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Transcript
PHYS3610/6610 Electronics I – Final – Thursday December 10th 1) The behavior of the transistor used the circuit shown on the left of the figure is described by the characteristic curves on the right of the figure. We have V B=5V , V CC =10V , R B=220k and RC =1k . Take the voltage drop across a forward biased PN junction to be 0.6V . a) [16 points] What is the value of the base current IB ? The base voltage is 5V−0.6V=4.4V . The base current I B =4.4V /R B =4.4V/ 22000=20 A b) [16 points] What are the values of the collector current I C and voltage V C ? The load line is set by the points V CE =10V , I C =0A and V CE =0V , I C =10V/1k =0.01A it intersect the characteristic curve for I B =20 A for V CE ≈7V , I C ≈0.003A=3mA c) [16 points] What is the value of the DC current gain DC ? I C 3×10−3 =150 By definition DC = = I B 20×10−6 d) [17 points] What is the minimal value of R B so the transistor functions as a switch in this circuit with I C =0.01A when the transistor is on? From the graph we see the base current I B must be greater than 60 A so we must have R B5V−0.6V/60×10−6=73.3k PHYS3610/6610 Electronics I – Final – Thursday December 10th 2) Consider the circuit in the figure with R=60 , C g=8 F and C f =0.22nF . Note: this exercise does not involve any complicated complex number calculation. a) [14 points] What is value of the gain at very low frequencies? At very low frequencies the impedances of the 4R capacitors is infinite so A DC =− =−4 R b) [14 points] What is value of the gain A=V OUT /V IN at very high frequencies? At very high frequencies the impedances of the capacitors is negligible in comparison to 4R // 12R 3R =− =−12 the resistor values so A∞ =− R // R /3 R/ 4 c) [14 points] What is the value of the frequency f g for which the input capacitor impedance ∣Z C ∣=R/3 ? What is the value of the frequency f f for which the feedback capacitor impedance ∣Z C ∣=12R ? g f 1 ∣Z C ∣=R/3 ⇒ 2 C g g fg 1 ∣Z C ∣=12R ⇒ 2 C f f f =R / 3⇒ f g = 1 1 = ≈1kHz 2 C g R/ 3 2 8×10−6 60 /3 =12R ⇒ f f = f 1 1 = ≈1MHz 2 C f 12R 2 0.22×10−9 12×60 d) [14 points] What is the value of the gain f g and f f ? A Mid =V OUT /V IN for frequencies between At a frequency f such that f g ≪ f ≪ f f , the input impedance is Z g ≈ R // R/ 3=R/ 4 and the feedback impedance is Z f ≈4R so the gain is 4R A Mid =− =−16 R/ 4 e) [14 points] Calculate the dB values of ∣ ADC∣ , ∣ AMid∣ and ∣ A∞∣ and trace a graph representing the dB gain of the circuit as a function of the frequency on the back of the page. ∣ ADC∣dB=20 log10 4=12.0dB ∣ AMid∣dB=20 log10 16=24.0dB ∣A∞∣dB=20 log 10 12=21.6dB PHYS3610/6610 Electronics I – Final – Thursday December 10th PHYS3610/6610 Electronics I – Final – Thursday December 10th 3) Consider the three inputs (A,B,C), one output (R) digital circuit in the figure. a) [16 points] Fill in the truth table of this circuit. b) [22 points] Fill in the Karnaugh table of this circuit. A B C R 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 A B R C 0 0 0 1 1 1 1 0 0 0 0 1 0 1 1 1 1 1 c) [16 points] From the Karnaugh table write an expression for R as a sum of minterms. We can identify two overlaping groups of conditions for which corresponding expression is R= A⋅BC R=1 and the d) [11 points] Manipulate the expression you obtained for R so it can be implemented with NAND gates only. Provide a schematic of the NAND implementation. Applying de Morgan relations we have R=A.B+C = A.B . C