Quèstion 24 Determine the Damping Factor for Stage Two of the low pass filter below. C1...
Question 27 19 kQ. Assume R6 Optimize this filter for a quality of 63. C1 C2 0.22 uF R1 R2 R3 - R4 R7=4.7 k2, Do not type the unit. Enter your results for R5 in MQ. o Vou (BP) C2 oVout HP C R3 RY R2 RA R7 W oVourLP Rs Summing amplifier Integrator Integrator Re w Question 27 19 kQ. Assume R6 Optimize this filter for a quality of 63. C1 C2 0.22 uF R1 R2 R3 -...
Assume R6=20 kQ. Optimize this filter for a quality of 70. C1 C2 -0.22 uF R1 R2 - R3 R4-R7= 4.7 k, Do not type the unit. Enter your results for R5 in MQ. oVr C R7 oaP Rs Summing amplifier Integrator Integrator Re r Assume R6=20 kQ. Optimize this filter for a quality of 70. C1 C2 -0.22 uF R1 R2 - R3 R4-R7= 4.7 k, Do not type the unit. Enter your results for R5 in MQ. oVr...
Using nodal analysis, calculate the transfer function of notch filter. Use the transfer function to calculate the expected gain (using the actual resistor and capacitor values that were measured for your experiment-See below). Hint: You need to solve the circuit with nodal analysis, using the impedance of a capacitor as - j / (2 * pi * C). The amplifier at the end is just a unity gain amplifier with a gain of 1, so it won't enter into the...
Calculate the voltage gain (Av) for the loaded Common Emitter Amplifier below: BAC 200 for all transistors. Assume re 15 Q for the CE Amplifier. Ignore re' for the DP Amplifier Vcc 12 V, R1 56 K, R2 10 KQ, R4 22 KQ R3 5.6 KQ, RL 8 RE(CE) 628 0, RE(CC) 330 Rc 3.5 KQ, Enter your results. No units. SR3 R1 RC C3 Vcc C1 Beta Q2a HH Vin Beta Q1 Beta Q2b C4 R4 Vs R2 C2...
a) Design a low-pass filter using the given circuitry with a cut-off value of 1 kHz and plot the frequency response curve on the given axes 1.0 0.7 0.5 in out 0.0 101 102 103 104 10s Hz b) Design a band-pass filter using the given circuitry with a bandwidth of 500 Hz and a lower cut-off value of 100 Hz, and draw the frequency response curve. Keep all resistors at the same value (i.e. Ri-R-R3-R4). 1.0 0.7 0.5 0.0...
NAME u (33 PTS.) A FET two-stage amplifier is shown below. (a) Draw the small-signal equivalent circuit oe entire amplifier. (b) Find the voltage gain A,-./. (c) Find the input resistanc resistance Ro. Assume that Id = 1mA. e R, and the output Kn 31.6mAN2 V-1.73 Lambda0 R5 R4 C2 M2 C3 Ro R2 C1 00uF RL 50k Ri C4 R3 R10 1.8K
l (33 PTS.) A MOSFET two-stage amplifier is shown below a) Determine gm for both MOSFETS )ndicate the configuration (topology) of each amplifier stage raw the small-signal equivalent circuit of the entire amplifier (with transistors) d) Find Av Vou/vs, R and R Note: lo 1mA, Kn-31 6mAN2 VM-1.73 Lambda0 Mi Config. Mz Config.- RS sk R4 100 C2 OuF Ro R2 50 C1 M1 100uF 50% Ri Vs R3 50k 000uF R10 1.8k
R1 = 49.7 Ohms, C1 = 1.60 uF Using the circuit shown in (Figure 1), design a narrow band bandreject filter having a center frequency of 4 kHz and a quality factor of 10. Base the design on igure 1 of 2 (> (1-o)R Figure 2 of 2 l6 RI R3 RC 2 Rs Part C Determine the resistance R2 in the filter Express your answer to three significant figures and include the appropriate units R2Value Units Submit Request Answer...
Question I : Consider the amplifier circuit shown below (p-150 for both transistors) (18 marks) +12V R4 1k8 R1 15k Q1 2N3904 C2+ 10μ R2 6V 4k7 2mA Out Q2 2N3904 C1 R3 10k R5 1k8 In 10H (i) Perform DC analysis and prove that the indicated voltages and currents in the figure are correctly calculated. Find the operating point of Q1 and Q2 (5 marks) (ii) Calculate the gain of this amplifier (5 marks) (iii) In the lab, only...
4. The switched-capacitor filter implementation of a two-integrator loop is shown below. Assume a clock frequency of fe 100 kHz, and C-C2-5 pF Ca (by (a) An optimally flat low-pass response is realized when Q 1/V2 and wan w.Design the circuit so the 2d integrator's output is an optimally flat low-pass function where w1000 and the DC gain is 1. (Hint: see section 17.11.2 in the text) 4. The switched-capacitor filter implementation of a two-integrator loop is shown below. Assume...