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Semiconductor Electronics: Materials, Devices and

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2 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

A potential barrier of 0.4V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0 * 105. The speed with which electron enters the p side will be x 3 * 1 0 5 m s − 1  the value of x is__________.

(Given mass of electron = 9 * 10-31 kg, charge on electron = 1.6 * 10-19 C.)

 
0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

According to Work energy theorem, we can write

K f − K i = W E l e c t r i c     F o r c e ⇒ 1 2 m v 2 − 1 2 m v 0 2 = − e V ⇒ v 2 = v 0 2 − 2 e V m

⇒ v 2 = ( 6 . 0 * 1 0 5 ) 2 − 2 * 1 . 6 * 1 0 − 1 9 9 * 1 0 − 3 1 = 3 2 4 − 1 2 8 9 * 1 0 1 0 = 1 9 6 9 * 1 0 1 0 ⇒ V = 1 4 3 * 1 0 5 m / s

New answer posted

2 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

A potential barrier of 0.4V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0 * 105. The speed with which electron enters the p side will be x 3 * 1 0 5 m s − 1  the value of x is__________.

(Given mass of electron = 9 * 10-31 kg, charge on electron = 1.6 * 10-19 C.)
0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

According to Work energy theorem, we can write

K f − K i = W E l e c t r i c     F o r c e ⇒ 1 2 m v 2 − 1 2 m v 0 2 = − e V ⇒ v 2 = v 0 2 − 2 e V m  

⇒ v 2 = ( 6 . 0 * 1 0 5 ) 2 − 2 * 1 . 6 * 1 0 − 1 9 9 * 1 0 − 3 1 = 3 2 4 − 1 2 8 9 * 1 0 1 0 = 1 9 6 9 * 1 0 1 0 ⇒ V = 1 4 3 * 1 0 5 m / s  

New answer posted

3 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10µA and the collector current changes by 1.5 mA. The load resistance is 5 kW. The voltage gain of the transistor will be ________.
0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Ib = 10 µA

IC = 1.5 mA

RL = 50 kW or (Rc)

Base – emitter voltage = 10 mv

R B = V B I B = 1 0 * 1 0 − 3 1 0 * 1 0 − 6 = 1 0 3 Ω  

A v = ( Δ l C Δ l B ) * ( R C R B )

=  1 . 5 * 1 0 − 3 1 0 * 1 0 − 6 * 5 * 1 0 3 1 0 3

Av = 750

New answer posted

3 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

Drawing power form a 12V car battery, a 9V stabilized DC voltage is required to power a car stereo system, attached to the terminals A and B, as shown in the figure. If a Zener diode with ratings, Vz = 9V and Pmax = 0.27 W, is connected as shown in the figure, for the above purpose, the minimum series resistance Rs must be:-
0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

i m a x = P m a x V z = 3 1 0 0

∴ R s m i n = V R S i m a x

= 3 3 / 1 0 0 = 1 0 0 Ω

New answer posted

3 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

Choose the correct waveform that can represent the voltage across R of the following circuit, assuming the diode is ideal one:
0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Diode, in forward biased condition only, will allow current to flow through it.

Pot. different across resistor is

Δ V = ( 1 0 s i n ω t − 3 ) v o l t  

But in reverse biased condition of diode,

Δ V = 0 (across diode)

New answer posted

3 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

In the given circuit find the current through the Zener diode in mA.
0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

    Potential difference across R?
(R? / (R? +R? ) * V is greater than zenor voltage
⇒ i? = V_z/R? = 20/1000 A = 20 mA
Current through R? , i? = (40-V_z)/R? = 20/500 = 40 mA


Current through zener diode = i? - i? = 20 mA

New question posted

3 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

The truth table for the circuit given in the figure is :
0 Follower 2 Views

New answer posted

3 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

Calculate the value of output voltage V0 (in V) if the Si diode and  Ge diode conduct at 0.7V and 0.3V respectively as shown in figure.

 
0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  I = 1 2 − 0 . 3 5 * 1 0 3 = 2 . 3 4 m A

V 0 = I R = ( 2 . 3 4 * 1 1 − 3 ) ( 5 * 1 0 3 ) = 1 1 . 7 V  

            

New answer posted

3 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

Consider a p-n junction diode which has a potential drop of 0.5 V which is to be taken independent of current (under forward bias). If we want to use 1.5 V cell to forward bias the diode then what should be the value of resister (in Ω) used in series with the diode so that current may not exceed 10 mA, and hence may work safely.
0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

1R + 0.5 = 1.5
1R = 1
R = 1/10 = 100Ω

New answer posted

3 weeks ago

Semiconductor Electronics: Materials, Devices and Class 12th

A common transistor ratio set requires 12V (D.C) for its operation. The D.C source is constructed by using a transformer and a rectifier circuit, which are operated at 220V (A.C) on standard domestic A.C, supply. The number of turns of secondary coil are 24, then the number of turns of primary are___________.
0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  N p N s = V p V s ⇒ N p = V p V s * N s = 2 2 0 1 2 * 2 4 = 4 4 0

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