MOSFET (10%) Transfer ID−VG characteristic of a MOSFET at 360 K is shown in figure below. (a) What is the subthreshold swing? (4%) (b) Decrease the temperature to 240 K, please plot is ID-VG characteristic against the curve below (compare with the one measured at 360 K), what is the subthreshold swing at 240 K? ( 6%)
The feedback transresistance amplifier utilizes two identical MOSFETs biased by ideal current sources I. The MOSFETs have the parameters, gm1, ro1 for Q1 and gm2, ro2 for Q2, respectively. The feedback resistance RF. (a) Provide the A circuit and derive an expression for A in terms of gm1, ro1, gm2, ro2, and RF. (b) What is β ? (c) Derive an expression for Af = Vo/Is. (d) Derive expressions for Rin and Rout . When the current sources I = 0.4 mA, the MOSFETs are sized to operate at VOV = 0.2 V and have Vt = 0.5 V and VA = 16 V. The feedback resistance RF = 10 kΩ. (e) If IS has a zero de component, find the dc voltage at the input, and at the drain of Q1. (f) Find gm and ro of Q1 and Q2. (g) Evaluate Af, Rin and Rout for the component values given.
The subthreshold current in a MOSFET is given by ID = I0 exp(VGS nVth) Determine the change in applied VGS for a factor of 10 increase in ID for (a) n = 1, (b) n = 1.5, and (c) n = 2.1. Note that Vth is given as 0.0259 V
An N-channel transistor has the threshold voltage Vt = 0.34 V and the subthreshold slope S = 80 mV/dec, In temperature, I0 = 100 nA, W = 10 μm and L = 50 nm. i. Estimate Ioff in room temperature. ii. Estimate Ids at Vg = 0.17 V in room temperature. iii. Resolve the question i and ii when the temperature change to 200 K and 400 K. Typical value of temperature coefficient for threshold voltage α = −2 mV/K. Assume the activation energy Ea = Eg/2, and Eg keeps unchanged in this question. Hint: Vt(T) = Vt(T0) − α×(T − T0) I0(T) = I0(T0)×exp(−Eak(1 T − 1 T0))
The circuit below represents the implementation of function OUT. The sizes of the transistors in the pull up network is shown in red color beside each transistor. What is the worst case delay of the pull up network in terms of Rn given that the mobility of the electrons is 4 times the mobility of the holes. Note: Round the number to the nearest integer.
Problem 2 (3 points) All transistors are identical and they are arranged in series (they share source/drain regions; their threshold voltage is VT = 1 V; the substrate is grounded. The total capacitance of the gate electrode is Ctot = Cox + Cgs+Cgd+Cpar. Assume that for parasitic capacitance Cpar = 0.1Ctot . A) What is the floating voltage value of the channel regions in transistors 2 and 3 ? B) What is the potential of source and drain region of transistor 5?
For the following circuits, assume VDD = +12 V. The n-channel enhancement-mode MOSFET has parameters Kn = 1 mA/V2 and VTN = 0.75 V. Find the Q-points (ID, VDS) of the circuits. What are the operating regions of the transistors? (a) (b) (c)
What should be the value of the current source IS so that the PMOS transistor is operating in saturation with VSG = VSD = 1 V? VSS = 2.5 V, |Vt| = 0.5 V, kp = 4 mA/V2. Neglect the channel-length modulation (that is, λ = 0 ).
Determine the value of the source voltage VS in the circuit shown in the figure. Assume VDD = 5 V, VSS = −5 V, RS = 100 kΩ. The NMOS transistor has Vt = 0.8 V, kn = 0.5 mA/V2. Neglect the channel-length modulation (that is, λ = 0).
Consider a Class A power amplifier circuit shown in Figure Q.1. Given that β = 120, VBE = 0.7 V, IBQ = 5.3 mA and vce,sat = 0.5 V. Figure Q. 1: Class A power amplifier (a) Sketch with complete labelling the DC and AC load lines. (12 marks) (b) Determine maximum output peak voltage, voutp, without distortion. (4 marks) (c) Calculate the efficiency of the power amplifier at condition of maximum voutp. (4 marks) (d) Examine the expected change in the AC load line when RL value increases. Describe how it affect the maximum output voltage swing. (3 marks)
For the circuit of the figure shown below, given that Vm = 9.8 V, R1 = 7.2 Ω, R2 = 13.5 Ω, ZL1 = j5.3 Ω, ZL2 = j18 Ω, ZM = j5.7 Ω and ZC = −j24 Ω, then the mesh equation (2) is: a. −j5.7 I1 + (13.5 − j42.0) I2 = 0 b. -j5.3 I1 + (13.5 - j6.0) I2 = 0 c. j5.3 I1 + (13.5 + j6.0) I2 = 0 d. -j5.7 I1 + (13.5 − j6.0) I2 = 0 e. j5.7 I1 + (13.5 − j6.0) I2 = 0