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Tutorial - 3: Questions on PN junction Resistances and Capacitances (EDC)

Electronic Devices & Circuits


Unit – 1 : Tutorial – 3 (11th August 2015)
Topic: V-I characteristics 
           Static & Dynamic resistances
           Diffusion & Transition Capacitance
           Energy band Diagram
Submission Date : 20th August 2015 (before 12:30 pm)

Theory Questions:
1. Define Cut-in voltage and draw V-I characteristics of Silicon and Germanium PN junction separately.
2. Why reverse saturation current of Germanium is more than silicon at any temperature?
3. Explain importance of modeling and in how many ways a diode can be modeled.
4. Derive mathematical expression for dynamic resistance.
5. Why Diffusion capacitance dominates in forward bias and Transition capacitance dominates in reverse bias?
6. Derive the expression for transition capacitance of abrupt P+N junction.
7. Find the expression for built in potential barrier energy (Eo) with the help of energy band diagram under open circuit PN junction.
8. Draw Energy Band diagram of PN junction under forward and reverse bias.
9. Explain how Diffusion capacitance is formed in forward bias and give the expression for it.
10. Define Mean lifetime (τ) and Diffusion length (L) of minority carrier in a PN junction.

Numerical Questions:
1. Find the voltage at “C” in both circuits, when A and B are binary signals, which can hold either 0 volts (LOW) or VCC (HIGH). Write comments on the results obtained.

2. Find V0, I, I1 and I2 when (a) the diodes are ideal (b) when D1 has cut-in voltage of 0.2 volts and D2 has cut-in voltage of 0.7 voltage.


3.Find VD1, VD2 and IO when (a) the diodes are ideal (b) when D1 has cut-in voltage of 0.2 volts and D2 has cut-in voltage of 0.7 volts.


4. Find the dynamic resistance of a silicon PN junction for a forward bias of 0.6 volts. Assume reverse saturation current is 50 nA at room temperature.

5. A silicon P+N junction has Diffusion capacitance of 1 µF, when carrying a current of 1 mA. Calculate Mean life time of holes.

6. A silicon PN junction under reverse bias has depletion region width of 10 µm. Find the depletion capacitance of the diode per square meter.

7. consider an Abrupt PN junction with a built in potential of 0.8 volts. If the transition capacitance is 1 pF when the reverse bias voltage is 1.2 volts. Find the transition capacitance when the reverse voltage is 7.2 votls.

8. Calculate the barrier capacitance of a Germanium PN junction, whose area is 0.5 x 0.5 mm2 and space charge thickness is 3 µm.

9. The transition capacitance of an abrupt PN junction is 10 pF at barrier voltage (VB) of 4 volts. Calculate the change in capacitance for a 1 volt increase in barrier voltage.

10. (a) calculate the factor by which the reverse saturation current is multiplied, when the temperature is increased from 27oC to 100oC.
     (b) A silicon PN junction has a reverse saturation current of 1 µA at 27oC. Find the dynamic resistance at 27oC and at 100oC.

Solutions for Tutorial - 2 on PN Junction (Diode) Electronic Devices and Circuits (EDC)

Unit – 1 : Solutions to Tutorial – 2 (22nd July 2015)
Topic: PN junction under open circuit condition
Submission Date : 31st July 2015 (before 12:30 pm)

Short Questions:
1. What is the significance of arrow direction on symbol of PN junction?
Ans: The arrow direction on the diode symbol indicates the direction of conventional current when the diode is forward biased.

2. Draw the V – I characteristics of Ideal diode and find the resistance in forward bias and reverse bias from the characteristics
Ans:
A diode is said to be in forward bias, when voltage on p-side is more positive than voltage on n-side. For Ideal diode, even if the voltage on p-side is slightly larger (by a minute amount like nano volts or pico volts), then it comes into forward bias and a large current flows from P to N.

Resistance in forward bias of ideal diode = RF = VF/IF ≈ 0/IF ≈ 0 ohms i.e. Ideal diode acts as short circuit in forward bias.

A diode is said to be in reverse bias, when voltage on n-side is more positive than p-side. In practical diode, a small current flows in reverse bias due to movement of minority carriers. But in Ideal diode, no current flows in reverse bias i.e. reverse current is zero.

Resistance in reverse bias of ideal diode = RR = VR/IR ≈ VR/0 ≈ ∞ (infinity) i.e. Ideal diode acts as open circuit in reverse bias.

3. Define Diffusion phenomenon. Discuss whether diffusion occurs due to majority or minority carriers in PN junction.
Ans:
Diffusion is the process of movement of charge carriers from higher concentration region to lower concentration region to make both regions equal, even in the absence of applied electric field.
Example: when a purple drop is placed in pure water, initially it will be at one place, which can assume higher concentration region and surrounding pure water is absence of purple color is assumed as lower concentration region. Under steady state (after some time), as a process of diffusion, entire glass of water will be of same purple color i.e. purple has diffused from higher concentration region to lower concentration region to get equal distribution.
We know in p-type semiconductor, holes are majority carriers and electrons are minority carriers because electrons are less in number compared to holes. Similarly in n-type semiconductor, electrons are majority carriers and holes are minority carriers.
A PN junction is formed when one half of intrinsic semiconductor is doped with trivalent impurities (to get p-region) and other half is doped with pentavalent impurities (to get n-region).
In PN junction, initially holes are more in p-side than n-side and electrons more in n-side than p-side. So due to diffusion phenomenon, charge carriers move from higher concentration region to lower concentration i.e. holes move from P to N side and electrons move from N to P side, which are majority carriers in their respective regions, Diffusion results only due to majority carriers in PN junction.
Note: Movement of charge carriers results in current, so a current results due to diffusion of majority carriers from one side to other in a PN junction, which is called Diffusion current. It is always due to movement of majority carriers and is always flow from P-side to N-side.

4. Define Drift phenomenon. Discuss whether drift occurs due to majority or minority carriers in PN junction.
Ans:
Drift is, by definition, charged particle motion in response to an applied electric field.
When an electric field is applied across a semiconductor, holes will move in the direction of electric field and electrons will move in the opposite direction of electric field, which results in a current, called drift current.

In a PN junction, as a result of diffusion, majority carriers move from one side to other side and creates depletion region, consisting of negative ions on p-side and positive ions on n-side near the junction. Since ions are immobile, there is a fixed positive charge on n-side and equal amount of negative charge on p-side near the junction appear which results in built in electric field, originating in the direction from N to P region as shown. The function of this electric field is to oppose movement of majority carriers to cross the junction but it helps minority carriers to cross the junction by attracting them. Hence minority carriers cross the junction i.e. electrons from p-region enters n-region (opposite direction of electric field) and holes from n-region to p-region (same direction as electric field). Drift definition satisfies to the movement of minority carriers in PN junction. So current due to movement of minority is called drift current. Hence Drift is always due to minority carriers in PN junction.
Note: In a semiconductor, both currents, Drift and Diffusion currents can exist. In short Diffusion current depends on concentration gradient (non-uniform semiconductor) and Drift current depends on potential gradient (electric field).

5. Draw the diagram of PN junction at open circuit condition under equilibrium and indicate depletion region widths, open circuit voltage and electric field direction and neutral regions.
Ans:

Wn = width of depletion region on n-side from the junction
Wp = width of depletion region on p-side from the junction
W = Total width of depletion region
Vo = built in potential barrier voltage (or) contact potential (or) cut in voltage (or) Knee voltage

Long Questions:
6. Prove that the depletion region penetrates more into lightly doped side of PN junction than heavily doped.
Ans:

We know, charge present on p-side of depletion region is always equal to charge present on n-side of depletion region.
Charge on p-side = charge on n-side
volume charge density on p-side * volume on p-side = volume charge density on n-side * volume on n-side
ρvp * volume on p-side = ρvn * volume on n-side
qNA * Area * Wp = qND * Area * Wn
NAWp = NDWn - (1)
from equation (1), if NA = ND then Wp = Wn i.e. width of depletion region penetrates equally for PN junction with equal doping on both sides.
If NA ≠ ND then Wp ≠ Wn i.e. Depletion region penetrates unequally for PN junction with unequal doping on both sides.
If NA > ND (P+N junction) then Wp < Wn and if NA < ND (PN+ junction) then Wp > Wn
Hence we conclude that depletion region penetrates more into lightly doped region than heavily doped region of PN junction.


7. For a PN junction, calculate the change in contact potential if the doping on P-side is increased by a factor of 1000 and doping on N-side is unaffected.
Ans: 0.18 volts

8. For a Germanium diode with ND = 1016 donors/cm3 and NA = 3 x 1018 /cm3. Calculate for a PN junction under open circuit
(a) Built in potential (V
o)
(b) Width of potential barrier
(c) Maximum electric field

Ans:
(a) Vo = 0.138 v
(b) given NA > ND (P+N junction) then Wp < Wn , so W ≈ Wn = [(2ϵVo/q)* (1/ND)]1/2 = 2.4 μm
(c) εmax = [(qNDWn)/ϵ] = 434 V/m

9. Define forward and reverse bias. Mention unique characteristics of PN junction under forward and reverse bias.
Ans :

Forward Bias: A diode is said to be in forward bias, if the voltage on p-side is more positive than voltage on n-side.
In Forward bias,
effective electric field across the junction is reduced compared to electric field under open circuit, since the applied electric field under forward bias is in opposite direction of built in electric field, so net electric field will reduced across the junction.
Movement of majority carriers (Diffusion current) dominates over movement of minority carriers (Drift current), so net current in forward bias is only diffusion current due to movement of majority carriers and always in the direction from P to N .
since majority carriers are very large in number, forward current is large and is in the range of mA.
Because of large current in forward bias, very small resistance is present when the diode is in forward bias. The range of forward resistance is from 10 Ω – 200 Ω.
In forward bias, potential barrier of the diode reduces i.e. height and width of potential barrier decreases when compared to potential barrier under open circuit.
Reverse Bias : A diode is said to be in Reverse bias, if the voltage on n-side is more positive than voltage on p-side.
In Reverse bias,
effective electric field across the junction is increased compared to electric field under open circuit, since the applied electric field under reverse bias is in the same direction of built in electric field, so net electric field will increased across the junction.
Movement of minority carriers (Drift current) dominates over movement of majority carriers (Diffusion current), so net current in reverse bias is only Drift current due to movement of minority carriers and always in the direction from N to P.
As the name implies, minority carriers are small in number, so current due to minority carriers in reverse bias also small, in the range of µA for Germanium and nA for silicon.
We know minority carriers will be created only when covalent bonds break, which depends only on external energy like temperature or light, but not on applied voltage under reverse bias. Hence reverse current is constant for a given temperature, and is independent of applied reverse voltage, because of this reason, reverse current is frequently called as reverse saturation current.
In reverse bias, potential barrier of the diode increases i.e. height and width of potential barrier increases when compared to potential barrier under open circuit.


10. For a Step graded PN junction with NA > ND , under open circuit, Draw a neat sketch for volume charge distribution (ρv), electric field intensity (ε) and effective voltage for holes.
Ans:
from Poisson’s equation, d2v/dx2 = - (ρv/ϵ)
so electric field intensity, ε = -dv/dx = ∫(ρv/ϵ) dx
and potential barrier voltage, V = - ∫ ε dx




Note: The magnitude of Electric field will be maximum at the junction i.e. at x = 0.

Solutions to EDC Tutorial - 1 on Semiconductors

Electronic Devices & Circuits
Solutions to Tutorial – 1 
Topic : Semiconductors 

Short Questions:
1. Explain why every semiconductor is electrically neutral in nature?
Ans: Every atom is electrically neutral in nature as they consists of equal amount of protons (positive charge) and electrons (negative charge). We know the charge of an electron is -1.6 x 10-19coulombs and the charge of a proton is +1.6 x 10-19coulombs.
Intrinsic semiconductor is a material which consists of all identical atoms arranged in a particular dimension. Since atom is electrically neutral, so the intrinsic semiconductor.
Extrinsic semiconductor is formed when impurity is added to intrinsic semiconductor. The impurity may be either pentavalent impurity atom (to get N-type ) or trivalent impurity atom(to get P-type). Since every atom is neutral, impurity atom is also electrically neutral so extrinsic semiconductor is also electrically neutral.
Hence every semiconductor is electrically neutral.
Note: Ions are atoms with extra electrons (or) missing electrons.
Explanation Video:  https://www.youtube.com/watch?v=-fs1FkT51bM

2. Define free electron, bound (valence) electron, hole, electron current and hole current in a semiconductor?
Ans: electron in outermost orbit or valence band is called valence electron or bound electron, as it is in the control of parent atom. When external energy (like temprature or light) is applied, some covalent bonds break, so that many electrons will move from valence band to conduction band. The electrons in conduction band are called free electrons.
In semiconductor, current flows due to movement of both charge carriers, electrons and holes. Electron current is defined as movement of electrons in conduction band, where as hole current is defined as the opposite direction of current results from movement of electrons in valence band.
Note: hole is an empty energy level in valence band, which can't move itself. But as electron in valence band tries to occupy that vacancy, it looks like hole is moving in the direction opposite to the direction of electron in valence band.
Explanation Video:  https://www.youtube.com/watch?v=I_meqVX3log

3. why intrinsic concentration of germanium is more than silicon at any temperature?
Ans: Intrinsic concentration (ni) is referred as the number of covalent bonds break at a given temperature per unit volume. It is a function of temperature, so as temperature increases, ni increases.
We know, energy gap is defined as the difference between lowest energy level of conduction band and highest energy level of valence band, and is the minimum energy required to break a covalent bond so that an electron from valence band can reach conduction band. When a covalent bond breaks, an electron in conduction band (free electron) and a vacancy in valence band (hole) is created.
Germanium atom has less energy gap than Silicon atom, so at a given temperature ie. For the given thermal energy, more covalent bonds will break in germanium than in silicon, so ni is more in Ge than Si.
Explanation Video: https://www.youtube.com/watch?v=A6ZurAx90bE

4. Why mobility of electron is more than mobility of hole in a semiconductor?
Ans: we know, conduction band consists of only free electrons, remaining are all empty energy levels where as valence band consists of valence electrons and holes (empty energy levels in valence band). So more empty energy levels are available in conduction band than valence band so electron in conduction band can move easily than electron in valence band, that is the reason why mobility of electron is more than hole.
Explanation Video: https://www.youtube.com/watch?v=MzFc4WIlvIA

5. Justify : “Majority carriers in extrinsic semiconductor depends on doping concentration, while minority carriers depends on externally applied energy like temperature, light etc.. (but not on voltage)”.
Ans: Take N-type semiconductor to justify.
N-type semiconductor is formed when pentavalent impurity is added to pure semiconductor. Assume silicon crystal as intrinsic semiconductor and phosphorous as pentavalent impurity atom. Phosphorous has five valence electrons, four out of them covalently bonded with neighboring electrons of silicon and one electron is available as extra electron, which occupies the energy level just below lowest energy level of conduction band of silicon.
At T = 0ok, no thermal energy is applied, so no covalent bonds will break. Hence conduction band is completely empty and valence band is completely full.
no. of free electrons in conduction band = electron concentration (nn) = zero
no. of holes in valence band = hole concentration (pn) = zero
no. of extra electrons = ND
At T = 50ok, all extra electrons get excited as small amount of thermal energy is applied and they all moved into conduction band, so become free electrons. But no covalent bond will break at this temperature i.e. no EHPs are generated.
no. of free electrons in conduction band = electron concentration (nn) = ND
no. of holes in valence band = hole concentration (pn) = zero
no. of extra electrons = zero
At T = 300ok, some covalent bonds will break and that many EHPs will be generated. Say 'x' no. of covalent bonds are broken at 300ok, so x no. of free electrons in conduction band and x no. of holes in valence band gets created. Now
no. of free electrons in conduction band = electron concentration (nn) = ND+ x ≈ ND
no. of holes in valence band = hole concentration (pn) = zero + x = x
so nn depends directly on ND, impurity doping concentration and pn depends on EHPs generated which depends directly on applied external energy in the form of light or temperature.


Long Questions:
1. Find the electrid field required to move an electron from valence band to conduction bond? Assume Silicon atom.
Ans: ε = 45.6 kV/cm.
Solution: The energy of an electron having mass 'm' moving with an average drift velocity 'v' is
w = ½ mv2 = ½ m(μ.ε)2 joules
we know 1eV = 1.6 x 10-19 Joulesand Eg of Si at room temperature ≈ 1 eV = 1.6 x 10-19 Joules
equate the energy of electron to the energy gap of silicon, and find required electric field, ε.
after solving for ε, ε = 45.6 kV/cm.

2. Find the conductivity of intrinsic silicon at room temperature? If it is doped with phosphorous so that majority carrier concentration becomes 10 16 donors/cm3. Find the conductivity of doped semiconductor? Give your comments from the results obtained.
Ans:
Conductivity of intrinsic silicon = 4.32 x 10-6 (Ω-cm)-1
Conductivity of doped semiconductor = 2.08 (Ω-cm)-1
Hence after doping, conductivity is increased by around 5,00,000 times.
Note: Intrinsic semiconductor is not used practically because of its low conductivity, which is not sufficient for practical applications. So impurity is added to intrinsic semiconductor to improve conductivity i.e. extrinsic semiconductor has higher conductivity than intrinsic semiconductor, which is used in real world applications.

3. An N type silicon bar 0.1 cm long and µm2 in cross sectional area has a majority carrier concentration of 5 x 1020 m-3 and the carrier mobility is 0.13 m2/V-sec at 300oK. If the charge of an electron is 1.6 x 10-19 coulomb. Find the resistance of the bar..
Ans: Resistance of the N-type semiconductor bar = 96 MΩ

4. Define Mass Action Law and also find the minority carrier concentration of an n-type silicon, if it is doped with 1 phosphorous in every 1000 silicon atoms per cubic centimeter.

Mass Action Law: under thermal equilibrium, the product of total positive charge and total negative charge is always constant and is equal to square of intrinsic concentration.
n.p = ni2
where n = electron concentration in conduction band and p = hole concentration in valence band.
Ans: Minority carrier concentration of N-type semiconductor given (pn) = 4.5 holes/cm3
Note: Using Mass Action Law, we can find minority carrier concentration in a semiconductor if majority carrier concentration is known and vice versa.

5. Find Minority carrier concentration of p-type silicon sample, whose resistivity is 1 kΩ-cm.
Ans: Minority carrier concentration of P-type semiconductor given (np) = 5 x 1012 electrons/cm3