11midt2sol

11midt2sol - MIDTERM 2 EE 432 Fall 2011 Pledge: No aid...

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Unformatted text preview: MIDTERM 2 EE 432 Fall 2011 Pledge: No aid given, received, nor observed. Tintin Name: Signature: Milou Question # 1 / 6 FET Types (12 points) (a)Indicate below which device is a (J)FET, (MES)FET, (H)EMT, (MOS)FET (4 points). (b)Circle the channel in the diagram below (4 points). (c)Specify for each device whether we have a n or p channel (4 points). S S D G G G D S G ¦¡¥¡¡¡¡¦¥ ¦ ¦¥ ¦¥ ¦¥ ¦¥¡¦¡¥¡¥¡¥¡¡¥ ¥¡¥¡¦¥¡¦¥¡¦¥¡ ¦¡¦¡¡¡¡¦¥ %% ¥¡¥¡¦¡¦¡¦¡¥¡ ¡¡¡¡¡¦¡¥¥ &&&& %¡%¡%¡%¡ &¡&¡&¡&¡ %¡%¡%¡%¡%& &¡&¡&¡&¡%&¡ %¡%¡%¡%¡%&¡ ¡¡¡¡¡% N Depletion ¡¡¡¡¡¡¡  ¡¡¡¡¡¡¡ ¡¡¡¡¡¡¡ ¡¡¡¡¡¡¡  ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¡¡¡¡¡¡ ¡¡¡¡¡¡¡ ©  ©¡©¡©¡©¡©¡©¡©¡ Depletion GaAs N Depletion AlGaAs N N GaAs N G (J)FET p-channel ¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡ ££££££££££££ ¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡ ¤¡¡¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤ £¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡ £¤¡£¡¡£¡£¡£¡£¡£¡£¡£¡£¡£¡ ¡¤¡£¤¡¡¡¡¡¡¡¡¡¡£¤£¤£ ¨ §¡§¡§¡§¡¨ ¨¨¨¨ §¡¨¡¨¡¨¡¨¡§¡ ¡¡¡¡¡¨¡¨ ¨¡ §¡ §¡ §¡ §¡ ¨¡ ¡¡¡¡¡ § §¨¡¡¡¡¡¡ §¨¨ ¡¡¡¡¡¡ ¨¨¨¨ ¡¡¡¡ §¡ §¡§¡§¡§¡§¡ ¨§¡¡¡¡¡ ¨¡§ ¨¡¨¡¨¡¨¡ §¡ ¡ ¡ ¡ ¡§¡ ¡¡¡¡¡ ¡§¨ §¡§¡§¡§¡ P D SI GaAs ¡!¡!¡!¡!¡!¡!¡!¡!¡!¡ """"""""" !"¡¡¡¡¡¡¡¡¡¡!" ¢¡¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¡¡¡¡¡¡¡¡¡¡¢ ¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢ ¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡¢¡ ¡¡¡¡ ¡¡¡¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¢ ¢¡¡¡¡¡¡¡¡¡¡ ¢ ¡¡¡¡ ¡¡¡¡ $$$$$$$ #¡#¡#¡#¡#¡#¡#¡ ¡¡¡¡¡¡¡#$ S N Depletion SiO 2 GaAs N N i AlGaAs 2DEG i GaAs (H)EMT n-channel (MES)FET n-channel IC D CHANNEL 1 Depletion Channel S D CHANNEL 4 Depletion Channel Depletion K I J ∆V DS D L D VDS • at the IV point I: 1 B (MOS)FET n-channel IC Depletion CHANNEL 2 S N Depletion P Si Question # 2 / 6 Channel State (12 points) Consider the JFET IV below. Specify which CHANNEL state # applies VGD = VT D CHANNEL 5 Depletion Channel S Depletion CHANNEL 3 Depletion Channel S Depletion D D Depletion Channel Depletion • at the IV point J: 2 F S ∆V DS E CHANNEL 6 Depletion Channel S Depletion S=Source D=Drain L' Velocity saturation VDS L' 4 at the IV point E: 5 at the IV point F: • at the IV point K: 3 at the IV point G: 6 • at the IV point L: 2 at the IV point H: Circle the channel regions where the voltage ∆VDS is dropped if any. Indicate the new effective gate length L in CHANNEL 3 and 6. 1 G H 5 Question # 3 / 6 MOS Capacitor (12 points) For each band diagram: ε (V = 0) V>0 V>0 E V=0 ε Ec c q φm q φs ( 23(23 3((2 32(23 ) 6 2((2 3(23 ()()()()()( 2(23( '('('('('( ((22 (((((')') ) '( )('( '()() ) ) ) ) ('('('('('('( Metal 1 0( 1( 0(01 5(5(5(5(5(5(5( 01 1( 0( 5(4(4(4(4(4(4(5 01 1( 0( 7 5 7 5 7 5 7 5 7 5 7 5 1( 7( 54(5(5(5(5(5(5(45 018701 8(8(8(8(8(8(0( ((((((( 8( 4(4(4(4(4(4(4(45(4 4(4(4(4(4(4(4(4( ((((((((44 EFm Oxide 1 Ev Semiconductor EFs EFs Ei EFs Ei Ei Ec −qV v EFm M O E Ev −qV ε V<0 EFm Ec Ei −qV EFs Ev EFm S M O S M O S 4 2 3 A I D E Specify whether we have (E)quilibrium, (A)ccumulation, (D)epletion and (I)nversion. Give the polarity of the voltage VGS applied on the metal gate relative to the semiconductor. Specify whether the gate to semiconductor CGS capacitance is CGS (1) = Ci = i /di or CGS (2) = Ci Cd /(Ci + Cd ) with Cd the depletion capacitance when the device is in: Inversion (MOS low frequency) CGS = 1 , Accumulation CGS = 1 , 2 2 Inversion (MOS high frequency) CGS = , Depletion CGS = , 1 Inversion (MOSFET) CGS = . Note that in the flat band case the depletion capacitance is infinite Cd = ∞. Question # 4 / 6: P-Channel M-O-S(N-type) Capacitor (25 Points) Draw the band-diagram of P-channel MOS capacitor at the threshold of inversion (VGS = VT ). Assume that the metal and semiconductor have DIFFERENT work-functions: qφm = 2.6 eV AND qφs = 2.3 eV. The semiconductor band gap is 1 eV. The Fermi level is located 0.3 eV below the conduction band edge Ec . For simplicity assume that the voltage Vi dropped across the insulator is HALF the voltage VS dropped across the semi-conductor. Hint: start from the semi-conductor band-diagram located on the right side. The band diagram must be drawn to scale and should include: • the Fermi levels (metal EF m and N-type semiconductor EF s ) (4 points) • the local vacuum level: −qV (x). (3 points) • the depletion charge (+ or -) in the forbidden band(3 points) • the charge (+ or -) developed in the channel due to inversion (3 points) • label qφF = EF s − Ei in the semiconductor (3 points) • the conduction band edge (2 points) • the valence band edge (2 points) • the intrinsic energy level (assumed in the center of the forbidden band) (2 points) • Give the threshold voltage VT relative to the semiconductor (3 points) 2 Metal Oxide N−type semiconductor 0 Vacuum level −1 qφs VT= -0.3V - + - + + + Ec + −2 -qVT = 0.3 eV E Ei + Ev x Metal W+d i Insulator W+ W − N−type semiconductor W Oxide 0 Depletion 3 −3 −4 Fs Question # 5 / 6:P-Channel MOS Threshold Voltage (27 Points) In this question we derive the threshold voltage of a P-channel MOS capacitor. Assume that the metal and P-semiconductor have the same work-function. Further assume that semiconductor is doped with donors and acceptors such that the donor and acceptor concentrations verify: Nd > Na . This problem uses the x-axis of Question 4 shown on the previous page. a) Calculate the potential qφF = EF s − Ei (0) in terms of the acceptor doping Na , ni (intrinsic carrier concentration) and kT . EF s is the Fermi level in the semiconductor. Start from: n = Nd − Na = ni exp EF s − EF i (∞) kT qφF = EF s − Ei (0) = kT log[(Nd-Na)/ni] b) Give the potential drop VS (inv) (which arises in the semiconductor in inversion) in terms of φF . Hint: VS (inv) < 0 and φF > 0. VS (inv) = V (W ) − V (0) = c) We wish to integrate the Poisson equation q (Nd − Na ) d2 V =− 2 dx s in the depletion region of width W to obtain the surface potential VS (inv) in terms of W . We use the solution: V (x) = (g/2)x2 + f x. At x = 0 the reference for the potential V (x) is V (x = 0) = 0 and the electric field verifies: E (x = 0) = − dV (x) = 0. See the previous dx x=0 page for the x axis definition. Express the first two derivatives of the potential V (x) in terms of the constants g and f . dV (x) =0 = f dx x=0 d2 V q (Nd − Na ) =− =g 2 dx s d) Give the resulting potential VS (inv) in terms of W and conversely give W first in terms of VS (inv) and second in terms of φF VS (inv) = V (x = W ) = W (VS (inv)) = W (φF ) = 4 d) Express the depletion charge Qd per unit area Qd (inv) first in terms of the depletion width W and the compensated doping concentration Nd − Na . Qd (inv) = e) Gauss’s law in the depletion region of the semiconductor dE = q Nd −Na is used to establish dx s the relation given below between the electric field E (W − ) and the depletion charge Qd (inv) and s . Specify the range of integration used on the two integrals to complete this derivation: W E (W − ) = s dE = q (Nd − Na )dx = Qd (inv) (1) s 0 0 f) We assume the device is in inversion with a two dimensional hole surface charge Qp located at x = W . Using the discontinuity of the displacement field at x = W in the presence of a surface charge and Equ (1): i E (W + )= s E (W − ) + Qp = Qd (inv) + Qp derive an expression giving the potential Vi in terms of the depletion charge Qd (inv), the holes surface charge Qp and the oxide capacitance per unit area Ci = i /di . Simply start from the identity below and substitute the above expression for E (W + ): Vi = V (W +di )−V (W ) = −E (W + )di = g) We shall now derive the relation between the hole channel surface charge Qp (VGS ) and the gate voltage VGS in inversion. For this purpose we assume the depletion width and thus the surface potential VS (inv) remain constant in inversion and we write: VGS = Vi + VS (inv) = Replace Vi in the above equation by the result in question f) and solve for Qp as a function of VGS . Qp (VGS ) = h) Identify the threshold voltage as VGS yielding Qp = 0: VT will be expressed in terms of Ci , Qd (inv) and VS (inv). The signs are critical. VT = i) Give the VGS range of validity for the formula derived in question g). 5 Question # 6 / 6 Long/Short Channel MOSFET (12 points) ID V GS =7 V GS =6 V GS =5 V GS=4 V GS=3 V GS=2 V GS 1 = V V V V V V V V GS= 0=V T V DS • Label below the IV characteristics associated with 1) “dashed lines” or 2) “plain lines”. LONG CHANNEL SHORT CHANNEL answer 1 2 • Current saturation originates from 1) pinchoff (VGD = Vt ). or 2) velocity saturation LONG CHANNEL SHORT CHANNEL answer 1 2 • The saturation current varies 1) quadratically (VGS − VT )2 or 2) linearily (VGS − VT ). LONG CHANNEL SHORT CHANNEL answer 2 1 • The transconductance current is 1) varies linearily (VGS − VT ). or 2) constant LONG CHANNEL SHORT CHANNEL answer 2 1 • The unity current gain cutoff frequency varies as 1) 1/L or 2) 1/L2 with L the gate length (source to drain) LONG CHANNEL SHORT CHANNEL answer 2 1 6 ...
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