input characteristics: Connect circuit as shown in the circuit diagram for input characteristics. Connect variable power supply 0-30V (VEE) at emitter base circuit and another power supply 0-30V at collector base circuit (Vcc). Keep Vcc fix at 0V (Or do not connect Vcc). Increase VEE from 0V to 20V, note down readings of emitter current Ie and emitter to base voltage Veb in the observation table. Repeat above procedure for Vcc = +5V and Vcc = +10V. Draw input characteristics curve. Plot Veb on X axis and Ie on Y axis. For output characteristics: Connect circuit as shown in the circuit diagram for output characteristics Connect variable power supply 0-30V at emitter circuit and collector circuit. Keep emitter current fix (Initially 0) Increase VCC from 0V to 30V, note down readings of collector current Ic and collector to base voltage Vcb in the observation table. Repeat above procedure for base currents Ie = 1mA, 5 ma and 10mA. Increase emitter current by increasing VEE. Draw output characteristics curve. Plot Vcb on X axis and Ic on Y axis. OBSERVATION TABLE: Input Characteristics:
Output Characteristics: CONCLUSION:
EXPERIMENT - 8 AIM: To obtain characteristic of transistor as a switch circuit. THEORY: Transistor can be operated in three region: cut-off region, active region and saturation region. While using transistor in amplifier circuit, we are using active region. If transistor operated in cut-off and saturation region in amplifier, clipping of waveforms will occur. When we use transistor as a switch, only two regions cut-off and saturation are used. In saturation region transistor acts as ON switch. In cut-off region, transistor acts as OFF switch. We are using only two points of DC load line while using transistor as a switch. CIRCUIT DIAGRAM: PROCEDURE: Connect circuit as shown in the circuit diagram for input characteristics Adjust collector supply Vcc = +12V and base supply VBB=+5V You may use base voltage supply switch instead of switch shown in the circuit diagram. Measure base current when switch is OFF (It will be zero). Measure collector current and voltage between collector and emitter. LED is off because transistor is in cut-off region, Apply base voltage +5V, LED will be ON. Measure collector current and collectoremitter voltage. Transistor is in saturation region. OBSERVATION: [1] Switch off (VBB=0): Ib = VCE = Ic = [2] Switch ON (VBB=+5V): Ib = VCE = Ic = CONCLUSION:
EXPERIMENT - 9 AIM: To obtain the transfer characteristics of FET. THEORY: The Field Effect Transistor (FET) is a three terminal device. Three terminals are Drain (D), Source (S) and Gate (G). In FET, current flow is due to only one type of charge particles, either electrons or holes. So FET is known as unipolar device. The name field effect is derived from the fact that the current is controlled by an electric field set up in the device by an externally applied voltage. Thus FET is a voltage controlled device while bipolar transistor is current controlled device. The Field Effect Transistor (FET) can be broadly classified into following categories: In this experiment we will obtain output characteristics of N-channel FET using CS (Common source) Configuration. It is also known as drain characteristics. Basic construction of N-channel FET and its symbol are shown in the following figure. When gate to source voltage VGS is zero, N type channel is open so drain current will flow through it. As we increase negative voltage on the gate terminal, VGS=-1V, -2V, -3V etc., drain current reduces. The reduction in drain current is due to reduction in width of channel. As we increase negative gate voltage, width of depletion region spreads in the channel. Depletion region (generated field due to reverse bias) does not have charge carriers so width of channel will reduce. As we increase negative value of VGS, penetration of depletion region (field) will be more and more due to which channel becomes narrower. At one point drain current reduces to zero when entire channel will be closed due to penetration of depletion region. The value of VGS at which drain current reduces to zero is called cut-off voltage VGS (off). Normally Drain current reduces to zero at VGS=-Vp. Thus VGS (off) = -VP where VP is pinch-off voltage. Pinch-off voltage VP is the value of voltage VDS at which drain current becomes constant.
CIRCUIT DIAGRAM: PROCEDURE: Connect circuit as shown in the circuit diagram for output (drain) characteristics. Connect variable power supply 0-10V at gate circuit and 0-12V at drain circuit. Keep gate to source voltage zero (VGS=0). Increase drain supply Vdd from 0V to 12V, note down readings of drain current Id and drain to source voltage Vds in the observation table.
Repeat above procedure for different gate to source voltages VGS = -1, -2, -3, -4 etc. Note down reading of Gate to source voltage at which drain current remains zero. This is cutoff voltage VGS(off). Note down pinch-off voltage for all values of VGS. Draw output characteristics curve. Plot Vds on X axis and Id on Y axis. OBSERVATION TABLE: CONCLUSION:
EXPERIMENT - 10 AIM: To obtain frequency response of single stage transistor amplifier. THEORY: The circuit consists of voltage divider bias with bypass capacitor Ce and two coupling capacitors C1 and C2, one at the base of the transistor and another at the collector of the transistor. The input resistance Rs is connected to the transistor base via coupling capacitor C1. Capacitor C2 couples external load resistor Rl to the transistor collector. Coupling capacitors are connected to maintain the stability of bias condition. Coupling capacitors acts as open circuit to dc thus maintaining stable biasing condition even after connection of Rs and Rl. The advantage of connecting C1 is that any dc component in the signal is blocked and only ac signal is routed to the transistor amplifier. The emitter resistance Re is one of the component which provides bias stabilization. But it also reduces the voltage swing at the output. The emitter bypass capacitor Ce provides a low reactance path to the amplified ac signal increasing the output voltage swing. During positive half cycle of input signal as the input voltage increases, it increases forward bias on basetimes the base current, the collector current will also increase. This increases voltage drop across Rc which in turn decrease collector voltage Vc (Vc=Vcc-IcRc). Thus, as Vi increase in a positive direction, Vo goes in a negative direction and it results in negative half cycle of output voltage for positive half cycle of input voltage. Therefore, there is a phase shift of 1800 between input and output voltage. CE amplifier provides moderate voltage and current gain. It has low input impedance and high output impedance. CIRCUIT DIAGRAM:
PROCEDURE: Connect the circuit as per the circuit diagram. Set the function generator to give 100mVp-p sine wave to the amplifier. Observe the output on CRO. Keeping the input voltage constant, vary the frequency from 100 Hz to 100 KHz and note down the corresponding output. Plot the graph: gain (db) v/s frequency. For finding input impedance, connect decade resistance box (DRB) in series with the input. Keep zero resistance in DRB. Connect CRO at the input. Set the input to give 2 Vp-p outputs. Change the resistance of DRB in order to get half i.e. 1V output. The resistance of DRB will be the input impedance. For finding output impedance, connect decade resistance box (DRB) in parallel with the output. Keep maximum resistance in DRB. Set the input to give 2 Vp-p output. Change the resistance of DRB in order to get half i.e. 1V output. The resistance of DRB will be the output impedance. After plotting the graph, mark the points which are 3dB lower than midband gain. Make the projection of these points on frequency axis. The difference of two points indicates bandwidth of the amplifier. OBSERVATION TABLE: Input Voltage: V1 = Sr. No. 1 2 3 4 5 6 7 8 9 10 Frequency at the input Output Voltage (Vo) Output Voltage (V3) Gain A = Vo/Vi Gain(dB) = 20 log (Vo/Vi) CONCLUSION: