Microelectronics: Circuit Analysis and Design
Microelectronics: Circuit Analysis and Design
4th Edition
ISBN: 9780073380643
Author: Donald A. Neamen
Publisher: McGraw-Hill Companies, The
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Chapter 17, Problem 17.9EP

The parameters of the TIL NAND circuit in Figure 17.24 are: R 1 = 12 k Ω , R 2 = 4 k Ω , R B = 2 k Ω , and R C = 6 k Ω . Assume β F β = 25 and β R = 0.1 (for each input emitter). For a no−load condition, determine the base and collector currents in each transistor for: (a) υ X = υ Y = 0.1 V and (b) υ X = υ Y = 5 V . (Ans. (a) i 1 = i B 1 = 0.342 mA , i C 1 0 , i B 2 = i C 2 = 0 , i B o = i C o = 0 ; (b) i 1 = i B 1 = 0.225 mA , i B 2 = | i C 1 | = 0.27 mA , i 2 = i C 2 = 1.025 mA i B o = 0.895 mA , i C o = 0.8167 mA )

(a)

Expert Solution
Check Mark
To determine

The value of the base and the collector current in each of the transistor.

Answer to Problem 17.9EP

The value of the currents are iC1 is 0 , iC2 is 0 , iB2 is 0 , iBO is 0 , iCO is 0 , iB1 is 0.342mA and i1 is 0.342mA ,

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1.

  Microelectronics: Circuit Analysis and Design, Chapter 17, Problem 17.9EP

The expression for the voltage vB2 is given by,

  vB2=vX+vCE(sat)

Substitute 0.1V for vX and 0.1V for vCE(sat) in the above equation.

  vB2=0.1V+0.1V=0.2V

The second transistor and the transistor at the output are both cut off, then the base voltage vB1 is given by,

  vB1=vX+vBE(sat)

Substitute 0.1V for vX and 0.8V for vBE(sat) in the above equation.

  vB1=0.1V+0.8V=0.9V

The expression for the value of the current i1 is given by,

  i1=5VvB112kΩ

Substitute 0.9V for vB1 in the above equation.

  i1=5V0.9V12kΩ=0.342mA

The expression to determine the value of the base current iB1 is given by,

  iB1=i1

Substitute 0.342mA for i1 in the above equation.

  iB1=0.342mA

The transistor two and the output transistor are cutoff, therefore the value of the current are given by,

  iC1=0iC2=0iB2=0iBO=0

The expression for the value of the current iCO is given by,

  iCO=0 .

Conclusion:

Therefore, the value of the currents are iC1 is 0 , iC2 is 0 , iB2 is 0 , iBO is 0 , iCO is 0 , iB1 is 0.342mA and i1 is 0.342mA

(b)

Expert Solution
Check Mark
To determine

The value of the base and the collector current in each of the transistor.

Answer to Problem 17.9EP

The value of the currents are iC1 is 0.27mA , i2 is 1.025mA , iC2 is 1.025mA , iB2 is 0.27mA , iBO is 0.895mA , iCO is 0.8167mA , iB1 is 0.255mA and i1 is 0.255mA .

Explanation of Solution

Calculation:

The input transistor is biased with inverse active mode when the input voltage is 5V .

The expression for the voltage vB1 is given by,

  vB1=vBE(sat)QO+vBE(sat)Q1+vBE(sat)Q1

Substitute 0.7V for vBE(sat)Q1 , 0.8V for vBE(sat)QO and 0.8V for vBE(sat)Q2 in the above equation.

  vB2=0.8V+0.8V+0.7V=2.3V

The expression to determine the value of the collector voltage is given by,

  vC2=VBE(sat)QO+VCE(sat)Q2

Substitute 0.1V for VCE(sat)Q2 and 0.8V for vBE(sat)QO in the above equation.

  vC2=0.1V+0.8V=0.9V

The expression to determine the value of the current i1 is given by,

  i1=5VvB112kΩ

Substitute 2.3V for vB1 in the above equation.

  i1=5V2.3V12kΩ=0.255mA

The expression to determine the value of the base current iB1 is given by,

  iB1=i1

Substitute 0.255mA for i1 in the above equation.

  iB1=0.255mA

The expression to determine the value of the current iB2 is given by,

  iB2=(1+2βR)i1

Substitute 0.1 for βR and 0.225mA for i1 in the above equation.

  iB2=(1+2( 0.1))(0.225mA)=0.27mA

The expression to determine the value of the collector current iC1 is given by,

  iC1=iB2

Substitute 0.27mA for iB2 in the above equation.

  iC1=0.27mA

The expression to determine the value of the current i2 is given by,

  i2=5VvC24kΩ

Substitute 0.9V for vC2 in the above equation.

  i2=5V0.9V4kΩ=1.025mA

The expression to determine the value of the base current iC2 is given by,

  iC2=i2

Substitute 1.025mA for i2 in the above equation.

  iC2=1.025mA

The expression to determine the value of the base current iE2 is given by,

  iE2=i2+iB2

Substitute 1.025mA for i2 and 0.27mA for iB2 in the above equation.

  iE2=0.27mA+1.025mA=1.295mA

The expression to determine the value of the current i4 is given by,

  i4=VBE(sat)2kΩ

Substitute 0.8V for VBE(sat) in the above equation.

  i4=0.8V2kΩ=0.4mA

The expression to determine the value of the current iBO is given by,

  iBO=iE2i4

Substitute 1.295mA for iE2 and 0.4mA for i4 in the above equation.

  iBO=1.295mA0.4mA=0.895mA

The expression to determine the value of the current iCO is given by,

  iCO=5VVCE(sat)6kΩ

Substitute 0.1V for VCE(sat) in the above equation.

  iCO=5V0.1V6kΩ=0.8167mA

Conclusion:

Therefore, the value of the currents are iC1 is 0.27mA , i2 is 1.025mA , iC2 is 1.025mA , iB2 is 0.27mA , iBO is 0.895mA , iCO is 0.8167mA , iB1 is 0.255mA and i1 is 0.255mA .

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Chapter 17 Solutions

Microelectronics: Circuit Analysis and Design

Ch. 17 - The ECL circuit in Figure 17.19 is an example of...Ch. 17 - Consider the basic DTL circuit in Figure 17.20...Ch. 17 - The parameters of the TIL NAND circuit in Figure...Ch. 17 - Prob. 17.10EPCh. 17 - Prob. 17.5TYUCh. 17 - Prob. 17.6TYUCh. 17 - Prob. 17.7TYUCh. 17 - Prob. 17.8TYUCh. 17 - Prob. 17.11EPCh. 17 - Prob. 17.12EPCh. 17 - Prob. 17.9TYUCh. 17 - Prob. 17.10TYUCh. 17 - Prob. 17.11TYUCh. 17 - Prob. 1RQCh. 17 - Why must emitterfollower output stages be added to...Ch. 17 - Sketch a modified ECL circuit in which a Schottky...Ch. 17 - Explain the concept of series gating for ECL...Ch. 17 - Sketch a diodetransistor NAND circuit and explain...Ch. 17 - Explain the operation and purpose of the input...Ch. 17 - Sketch a basic TTL NAND circuit and explain its...Ch. 17 - Prob. 8RQCh. 17 - Prob. 9RQCh. 17 - Prob. 10RQCh. 17 - Explain the operation of a Schottky clamped...Ch. 17 - Prob. 12RQCh. 17 - Prob. 13RQCh. 17 - Sketch a basic BiCMOS inverter and explain its...Ch. 17 - For the differential amplifier circuit ¡n Figure...Ch. 17 - Prob. 17.2PCh. 17 - Prob. 17.3PCh. 17 - Prob. 17.4PCh. 17 - Prob. 17.5PCh. 17 - Prob. 17.6PCh. 17 - Prob. 17.7PCh. 17 - Prob. 17.8PCh. 17 - Prob. 17.9PCh. 17 - Prob. 17.10PCh. 17 - Prob. 17.11PCh. 17 - Prob. 17.12PCh. 17 - Prob. 17.13PCh. 17 - Prob. 17.14PCh. 17 - Prob. 17.15PCh. 17 - Prob. 17.16PCh. 17 - Prob. 17.17PCh. 17 - Prob. 17.18PCh. 17 - Consider the DTL circuit shown in Figure P17.19....Ch. 17 - Prob. 17.20PCh. 17 - Prob. 17.21PCh. 17 - Prob. 17.22PCh. 17 - Prob. 17.23PCh. 17 - Prob. 17.24PCh. 17 - Prob. 17.25PCh. 17 - Prob. 17.26PCh. 17 - Prob. 17.27PCh. 17 - Prob. 17.28PCh. 17 - Prob. 17.29PCh. 17 - Prob. 17.30PCh. 17 - Prob. 17.31PCh. 17 - Prob. 17.32PCh. 17 - Prob. 17.33PCh. 17 - For the transistors in the TTL circuit in Figure...Ch. 17 - Prob. 17.35PCh. 17 - Prob. 17.36PCh. 17 - Prob. 17.37PCh. 17 - Prob. 17.38PCh. 17 - Prob. 17.39PCh. 17 - Prob. 17.40PCh. 17 - Prob. 17.41PCh. 17 - Prob. 17.42PCh. 17 - Prob. 17.43PCh. 17 - Prob. 17.44PCh. 17 - Design a clocked D flipflop, using a modified ECL...Ch. 17 - Design a lowpower Schottky TTL exclusiveOR logic...Ch. 17 - Design a TTL RS flipflop.
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