EE 40: Introduction to Microelectronic CircuitsSpring 2008: HW 9(due 4/28, 5 pm)Venkat AnantharamApril 21, 2008Referenced problems from Hambley, 4th edition.1. P10.832. Design a clipper circuit for a negative limit of −2.1V and a positive limit of 2.2V . The input voltageis peak limited to ±5V .You are allowed to use two diodes, two d.c. voltage sources and a resistor.The maximum allowable current through each diode is 2.5mA and the threshold voltage of each diodeis 0.7V . Use an ideal voltage with threshold, ignoring breakdown.Hint: See Figure 10.32 in textbook.3. A voltage source produces a periodic voltage vin(t) with perio d T = 2s with waveform as in Figure 1.The positive peaks are of 6V and the negative peaks are at −2V . Note that−2 −1 0 1 2 3−2−10123456vin(t) in Vt in sFigure 1: Waveform Input Voltagevin(t) = vd+ va(t)where vd= 2V and va(t) is a periodic signal with zero dc component.It is desired to clamp vin(t) to a positive peak of 10V . You are allowed to use at most one of each ofthe following:1(a) an ideal diode.(b) an ideal Zener diode of arbitrary breakdown voltage.(c) a resistor of size 1kΩ.(d) a dc voltage source of arbitrary value.(e) a capacitor of arbitrary value.You may assume implicitly that there is a small resistance in series with the voltage source vinwhichis so small compared to 1kΩ that it can be neglected.Design a clamp circuit to perform the desired task, giving some guidelines for the choice of values forthe elements (b), (d), and (e).4. In the circuit in Figure 2, the op-amp is assumed to be ideal. The circuit is called a precision rectifier.Here ±V denotes the supply voltages. Take V = 12V . Note that vout(t)is defined at the − input of−vin(t)+−+R+vout(t)−+V−VFigure 2: Precision Rectifierthe op-amp.(a) Let vin(t) = Vmsin(ωt). Let Vmbe significantly smaller than V , e.g. Vm= 6V . D etermine vout(t)under the assumption that the diode is ideal.(b) Now, suppose that the diode is modelled as having a threshold voltage vth= 0.7V , being an idealshort circuit at threshold voltage and an ideal open circuit below the threshold voltage. Again,dtermine vout(t).(c) Why is the circuit called a precision rectifier?5. Consider the circuit in Figure 3. This circuit is known as an inverting precision rectifier.Assume that the op-amp has supply voltages at ±12V and that vin(t) = 6V sin(ωt). Further assumethat the diode has a threshold voltage of 0.7V but is otherwise ideal.Determine the voltage vout(t).2−+vin(t) = 6V sin(ωt)+vout(t)−−++12V−12VFigure 3: Inverting Precision Rectifier6. For this problem, note that the intrinsic carrier concentration of both electrons and holes in pure Si atroom temperature (≈ 300K) can be taken to be 1010cm−3, while that in pure Ge can be taken to be2 ∗ 1013cm−3.Identify the majority carrier and find the electron and hole concentrations at room temperature in thefollowing semiconductors.(a) Silicon doped with phosphorus at a doping concentration of 1016cm−3.(b) Silicon simultaneously doped with arsenic at a concentration of 5 ∗ 1017cm−3and with b oron at aconcentration of 5.1 ∗ 1017cm−3.(c) Germanium doped with boron at a concentration of 2 ∗ 1015cm−3.Consider a slat of silicon as shown in Figure 4. We focus on the x direction and assume that there is noFigure 4: A slat of Siliconvariation in planes transverse to the x-direction.Recall that Gauss’s law tells us that if there is a charge density profile ρ(x) in the slat (measured inCcm3)then the associate electric field E measured inVcmand with reference as pointing in the x direction is givenbydEdx(x) =ρ(x)ǫHere, ǫ is the permittivity of the material (here, the material is silicon, for which you can take ǫ = 11.7ǫ0,where ǫ0, the permittivity of vacuum, is 8.85 ∗ 10−14Fcm).Further, the associated electrostatic potential Φ(x), measure in V , satisfiesE(x) = −dΦdx(x)which, together with Gauss’s law, gives Poisson’s equationd2Φdx2(x) = −ρ(x)ǫ3This determines potential differences: the actual value of the potential depends on a choice of reference.This discussion is relevant to the following three problems.7. Suppose the charge density in a slat of silicon (variations only in the x direction) is given by the graphin Figure 5.−120 −100 −80 −60 −40 −20 0 20 40 60 80−4−3−2−1012ρ(x) in mC/cm3x in nmFigure 5: Charge Density Profile(a) Verify that the sample as a whole is electrically neutral.(b) Use Gauss’s law to determine the electric field as a function of x.(c) Explain why the direction of the electric field (which is determined by its sign) does not changethroughout the range −120nm < x < 80nm.(d) Determine the potential ρ(x) as a function of x, assuming as a reference that ρ0 = 0, thus solvingPoisson’s equation.8. The electric field in a slat of silicon (variations only in the x direction) is given by the graph in Figure6.(a) Determine the associated charge density ρ(x).(b) Assuming as a reference that Φ(0) = 0, solve for the corresponding potential Φ(x).(c) Verify that the sample as a whole is electrically neutral.In addition to the discussion of Gauss’s law and Poisson’s equation, the following two problems refer to thedepletion approximation for pn diodes.In addition to the supplementary reader, you should also recall, as discussed in class, that, in the absenceof an eventually applied bias, the potential in the bulk of the n-type material (away from the depletionregion) can be taken to be VTlnNdniand that in the bulk of the p-type material (away from the depletionregion) can be taken to be VTlnniNa. Here VT=kTqdenotes the thermal voltage, nithe intrinsic carrierconcentration of holes and of electrons in pure silicon, and Naand Ndrespectively denote the doping densitiesof acceptors and donors (in the p-type and n-type materials, respectively).The potentials are thought of as referenced to an “intrinsic” situation - if one solves Poisson’s equationexactly with these boundary conditions, the potential will be zero precisely when the electron density andthe hole density are equal (each equal to the intrinsic carrier concentration).40 100 200 300 400 500 600 700 800−1.5−1−0.500.511.522.533.5E(x) in kV/cmx in nmFigure 6: Profile of Electric Field9. Consider a Silicon p-n junction in thermal equilibrium at room temperature with no externally appliedbias. Assume that Na= 1019cm−3and Nd= 1017cm−3.Using the depletion approximation, determine
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