Explanation Calculation: The given diagram is shown in Figure 1. The expression for the voltage v B 1 is given by, v B 1 = V B E ( sat ) Q O + V B E ( sat ) Q 2 + V B E ( sat ) Q 1 Substitute 0.8 V for V B E ( sat ) Q O , 0.8 V for V B E ( sat ) Q 2 and 0.7 V for V B E ( sat ) Q 1 in the above equation. v B 1 = 0.8 V + 0.8 V + 0.7 V = 2 .3 V The expression to determine the collector voltage v C 2 is given by, v C 2 = V B E ( sat ) Q O + V C E ( sat ) Q 2 Substitute 0.8 V for V B E ( sat ) Q O and 0.1 V for V C E ( sat ) Q 2 in the above equation. v C 2 = 0.8 V + 0.1 V = 0 .9 V The expression to determine the current i 1 is given by, i 1 = 5 V − v B 1 12 k Ω Substitute 2 .3 V for v B 1 in the above equation. i 1 = 5 V − 2 .3 V 12 k Ω = 0.225 mA The expression for the current i B 2 is given by, i B 2 = ( 1 + 2 β R ) i 1 Substitute 0.1 for β R and i 1 is 0.225 mA in the above equation. i B 2 = ( 1 + 2 ( 0.1 ) ) ( 0.225 mA ) = 0.27 mA The expression for the current i 2 is given by, i 2 = 5 V − v C 2 4 k Ω Substitute 0

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Ch. 17 - Consider the differential amplifier circuit in...Ch. 17 - Prob. 17.2EP Ch. 17 - The reference circuit in Figure 17.5 is to be...Ch. 17 - Assume the maximum currents in Q3 and Q4 of the...Ch. 17 - Prob. 17.5EP Ch. 17 - Prob. 17.6EP Ch. 17 - Prob. 17.1TYU Ch. 17 - Prob. 17.2TYU Ch. 17 - Prob. 17.7EP Ch. 17 - Prob. 17.3TYU

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