Question Topics :
All Questions
Answer: Fermi-Dirac function specifies the fraction of all energy states at energy E occupied when the system(the crystal) is in thermal equilibrium. According to quantum statistics, the Fermi-Dirac probability function, f(E), is expressed as, f(E) = 1 /(1+ exp(E-Ef)/kT where, k = Boltzmann constant, T = Temperature ( degrees kelvin ), Ef = Fermi level or characteristic energy for the crystal in eV.For semiconductors,it's value is 1/2,meaning that fermi energy is that energy level for which the probability of occupancy is half. We have talked about Fermi energy in semiconductors in the lectures. More about Fermi-Dirac function can be read in any book on Solid State Physics. I hope ,it is sufficient for your current query. D C Dube
Answer: No. Fermi level is not always in the middle of valence and conductance bands. Only for pure ( undopped) semiconductor it is in the middle of the two bands and that is true only under the assumption of equal effective masses for electron and hole. Useful phenomenon of conduction in semiconductors occur only in the outer most quantum shells. The usefulness and significance of Fermi level can be better appreciated through p-n junction.When p- and n- regions co-exist in a semiconductor crustal, majority holes diffuse towards hole defficient n- region. Similarly, electrons diffuse towards p- region.Till how long this charge movement across the junction continues? It stops as soon as Fermi levels on the p- and n- sides allign and become one. And this allignment gives rise to potential barrier at the junction.
Answer: The direction of externally applied voltage creats field at the p-n juction in the same direction as the built-in field.So the two fields are added up and the potential barrier increases as compared to unbised junction.This is possible only when fermi levels move apart from their alligned position of unbised junction. In the forward bias case,the movement of fermi levels on the two sides is opposite and much smaller. This is so, because in forward bias high current flows and as a result, large voltage drop occurs across the resistor used in the circuit and not across the junction.
Answer: In reverse bias since the sign is negative,so applied voltage and built-in voltage get added up. But in forword bias, as soon as the applied voltage tries to build-up and is close to built-in voltage,the current rises exponentially and gives large voltage drop acsoss the resistor in he circuit,therefore,(VB - V) is not positive.
Answer: The diode in the biasing cercuit provides trckel current required for class AB operation. The characteristics of this diode are identical to that of emitter juction of the transistor. If the temp. changes, emitter juction voltage drops and trickel current will also fall. But drop at biasing diode will also drop and that will increase the collector current. (Look at the expression for collector current). This way operating point is stablyzed.
Answer: Voltage drop across a forward biased diode is not equal to applied voltage, but it is 0.7V for silicon diode. Look at the input I-V charactristics of a p-n diode.A resistance in series of the diode is always connected, and this resistor limits the current in the circuit. In the absence of this resistor, current will rise expotentially as soon as applied voltage in forward bias goes above 0.7 volt spoiling the diode. With the resisitor,voltage drop in excess of 0.7V,drops across the external resistor.The net voltage across diode is ( Vb -V) but here V should be viewed as applied voltage minus drop across the resistor, such that (Vb-V) is always positive. In the reverse bias,since voltage is negative, and the total potential across the diode is (Vb + V).In that case, no current flows in the circuit and drop at the resistor is zero.
Answer: Current flow requires flow of charges. In an unbised p-n diode,charges are prevented from flowing by the potential barrier. Connecting a resistor acrooss the diode does not affect the barrier height.Further,when we connect the metal electrodes ( terminals of resistor )the fermi levels on the two sides are compensated, preventing any current to flow.
Answer: Holes in N- type semiconductor are there as the semiconductor is at a finite temperature, say at room temperature ( 300 K ). The heat available breaks some bonds and puts some electrons in conduction band leaving corresponding holes in the valence band. This way holes are created in the N-material. The doping atoms added to make material N-type just push electrons in the conduction band,leaving behind ionized impurity atoms at the positon of substitution in the lattice.
Answer: Moblity of a charge carrier is defind as the velocity per unit field.Thus,higher mobility implies operation of the device upto higher frequency. In germanium or silicon,mobility is much higher for electrons as compared to holes.PNP or NPN,both type are alright but NPN will work satisfactorily up to higher frequency as compared to PNP. In NPN transistor,electrons play the major role in transistor operation.
Answer: Schottky diode is quite different from a p-n diode.Schottky diode is formed by having metal wire ( silver,gold etc.) on one side and silicon ( normally n-type) on the other side.Electrons are charge carrier. There is no depletion layer, therefore this diode is free from the problem of reverse recovery time (which limits the frequency of operation of a p-n diode).A schottky diode can work up to the frequency of several hundred MHz.
Answer: The depletion width is a function of doping. When doping is increased,obviously,charge density increases.At the junction,these charges are also recombination centers. When charge carriers diffuse from both sides of the junction,recombination occurs close to the junction only due to higher charge density,resulting in decreased depletion width.
Answer: In place of resistance, it is better to talk in tems of resistivity or conductivity (which is inverse of resistivity).One can make either p-type or n-type semiconductor of almost any resistivity. Resistivity of a material depends on three factors:movile charge carrier concentration,movility of charge carriers and on the charge of the carrier. Charge of an electron and hole are same.Electron movility is nearly three times of that of hole in silicon semiconductor. Normally the extent of doping ( which gives carrier concertration ) in a semiconductor desides the magnitude of resistivity.In devices, one of the dopings ( either n- or p-) is very large compared to the other one. One can make p-type semiconductor more resistive by doping it very lightly.
Answer: Have you observed movement of a bubble in a beaker containing water? What is bubble? It is absence of water /fluid at that place. And movement of bubble is actually movement of fluid in the opposite direction. This way, you see that bubble gives rise to movement of fluid. Simlilarly, hole ( which is a vacant place in otherwise filled states -filled with valence electrons) gives rise to movement of valence electrons. If all energy states are filled in the valence band meaning there are no holes in the valence band-then this eletron movement will not be possible. And contribution of holes in the conductivity of semiconductor will not be there. In general,conduction electrons in the conduction band and holes in the valence band in a semiconductor, both are present and both contribute to the total conductivity of semiconductor.
Answer: During the formation of p-n junction,electrons fron n-side diffuse towards p-side of semiconductor.Similarly,holes diffuse from p-side to n-semiconductor.At the place where n- and p- semiconductors meet, these diffusing electrons and holes recombine and elctrons and holes both disappear from that narrow region . This creates space charge region which contains negative ions( IVth group acceptor impurities which become negatively inozed after recieving extra electron)in the p-region and positive ions on n-side ( Vth group donor impurity atoms which become positive ions after releasing their Vth electron)in the n-region.This space charge(pesence of net negative and positive charges in the space region) gives rise to an electric field that opposes the further diffusion of charges. An equilibrium is established when diffusion and drift currents are equal and space charge is thus created.
Answer: Carbon though belongs to group IV ,but it is not a semiconductor. In the crystalline form, for example, it has band gap of around 6 electron volt, which is very high for the material to be a semiconductor. Crystal carbon( diamond)is the best insulator.
Answer: Germanium has a band gap of around 0.7ev while silicon has around 1.1ev. Therefore, germanium devices are more sensitive to temperature changes. Device characteristics change significantly with temperature rise of say 20-30 degrees. This is one reason for lesser use of Ge. In IC formation, Ge is not used because its oxide is difficult to be formed. Oxide formation is an important step in IC fabrication.
Answer: The frequency of operation of a transistor depends on how quickly the charge carriers respond to changing voltages. Higher the charge mobility, quicker will be response and higher will be frquency of operartion.
Answer: The values of built-in voltage and depletion width can be calulated if doping densities on p- and n- side are known. Expressions for these quantities can be found in any book on electrocs.
Answer: Look at the output characteristics of transistor.These characteristics may be divided into three regions,saturation region,cutoff region and active region.In the active region,the collector current ( output current) varies almost linearly with input current( base current in common emitter setup) and this is a requirement if the transistor has to function as an amplfier.The cutoff and saturation regions are used when transistor works as a switch. Common nemitter configuration is most widely used as this is the only fonfiguration which shows current gain as well as valtage gain and also it exhibits input and output impedances of the same order which is helpful in multi stage connection.helps
Answer: This has been answered in the response to your other question.
Answer: You have not followed what I have said in the lecture. Look, there are two types of contacts,viz.ohmic and rectifying. Take a crystal of a semiconductor , we need to put contacts on both the sides of semiconductor crystal so that voltage can be applied.If the contacts are ohmic,current through the semiconductor will remain same on changing the polarity of the applied voltage across the semiconductor.If the contacts are of rectifying type,current will change. The transistors are made by the manufacturer with ohmic contacts on emitter,base and collector regions. In an amplifying circuit,emitter-base junction is forward biased and collector-base is reverse biased always. you should not expect same behavior of transistor if you change the polarity of bias at two junctions.
