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I.

Question: 1. A n-type piece of silicon experiences an electric ...​

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1. A n-type piece of silicon experiences an electric field equal to 0.1 V/m.
a. Calculate the velocity of electrons and holes in the material.
b. What doping level is necessary to provide a current density of 1 mA/m^2 under these
conditions? Assume the hole concentration is negligible.

2. A n-type piece of silicon with a length of 0.1 m and cross section area of 0.05 m x 0.05 m sustains a voltage difference of 1 V. If the doping level is 10^17cm^-3, calculate the total
current flowing through the device T=300K.

3. The figure shows a p-type bar of silicon that is subjected to electron injection from left and
hole injection from right. Determine the total current flowing through the device if the cross
section area is equal to 1 m x 1m.

4. Due to a manufacturing error, the p-side of a pn junction has not been doped. If
ND=3X10^13cm^-3, calculate the built-in potential barrier at T=300K.

5. A junction employs ND= 5 x 10^17cm^-3, and NA= 4 x 10^16cm^-3,
a. Determine the majority and minority carrier concentrations on both sides.
b. Calculate the built in potential at 250K, 300K, and 350K.

6. We have received the circuit shown in the figure and wish to determine R-sub1 and I-sub-s.
We note that Vx=1v-->Ix=0.2mA and Vx=2V-->Ix=0.5mA. Calculate R 1and I.s
7. IfIs= 3 x 10^16A, calculate VD1 for IX=1 mA, 2mA, and 4mA.


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  1.
  see the required equation is
  Vd = u*E where Vd is the drift velocity and u is the mobility, E is the field.
  here mobility is the dominant factor because it depends on a lot of things. which must be given. othwerwise you can't guess the velocity.
  although in general in undoped silicon velocity is higher than in doped silicon for both electron and holes.
  
  b)
  here the current density is given by
  J = n*e*Vd
  where n is the carrier concentration, e is the electronic charge and Vd is the drift velocity.
  from the above calculated velocity we can find out n.
  
  
  2. voltage is given so field accross the bar can be foun out
  E = V/d , where d is the length of the bar.
  = 10 V/m
  now as we know
  I = J*A = n*e*A*Vd
  where n is electron charge and A is the area. just put values and get the current value.
  remember Vd = u*E where u is the mobilty which must be mentioned. mobility alos depends on temperature.
  
  3. here the curren is due to diffusion. the formula for this
  A = A*D*e*dQ/dx
  where A area, D is diffusion constant, dq/dx is the carrier concentration gradient.
  the above values need to be give for calculation.
  if the carries injection is done from a battery souce then there will be significant current.
  if it is only charge bombarding then at some point there will be equillibrium and current will be zero.
  
  4. builtin potential is given by,
  Vbi = (kT/q)*(ln(NA*ND/ni^2))
  where k is the boltzman constant, q the electron charge, NA acceptor ion concentration, ND donar ion concentration, ni is the intrinsic concentration or undoped silicon concentration.
  here as NA = ni as there is no P type doping so
  Vbi = (kT/q)*(ln(NA/n)) ; kT/q =0.026 V put the value and get theresult.
  
  5. ND= 5 x 10^17cm^-3, and NA= 4 x 10^16cm^-3 and from mass law action we know
  NA*ND = ni^2
  from the information of the ni in each side calculate NA in the n side and ND in the p side.
  which is greater in magnitude that is the majority carrtier and the other one is the mainority carrier.
  fror the biult in potential use the formula as
  Vbi = (kT/q)*(ln(NA*ND/ni^2))
  6.
  there are two governing equation here,
  (1-Vd1)/R1 = Is*exp(q*Vd1/kT) = 0.2 m--------------------------1
  (2-Vd2)/R1 = Is*exp(q*Vd2/kT) = 0.5 m--------------------------2
  solving the two equationds we get
  2.5 Vd1 - Vd2 = 0.5 and Vd2 - Vd1 = 0.0238
  or Vd1 = 0.373 V and Vd2 = 0.3492 V
  put this vaues in the corresponding above equation you get, Is = 2.938 *10^-10 A and R1 = 3.254 K ohm.
  
  7. t he governing formula is
  Ix = Vd/R1 + Is*exp(q*Vd/k*T) whre k is the boltzman costant and T is the teperature.
  put all the values you will get equations on which you need to do iterative process to find out Vd's. because the equations are nonlinear.
  
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II.

Question: Consider a pn junction in forward bias: To obtain a current of 1 mA with a voltage of 750 mV, ho...​

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III.

Question: Two identical pn junctions are placed in series. a) prove that this combination can be view...​

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Two identical pn junctions are placed in series.
a) prove that this combination can be viewed as a single two-terminal device having an exponential characteristic.
b) For a tenfold change in the current, how much voltage change does such a device require?

please write it clearly!!

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pa unlock po boss

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Question: 1. Given the graph G as shown, give the vertex set and edge set of G. Also, find the degree of ea...​

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  1.)
  Vertex set of G = { v1, v2, v3, v4, v5, v6, v7, v8 }
  Edge set of G = { (v1,v2), (v1, v5), (v2, v3), (v2, v4), (v2, v5), (v3, v5), (v3, v4), (v4, v5), (v4, v6), (v5, v6), (v6, v7), (v7, v8)}
  (remember the edges are undirected, hence (v1,v2) doesn't mean an edge from v1 to v2 but it just means that an edge between v1 and v2
  The degree of each vertex :
  v1 - 2 ,
  v2 - 4 ,
  v3 - 3 ,
  v4 - 4 ,
  v5 - 5,
  v6 - 3,
  v7 - 2,
  v8 - 1
  The order of graph G = 8
  The size of graph G = 12
  2.) Graph will be -
  
  If person 1 is John, then John knows 3 persons
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[Unkown User]

Done!

malaki masyado yung file di maupload dito kaya sa dropmb ko nalang i uupload download mo nalang doon.

Eto yung link:​

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[jonbinsdesu]

Thank you sir

 

[jonbinsdesu]

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Sir pwede pa unlock po ule? Ito po kasi yung answer sa number 3 eh pero wala dun sa pinaunlock ko nung una huhuhuhu salamat po ng marami

 

[Unkown User]

Here,

Question: Draw a flow chart that displays the logical flow possibilities in a program. One execution sequen...​

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Draw a flow chart that displays the logical flow possibilities in a program. One execution sequence that starts at the first vertex A and
ends at the last vertex H. That is, A goes to B, B goes to C, D, and E, C goes to E, D goes to F, E goes to F, F goes to G, G goes to B, and G
goes to H. What is the length of an execution sequence with the least number of steps (or vertices). Also, give a possible execution
sequence with 15 steps (or vertices)

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Pwede po paunlock? Salamat po!

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[XX='Elo Max, c: 1229864, m: 871424'][/XX]
[XX='Elo Max, c: 1229864, m: 871424'][/XX] slamaat po

 

[BototoyTheGreat24]

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sana mapansin ts. thank you in advance!

 

[Unkown User]

Done idol,

 

[Unkown User]

Done idol,

 

[aLpha1267]

Elo Max
Maraming3x salamat po master Godbless po

 

[aLpha1267]

Patulong po ulet master Elo Max
sa pag unlock po
Maraming salamat po..

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paunlock ts

 

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Elo Max pa unlock po ako lods nitong chegg​

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pa unlock po ulit ako, super need lang po.

 

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Question: Example 16.1. A single-effect evaporator is to concentrate 20,000 lb/h (9070 kg/h) of a 20 percen...​

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  Feed = 9070 kg/h
  NaOH in Feed = 0.2
  NaOH in Product = 0.5
  Steam Pressure = 1.37 atm = 138.815 kPa
  Vap Pressure = 100 mmHg = 13.33 kPa
  U = 1400 W/m2°C
  Feed Temperature, Tf = 37.8°C
  Overall Material Balance, F = L + V = 9070 = L+V ....(1)
  Component balance on NaOH, Fxf = Lxl
  9070*0.2 = L *0.5
  L = 3628 kg/h
  V = 5442 kg/h (from Eq (1))
  To determin T1 = Tsat + BPR of 50% concentrated NaOH product, first we obtain Tsat of Pure water from Steam Table. At 13.33 kPa, Tsat = 52°C
  From Duhring chart for Tsat = 41.06 & 50% NaOH, boiling point of the solution is, T1 = 93.33°C
  BPR = T1-Tsat = 93.33-52 = 41.33°C
  From enthalpy concentration chart for superheated steam, Tf = 37.8 °C & xf = 0.2, then hf = 150 kJ/kg
  T1 = 93.33°C & Xl = 0.5 then hf = 545 kJ/kg
  For saturated steam at 1.37 atm from steam table we get,
  Ts = 109.32°C and = 2231.9 kJ/kg
  To get Hv for superheated vapor, first we obtain the enthalpy at T sat = 52°C & P1 = 13.33 kPa, get Hsat = 2594.04kJ/kg
  Then using heat capacity of 1.884 kJ/kg K of superheated steam, Hv = Hsat + Cp BPR
  Hv = 2594.04 + 1.884 *41.33 = 2671.9 kJ/kg
  Substituting in heat balance equation and solving for S,
  Fhf + S lambda = Lhl + VHv
  9070*150 + S*2231.9 = 3628*545 + 5442*2671.9
  S = 6791.196 kg steam / hr
  q = S lambda
  q = 6791.18*2231.9*1000/3600 = 4210344 W
  q = UA(Ts-T1)
  A = 4210344 /(1400* (109.32-93.33)
  A = 188 m2
  Economy = 5442/6791.196
  Economy = 0.8
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I.

Question: 1. Consider the grammar given below (5 points) = [, ] | := |( ) =:= x | y | z (a) What are ...​

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  (C) part question is not given.
  
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II.

Question: 1. Describe in English the language defined by the following grammar: <S> -> <A> &...​

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1. Describe in English the language defined by the following grammar:
<S> -> <A> <B> <C>
<A> -> a <A> | a
<B> -> b <B> | b
<C> -> c <C> | c
2. Consider the following grammar:
<S> -> a <S> c <B> | <A> | b
<A> -> c <A> | c
<B> -> d | <A>
Which of the following sentences are in the language defined by the grammar? Justify your answer by showing derivations or parse trees.
(a) abcd
(b) acccbd
(c) acd
(d) accc
3. Either prove or disprove the ambiguity of the following grammar.
<S> -> <E>
<E> -> <E> * <E> | <id>
<id> -> X | Y | Z
4. Rewrite the following grammar to give + precedence over * and force + to be right associative:
<assign> -> <id> = <expr>
<id> -> A | B | C
<expr> -> <expr> + <term> | <term>
<term> -> <term> * <factor> | <factor>
<factor> -> (<expr>) | <id>
5. Using the your rewritten grammar of Problem 4 above, show a parse tree and a leftmost derivation of the following statement:
A = B * (C * (A + B))
6. Convert the following EBNF to BNF:
<identifier> -> <letter> { [ _ | $ ] <letter_digit>}
<letter_digit> -> <letter> | <digit>
<letter> -> <upper_case_letter> | <lower_case_letter>
<upper_case_letter> -> A | B | C | D | E | F | G
<lower_case_letter> -> a | b | c | d | e | f | g
<digit> -> 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
7. You are to define the syntax of BNF itself.
(a) Write a BNF grammar describing the syntax of BNF; i.e., define the syntax of BNF itself using BNF. Use a pair of single quotes to use meta symbols like | and -> as terminal symbols, e.g., '|' and '->'. You may assume that non-terminals such as <identifier> and <special-symbol> are already defined.
(b) Is the following a sentence of the language you defined in (a) above? Justify your answer by stating the reason.
<expr> -> <expr> + <term> | <term>
<term> -> 5
(c) Based on your answer to (a) above, draw a parse tree of the following sentence.
<expr> -> <expr> <opr> <expr>
<opr> -> + | -
8. Define the syntax of the Connect Four Web service APIs and their JSON outputs. The Web service provides three APIs: info, new, and play. For each service API, define the syntax of:

• calling it, i.e., URL including its query string, e.g., You do not have permission to view the full content of this post. Log in or register now.
• JSON outputs (normal and error/exceptional responses), e.g., {"width":7,"height":6,"strategies":["Smart","Random"]}

Use the Java client (c4-web.jar) available from the course website and/or your favorite Web browser to learn about the APIs of the Web service and their sample outputs.
Refer to the JSON website (You do not have permission to view the full content of this post. Log in or register now.) for the BNF syntax of JSON.
9. Consider the following BNF grammar that describes some of the "technical elective" courses allowed by the CS degree plan.
<tech-electives> -> <tech-courses>
<tech-courses> -> <tech-course> | <tech-course> <tech-courses>
<tech-course> -> CS4390 | CS4371 | CS4373 | CS4330 | CS4317 | CS4339 | CS4342
You need to take at least five technical elective courses. However, no more than two courses of CS4390 (Special Topics), CS4371 (Independent Studies) and CS4373 (Internship) in any combination can count for technical electives.
CS4390 CS4330 CS4317 CS4339 CS4342 -- ok: one special course
CS4390 CS4390 CS4317 CS4339 CS4342 -- ok: two
CS4390 CS4371 CS4317 CS4339 CS4342 -- ok: two
CS4390 CS4390 CS4390 CS4330 CS4317 CS4339 -- ok: why?
CS4390 CS4330 CS4317 -- not okay: < five
CS4390 CS4390 CS4373 CS4330 CS4342 -- not okay: three
Extend the given BNF grammar to an attribute grammar to specify the above rules.

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