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 $V_{\gamma }=0.6\text{V}$ for each diode. Determine the outputs $V_{O1}$ and $V_{O2}$ for: (a) $V_1=V_2=5\text{V}$ ; (b) $V_1=0$ , $V_2=5\text{V}$ ; and (c) $V_1=V_2=0$ . What can be said about the relative values of $V_{O1}$ and $V_{O2}$ in their “low” state?

Figure P2.62

To determine

Values of output voltages $V_{O1}\text{ and }{V}_{O2}$ for given voltages in all parts.

Answer to Problem 2.62P

The values of output voltages are $V_{O1}=5\text{ V and }{V}_{O2}=5\text{ V}$ .

The values of output voltages are $V_{O1}=0.6\text{ V and }{V}_{O2}=1.2\text{ V}$

The values of output voltages are $V_{O1}=0.6\text{ V and }{V}_{O2}=1.2\text{ V}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Voltages given:

  $V_1=V_2=5\text{ V}$

  $V_1=0,V_2=5\text{ V}$

  $V_1=V_2=0$

Calculation:

(i) Assume voltage $V_y=0.6\text{ V}$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V_1$ & $V_2$ 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, $V_{O1}$ is LOW if any one of the inputs, $V_1$ and $V_2$ is LOW

Hence, if the inputs of circuit are $V_1=V_2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V_1=0,V_2=5\text{ V}$ then the output is in LOW state, which means that $D_1$ is ON.

Similarly, if the inputs of circuit are $V_1=5\text{ V},V_2=0$ then the output is in LOW state, which means that $D_2$ is ON.

Similarly, if the inputs of circuit are $V_1=V_2=5\text{ V}$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $V_{O1}$ from all the above combinations.

Table 1

$V_1$	$V_2$	$V_{O1}$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $V_{O1}$ is

  $V_{O1}=V_1\text{ AND }{V}_2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $V_{O1}\text{and }{V}_x$ values is LOW then the output, $V_{O2}$ is LOW. Also, the voltage $V_x$ is always HIGH. So, the output $V_{O2}$ depends on $V_{O1}$ .Since, if $V_{O1}$ is HIGH the output voltage $V_{O2}$ is also HIGH.

So the expression for the output voltage $V_{O2}$ for ideal diode can be,

  $V_{O2}=V_{O1}$

Now for the given inputs $V_1=V_2=5\text{ V}$ .

The diodes $D_1{\text{and D}}_2$ are in the forward biased or ON state.

Hence, from the expression for ideal diode output voltages are $V_{O1}=5\text{ V and }{V}_{O2}=5\text{ V}$ .

(ii) Let’s assume voltage $V_y=0.6\text{ V}$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V_1$ & $V_2$ 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, $V_{O1}$ is LOW if any one of the inputs, $V_1$ and $V_2$ is LOW

Hence, if the inputs of circuit are $V_1=V_2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V_1=0,V_2=5\text{ V}$ then the output is in LOW state, which means that $D_1$ is ON.

Similarly, if the inputs of circuit are $V_1=5\text{ V},V_2=0$ then the output is in LOW state, which means that $D_2$ is ON.

Similarly, if the inputs of circuit are $V_1=V_2=5\text{ V}$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $V_{O1}$ from all the above combinations.

Table 1

$V_1$	$V_2$	$V_{O1}$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $V_{O1}$ is,

  $V_{O1}=V_1\text{ AND }{V}_2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $V_{O1}\text{and }{V}_x$ values is LOW then the output, $V_{O2}$ is LOW. Also, the voltage $V_x$ is always HIGH. So, the output $V_{O2}$ depends on $V_{O1}$ .Since, if $V_{O1}$ is HIGH the output voltage $V_{O2}$ is also HIGH.

So the expression for the output voltage $V_{O2}$ for ideal diode can be,

  $V_{O2}=V_{O1}$

Now for the given inputs, $V_1=0,V_2=5\text{ V}$ .

The diodes $D_1$ is in the reverse biased while diode $D_2$ is in forward biased.

Calculating the value of output voltage, $V_{O1}$ is,

  $\begin{array}{l}{V}_{O1}=V_1+V_y\\ =0+0.6\\ =0.6\text{ V}\end{array}$

Calculating the value of output voltage, $V_{O2}$ is,

  $\begin{array}{l}{V}_{O2}=V_{O1}+V_y\\ =0.6+0.6\\ =1.2\text{ V}\end{array}$

Hence, output voltages are $V_{O1}=0.6\text{ V and }{V}_{O2}=1.2\text{ V}$ .

(iii) Let’s assume voltage $V_y=0.6\text{ V}$ across each diode in the circuit.

The entire circuit can be divided into two parts.

Now considering the first part, $V_1$ & $V_2$ 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, $V_{O1}$ is LOW if any one of the inputs, $V_1$ and $V_2$ is LOW

Hence, if the inputs of circuit are $V_1=V_2=0$ then the output is in LOW state, which means that the diodes are ON.

Similarly, if the inputs of circuit are $V_1=0,V_2=5\text{ V}$ then the output is in LOW state, which means that $D_1$ is ON.

Similarly, if the inputs of circuit are $V_1=5\text{ V},V_2=0$ then the output is in LOW state, which means that $D_2$ is ON.

Similarly, if the inputs of circuit are $V_1=V_2=5\text{ V}$ then the output is in HIGH state, which means that diodes are in OFF.

Let’s make the truth table for $V_{O1}$ from all the above combinations.

Table 1

$V_1$	$V_2$	$V_{O1}$
0 0 5 5	0 5 0 5	0 0 0 5

On considering, the Boolean expression for $V_{O1}$ is,

  $V_{O1}=V_1\text{ AND }{V}_2$

Now, taking considerations from the second part, as diodes are connected in reverse bias it is concluded that if any one of the inputs, $V_{O1}\text{and }{V}_x$ values is LOW then the output, $V_{O2}$ is LOW. Also, the voltage $V_x$ is always HIGH. So, the output $V_{O2}$ depends on $V_{O1}$ .Since, if $V_{O1}$ is HIGH the output voltage $V_{O2}$ is also HIGH.

So the expression for the output voltage $V_{O2}$ for ideal diode can be,

  $V_{O2}=V_{O1}$

Now for the given inputs $V_1=V_2=0$ .

The diodes $D_1{\text{and D}}_2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $V_{O1}$ is

  $\begin{array}{l}{V}_{O1}=V_2+V_y\\ =0+0.6\\ =0.6\text{ V}\end{array}$

Calculating the value of output voltage, $V_{O2}$ is

  $\begin{array}{l}{V}_{O2}=V_{O1}+V_y\\ =0.6+0.6\\ =1.2\text{ V}\end{array}$ .

Hence, output voltages are $V_{O1}=0.6\text{ V and }{V}_{O2}=1.2\text{ V}$

Therefore, in their LOW states the value of $V_{O2}\text{ is 0}\text{.6 V greater than }{V}_{O1}$

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.

  

Circuit diagram:

  

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

The diodes $D_1{\text{and D}}_2$ are in the reversed biased or OFF state.

Calculating the value of output voltage, $V_{O1}$ is,

  $\begin{array}{l}{V}_{O1}=V_2+V_y\\ =0+0.6\\ =0.6\text{ V}\end{array}$

Calculating the value of output voltage, $V_{O2}$ is,

  $\begin{array}{l}{V}_{O2}=V_{O1}+V_y\\ =0.6+0.6\\ =1.2\text{ V}\end{array}$ .

Hence, output voltages are $V_{O1}=0.6\text{ V and }{V}_{O2}=1.2\text{ V}$ .