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denpristay [2]
3 years ago
9

(a) In your own words, explain how donor impurities in semiconductors give rise to free electrons in numbers in excess of those

generated by valence band–conduction band excitations.
(b) Also, explain how acceptor impurities give rise to holes in numbers in excess of those generated by valence band–conduction band excitations.
Engineering
1 answer:
Kazeer [188]3 years ago
8 0

Answer:

A. N type impurities

B. P type impurities

Explanation:

A. The impurities contribute free electrons and changing the conducting property of the semi conductor. When a pentavalent impurities in a semi conductor( impurities with five valence electron) , the impurity atom replace some of the semi conductor atoms in the crystal structure where 4 of the valence electron would be involved in bonding of 4 neighbouring semiconductor while leaving the fifth electron to be free(negative charge carrier) which is available for detachment.

B. When a trivalence impurity is added to semiconductor, instead of excess electron, there will be excess hole created by crystals. Reason for this attribute is the trivalence atom will replace some tetra valence semiconductor atom, when three valence electrons of the 3 valence electrons of the trivalent impurity atom make bond with three neighbouring semiconductor which gives rise to lack of electron in the bond of the fourth neighbouring semiconductor which contribute a whole to the crystalline since trivalent impurity contribute excess holes to the crystal of semi conductor, this holes can accept electrons.

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posledela

Answer:

Explanation:

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3 0
3 years ago
A technician needs to check the heating operation of a heat pump that has no gauge access ports. The technician should start by:
OverLord2011 [107]

The technician should start by checking the temperature rise across the indoor coil.

<h3>Who is a technician?</h3>

This is a person who has skill in a particular area of job. A technician is responsible for repairs and also ensure different equipment and systems are working perfectly.

Hence, the technician should start by checking the temperature rise across the indoor coil.

Learn more about technician here : brainly.com/question/13315405

#SPJ1

7 0
2 years ago
Water flows with a velocity of 3 m/s in a rectangular channel 3 m wide at a depth of 3 m. What is the change in depth and in wat
strojnjashka [21]

Answer: new depth will be 3.462m and the water elevation will be 0.462m.

The maximum contraction will be achieved in width 0<w<3

Explanation:detailed calculation and explanation is shown in the image below

8 0
3 years ago
Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh
Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

R_a=e^{-\left ( \frac{T-r}{n}\right )B}

R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}

R_a=e^{-4.4}=0.01227

For B

R_b=e^{-\left ( \frac{300-336}{100}\right )3}

R_b=e^{1.08}=2.944

B is better for 100 hours

(b)For 750 hours

R_a=e^{-0.5866}=0.55621

R_b=e^{0.144}=1.154

So here B is more Reliable.

3 0
3 years ago
The following C program asks the user for two input null-terminated strings, each stored in uninitialized 100-byte buffer, and c
marissa [1.9K]

Answer:

Code is given below:

Explanation:

.data  

str1: .space 20  

str2: .space 20  

msg1:.asciiz "Please enter string (max 20 characters): "  

msg2: .asciiz "\n Please enter string (max 20 chars): "  

msg3:.asciiz "\nSAME"  

msg4:.asciiz "\nNOT SAME"  

.text

.globl main

main:  

   li $v0,4        #loads msg1  

   la $a0,msg1  

   syscall

   li $v0,8

   la $a0,str1

   addi $a1,$zero,20

   syscall          #got string to manipulate

   li $v0,4        #loads msg2

   la $a0,msg2

   syscall

   li $v0,8

   la $a0,str2

   addi $a1,$zero,20

   syscall         #got string  

       la $a0,str1             #pass address of str1  

   la $a1,str2         #pass address of str2  

   jal methodComp      #call methodComp  

   beq $v0,$zero,ok    #check result  

   li $v0,4

   la $a0,msg4

   syscall

   j exit

ok:  

   li $v0,4  

   la $a0,msg3  

   syscall  

exit:  

   li $v0,10  

   syscall  

methodComp:  

   add $t0,$zero,$zero  

   add $t1,$zero,$a0  

   add $t2,$zero,$a1  

loop:  

   lb $t3($t1)         #load a byte from each string  

   lb $t4($t2)  

   beqz $t3,checkt2    #str1 end  

   beqz $t4,missmatch  

   slt $t5,$t3,$t4     #compare two bytes  

   bnez $t5,missmatch  

   addi $t1,$t1,1      #t1 points to the next byte of str1  

   addi $t2,$t2,1  

   j loop  

missmatch:    

   addi $v0,$zero,1  

   j endfunction  

checkt2:  

   bnez $t4,missmatch  

   add $v0,$zero,$zero  

endfunction:  

   jr $ra

3 0
3 years ago
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