Q1 (50p). Consider the differential amplifier and the Wilson current mirror and answer the questions given below. Assume all MOS and BJT devices are exactly matched. [ β = 10.000, μnC0 K = 8 mA/V2, VA = 50 V, VBE = 0.7 V] Wilson Current Mirror a) Calculate Rid and Rod. (6p) b) Calculate Ad1 = Vo1/Vid, Ad2 = Vo1/Vid, and Ad = Vod/Vid by using half circuit equivalent model. (12p) c) Calculate Acmsingle and Acmdiff by using half circuit model when a common-mode voltage Vcm is applied to the inputs. (8 p) d) Calculate CMRRsingle and CMRRdiff in dB. (6p) e) Design the Wilson current mirror that generates the bias current of 1 b = 2 mA. (8p) f) Replace the ideal current source by the Wilson current mirror and repeat c and d. (10p)
Question 6. Consider the Common Emitter (CE) amplifier in figure B3 below. The capacitor Cin = 1 μF, VCC = 12 V, βF = 150, and the low cut-off frequency is 100 Hz. Assume VBE = 0.7 V. VCE−5 AT for this BJT transistor is 0.2 V. RC = 800 Ω, and RE = 120 Ω. The quiescent collector current is 5 mA Figure B3: Common Emitter Amplifier circuit (I) Draw the small signal model for the CE amplifier [3 marks] (II) What is the transconductance gain of the amplifier? [2 marks] (III) What is the voltage gain of the amplifier - with and without a bypass capacitor (CE)? [4 marks] (IV) Determine a suitable value for the bypass capacitor CE. [3 marks] (V) Determine a suitable value for the total bias resistance (RB1+RB2). [5 marks] (VI) Determine specific values for bias resistors RB1, RB2. [5 marks] (VII) Figure 4 below shows a common-source amplifier constructed using a MOSFET transistor. igure 4: common-source amplifier circuit This circuit also uses two bias resistors to control the operating point of the amplifier. What advantage does this circuit have over the common emitter circuit in terms of selecting component values for the bias resistors to reduce power consumption? [3 marks]
Consider the circuit shown below. Use the following parameters for your calculations. VDD = −VSS = 5 V, Kn = 0, 7 mA/V2, VTN = 0, 8 Vλ = 0, R1 = 237 kΩ, R2 = 63 kΩ, RD = 3 kΩ, RS = 1 kΩ, a-Please draw AC equivalent circuit and small signal equivalent circuit. (6 points) b-Find VGS (6 points) c-Find operating point Current loa and Voltage VDSQ (8 points) d-Find Av Voltage Gain (5 points)
A. ORCAD Experiment - NMOS Differential Amplifier Set up and simulate the given circuit in Fig. 1. with VDD = 12 V, Ro = 10 kΩ, RD = 3.9 kΩ. Use the NMOS parameters you have used previously in the labs. Use appropriate simulation settings you have learned so far in the labs to get the required measurements. Figure 1. NMOS differential amolifier. Set v1 = v2 = 0. Fill up the following table: If necessary, make necessary modifications to resistors to set IΩ = 1 mA, VDSQ = 8 V. Apply v1 and v2 a sinusoidal signal with Vpp = 1 V and f = 1 kHz to find Acm2 = Change v2 = 0 to find Adm2 = Calculate CMRR = 6. Based on your results, discuss the significance of the circuit in Fig. 1 with your lab partner and summarize your discussion in 3 sentences below:
Consider the circuit in the figure below, with values: K1 = 506 mA/V2 Vt1 = 0.5 V Vt2 = 0.6 V Calculate the transconductance, gm,T1, of the transistor T1. The Early effect of the transistor can be neglected. MOS transistor equation: I = K2(Vgs − Vt)2 gm,T1 = …mA/V
Q2. The circuit shown in Figure Q2 is a common source MOSFET amplifier, where Rs = 500 Ω, R1 = 30 kΩ, R2 = 50 kΩ, Rd = 10 kΩ, RL = 10 kΩ, Id = 1 mA, Kn(W/L) = 30 mA/V2, Vth = 2 V, λ = 0.005 V−1( = 1/VA), VDD = 12 V, C1 = 1 μF, C2 = 1 μF. Figure Q2 a) Perform the DC analysis of the circuit, stating any assumptions that you make. [25%] b) Perform the AC analysis of the equivalent circuit and draw the small signal circuit. [25%] c) Determine the input and output impedances of the circuit. [25%] d) Determine the lower cut-off frequencies of the amplifier. [25%]
Q7) Consider the ac equivalent circuit of a CMOS common-source amplifier shown in the figure below. The parameters of the NMOS transistors are VTN = 0.5 V, kn' = 85 μA/V2, (W/L)2,3 = 50 and λ = 0.05 V−1. The parameters of the PMOS transistor are VTP = −0.5 V, kp' = 40 μA/V2, (W/L)1 = 50 and λ = 0.075 V−1. Determine the small-signal voltage gain.