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Answer Part A Solution Q1 a) 20K i 10K 1 50 _ Vs + _ 2M + _ Vd O AVd + Figure 1a b) We solve the circuit in Figure 1a above by using nodal analysis. At node 1, KCL gives: vs v1 v1 v v 1 o3 3 3 10 x10 2000 x10 20 x10 Multiplying through by 2000x10 3 , we obtain 200vs 301v1 100vo or 2v vo v1 s (1) 3 At node 0, v1 vo vo Avd 20 x103 50 But vd v1 and A 200,000v1 . Then v1 vo 400(vo 200,000v1 ) (2) Substituting v1 from Eq (1) into Eq (2) gives, 0 26,667,067vo 53,333,333vs vo 20 log( 1.9999699) = 6.02 dB vs c) vo 2 x(1.9999699) 3.9999398V from Eq (1) we obtain v1 20.066667V Thus i v1 vo 0.1999mA 20 x103 Q2. Ideal Operational amplifier shown in Figure 1 below with input voltage is vin= 5 sin 3t mV. Rf i R1 _ va i vb + vout vin Figure 2 a. Name the type of Op Amp configuration, shown in Figure 2. b. Derive the formula of close-loop gain, ACL. c. Given vin= 5 sin 3t mV, R1 = 4.7 k and Rf = 47 k . Determine and draw the output voltage waveform. Solution Q2 a) Non Inverting Op-Amp b) Used nodal analysis at node va : v v v 0 a a out R1 Rf at node vb : vb vin at node a va vin v v v 0 in in out R1 Rf R vout (1 f ) vin R1 c) Output voltage vout (1 47 )vin = 55 sin 3t mV 4 .7 vout 55mV 5mV vin wt Answer Part B Q1 i) Derive the formula to calculate Vout. Applying KCL at inverting terminal: I1 + I2 + I3 = If V1 0 V2 0 V3 0 0 Vout R1 R2 R3 Rf V1 V2 V3 Vout R1 R2 R3 Rf Vout V1 V2 V3 Rf R1 R2 R3 V V V Vout R f 1 2 3 R1 R2 R3 ii) Repeat part (i) using R1 = R2 = R3 = R. V V V Vout R f 1 2 3 R1 R2 R3 R1 R2 R3 R iii) V V V Vout R f 1 2 3 R R R Rf Vout V1 V2 V3 R Calculate the output voltage, Vout if V1= 6V, V2= 5V, V3= 8V and Rf = 50kΩ, R1= 70kΩ, R2= 80kΩ, R3= 100kΩ. V V V Vout R f 1 2 3 R1 R2 R3 5V 8V 6V Vout 50k 70k 80k 100k Vout 11.4107V iv) Calculate the output voltage, Vout if R1 = R2 = R3 = 50kΩ. V V V Vout R f 1 2 3 R R R Rf Vout V1 V2 V3 R 50k 6V 5V 8V 50k 19V Vout Vout v) Find the value of R1, R2, R3 if the output voltage, Vout is equal to the average value of the inputs. Assume Rf = 35kΩ. Averaging Amplifier Rf R 1 ; n = no of inputs n R n Rf R n 35k R 3 35k R 105k Q2 Q3. Figure 3 Answer the questions below according to Figure 3. i) Derive the formula to calculate the rate of change of the output, Q Ic t Vout . t Q C Vc C Vc I c t I Vc c t C Vout Vc I Vout c t C I c I in Vin 0 R V I in in I c R I in ii) Vout Vout Vin R t C V in t RC Vout V in t RC Vout during the time that t capacitor charging and discharging. Assume an input to the circuit is a square wave with maximum value of ±5V. The value of capacitor, C and resistor, R are 15nF and 20kΩ. Determine the output voltage’s rate of change, Vout V in t RC 5V (5V ) Vout 10V 33.33mV / S t 300 S 20k 15 F Vout V in t RC 5V (5V ) Vout 10V 33.33mV / S t 20k 15 F 300 S iii) Calculate input and output frequencies when the output, Vout value is 10 Vp-p. Draw the input and output waveforms. Vout 33.33mV / S t Vout t 33.33mV / S 10V thalfcycle 300.03 S 33.33mV / S 1 1 1 freq 1.6665kHz t fullcycle thalfcycle 2 300.03 S 2 Q4. Figure 4 Answer the questions below based on Figure 4. i) Derive the formula to calculate Vout. I Vc c t C V Ic c C t Vc Vin I c I in V I in in C t V Vout RI c RI in R in C t V V Vout RC in RC in t t ii) Determine the output voltage, Vout and the feedback resistance, R. Assume an input to the circuit is a triangular wave with maximum value of ±4V and frequency of 100 kHz. The value of capacitor, C is 0.0011 µF and time constant, τ is 3.3 µS. freq 1 2 thalfcycle thalfcycle 1 1 5 S 2 freq 2 100kHz V V Vout RC in RC in t t V 4V (4V ) slope _ when _ ch arg ing in 1.6V / S 5 S t V Vout RC in t 3.3 S 1.6V / S 5.28V V 4V (4V ) slope _ when _ disch arg ing in 1.6V / S 5 S t V Vout RC in t 3.3 S 1.6V / S 5.28V time _ cons tan t , RC 3.3 S R 3k C 0.0011 F iii) Draw the input and output waveforms.