Semiconductor Electronics

Get insights from 57 questions on Semiconductor Electronics, answered by students, alumni, and experts. You may also ask and answer any question you like about Semiconductor Electronics

Follow Ask Question
57

Questions

0

Discussions

6

Active Users

0

Followers

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 F=μ0*24πi1i2dl

105=107*2*5*5d=10100

d=5*102m

d = 5 cm

New answer posted

2 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

A portion of the output power is returned back to the input.

New answer posted

2 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

g r 0 r R

g 1 r 2 r > R

New question posted

2 months ago

0 Follower 2 Views

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Power gain = ( Δ i c Δ i b ) 2 * R 0 R i

= ( 1 0 * 1 0 3 1 0 0 * 1 0 6 ) 2 * 2 1

= 2 * 104 = x * 104

= 2

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Voltage gain = ΔlCΔlB*RCRB

5*103100*106*20.5

= 200

New answer posted

2 months ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

l=90? 304000=15mA

I1=305000=6mA&l2=9mA

New answer posted

2 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

New answer posted

2 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

τ? =μ? *B?

0.018=μ (0.06)sin? 30? μ=0.6 Work =Uf-Ui2μB7.2*10-2J.

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

D1 forward biased

D2 Reversed biased

So, current will flow through only  D1

l = ( 1 0 . 6 ) v ( 6 0 + 4 0 ) Ω = 0 . 4 1 0 0                

= 4 * 10-3 A

= 4mA

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.