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Semiconductor Electronics: Materials, Devices and

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5 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

In the given circuit, the magnitude of V_L and V_C are twice that of V_R. Given that f = 50 Hz. the inductance of the coil is $\frac{1}{K π} m H .$ The value of K is_________.

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alok kumar singh

Contributor-Level 10

Given

V_L = V_C= 2V_R, f = 50Hz

$L = \frac{1}{k π} m H$

since, V_L = V_C.

then

$V_{n e t} = \sqrt[]{{(V_{L} - V_{C})}^{2} + {(V_{R})}^{2}}$

V_R = V_Net= 220V

I = $\frac{V_{N e t}}{Z} = \frac{220}{\sqrt[]{{(X_{L} - X_{C})}^{2} + 1 2^{2}}}$

$I = 44 A$

$V_{L} = I x_{L} = 2 v_{R}$

= I _xL = 440

x_L = 10

WL = 10

2πfL = 10

$L = \frac{1}{\frac{1}{100} π} k = \frac{1}{100} = 0.01 \approx 0$

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5 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

In the given circuit the input voltage V_in is shown in figure. The cut-in voltage of p-n junction diode (D₁ or D₂) is 0.6 V. Which of the following output voltage (V₀) waveform across the diode is correct?

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alok kumar singh

Contributor-Level 10

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8 months ago

Semiconductor Electronics: Materials, Devices and NCERT Solutions

Why is a p-type semiconductor electrically neutral despite having more holes than electrons?

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Aayush Kumari

Beginner-Level 5

A p-type semiconductor is electrically neutral despite having more holes, because the number of positively charged holes is exactly balanced/equal by the number of negatively charged acceptor ions introduced during doping. so practically untill any volatage is applied the semiconductor remains chargeless in other words doesn't produce any current even after doping.

As per the NCERT Textbooks information"Although the number of holes is more than the number of electrons in a p-type semiconductor, the material as a whole is electrically neutral because the charge of holes is balanced by the negatively charged acceptor ions.”

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8 months ago

Semiconductor Electronics: Materials, Devices and Semiconductor Electronics

How semiconductor diode get damaged under high current?

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Nishtha Datta

Beginner-Level 5

As per NCERT Textbboks"If forward current is too large, it can produce large heating and damage the junction. So, a resistor is used in series with the diode to limit the current in the circuit.”

It means when a semiconductor diode is connected to a source under high current it causes excess heat due to more electrical energy. This excess heat is responsible to damage the junction in semiconductor diode permanently. Student can check out NCERT Solutions for Semicondutor Electronics of class 12 physics.

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8 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

14.15 Write the truth table for the circuits given in Fig. 14.40 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

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Payal Gupta

Contributor-Level 10

14.15 A acts as two inputs of the NOR gate and Y is the output. As shown in the following figure. Hence the output of the circuit is $\overset{?}{A + A}$ = $\overset{?}{A}$

The truth table for the same is given as:

A	Y = ( $\overset{?}{A}$ )
0	1
1	0

This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.

A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates are $\overset{?}{A}$ and $\overset{?}{B}$ , as shown in the following figure

Above is given the inputs for the last NOR gate.

Hence, the output for the circuit can be written as:

Y = $\overset{?}{A + B}$ = $\overset{?}{\overset{?}{A} . \overset{?}{B}}$ = A.B

The truth table for the same can b

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14.15 A acts as two inputs of the NOR gate and Y is the output. As shown in the following figure. Hence the output of the circuit is $\overset{?}{A + A}$ = $\overset{?}{A}$

The truth table for the same is given as:

A	Y = ( $\overset{?}{A}$ )
0	1
1	0

This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.

Above is given the inputs for the last NOR gate.

Hence, the output for the circuit can be written as:

Y = $\overset{?}{A + B}$ = $\overset{?}{\overset{?}{A} . \overset{?}{B}}$ = A.B

The truth table for the same can be written as:

A	B	Y(=A!B)
0	0	0
0	1	0
1	0	0
1	1	1

This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.

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New answer posted

8 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

14.14 Write the truth table for circuit given in Fig. 14.39 below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)

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Payal Gupta

Contributor-Level 10

14.14 A and B are the inputs of the given circuit. The output of the first NOR gate is $\overset{?}{A}$ + $\overset{?}{B}$ . It can be observed from the following figure that the inputs of the second NOR gate become the output of the first one.

Hence, the output of the combination is given as:

Y = $\overset{?}{\overset{?}{A + B} + \overset{?}{A + B}}$ = $\overset{?}{\overset{?}{A} . \overset{?}{B}}$ + $\overset{?}{\overset{?}{A} . \overset{?}{B}}$ = = $\overset{?}{\overset{?}{A} . \overset{?}{B}}$ = $\overset{?}{A}$ + $\overset{?}{B}$ = $\overset{?}{A + B}$

The truth table for this operation is given as:

This is the truth table of an or gate. Hence, this circuit functions as an or gate.

A	B	Y ( = A + B)
0	0	0
0	1	1
1	0	1
1	1	1

This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.

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8 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

14.13 You are given two circuits as shown in Fig. 14.38, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

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Payal Gupta

Contributor-Level 10

14.13 The output of the left NAND gate will be $\overset{?}{A . B}$ , as shown in the following figure:

Hence, the output of the combination of two NAND gates is given as:

Y = ( $\overset{?}{A . B}$ ).( $\overset{?}{A . B}$ ) = $\overset{?}{A . B}$ + $\overset{?}{A . B}$ = AB

Hence the circuit functions as an AND gate.

$\overset{?}{A}$ is the output of the upper left of the NAND gate and $\overset{?}{B}$ is the output of the lower half of the NAND gate, as shown in the following figure.

Hence, the output of the combination of the NAND gates will be given as:

Y = $\overset{?}{A}$ . $\overset{?}{B}$ = $\overset{?}{A}$ + $\overset{?}{B}$ = A + B

Hence, this circuit functions as an OR gate.

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8 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

14.12 Write the truth table for a NAND gate connected as given in Fig. 14.37.

\overset{̅}{A + A}

Hence identify the exact logic operation carried out by this circuit.

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Payal Gupta

Contributor-Level 10

14.12 A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.

Hence, the output can be written as:

Y = $\overset{?}{A + A}$ = $\overset{?}{A}$ + $\overset{?}{A}$ = $\overset{?}{A}$ ……………(i)

The truth table for equation (i) can be drawn as:

A	Y = ( $\overset{?}{A}$ )
0	1
1	0

This circuit functions as a NOT gate. The symbol for this logic circuit is as shown below:

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New answer posted

8 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

14.11 You are given the two circuits as shown in Fig. 14.36. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.

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Payal Gupta

Contributor-Level 10

14.11 A and B are the inputs and Y is the output of the given circuit. The left half of the given figure acts as the NOR gate, while the right half acts as the NOT gate

Hence the output of the NOR gate is $\overset{?}{A + B}$

This will be input for the NOT gate. Its output will be $\overset{?}{A + B}$ = A + B

So Y = A + B

Hence, this circuit functions as an OR gate.

A and B are the inputs and Y is the output of the given circuit. It can be observed from the following figure that the inputs of the right half NOR gate are the outputs of the two NOT gates.

Hence, the output of the given circuit can be written as:

Y = $\overset{?}{A + B}$ = $\overset{?}{A + B}$ = A + BHence this circuit functions as an AND gate.

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New answer posted

8 months ago

Semiconductor Electronics: Materials, Devices and Class 12th

14.10 In a p-n junction diode, the current I can be expressed as

where I0 is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, $k_{B}$

is the Boltzmann constant (8.6*10–5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 * 10–12 A and T = 300 K, then

(a) What will be the forward current at a forward voltage of 0.6 V?

(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?

(c) What is the dynamic resistance?

(d) What will be the current if revers

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14.10 In a p-n junction diode, the current I can be expressed as

is the Boltzmann constant (8.6*10–5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 * 10–12 A and T = 300 K, then

(a) What will be the forward current at a forward voltage of 0.6 V?

(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?

(c) What is the dynamic resistance?

(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

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0 Follower 5 Views

Payal Gupta

Contributor-Level 10

14.10 In a p-n junction diode, the expression for current is given as:

$I = I_{0} e x p (\frac{e V}{2 k_{B} T} - 1)$

Where, $I_{0}$ = Reverse saturation current = 5 $* 10^{- 12}$ A

T = Absolute temperature = 300 K

$k_{B}$ = Boltzmann constant = 8.6 $* 10^{- 5}$ eV/K = 8.6 $* 10^{- 5} * 1.6 * 10^{- 19}$ J/K = 1.376 $* 10^{- 23} J / K$

V = voltage across the diode

e = charge of an electron = 1.6 $* 10^{- 19}$ C

Forward voltage, V = 0.6V

Current, $I = 5 * 10^{- 12} * e x p (\frac{1.6 * 10^{- 19} * 0.6}{1.376 * 10^{- 23} * 300} - 1)$ = $5 * 10^{- 12} * e x p ? (22.256)$

= 0.02315 A

For forward voltage, V = 0.7V, we can write

Current, $I' = 5 * 10^{- 12} * e x p (\frac{1.6 * 10^{- 19} * 0.7}{1.376 * 10^{- 23} * 300} - 1)$ = $5 * 10^{- 12} * e x p ? (26.132)$

= 1.117 A

Hence increase in the current, ΔI = I'-I = 1.117 – 0.02315 = 1.0934 A

Dynamic resistance = $\frac{C h a n g e i n v o l t a g e}{C h a n g e i n C u r r e n t}$ = $\frac{0.7 - 0.6}{1.0934} = 0.091 Ω$

If the reverse bias voltage is changed from 1 V to 2 V, the current will remain same as $I_{0}$ will be equal in both cases. Therefore

...more

14.10 In a p-n junction diode, the expression for current is given as:

$I = I_{0} e x p (\frac{e V}{2 k_{B} T} - 1)$

Where, $I_{0}$ = Reverse saturation current = 5 $* 10^{- 12}$ A

T = Absolute temperature = 300 K

$k_{B}$ = Boltzmann constant = 8.6 $* 10^{- 5}$ eV/K = 8.6 $* 10^{- 5} * 1.6 * 10^{- 19}$ J/K = 1.376 $* 10^{- 23} J / K$

V = voltage across the diode

e = charge of an electron = 1.6 $* 10^{- 19}$ C

Forward voltage, V = 0.6V

Current, $I = 5 * 10^{- 12} * e x p (\frac{1.6 * 10^{- 19} * 0.6}{1.376 * 10^{- 23} * 300} - 1)$ = $5 * 10^{- 12} * e x p ? (22.256)$

= 0.02315 A

For forward voltage, V = 0.7V, we can write

Current, $I' = 5 * 10^{- 12} * e x p (\frac{1.6 * 10^{- 19} * 0.7}{1.376 * 10^{- 23} * 300} - 1)$ = $5 * 10^{- 12} * e x p ? (26.132)$

= 1.117 A

Hence increase in the current, ΔI = I'-I = 1.117 – 0.02315 = 1.0934 A

Dynamic resistance = $\frac{C h a n g e i n v o l t a g e}{C h a n g e i n C u r r e n t}$ = $\frac{0.7 - 0.6}{1.0934} = 0.091 Ω$

If the reverse bias voltage is changed from 1 V to 2 V, the current will remain same as $I_{0}$ will be equal in both cases. Therefore the dynamic resistance in the reverse bias will be $\frac{1}{0} = \infty$

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