Circuits and systems

1 In the circuit as shown in Fig Q.1, the Zener diode has a reverse-biased voltage V Z = 2.5 V. Calculate the values of the load voltage V L and its load current I L if R L = 1 k ΩR5RLVL -+Fig. Q.1The load voltage V L is calculated to be(a) 1.62V (b) 2.22V (c) 2.50V (d) 0 V2 In the circuit as shown in Fig Q.1, find the load current I L for the condition as described in Q.1.(a) 0 A(b) 1.62 mA (c) 2.22 mA (d) 6.0 mA3. In the circuit as shown in Fig Q.1, find the zener diode current I Z for the condition as described in Q.1. (a) 0 A (b) 1.62 mA (c) 2.22 mA (d) cannot be determined, required non-linear diode characteristics4. For the circuit in Fig Q.4, the BJT (Q2) has current gain β = 150 , output resistance r o = 20 k Ω , and emitter resistance r e = 15Ω (using T-model representation, operating at the operating temperature).VoutVinFig Q.4When the circuit in Fig Q.4 is used to amplify small a.c. signals, determine the value for the resistance r π for the equivalent a.c. small signal equivalent circuit(a) 20 k Ω (b) 2.25 k Ω (c) 2.265 k Ω (d) 22.5 k Ω5.For the circuit in Fig. Q.4, which one of the following model is NOT a correct representation of the small signal equivalent circuit for the bipolar transistor Q2:Fig 5(a) Fig 5(b)(a) Fig 5 (a) (b) Fig 5 (b) (c) Fig 5 (c) (d) Fig 5 (d)β iFig 5(c) Fig 5(d)6.For the circuit in Fig Q.4, which of the following is the equivalent a.c. small signal circuit representation of the whole circuit:RC 3.9kbeta iRC 3.9kFig 6(a) Fig 6(b)VoutRC 3.9kVinbeta ibeta iRC 3.9kFig 6(c) Fig 6(d)(a) Fig 6 (a) (b) Fig 6 (b) (c) Fig 6 (c) (d) Fig 6 (d)Small signal a.c. equivalent circuit:7. For the circuit in Fig Q.4 and for the conditions as described in Q.4, determine the inputimpedance as seen by the signal source, Z i(a) 2.265 k Ω (b) 2.254 k Ω (c) 470 k Ω (d)1.99 k Ω8. For the circuit in Fig Q.4 and for the conditions as described in Q.4, determine the outputimpedance Z o seen by the load when connected to the terminal V out(a) 3.9 k Ω (b) 20 k Ω(c) 1.322 k Ω (d) 3.264 k Ω9. For the circuit in Fig. Q.4 and for the conditions as described in Q4, determine the voltage gainA v of the amplifier circuit.(a) 217.6 (b) -217.6 (c) 258.3 (d) -258.310. Find an expression for the output voltage V o of the circuit as shown in Fig. Q.10 , relating to theinput voltages V i .R2Fig Q.10(a)42311o i R R V V R R ⎛⎫=+ ⎪⎝⎭(b)32141o i R R V V R R ⎛⎫=+ ⎪⎝⎭(c)423141o i R R V V R R R ⎛⎫=+ ⎪+⎝⎭(d) 214311o iR V V R R ⎛⎫+⎪⎝⎭=⎛⎫+ ⎪⎝⎭11. Using the circuit as shown in Fig. Q.10, calculate the voltage gain (Vout/Vin) using the numericalvalues for the resistors. Given R 1 = 10 k Ω, R 2 = 20 k Ω, R 3 = 15 k Ω, R 4 = 5 k Ω(a) 1 (b) 9(c) 0.75 since 5201025307551510v k k A ..k k k ⎛⎫⎛⎫=+=⨯= ⎪⎪+⎝⎭⎝⎭(d) 2.2512. Find the relationship Vo/V_in for the circuit as shown in Fig. Q.12. Express the relationship interms of the resistance R 1 and R 2 and hence evaluate the numerical value.R2Fig Q.12(a) 2_1oin R V V R =, hence, Vo = 10 V _in (b) 2_1o in RV V R =-, hence, Vo = -10 V _in(c) 2_11o in R V V R ⎛⎫=+ ⎪⎝⎭ , hence Vo = 11 V _in(d) 2_11o in R V V R ⎛⎫=-+ ⎪⎝⎭ , hence Vo = -11 V _in13. Find the relationship V_out/V_in for the circuit as shown in Fig. Q.12. Express the relationship interms of the resistance R 1, R 2, R 3 and R 4 and then evaluate the numerical value.(a) 42__31outin R R V V R R ⎛⎫-=⎪⎝⎭ , hence V _out = -25 V _in (b)42__31out in R R V V R R ⎛⎫⎛⎫= ⎪⎪⎝⎭⎝⎭ , hence V _out = 25 V _in(c) 42__311outin R R V V R R ⎛⎫-=+ ⎪⎝⎭ , hence V _out = -27.5 V _in (d) 2242__3111outin R R V V R R ⎛⎫⎛⎫=++ ⎪⎪⎝⎭⎝⎭ , hence V _out = 38.5 V _in14. Find the relationship Vo/Vs for the circuit as shown in Fig. Q.14. Express the relationship interms of the resistance value R 1 and R 2.R2Fig Q.14(a)21o S R V V R =-(b)121o S R V V R ⎛⎫=+⎪⎝⎭(c) 211o S R V V R ⎛⎫=+ ⎪⎝⎭(d) 12o S RV V R =-15. Suggest numerical values for R2 if the voltage gain Vo/Vs is the same as the voltage gain(V_out/V_in) obtained in Q.13. Given R1 = 5 k Ω(a) 125 k Ω (b) 120 k Ω (c) 132.5 k Ω (d) 187.5 k ΩThe op amp is connected in the non-inverting mode, henceVoltage gain is : 211V RA R =+Given Av = 25, R 1 = 5 k Ω, then R 2 = 24 R 1 = 120 k Ω16. For the circuit as shown in Fig Q.16, express the output voltage Vo as a function of the threeinput voltages V1, V2 and V3.R2Fig Q.16(a) Vo = -10 V1 + 5 V2 – 4 V3 (b) Vo = 10 V1 – 5 V2 + 4 V3 (c) Vo = 11 V1 – 5 V2 + 5 V3 (d) Vo = -11 V1 + 5 V2 – 5 V3Consider the first stage of the inverting amplifier,Consider the second stage of the inverter amplifier,17. For the circuit as shown in Fig Q.16, suggest what value of resistance you would change if theoutput voltage is to be expressed as the following : Vo = 5 V1 – 5V2 + 2 V3(a) Change R5 from 100k to 50 k(b) Change R2 from 100k to 50 k(c) Change R6 from 25k to 50 k(d) Change R3 from 100k to 200k - to replaced by : Change R1 from 10k to 20kThis question has two possible answers as it stands. It should have only one correct answer, and if (d) is replaced by Change R1 from 10k to 20k, then the correct answer will be (b)only18. Consider the emitter-follower amplifier as shown in Fig Q.18.Fig Q.18Calculate the d.c. bias current for the collector, I CQ(a) 2.383 mA(b) 6.415 mA(c) 11.33 mA(d) 13.50 mA19. For the circuit as shown in Q.18 (Fig Q.18), calculate the d.c. bias collector to emitter voltage,V CEQ for the transistor(a) 1.37 V(b) 8.52 V(c) 13.87 V(d) 14.7 V()()()212126310157510101010575076415151011641511510164151010852B CCBB BEQBQB EBQCQBQCCCEQ ER kV V.VR R k kR R R k k kV V..I.AR R k kI I.mAV V I R..V-==⨯=++===Ω--===μ+β++⨯=β==-β+=-⨯⨯⨯=20. For the circuit as shown in Q.18 (Fig Q.18), draw the small signal a.c. equivalent circuit for mid-band operation. Assuming that the circuit is operating in the mid-band region for which the coupling and bypass capacitors are short circuitsr_piRLr_piRLFig Q20aFig Q20bRLRLFig Q20cFig Q20d(a) Fig Q20 a (b) Fig Q20 b (c) Fig Q20 c (d) Fig Q20 d21. For the circuit as shown in Q. 18 (Fig Q.18), find the value of r π (small signal a.c. base to emitterresistance). Given VT = 26 mV.(a) 1.09 k Ω(b) 405.3 Ω (c) 229.5 Ω (d) 192.6 Ω22. For the circuit as shown in Q.18 (Fig Q.18), calculate the voltage gain, A V .(a) 123.4 (b) 82.2 (c) 3.01 (d) 0.98Using the equivalent circuit to determine Vo/Vi :23. For the circuit as shown in Q.18 (Fig Q.18), find the input impedance Z i .(a) 5 k Ω (b) 4.36 k Ω (c) 1 k Ω (d) 34.07 k ΩTo find the input impedance, ()()()()11111140531013333340753407436in B it''L b b in it L b b in Z R Z i r i R V Z r R ...k i i Z k .k .k ππ=+β+===+β+=+⨯=Ω==Ω24.For the circuit as shown in Q.18 (Fig Q.18), find the output impedance Zo .(a) 12.1 Ω (b) 333.3 Ω (c) 500 Ω (d) 1 k ΩTo find the output impedance,()()()()()()()()1158333111111114053833311211101x out x''xx S S B x S b b E'x x S x x b b b 'E E ES x 'out x E S 'S out E V Z i V i i ;R R R k k .;V i r R R V i i i r R i i V R R R r R i Z V R r R r R ..Z R k .πππππ=β++====Ω=-+⎛⎫β+∴β++=-+⇒=-β+=+ ⎪+⎝⎭⎛⎫β+∴==+ ⎪+⎝⎭⎛⎫++⎛⎫∴=== ⎪ ⎪β+⎝⎭⎝⎭Ω25. For the circuit as shown in Q.18 (Fig Q.18), find the current gain, A i .(a) -41.4 (b) 41.4 (c) 8.61 (d) 22To find current gain, and power gain :()43609888615000988861851out L o in i V i in in L V i V Z iZ .k A A ..i Z G A A ...====⨯===⨯=。