Q1) What is a semiconductor?
A1) A semiconductor is a substance whose resistivity lies between the conductors and insulators. The property of resistivity is not the only one that decides a material as a semiconductor, but it has few properties as follows.
Q2) Explain intrinsic and extrinsic semiconductors? A2) Intrinsic Semiconductors A Semiconductor in its extremely pure form is said to be an intrinsic semiconductor. The properties of this pure semiconductor are as follows −
In order to increase the conduction capability of intrinsic semiconductor, it is better to add some impurities. This process of adding impurities is called as Doping. Now, this doped intrinsic semiconductor is called as an Extrinsic Semiconductor. Extrinsic Semiconductor An impure semiconductor, which is formed by doping a pure semiconductor is called as an extrinsic semiconductor. There are two types of extrinsic semiconductors depending upon the type of impurity added. They are N-type extrinsic semiconductor and P-Type extrinsic semiconductor.
Q3) Explain diode equivalent circuits?
A3) Diode is an active element in electrical circuit it has some resistance and some value of junction capacitance when the diode in circuit is replaced by same value of the resistance and capacitance its called as the Diode equivalent circuit. To define the dc diode model, we look at the characteristics of an ideal diode and the modifications that were required due to practical considerations. To review: ¾ Ideal diode: VON = 0, Rr = ∞ and Rf = 0. In other words, the ideal diode is a short in the forward bias region and an open in the reverse bias region. ¾ Practical diode (silicon): VON = 0.7V, Rr < ∞ (typically several MΩ), Rf ≈ rd (typically < 50 Ω). The general representation for a practical diode under dc operating conditions is shown below. Recall the diode is a two-terminal device, which simply means that it connects to other circuit elements at connection points labeled a and b in the circuit below with the voltage Vab applied across the diode. Note that the terminal voltage Vab is the same as the voltage applied across the diode, vD – same animal, different notation. This model may be simplified if we can define the operating region as forward or reverse bias. For the forward bias region (vD ≥ 0.7 V for silicon), the ideal diode is a short and the terminal characteristics of the model above reduce to the parallel combination of Rr and Rf. Since Rr >> Rf, Rr ||Rf ≈ Rf. Likewise, when the voltage applied to the diode is less than VON (vD < 0.7 V for silicon), the ideal diode is an open and the resistance between terminals a and b is Rr. These two cases are illustrated below for reference
Forward bised DC model Reverse biased DC model
Even though the same, they behave very differently! Remember the values of the resistances in the different regions and you can see that diode characteristics in the forward and reverse bias regions are quite distinct.
Q4) Explain diode characteristics? A4) Basic static characteristics of diodes are the forward voltage VF and forward current IF, and the reverse voltage and current VR and IR. The area surrounded by the orange dashed line in the diagram on the right indicates the usable area of rectifying diodes. Specifically, this is the area within the range of allowable IF, and within the breakdown voltage range in the reverse direction. It should be noted that the area enclosed by the green dashed line is the usable area of Zener diodes, although these are not discussed in this chapter. This area isn’t usable for other diodes, and if this area is entered without any limits on the IR, device failure may occur.
Q5) Explain diode as a switch?
A5) Whenever a specified voltage is exceeded, the diode resistance gets increased, making the diode reverse biased and it acts as an open switch. Whenever the voltage applied is below the reference voltage, the diode resistance gets decreased, making the diode forward biased, and it acts as a closed switch. The following circuit explains the diode acting as a switch.
A switching diode has a PN junction in which P-region is lightly doped and N-region is heavily doped. The above circuit symbolizes that the diode gets ON when positive voltage forward biases the diode and it gets OFF when negative voltage reverse biases the diode.
As the forward current flows till then, with a sudden reverse voltage, the reverse current flows for an instance rather than getting switched OFF immediately. The higher the leakage current, the greater the loss. The flow of reverse current when the diode is reverse biased suddenly may sometimes create few oscillations, called RINGING.
This ringing condition is a loss and hence should be minimized.
Q6) Explain half wave rectifier?
A6) A complete half-wave rectifier circuit consists of 3 main parts:
A half wave rectifier circuit diagram looks like this:
First, a high AC voltage is applied to the to the primary side of the step- down transformer and we will get a low voltage at the secondary winding which will be applied to the diode.
During the positive half cycle of the AC voltage, the diode will be forward biased and the current flows through the diode. During the negative half cycle of the AC voltage, the diode will be reverse biased and the flow of current will be blocked. The final output voltage waveform on the secondary side (DC) is shown in figure above. If we replace the secondary transformer coils with a source voltage, we can simplify the circuit diagram of the half-wave rectifier as:
For the positive half cycle of the AC source voltage, the equivalent circuit effectively becomes:
This is because the diode is forward biased, and is hence allowing current to pass through. So, we have a closed circuit. But for the negative half cycle of the AC source voltage, the equivalent circuit becomes:
Because the diode is now in reverse bias mode, no current is able to pass through it. As such, we now have an open circuit. Since current cannot flow through to the load during this time, the output voltage is equal to zero. This all happens very quickly – since an AC waveform will oscillate between positive and negative many times each second (depending on the frequency). Here’s what the half wave rectifier waveform looks like on the input side (Vin), and what it looks like on the output side (Vout) after rectification (i.e. conversion from AC to DC):
The graph above actually shows a positive half wave rectifier. This is a half-wave rectifier which only allows the positive half-cycles through the diode,and blocks the negative half-cycle. The voltage waveform before and after a positive half wave rectifier is shown in figure 4 below.
Conversely, a negative half-wave rectifier will only allow negative half-cycles through the diode and will block the positive half-cycle. The only difference between a posive and negative half wave rectifier is the direction of the diode. As you can see in figure 5 below, the diode is now in the opposite direction. Hence the diode will now be forward biased only when the AC waveform is in its negative half cycle.
ADVANTAGES:- 1)Simple Construction. 2) Component required less. 3)Small size.
APPLICATION:- Walkman, low-cost power supply.
TRANSFORMER UTILISATION FACTOR(TUF): It indicates how well the ilp transformer is being utilized TUF= DC O/P Power / AC power rating of the transformer
Q7) Explain full wave rectifier?
A7) Full wave rectifier 1) A center-tapped rectifier is a type of full-wave rectifier that uses two diodes connected to the secondary of a center-tapped transformer.
Center tapped full wave rectifier operation:- I) During the positive half cycle of the i/p ac supply. diagram
D1.Is in forward biased & D2 is in reverse biased. II)During -ve half cycle.
D1 Reverse biased D2 Forward biased
ADVANTAGES
DISADVANTAGES PIV of the diode is 2vm, more size costly.
APPLICATION I) Battery charges. 2)power supply at the laboratory, high current, electronic ckt.
Q8) Explain the use of capacitor filter?
A typical capacitor filter circuit diagram is shown below. The designing of this circuit can be done with a capacitor C as well as load resistor (RL). The rectifier’s exciting voltage is given across the terminals of a capacitor. Whenever the voltage of the rectifier enhances then the capacitor will be charged as well as supplies the current to the load.
At the last part of the quarter phase, the capacitor will be charged to the highest rectifier voltage value that is denoted with Vm, and then the voltage of the rectifier starts to reduce. As this happens, the capacitor starts discharging through the voltage across it and load. The voltage across the load will reduce little only because the next peak voltage occurs instantaneously to charge the capacitor. This procedure will repeat many times and the output waveform will be seen that very slight ripple is missing in the output. Furthermore, the output voltage is superior because it remains significantly close to the highest value of the output voltage of the rectifier.
Q9) Compare half-wave and full-wave rectifier?
Q10) Explain breakdown mechanism?
The following two processes cause junction breakdown due to the increase in reverse bias voltage. (i) Zener Breakdown Zener Breakdown The Zener Breakdown is observed in the Zener diodes having Vz less than 5V or between 5 to 8 volts. When a reverse voltage is applied to a Zener diode, it causes a very intense electric field to appear across a narrow depletion region. Such an intense electric field is strong enough to pull some of the valence electrons into the conduction band by breaking their covalent bonds .these electrons then become free electrons which are available for conduction. A large number of such free electrons will constitute a large reverse current through the Zener diode and breakdown is said to have occurred due to the Zener effect. Characteristics of Zener Breakdown is shown below:
Zener Breakdown Characteristics A current limiting resistance should be connected in series with the Zener diode to protect it against the damage due to excessive heating. In Zener breakdown, the breakdown voltage depends on the temperature of P-N junction.The breakdown voltage decreases with increase in the junction temperature. Avalanche Breakdown The avalanche breakdown is observed in the Zener Diodes having Vz having than 8 V. In the reverse biased condition, the conduction will take place only due to the minority carriers. As we increase the reverse voltage applied to the Zener diode, these minority carriers tend to accelerated. Therefore, the kinetic energy associated with them increases. While travelling, these accelerated minority carriers will collide with the stationary atoms and impart some of the kinetic energy to the valence electrons present in the covalent bonds. Characteristics of Avalanche Breakdown is shown below:
Avalanche Breakdown Characteristics Due to this additionally acquired energy, these valence electrons will break their covalent bonds and jump into the conduction bond to become free conduction. Now these newly generated free electrons will get accelerated. They will knock out some more valence electrons by means of collision. This phenomenon is called as carrier multiplication.
Q11) Explain the operation of Zener diode?
A zener diode can be forward biased or reverses biased. its operation in the forward biased mode is same as that of a p-n junction diode but its operation in the reverse biased mode is sustainably deferent.
Q12) Explain the characteristics of Zener diode?
Q13) Explain photodiode ?
The photodiode is a p-n junction semiconductor diode which is always operated in the reverse biased condition.
The light is always focused through a glass lens on the junction of the photodiode As the photodiode is reverse biased the depletion region is quite wide, penetrated on both sides of the junction. The photons incident on the depletion region will impact their energy to the Ions present in the depletion region and generates e hole pairs. The photons incident on the depletion region, so the number of electron hole pairs will be generated, depends on the intensity of light [number of photons] These and holes will be attracted towards the +ve & -ve terminals respectively of the photocurrent. With an increase in the light intensity more number of e hole pairs are generated and the photocurrent increases thus the photocurrent is proportional to the light intensity.
Q14) Explain the construction and operation of LED?
An LED emits light when electrical energy is applied to it. for proper operation, it is necessary to forward bias the LED.
Construction of LED:-
To make the emission of light in one direction cup type construction is used for LED.
PRINCIPLE LED OPERATION When the led is forward biased the electrons in the n-region will cross the junction and recombine with the holes in the p-type material.
These free e- reside in the conduction band & hence at a higher energy level than the holes in the valence band
When recombination takes place this e- return peak to the valence band which is at a lower energy level than the conduction band. while returning back the recombining e-give away the excess energy in the form of light. This process is called electroluminescence. in this way an LED emits light.
Q15) Explain the CB operation of BJT?
Common Base CB The name itself implies that the Base terminal is taken as common terminal for both input and output of the transistor. The common base connection for both NPN and PNP transistors is as shown in the following figure.
For the sake of understanding, let us consider NPN transistor in CB configuration. When the emitter voltage is applied, as it is forward biased, the electrons from the negative terminal repel the emitter electrons and current flows through the emitter and base to the collector to contribute collector current. The collector voltage VCB is kept constant throughout this. In the CB configuration, the input current is the emitter current IE and the output current is the collector current IC. Current Amplification Factor α The ratio of change in collector current ΔIC to the change in emitter current ΔIE when collector voltage VCB is kept constant, is called as Current amplification factor. It is denoted by α. α=ΔIC/ΔIE atconstantVCB Expression for Collector current Along with the emitter current flowing, there is some amount of base current IB which flows through the base terminal due to electron hole recombination. As collector-base junction is reverse biased, there is another current which is flown due to minority charge carriers. This is the leakage current which can be understood as Ileakage. This is due to minority charge carriers and hence very small. The emitter current that reaches the collector terminal is αIE Total collector current IC=αIE+Ileakage If the emitter-base voltage VEB = 0, even then, there flows a small leakage current, which can be termed as ICBO collector−basecurrentwithoutputopencollector−basecurrentwithoutputopen. The collector current therefore can be expressed as IC=αIE+ICBO IE=IC+IB IC=α(IC+IB)+ICBO IC(1−α)=αIB+ICBO IC=(α1−α)IB+(ICBO1−α) IC=(α1−α)IB+(11−α)ICBO Hence the above derived is the expression for collector current. The value of collector current depends on base current and leakage current along with the current amplification factor of that transistor in use. Characteristics of CB configuration
η=ΔVEB/ΔIE atconstantVCB
ro=ΔVCB/ΔIC atconstant lE
Q16) Explain the operation of CE configuration?
The name itself implies that the Emitter terminal is taken as common terminal for both input and output of the transistor. The common emitter connection for both NPN and PNP transistors is as shown in the following figure.
Just as in CB configuration, the emitter junction is forward biased and the collector junction is reverse biased. The flow of electrons is controlled in the same manner. The input current is the base current IB and the output current is the collector current IC here. Base Current Amplification factor β The ratio of change in collector current ΔIC to the change in base current ΔIB is known as Base Current Amplification Factor. It is denoted by β β=ΔICΔ/IB Relation between β and α Let us try to derive the relation between base current amplification factor and emitter current amplification factor. β=ΔIC/ΔIB α=ΔIC/ΔIE IE=IB+IC ΔIE=ΔIB+ΔIC ΔIB=ΔIE−ΔIC We can write β=ΔIC/ΔIE−ΔIC Dividing by ΔIE
β=ΔIC/ΔIE/ΔIE/ΔIE−ΔIC/ΔIE We have α=ΔIC/ΔIE Therefore, β=α/1−α From the above equation, it is evident that, as α approaches 1, β reaches infinity. Hence, the current gain in Common Emitter connection is very high. This is the reason this circuit connection is mostly used in all transistor applications. Expression for Collector Current In the Common Emitter configuration, IB is the input current and IC is the output current. We know IE=IB+IC And IC=αIE+ICBO =α(IB+IC)+ICBO IC(1−α)=αIB+ICBO IC=α/1−α IB+1/1−α ICBO If base circuit is open, i.e. if IB = 0, The collector emitter current with base open is ICEO ICEO=1/1−αICBO Substituting the value of this in the previous equation, we get IC=α/1−α . IB+ ICEO IC=βIB+ICEO Hence the equation for collector current is obtained. Knee Voltage In CE configuration, by keeping the base current IB constant, if VCE is varied, IC increases nearly to 1v of VCE and stays constant thereafter. This value of VCE up to which collector current IC changes with VCE is called the Knee Voltage. The transistors while operating in CE configuration, they are operated above this knee voltage. Characteristics of CE Configuration
ri=ΔVBE/ΔIB at constant VCE
ro=ΔVCE/ΔIC at constant IB
Q17) Explain the load line analysis?
When the transistor is given the bias and no signal is applied at its input, the load line drawn at such condition, can be understood as DC condition. Here there will be no amplification as the signal is absent. The circuit will be as shown below.
The value of collector emitter voltage at any given time will be VCE=VCC−ICRC As VCC and RC are fixed values, the above one is a first degree equation and hence will be a straight line on the output characteristics. This line is called as D.C. Load line. The figure below shows the DC load line.
To obtain the load line, the two end points of the straight line are to be determined. Let those two points be A and B. To obtain A When collector emitter voltage VCE = 0, the collector current is maximum and is equal to VCC/RC. This gives the maximum value of VCE. This is shown as VCE=VCC−ICRC 0=VCC−ICRC IC=VCC/RC This gives the point A (OA = VCC/RC) on collector current axis, shown in the above figure. To obtain B When the collector current IC = 0, then collector emitter voltage is maximum and will be equal to the VCC. This gives the maximum value of IC. This is shown as VCE=VCC−ICRC =VCC (As IC = 0) This gives the point B, which means (OB = VCC) on the collector emitter voltage axis shown in the above figure. Hence, we got both the saturation and cutoff point determined and learnt that the load line is a straight line. So, a DC load line can be drawn.
Q18) Explain voltage divider bias configuration?
Voltage Divider Bias
Approximate analysis
VB = R2 Vcc/ R1 + R2
IE = V E / RE
Ri = (β + 1) RE ͌ β RE
β RE ≥ 10 R2
ICQ = IE
VE = VB – VBE
VCEQ = Vcc – Ic(RC + RE)
Transistor Analysis
IC sat = IC max = Vcc/ Rc + RE
Load Line Analysis
Ic = Vcc/ Rc + RE | Vce =0V
VCE = VCC| Ic =0mA
Determine the levels of ICQ and VCEQ for voltage divider configuration using exact and approximate techniques and compare solutions.
β . RE ≥ 10 R2
(50)(1.2 k Ω) ≥ 10(22 k Ω)
Rth = R1 || R2 82 kΩ || 22 kΩ = 17.35 kΩ
Eth = R2 Vcc/ R1 + R2 = 22kΩ (18V)/ 82kΩ + 22kΩ = 3.81 V IB = Eth – VBE/ Rth + (β +1) RE = 3.81 V – 0.7V / 17.35 kΩ + (51 )(1.2kΩ) = 3.11/ 78.55 kΩ = 39.6 μ A
ICQ = β IB = (50)(39.6μA) = 1.98 mA
VCEQ = Vcc – IC(RC + RE)
= 18V – (1.98mA)(5.6kΩ + 1.2 kΩ)= 4.54V
Q19) Explain transistor as an amplifier?
The transistor has three terminals namely emitter, base and collector. The emitter and base of the transistor are connected in forward biased and the collector base region is in reverse bias. The forward bias means the P-region of the transistor is connected to the positive terminal of the supply and the negative region is connected to the N-terminal and in reverse bias just opposite of it has occurred. Vee is applied to the input circuit along with the input signal to achieve the amplification. The DC voltage VEE keeps the emitter-base junction under the forward biased condition regardless of the polarity of the input signal and is known as bias voltage. When a weak signal is applied to the input, a small change in signal voltage causes a change in emitter current this change is almost the same in collector current because of the transmitter action. In the collector circuit, a load resistor RC of high value is connected. When collector current flows through such a high resistance, it produces a large voltage drop across it. Thus, a weak signal (0.1V) applied to the input circuit appears in the amplified form (10V) in the collector circuit. Input Resistance When the input circuit is forward biased, the input resistance will be low. The input resistance is the opposition offered by the base-emitter junction to the signal flow. Hence, it is the ratio of small change in base-emitter voltage (ΔVBE) to the resulting change in base current (ΔIB) at constant collector-emitter voltage. Input resistance, Ri=ΔVBE/ΔIb Where Ri = input resistance, VBE = base-emitter voltage, and IB = base current.
Output Resistance The output resistance of a transistor amplifier is very high. The collector current changes very slightly with the change in collector-emitter voltage. The ratio of change in collector-emitter voltage (ΔVCE) to the resulting change in collector current (ΔIC) at constant base current. Output resistance = Ro=ΔVCE/ΔIC Where Ro = Output resistance, VCE = Collector-emitter voltage, and IC = Collector-emitter voltage. Current gain It is the ratio of change in collector current (ΔIC) to the change in base current (ΔIB). Current gain, β=ΔIC/ ΔIB Voltage Gain It is the ratio of change in output voltage (ΔVCE) to the change in input voltage (ΔVBE).
Voltage gain, AV=ΔVCE/ΔVBE
= ΔIC x RAC / ΔIB x Ri = ΔIC / ΔIB x RAC / Ri = β x R AC/ Ri
Power Gain It is the ratio of output signal power to the input signal power. Power gain Ap = (ΔIC) 2 x RAC / (ΔIB ) 2 x Ri = (ΔIC / ΔIB) x ΔIC x RAC / ΔIB x Ri = current gain x voltage gain |
