Consider the circuit in Figure P2.62. The output of a diode AND logic gate is connected to the input of a second diode AND logic gate. Assume V γ = 0.6 V for each diode. Determine the outputs V O 1 and V O 2 for: (a) V 1 = V 2 = 5 V ; (b) V 1 = 0 , V 2 = 5 V ; and (c) V 1 = V 2 = 0 . What can be said about the relative values of V O 1 and V O 2 in their “low” state? Figure P2.62

Question

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Answer 1

Textbook Question

Chapter 2, Problem 2.62P

Consider the circuit in Figure P2.62. The output of a diode AND logic gate is connected to the input of a second diode AND logic gate. Assume $V γ = 0.6 V$ for each diode. Determine the outputs $V O 1$ and $V O 2$ for: (a) $V 1 = V 2 = 5 V$ ; (b) $V 1 = 0$ , $V 2 = 5 V$ ; and (c) $V 1 = V 2 = 0$ . What can be said about the relative values of $V O 1$ and $V O 2$ in their “low” state?

Chapter 2, Problem 2.62P, Consider the circuit in Figure P2.62. The output of a diode AND logic gate is connected to the input
Figure P2.62

(a)

Expert Solution

To determine

Values of output voltages $VO1 and VO2$ for given voltages in all parts.

Answer to Problem 2.62P

The values of output voltages are $VO1=5 V and VO2=5 V$ .

The values of output voltages are $VO1=0.6 V and VO2=1.2 V$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 1

Voltages given:

$V1=V2=5 V$

$V1=0,V2=5 V$

$V1=V2=0$

Calculation:

(i) Assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=5 V$ .

The diodes $D1 and D2$ are in the forward biased or ON state.

Hence, from the expression for ideal diode output voltages are $VO1=5 V and VO2=5 V$ .

(ii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs, $V1=0,V2=5 V$ .

The diodes $D1$ is in the reverse biased while diode $D2$ is in forward biased.

Calculating the value of output voltage, $VO1$ is,

$VO1=V1+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

(iii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=0$ .

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$

Therefore, in their LOW states the value of $VO2 is 0.6 V greater than VO1$

(b)

Expert Solution

To determine

Relative values of VO1 and VO2 in their “low” state

Answer to Problem 2.62P

Logic “0” signal degrades as it goes through additional logic gates.

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 2

Circuit diagram:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 3

Logic “0” signal degrades as it goes through additional logic gates.

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is,

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

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Answer 2

Textbook Question

Chapter 2, Problem 2.62P

Consider the circuit in Figure P2.62. The output of a diode AND logic gate is connected to the input of a second diode AND logic gate. Assume $V γ = 0.6 V$ for each diode. Determine the outputs $V O 1$ and $V O 2$ for: (a) $V 1 = V 2 = 5 V$ ; (b) $V 1 = 0$ , $V 2 = 5 V$ ; and (c) $V 1 = V 2 = 0$ . What can be said about the relative values of $V O 1$ and $V O 2$ in their “low” state?

Chapter 2, Problem 2.62P, Consider the circuit in Figure P2.62. The output of a diode AND logic gate is connected to the input
Figure P2.62

(a)

Expert Solution

To determine

Values of output voltages $VO1 and VO2$ for given voltages in all parts.

Answer to Problem 2.62P

The values of output voltages are $VO1=5 V and VO2=5 V$ .

The values of output voltages are $VO1=0.6 V and VO2=1.2 V$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 1

Voltages given:

$V1=V2=5 V$

$V1=0,V2=5 V$

$V1=V2=0$

Calculation:

(i) Assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=5 V$ .

The diodes $D1 and D2$ are in the forward biased or ON state.

Hence, from the expression for ideal diode output voltages are $VO1=5 V and VO2=5 V$ .

(ii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs, $V1=0,V2=5 V$ .

The diodes $D1$ is in the reverse biased while diode $D2$ is in forward biased.

Calculating the value of output voltage, $VO1$ is,

$VO1=V1+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

(iii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=0$ .

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$

Therefore, in their LOW states the value of $VO2 is 0.6 V greater than VO1$

(b)

Expert Solution

To determine

Relative values of VO1 and VO2 in their “low” state

Answer to Problem 2.62P

Logic “0” signal degrades as it goes through additional logic gates.

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 2

Circuit diagram:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 3

Logic “0” signal degrades as it goes through additional logic gates.

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is,

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

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Answer 3

Textbook Question

Chapter 2, Problem 2.62P

Consider the circuit in Figure P2.62. The output of a diode AND logic gate is connected to the input of a second diode AND logic gate. Assume $V γ = 0.6 V$ for each diode. Determine the outputs $V O 1$ and $V O 2$ for: (a) $V 1 = V 2 = 5 V$ ; (b) $V 1 = 0$ , $V 2 = 5 V$ ; and (c) $V 1 = V 2 = 0$ . What can be said about the relative values of $V O 1$ and $V O 2$ in their “low” state?

Chapter 2, Problem 2.62P, Consider the circuit in Figure P2.62. The output of a diode AND logic gate is connected to the input
Figure P2.62

(a)

Expert Solution

To determine

Values of output voltages $VO1 and VO2$ for given voltages in all parts.

Answer to Problem 2.62P

The values of output voltages are $VO1=5 V and VO2=5 V$ .

The values of output voltages are $VO1=0.6 V and VO2=1.2 V$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 1

Voltages given:

$V1=V2=5 V$

$V1=0,V2=5 V$

$V1=V2=0$

Calculation:

(i) Assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=5 V$ .

The diodes $D1 and D2$ are in the forward biased or ON state.

Hence, from the expression for ideal diode output voltages are $VO1=5 V and VO2=5 V$ .

(ii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs, $V1=0,V2=5 V$ .

The diodes $D1$ is in the reverse biased while diode $D2$ is in forward biased.

Calculating the value of output voltage, $VO1$ is,

$VO1=V1+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

(iii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=0$ .

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$

Therefore, in their LOW states the value of $VO2 is 0.6 V greater than VO1$

(b)

Expert Solution

To determine

Relative values of VO1 and VO2 in their “low” state

Answer to Problem 2.62P

Logic “0” signal degrades as it goes through additional logic gates.

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 2

Circuit diagram:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 3

Logic “0” signal degrades as it goes through additional logic gates.

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is,

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

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Answer 4

Textbook Question

Chapter 2, Problem 2.62P

Consider the circuit in Figure P2.62. The output of a diode AND logic gate is connected to the input of a second diode AND logic gate. Assume $V γ = 0.6 V$ for each diode. Determine the outputs $V O 1$ and $V O 2$ for: (a) $V 1 = V 2 = 5 V$ ; (b) $V 1 = 0$ , $V 2 = 5 V$ ; and (c) $V 1 = V 2 = 0$ . What can be said about the relative values of $V O 1$ and $V O 2$ in their “low” state?

Chapter 2, Problem 2.62P, Consider the circuit in Figure P2.62. The output of a diode AND logic gate is connected to the input
Figure P2.62

(a)

Expert Solution

To determine

Values of output voltages $VO1 and VO2$ for given voltages in all parts.

Answer to Problem 2.62P

The values of output voltages are $VO1=5 V and VO2=5 V$ .

The values of output voltages are $VO1=0.6 V and VO2=1.2 V$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 1

Voltages given:

$V1=V2=5 V$

$V1=0,V2=5 V$

$V1=V2=0$

Calculation:

(i) Assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=5 V$ .

The diodes $D1 and D2$ are in the forward biased or ON state.

Hence, from the expression for ideal diode output voltages are $VO1=5 V and VO2=5 V$ .

(ii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs, $V1=0,V2=5 V$ .

The diodes $D1$ is in the reverse biased while diode $D2$ is in forward biased.

Calculating the value of output voltage, $VO1$ is,

$VO1=V1+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

(iii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=0$ .

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$

Therefore, in their LOW states the value of $VO2 is 0.6 V greater than VO1$

(b)

Expert Solution

To determine

Relative values of VO1 and VO2 in their “low” state

Answer to Problem 2.62P

Logic “0” signal degrades as it goes through additional logic gates.

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 2

Circuit diagram:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 3

Logic “0” signal degrades as it goes through additional logic gates.

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is,

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

Values of output voltages $VO1 and VO2$ for given voltages in all parts.

Answer to Problem 2.62P

The values of output voltages are $VO1=5 V and VO2=5 V$ .

The values of output voltages are $VO1=0.6 V and VO2=1.2 V$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 1

Voltages given:

$V1=V2=5 V$

$V1=0,V2=5 V$

$V1=V2=0$

Calculation:

(i) Assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=5 V$ .

The diodes $D1 and D2$ are in the forward biased or ON state.

Hence, from the expression for ideal diode output voltages are $VO1=5 V and VO2=5 V$ .

(ii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs, $V1=0,V2=5 V$ .

The diodes $D1$ is in the reverse biased while diode $D2$ is in forward biased.

Calculating the value of output voltage, $VO1$ is,

$VO1=V1+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

(iii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=0$ .

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$

Therefore, in their LOW states the value of $VO2 is 0.6 V greater than VO1$

(b)

Expert Solution

To determine

Relative values of VO1 and VO2 in their “low” state

Answer to Problem 2.62P

Logic “0” signal degrades as it goes through additional logic gates.

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 2

Circuit diagram:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 3

Logic “0” signal degrades as it goes through additional logic gates.

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is,

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

Answer 8

(a)

Expert Solution

To determine

Values of output voltages $VO1 and VO2$ for given voltages in all parts.

Answer to Problem 2.62P

The values of output voltages are $VO1=5 V and VO2=5 V$ .

The values of output voltages are $VO1=0.6 V and VO2=1.2 V$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 1

Voltages given:

$V1=V2=5 V$

$V1=0,V2=5 V$

$V1=V2=0$

Calculation:

(i) Assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=5 V$ .

The diodes $D1 and D2$ are in the forward biased or ON state.

Hence, from the expression for ideal diode output voltages are $VO1=5 V and VO2=5 V$ .

(ii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs, $V1=0,V2=5 V$ .

The diodes $D1$ is in the reverse biased while diode $D2$ is in forward biased.

Calculating the value of output voltage, $VO1$ is,

$VO1=V1+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

(iii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=0$ .

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$

Therefore, in their LOW states the value of $VO2 is 0.6 V greater than VO1$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Values of output voltages $VO1 and VO2$ for given voltages in all parts.

Answer 13

Answer to Problem 2.62P

The values of output voltages are $VO1=5 V and VO2=5 V$ .

The values of output voltages are $VO1=0.6 V and VO2=1.2 V$

Answer 14

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 1

Voltages given:

$V1=V2=5 V$

$V1=0,V2=5 V$

$V1=V2=0$

Calculation:

(i) Assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=5 V$ .

The diodes $D1 and D2$ are in the forward biased or ON state.

Hence, from the expression for ideal diode output voltages are $VO1=5 V and VO2=5 V$ .

(ii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs, $V1=0,V2=5 V$ .

The diodes $D1$ is in the reverse biased while diode $D2$ is in forward biased.

Calculating the value of output voltage, $VO1$ is,

$VO1=V1+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

(iii) Let’s assume voltage $Vy=0.6 V$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V1$ & $V2$ are the inputs.

On considering the connection of Diodes it is known that diodes are connected in reverse bias.

So, taking considerations from the first part, it is concluded that the output, $VO1$ is LOW if any one of the inputs, $V1$ and $V2$ is LOW

Hence, if the inputs of circuit are $V1=V2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V1=0,V2=5 V$ then the output is in LOW state, which means that $D1$ is ON.

Similarly, if the inputs of circuit are $V1=5 V,V2=0$ then the output is in LOW state, which means that $D2$ is ON.

Similarly, if the inputs of circuit are $V1=V2=5 V$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $VO1$ from all the above combinations.

Table 1

$V1$	$V2$	$VO1$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $VO1$ is,

$VO1=V1 AND V2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $VO1 and Vx$ values is LOW then the output, $VO2$ is LOW. Also, the voltage $Vx$ is always HIGH. So, the output $VO2$ depends on $VO1$ .Since, if $VO1$ is HIGH the output voltage $VO2$ is also HIGH.

So the expression for the output voltage $VO2$ for ideal diode can be,

$VO2=VO1$

Now for the given inputs $V1=V2=0$ .

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$

Therefore, in their LOW states the value of $VO2 is 0.6 V greater than VO1$

Answer 15

(b)

Expert Solution

To determine

Relative values of VO1 and VO2 in their “low” state

Answer to Problem 2.62P

Logic “0” signal degrades as it goes through additional logic gates.

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 2

Circuit diagram:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 3

Logic “0” signal degrades as it goes through additional logic gates.

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is,

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Relative values of VO1 and VO2 in their “low” state

Answer 20

Answer to Problem 2.62P

Logic “0” signal degrades as it goes through additional logic gates.

Answer 21

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 2

Circuit diagram:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.62P , additional homework tip 3

Logic “0” signal degrades as it goes through additional logic gates.

The diodes $D1 and D2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $VO1$ is,

$VO1=V2+Vy =0+0.6 =0.6 V$

Calculating the value of output voltage, $VO2$ is,

$VO2=VO1+Vy =0.6+0.6 =1.2 V$ .

Hence, output voltages are $VO1=0.6 V and VO2=1.2 V$ .

Answer 22

Ch. 2 - Repeat Example 2.1 if the input voltage is...Ch. 2 - Consider the bridge circuit shown in Figure 2.6(a)...Ch. 2 - Assume the input signal to a rectifier circuit has...Ch. 2 - The input voltage to the halfwave rectifier in...Ch. 2 - Consider the circuit in Figure 2.4. The input...Ch. 2 - The circuit in Figure 2.5(a) is used to rectify a...Ch. 2 - The secondary transformer voltage of the rectifier...Ch. 2 - Determine the fraction (percent) of the cycle that...Ch. 2 - The Zener diode regulator circuit shown in Figure...Ch. 2 - Repeat Example 2.6 for rz=4 . Assume all other...

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Answer to Problem 2.62P

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Answer to Problem 2.62P

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