The saturated output voltages are ± V P for the Schmitt trigger in Figure P15.47. (a) Derive the expressions for the crossover voltages V T H and V T L (b) if V P = 12 V , V REF = − 10 V , and R 3 = 10 k Ω , find R 1 and R 2 such that the switching point is V S = − 5 V and the hysteresis width is 0.2 V. (c) Sketch the voltage transfer characteristics. Figure P15.47

Question

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Answer 1

Textbook Question

Chapter 15, Problem 15.47P

The saturated output voltages are $± V P$ for the Schmitt trigger in Figure P15.47. (a) Derive the expressions for the crossover voltages $V T H$ and $V T L$ (b) if $V P = 12 V$ , $V REF = − 10 V$ , and $R 3 = 10 k Ω$ , find $R 1$ and $R 2$ such that the switching point is $V S = − 5 V$ and the hysteresis width is 0.2 V. (c) Sketch the voltage transfer characteristics.

Chapter 15, Problem 15.47P, The saturated output voltages are VP for the Schmitt trigger in Figure P15.47. (a) Derive the
Figure P15.47

(a)

Expert Solution

To determine

To find: The expression for given crossover voltages.

Answer to Problem 15.47P

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 1

Calculation:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 2

From the above circuit

The inverting node is, $v+=vi$

Applying Kirchhoff's current law at non-inverting node:

$v+R1+v+−voR2+v+−VREFR3=0$

$v+R1+v+R2−voR2+v+R3−VREFR3=0$

$v+[1R1+1R2+1R3]−voR2−VREFR3=0$

$v+[1R1+1R2+1R3]=VREFR3+voR2$

$v*=VREFR3+voR21R1+1R2+1R3$

Substitute $v+=v,$ in the above equation,

$vl=VRFFR3+voR21R1+1R2+1R3……$

When $vo=VP$ and $vI=VTH$ in equation-(1),

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

When $vo=−Vp$ and $vt=VTL$ in equation-(1)

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Conclusion:

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

(b)

Expert Solution

To determine

To find: The values of $R1 and R2$

Answer to Problem 15.47P

The required values are $R1=10.17kΩ$ and $R2=600kΩ$

Explanation of Solution

Given:

Saturated output voltage is, $VP=12V$

Reference voltage is, $VREF=−10V$

One resistor value is, $R3=10kΩ$

Switching point is, $Vs=−5V$

Hysteresis width is $VTH−VTL=0.2V$

Calculation:

Substitute $Vp,VREF$ and $R3$ in upper crossover voltage,

$VTH=−1010×103+12R21R1+1R2+110×103$

Hence, $VTH=−1010×103+12R21R1+1R2+110×103$

Substitute $Vp,VREF$ and $R3$ in lower crossover voltage,

$VTL=−1010×103−12R21R1+1R2+110×103$

Substitute $VTH$ and $VTL$ in equation $VTH−VTL=−1010×103+12R21R1+1R2+110×103−−1010×103−12R21R1+1R2+110×103$

$−10 10× 10 3 + 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 − −10 10× 10 3 − 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 =0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 −( −10 10× 10 3 − 12 R 2 ) )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 + 10 10× 10 3 + 12 R 2 )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( 24 R 2 )=0.2$

$24 R 2 =0.2( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$24 0.2 = R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$120= R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 ).......( 3 )$

Assuming $VH$ and $VL$ are symmetrical about zero. The switching voltage becomes,

$Vs=VREFR31R1+1R2+1R3−5=110×1031R1+1R2+110×1035=101R1+1R2+110×1031R1+1R2+110×103=1010×10351R1+1R2+110×103=15×103$

Hence, we get $1R1+1R2+110×103=15×103$

Substitute $1R1+1R2+110×103=15×103$ in equation- (3)

$120=R2(15×103)R2=120(5×103)R2=600×103$

Therefore, value of $R2$ is $R2=600kΩ$

$We have, 1R1+1R2+110×103=15×103$

Substitute $R2=600kΩ$ in the above equation

$1R1+1600×103+110×103=15×1031R1=15×103−1600×103−110×1031R1=120−1−60600×103$

Therefore, the value of $R1$ is $R1=10.17kΩ$

Conclusion:

Therefore, the required values are $R1=10.17kΩ$ and $R2=600kΩ$

(c)

Expert Solution

To determine

To sketch: The voltage transfer characteristics.

Answer to Problem 15.47P

The voltage transfer characteristics are shown in Figure 1.

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 3

Calculation:

The upper crossover voltage of Schmitt trigger is

$VTH=VREFR3+VPR21R1+1R2+1R3=−1010×103+12600×10310.17×103+1600×103+110×103=−1010+12600110.17++1600+110$

$=−1+0.020.09833+0.00167+0.1=−0.980.2VTH=−4.9$

Therefore, the upper crossover voltage of Schmitt trigger is $VTH=−4.9V$

The lower crossover voltage of Schmitt trigger is

$VTL=VREFR3−VPR21R1+1R2+1R3=−1010×103−12600×103110.17×103+1600×103+110×103=−1010−12600110.17+1600+110=10.09833+0.00167+0.1=−1.020.2VTL=−5.1$

Therefore, the lower crossover voltage of Schmitt trigger is $VTL=−5.1V$

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 4

Figure 1

Conclusion:

Therefore, the voltage transfer characteristics are shown in Figure 1.

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Answer 2

Textbook Question

Chapter 15, Problem 15.47P

The saturated output voltages are $± V P$ for the Schmitt trigger in Figure P15.47. (a) Derive the expressions for the crossover voltages $V T H$ and $V T L$ (b) if $V P = 12 V$ , $V REF = − 10 V$ , and $R 3 = 10 k Ω$ , find $R 1$ and $R 2$ such that the switching point is $V S = − 5 V$ and the hysteresis width is 0.2 V. (c) Sketch the voltage transfer characteristics.

Chapter 15, Problem 15.47P, The saturated output voltages are VP for the Schmitt trigger in Figure P15.47. (a) Derive the
Figure P15.47

(a)

Expert Solution

To determine

To find: The expression for given crossover voltages.

Answer to Problem 15.47P

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 1

Calculation:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 2

From the above circuit

The inverting node is, $v+=vi$

Applying Kirchhoff's current law at non-inverting node:

$v+R1+v+−voR2+v+−VREFR3=0$

$v+R1+v+R2−voR2+v+R3−VREFR3=0$

$v+[1R1+1R2+1R3]−voR2−VREFR3=0$

$v+[1R1+1R2+1R3]=VREFR3+voR2$

$v*=VREFR3+voR21R1+1R2+1R3$

Substitute $v+=v,$ in the above equation,

$vl=VRFFR3+voR21R1+1R2+1R3……$

When $vo=VP$ and $vI=VTH$ in equation-(1),

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

When $vo=−Vp$ and $vt=VTL$ in equation-(1)

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Conclusion:

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

(b)

Expert Solution

To determine

To find: The values of $R1 and R2$

Answer to Problem 15.47P

The required values are $R1=10.17kΩ$ and $R2=600kΩ$

Explanation of Solution

Given:

Saturated output voltage is, $VP=12V$

Reference voltage is, $VREF=−10V$

One resistor value is, $R3=10kΩ$

Switching point is, $Vs=−5V$

Hysteresis width is $VTH−VTL=0.2V$

Calculation:

Substitute $Vp,VREF$ and $R3$ in upper crossover voltage,

$VTH=−1010×103+12R21R1+1R2+110×103$

Hence, $VTH=−1010×103+12R21R1+1R2+110×103$

Substitute $Vp,VREF$ and $R3$ in lower crossover voltage,

$VTL=−1010×103−12R21R1+1R2+110×103$

Substitute $VTH$ and $VTL$ in equation $VTH−VTL=−1010×103+12R21R1+1R2+110×103−−1010×103−12R21R1+1R2+110×103$

$−10 10× 10 3 + 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 − −10 10× 10 3 − 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 =0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 −( −10 10× 10 3 − 12 R 2 ) )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 + 10 10× 10 3 + 12 R 2 )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( 24 R 2 )=0.2$

$24 R 2 =0.2( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$24 0.2 = R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$120= R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 ).......( 3 )$

Assuming $VH$ and $VL$ are symmetrical about zero. The switching voltage becomes,

$Vs=VREFR31R1+1R2+1R3−5=110×1031R1+1R2+110×1035=101R1+1R2+110×1031R1+1R2+110×103=1010×10351R1+1R2+110×103=15×103$

Hence, we get $1R1+1R2+110×103=15×103$

Substitute $1R1+1R2+110×103=15×103$ in equation- (3)

$120=R2(15×103)R2=120(5×103)R2=600×103$

Therefore, value of $R2$ is $R2=600kΩ$

$We have, 1R1+1R2+110×103=15×103$

Substitute $R2=600kΩ$ in the above equation

$1R1+1600×103+110×103=15×1031R1=15×103−1600×103−110×1031R1=120−1−60600×103$

Therefore, the value of $R1$ is $R1=10.17kΩ$

Conclusion:

Therefore, the required values are $R1=10.17kΩ$ and $R2=600kΩ$

(c)

Expert Solution

To determine

To sketch: The voltage transfer characteristics.

Answer to Problem 15.47P

The voltage transfer characteristics are shown in Figure 1.

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 3

Calculation:

The upper crossover voltage of Schmitt trigger is

$VTH=VREFR3+VPR21R1+1R2+1R3=−1010×103+12600×10310.17×103+1600×103+110×103=−1010+12600110.17++1600+110$

$=−1+0.020.09833+0.00167+0.1=−0.980.2VTH=−4.9$

Therefore, the upper crossover voltage of Schmitt trigger is $VTH=−4.9V$

The lower crossover voltage of Schmitt trigger is

$VTL=VREFR3−VPR21R1+1R2+1R3=−1010×103−12600×103110.17×103+1600×103+110×103=−1010−12600110.17+1600+110=10.09833+0.00167+0.1=−1.020.2VTL=−5.1$

Therefore, the lower crossover voltage of Schmitt trigger is $VTL=−5.1V$

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 4

Figure 1

Conclusion:

Therefore, the voltage transfer characteristics are shown in Figure 1.

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Answer 3

Textbook Question

Chapter 15, Problem 15.47P

The saturated output voltages are $± V P$ for the Schmitt trigger in Figure P15.47. (a) Derive the expressions for the crossover voltages $V T H$ and $V T L$ (b) if $V P = 12 V$ , $V REF = − 10 V$ , and $R 3 = 10 k Ω$ , find $R 1$ and $R 2$ such that the switching point is $V S = − 5 V$ and the hysteresis width is 0.2 V. (c) Sketch the voltage transfer characteristics.

Chapter 15, Problem 15.47P, The saturated output voltages are VP for the Schmitt trigger in Figure P15.47. (a) Derive the
Figure P15.47

(a)

Expert Solution

To determine

To find: The expression for given crossover voltages.

Answer to Problem 15.47P

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 1

Calculation:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 2

From the above circuit

The inverting node is, $v+=vi$

Applying Kirchhoff's current law at non-inverting node:

$v+R1+v+−voR2+v+−VREFR3=0$

$v+R1+v+R2−voR2+v+R3−VREFR3=0$

$v+[1R1+1R2+1R3]−voR2−VREFR3=0$

$v+[1R1+1R2+1R3]=VREFR3+voR2$

$v*=VREFR3+voR21R1+1R2+1R3$

Substitute $v+=v,$ in the above equation,

$vl=VRFFR3+voR21R1+1R2+1R3……$

When $vo=VP$ and $vI=VTH$ in equation-(1),

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

When $vo=−Vp$ and $vt=VTL$ in equation-(1)

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Conclusion:

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

(b)

Expert Solution

To determine

To find: The values of $R1 and R2$

Answer to Problem 15.47P

The required values are $R1=10.17kΩ$ and $R2=600kΩ$

Explanation of Solution

Given:

Saturated output voltage is, $VP=12V$

Reference voltage is, $VREF=−10V$

One resistor value is, $R3=10kΩ$

Switching point is, $Vs=−5V$

Hysteresis width is $VTH−VTL=0.2V$

Calculation:

Substitute $Vp,VREF$ and $R3$ in upper crossover voltage,

$VTH=−1010×103+12R21R1+1R2+110×103$

Hence, $VTH=−1010×103+12R21R1+1R2+110×103$

Substitute $Vp,VREF$ and $R3$ in lower crossover voltage,

$VTL=−1010×103−12R21R1+1R2+110×103$

Substitute $VTH$ and $VTL$ in equation $VTH−VTL=−1010×103+12R21R1+1R2+110×103−−1010×103−12R21R1+1R2+110×103$

$−10 10× 10 3 + 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 − −10 10× 10 3 − 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 =0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 −( −10 10× 10 3 − 12 R 2 ) )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 + 10 10× 10 3 + 12 R 2 )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( 24 R 2 )=0.2$

$24 R 2 =0.2( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$24 0.2 = R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$120= R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 ).......( 3 )$

Assuming $VH$ and $VL$ are symmetrical about zero. The switching voltage becomes,

$Vs=VREFR31R1+1R2+1R3−5=110×1031R1+1R2+110×1035=101R1+1R2+110×1031R1+1R2+110×103=1010×10351R1+1R2+110×103=15×103$

Hence, we get $1R1+1R2+110×103=15×103$

Substitute $1R1+1R2+110×103=15×103$ in equation- (3)

$120=R2(15×103)R2=120(5×103)R2=600×103$

Therefore, value of $R2$ is $R2=600kΩ$

$We have, 1R1+1R2+110×103=15×103$

Substitute $R2=600kΩ$ in the above equation

$1R1+1600×103+110×103=15×1031R1=15×103−1600×103−110×1031R1=120−1−60600×103$

Therefore, the value of $R1$ is $R1=10.17kΩ$

Conclusion:

Therefore, the required values are $R1=10.17kΩ$ and $R2=600kΩ$

(c)

Expert Solution

To determine

To sketch: The voltage transfer characteristics.

Answer to Problem 15.47P

The voltage transfer characteristics are shown in Figure 1.

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 3

Calculation:

The upper crossover voltage of Schmitt trigger is

$VTH=VREFR3+VPR21R1+1R2+1R3=−1010×103+12600×10310.17×103+1600×103+110×103=−1010+12600110.17++1600+110$

$=−1+0.020.09833+0.00167+0.1=−0.980.2VTH=−4.9$

Therefore, the upper crossover voltage of Schmitt trigger is $VTH=−4.9V$

The lower crossover voltage of Schmitt trigger is

$VTL=VREFR3−VPR21R1+1R2+1R3=−1010×103−12600×103110.17×103+1600×103+110×103=−1010−12600110.17+1600+110=10.09833+0.00167+0.1=−1.020.2VTL=−5.1$

Therefore, the lower crossover voltage of Schmitt trigger is $VTL=−5.1V$

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 4

Figure 1

Conclusion:

Therefore, the voltage transfer characteristics are shown in Figure 1.

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Answer 4

Textbook Question

Chapter 15, Problem 15.47P

The saturated output voltages are $± V P$ for the Schmitt trigger in Figure P15.47. (a) Derive the expressions for the crossover voltages $V T H$ and $V T L$ (b) if $V P = 12 V$ , $V REF = − 10 V$ , and $R 3 = 10 k Ω$ , find $R 1$ and $R 2$ such that the switching point is $V S = − 5 V$ and the hysteresis width is 0.2 V. (c) Sketch the voltage transfer characteristics.

Chapter 15, Problem 15.47P, The saturated output voltages are VP for the Schmitt trigger in Figure P15.47. (a) Derive the
Figure P15.47

(a)

Expert Solution

To determine

To find: The expression for given crossover voltages.

Answer to Problem 15.47P

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 1

Calculation:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 2

From the above circuit

The inverting node is, $v+=vi$

Applying Kirchhoff's current law at non-inverting node:

$v+R1+v+−voR2+v+−VREFR3=0$

$v+R1+v+R2−voR2+v+R3−VREFR3=0$

$v+[1R1+1R2+1R3]−voR2−VREFR3=0$

$v+[1R1+1R2+1R3]=VREFR3+voR2$

$v*=VREFR3+voR21R1+1R2+1R3$

Substitute $v+=v,$ in the above equation,

$vl=VRFFR3+voR21R1+1R2+1R3……$

When $vo=VP$ and $vI=VTH$ in equation-(1),

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

When $vo=−Vp$ and $vt=VTL$ in equation-(1)

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Conclusion:

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

(b)

Expert Solution

To determine

To find: The values of $R1 and R2$

Answer to Problem 15.47P

The required values are $R1=10.17kΩ$ and $R2=600kΩ$

Explanation of Solution

Given:

Saturated output voltage is, $VP=12V$

Reference voltage is, $VREF=−10V$

One resistor value is, $R3=10kΩ$

Switching point is, $Vs=−5V$

Hysteresis width is $VTH−VTL=0.2V$

Calculation:

Substitute $Vp,VREF$ and $R3$ in upper crossover voltage,

$VTH=−1010×103+12R21R1+1R2+110×103$

Hence, $VTH=−1010×103+12R21R1+1R2+110×103$

Substitute $Vp,VREF$ and $R3$ in lower crossover voltage,

$VTL=−1010×103−12R21R1+1R2+110×103$

Substitute $VTH$ and $VTL$ in equation $VTH−VTL=−1010×103+12R21R1+1R2+110×103−−1010×103−12R21R1+1R2+110×103$

$−10 10× 10 3 + 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 − −10 10× 10 3 − 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 =0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 −( −10 10× 10 3 − 12 R 2 ) )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 + 10 10× 10 3 + 12 R 2 )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( 24 R 2 )=0.2$

$24 R 2 =0.2( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$24 0.2 = R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$120= R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 ).......( 3 )$

Assuming $VH$ and $VL$ are symmetrical about zero. The switching voltage becomes,

$Vs=VREFR31R1+1R2+1R3−5=110×1031R1+1R2+110×1035=101R1+1R2+110×1031R1+1R2+110×103=1010×10351R1+1R2+110×103=15×103$

Hence, we get $1R1+1R2+110×103=15×103$

Substitute $1R1+1R2+110×103=15×103$ in equation- (3)

$120=R2(15×103)R2=120(5×103)R2=600×103$

Therefore, value of $R2$ is $R2=600kΩ$

$We have, 1R1+1R2+110×103=15×103$

Substitute $R2=600kΩ$ in the above equation

$1R1+1600×103+110×103=15×1031R1=15×103−1600×103−110×1031R1=120−1−60600×103$

Therefore, the value of $R1$ is $R1=10.17kΩ$

Conclusion:

Therefore, the required values are $R1=10.17kΩ$ and $R2=600kΩ$

(c)

Expert Solution

To determine

To sketch: The voltage transfer characteristics.

Answer to Problem 15.47P

The voltage transfer characteristics are shown in Figure 1.

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 3

Calculation:

The upper crossover voltage of Schmitt trigger is

$VTH=VREFR3+VPR21R1+1R2+1R3=−1010×103+12600×10310.17×103+1600×103+110×103=−1010+12600110.17++1600+110$

$=−1+0.020.09833+0.00167+0.1=−0.980.2VTH=−4.9$

Therefore, the upper crossover voltage of Schmitt trigger is $VTH=−4.9V$

The lower crossover voltage of Schmitt trigger is

$VTL=VREFR3−VPR21R1+1R2+1R3=−1010×103−12600×103110.17×103+1600×103+110×103=−1010−12600110.17+1600+110=10.09833+0.00167+0.1=−1.020.2VTL=−5.1$

Therefore, the lower crossover voltage of Schmitt trigger is $VTL=−5.1V$

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 4

Figure 1

Conclusion:

Therefore, the voltage transfer characteristics are shown in Figure 1.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

To find: The expression for given crossover voltages.

Answer to Problem 15.47P

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 1

Calculation:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 2

From the above circuit

The inverting node is, $v+=vi$

Applying Kirchhoff's current law at non-inverting node:

$v+R1+v+−voR2+v+−VREFR3=0$

$v+R1+v+R2−voR2+v+R3−VREFR3=0$

$v+[1R1+1R2+1R3]−voR2−VREFR3=0$

$v+[1R1+1R2+1R3]=VREFR3+voR2$

$v*=VREFR3+voR21R1+1R2+1R3$

Substitute $v+=v,$ in the above equation,

$vl=VRFFR3+voR21R1+1R2+1R3……$

When $vo=VP$ and $vI=VTH$ in equation-(1),

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

When $vo=−Vp$ and $vt=VTL$ in equation-(1)

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Conclusion:

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

(b)

Expert Solution

To determine

To find: The values of $R1 and R2$

Answer to Problem 15.47P

The required values are $R1=10.17kΩ$ and $R2=600kΩ$

Explanation of Solution

Given:

Saturated output voltage is, $VP=12V$

Reference voltage is, $VREF=−10V$

One resistor value is, $R3=10kΩ$

Switching point is, $Vs=−5V$

Hysteresis width is $VTH−VTL=0.2V$

Calculation:

Substitute $Vp,VREF$ and $R3$ in upper crossover voltage,

$VTH=−1010×103+12R21R1+1R2+110×103$

Hence, $VTH=−1010×103+12R21R1+1R2+110×103$

Substitute $Vp,VREF$ and $R3$ in lower crossover voltage,

$VTL=−1010×103−12R21R1+1R2+110×103$

Substitute $VTH$ and $VTL$ in equation $VTH−VTL=−1010×103+12R21R1+1R2+110×103−−1010×103−12R21R1+1R2+110×103$

$−10 10× 10 3 + 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 − −10 10× 10 3 − 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 =0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 −( −10 10× 10 3 − 12 R 2 ) )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 + 10 10× 10 3 + 12 R 2 )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( 24 R 2 )=0.2$

$24 R 2 =0.2( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$24 0.2 = R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$120= R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 ).......( 3 )$

Assuming $VH$ and $VL$ are symmetrical about zero. The switching voltage becomes,

$Vs=VREFR31R1+1R2+1R3−5=110×1031R1+1R2+110×1035=101R1+1R2+110×1031R1+1R2+110×103=1010×10351R1+1R2+110×103=15×103$

Hence, we get $1R1+1R2+110×103=15×103$

Substitute $1R1+1R2+110×103=15×103$ in equation- (3)

$120=R2(15×103)R2=120(5×103)R2=600×103$

Therefore, value of $R2$ is $R2=600kΩ$

$We have, 1R1+1R2+110×103=15×103$

Substitute $R2=600kΩ$ in the above equation

$1R1+1600×103+110×103=15×1031R1=15×103−1600×103−110×1031R1=120−1−60600×103$

Therefore, the value of $R1$ is $R1=10.17kΩ$

Conclusion:

Therefore, the required values are $R1=10.17kΩ$ and $R2=600kΩ$

(c)

Expert Solution

To determine

To sketch: The voltage transfer characteristics.

Answer to Problem 15.47P

The voltage transfer characteristics are shown in Figure 1.

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 3

Calculation:

The upper crossover voltage of Schmitt trigger is

$VTH=VREFR3+VPR21R1+1R2+1R3=−1010×103+12600×10310.17×103+1600×103+110×103=−1010+12600110.17++1600+110$

$=−1+0.020.09833+0.00167+0.1=−0.980.2VTH=−4.9$

Therefore, the upper crossover voltage of Schmitt trigger is $VTH=−4.9V$

The lower crossover voltage of Schmitt trigger is

$VTL=VREFR3−VPR21R1+1R2+1R3=−1010×103−12600×103110.17×103+1600×103+110×103=−1010−12600110.17+1600+110=10.09833+0.00167+0.1=−1.020.2VTL=−5.1$

Therefore, the lower crossover voltage of Schmitt trigger is $VTL=−5.1V$

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 4

Figure 1

Conclusion:

Therefore, the voltage transfer characteristics are shown in Figure 1.

Answer 8

(a)

Expert Solution

To determine

To find: The expression for given crossover voltages.

Answer to Problem 15.47P

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 1

Calculation:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 2

From the above circuit

The inverting node is, $v+=vi$

Applying Kirchhoff's current law at non-inverting node:

$v+R1+v+−voR2+v+−VREFR3=0$

$v+R1+v+R2−voR2+v+R3−VREFR3=0$

$v+[1R1+1R2+1R3]−voR2−VREFR3=0$

$v+[1R1+1R2+1R3]=VREFR3+voR2$

$v*=VREFR3+voR21R1+1R2+1R3$

Substitute $v+=v,$ in the above equation,

$vl=VRFFR3+voR21R1+1R2+1R3……$

When $vo=VP$ and $vI=VTH$ in equation-(1),

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

When $vo=−Vp$ and $vt=VTL$ in equation-(1)

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Conclusion:

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To find: The expression for given crossover voltages.

Answer 13

Answer to Problem 15.47P

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Answer 14

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 1

Calculation:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 2

From the above circuit

The inverting node is, $v+=vi$

Applying Kirchhoff's current law at non-inverting node:

$v+R1+v+−voR2+v+−VREFR3=0$

$v+R1+v+R2−voR2+v+R3−VREFR3=0$

$v+[1R1+1R2+1R3]−voR2−VREFR3=0$

$v+[1R1+1R2+1R3]=VREFR3+voR2$

$v*=VREFR3+voR21R1+1R2+1R3$

Substitute $v+=v,$ in the above equation,

$vl=VRFFR3+voR21R1+1R2+1R3……$

When $vo=VP$ and $vI=VTH$ in equation-(1),

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

When $vo=−Vp$ and $vt=VTL$ in equation-(1)

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Conclusion:

The upper crossover voltage of Schmitt trigger is $VTH=VREFR3+VPR21R1+1R2+1R3$

The lower crossover voltage of Schmitt trigger is $VTL=VREFR3−VPR21R1+1R2+1R3$

Answer 15

(b)

Expert Solution

To determine

To find: The values of $R1 and R2$

Answer to Problem 15.47P

The required values are $R1=10.17kΩ$ and $R2=600kΩ$

Explanation of Solution

Given:

Saturated output voltage is, $VP=12V$

Reference voltage is, $VREF=−10V$

One resistor value is, $R3=10kΩ$

Switching point is, $Vs=−5V$

Hysteresis width is $VTH−VTL=0.2V$

Calculation:

Substitute $Vp,VREF$ and $R3$ in upper crossover voltage,

$VTH=−1010×103+12R21R1+1R2+110×103$

Hence, $VTH=−1010×103+12R21R1+1R2+110×103$

Substitute $Vp,VREF$ and $R3$ in lower crossover voltage,

$VTL=−1010×103−12R21R1+1R2+110×103$

Substitute $VTH$ and $VTL$ in equation $VTH−VTL=−1010×103+12R21R1+1R2+110×103−−1010×103−12R21R1+1R2+110×103$

$−10 10× 10 3 + 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 − −10 10× 10 3 − 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 =0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 −( −10 10× 10 3 − 12 R 2 ) )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 + 10 10× 10 3 + 12 R 2 )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( 24 R 2 )=0.2$

$24 R 2 =0.2( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$24 0.2 = R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$120= R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 ).......( 3 )$

Assuming $VH$ and $VL$ are symmetrical about zero. The switching voltage becomes,

$Vs=VREFR31R1+1R2+1R3−5=110×1031R1+1R2+110×1035=101R1+1R2+110×1031R1+1R2+110×103=1010×10351R1+1R2+110×103=15×103$

Hence, we get $1R1+1R2+110×103=15×103$

Substitute $1R1+1R2+110×103=15×103$ in equation- (3)

$120=R2(15×103)R2=120(5×103)R2=600×103$

Therefore, value of $R2$ is $R2=600kΩ$

$We have, 1R1+1R2+110×103=15×103$

Substitute $R2=600kΩ$ in the above equation

$1R1+1600×103+110×103=15×1031R1=15×103−1600×103−110×1031R1=120−1−60600×103$

Therefore, the value of $R1$ is $R1=10.17kΩ$

Conclusion:

Therefore, the required values are $R1=10.17kΩ$ and $R2=600kΩ$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To find: The values of $R1 and R2$

Answer 20

Answer to Problem 15.47P

The required values are $R1=10.17kΩ$ and $R2=600kΩ$

Answer 21

Explanation of Solution

Given:

Saturated output voltage is, $VP=12V$

Reference voltage is, $VREF=−10V$

One resistor value is, $R3=10kΩ$

Switching point is, $Vs=−5V$

Hysteresis width is $VTH−VTL=0.2V$

Calculation:

Substitute $Vp,VREF$ and $R3$ in upper crossover voltage,

$VTH=−1010×103+12R21R1+1R2+110×103$

Hence, $VTH=−1010×103+12R21R1+1R2+110×103$

Substitute $Vp,VREF$ and $R3$ in lower crossover voltage,

$VTL=−1010×103−12R21R1+1R2+110×103$

Substitute $VTH$ and $VTL$ in equation $VTH−VTL=−1010×103+12R21R1+1R2+110×103−−1010×103−12R21R1+1R2+110×103$

$−10 10× 10 3 + 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 − −10 10× 10 3 − 12 R 2 1 R 1 + 1 R 2 + 1 10× 10 3 =0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 −( −10 10× 10 3 − 12 R 2 ) )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( −10 10× 10 3 + 12 R 2 + 10 10× 10 3 + 12 R 2 )=0.2$

$1 1 R 1 + 1 R 2 + 1 10× 10 3 ( 24 R 2 )=0.2$

$24 R 2 =0.2( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$24 0.2 = R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 )$

$120= R 2 ( 1 R 1 + 1 R 2 + 1 10× 10 3 ).......( 3 )$

Assuming $VH$ and $VL$ are symmetrical about zero. The switching voltage becomes,

$Vs=VREFR31R1+1R2+1R3−5=110×1031R1+1R2+110×1035=101R1+1R2+110×1031R1+1R2+110×103=1010×10351R1+1R2+110×103=15×103$

Hence, we get $1R1+1R2+110×103=15×103$

Substitute $1R1+1R2+110×103=15×103$ in equation- (3)

$120=R2(15×103)R2=120(5×103)R2=600×103$

Therefore, value of $R2$ is $R2=600kΩ$

$We have, 1R1+1R2+110×103=15×103$

Substitute $R2=600kΩ$ in the above equation

$1R1+1600×103+110×103=15×1031R1=15×103−1600×103−110×1031R1=120−1−60600×103$

Therefore, the value of $R1$ is $R1=10.17kΩ$

Conclusion:

Therefore, the required values are $R1=10.17kΩ$ and $R2=600kΩ$

Answer 22

(c)

Expert Solution

To determine

To sketch: The voltage transfer characteristics.

Answer to Problem 15.47P

The voltage transfer characteristics are shown in Figure 1.

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 3

Calculation:

The upper crossover voltage of Schmitt trigger is

$VTH=VREFR3+VPR21R1+1R2+1R3=−1010×103+12600×10310.17×103+1600×103+110×103=−1010+12600110.17++1600+110$

$=−1+0.020.09833+0.00167+0.1=−0.980.2VTH=−4.9$

Therefore, the upper crossover voltage of Schmitt trigger is $VTH=−4.9V$

The lower crossover voltage of Schmitt trigger is

$VTL=VREFR3−VPR21R1+1R2+1R3=−1010×103−12600×103110.17×103+1600×103+110×103=−1010−12600110.17+1600+110=10.09833+0.00167+0.1=−1.020.2VTL=−5.1$

Therefore, the lower crossover voltage of Schmitt trigger is $VTL=−5.1V$

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 4

Figure 1

Conclusion:

Therefore, the voltage transfer characteristics are shown in Figure 1.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

To sketch: The voltage transfer characteristics.

Answer 27

Answer to Problem 15.47P

The voltage transfer characteristics are shown in Figure 1.

Answer 28

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 3

Calculation:

The upper crossover voltage of Schmitt trigger is

$VTH=VREFR3+VPR21R1+1R2+1R3=−1010×103+12600×10310.17×103+1600×103+110×103=−1010+12600110.17++1600+110$

$=−1+0.020.09833+0.00167+0.1=−0.980.2VTH=−4.9$

Therefore, the upper crossover voltage of Schmitt trigger is $VTH=−4.9V$

The lower crossover voltage of Schmitt trigger is

$VTL=VREFR3−VPR21R1+1R2+1R3=−1010×103−12600×103110.17×103+1600×103+110×103=−1010−12600110.17+1600+110=10.09833+0.00167+0.1=−1.020.2VTL=−5.1$

Therefore, the lower crossover voltage of Schmitt trigger is $VTL=−5.1V$

Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.47P , additional homework tip 4

Figure 1

Conclusion:

Therefore, the voltage transfer characteristics are shown in Figure 1.

Answer 29

Ch. 15 - Design a twopole lowpass Butterworth filter with a...Ch. 15 - Consider the switchedcapacitor circuit in Figure...Ch. 15 - Prob. 15.3EP Ch. 15 - (a) Design a threepole highpass Butterworth active...Ch. 15 - Prob. 15.2TYU Ch. 15 - Prob. 15.3TYU Ch. 15 - Simulate a 25M resistance using the circuit in...Ch. 15 - Design the phaseshift oscillator shown in Figure...Ch. 15 - Design the Wienbridge circuit in Figure 15.17 to...Ch. 15 - Prob. 15.5TYU

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