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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Chap. 5 Field-effect transistors (FET) Widely used in VLSI used in some analog amplifiers - output stage of power amplifers (may have good thermal characteristics if designed properly) n-channel or p-channel structure FET - voltage controlled device BJT - current controlled device

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Physical structure of a n-channel device: Typically L = 0.35 to 10 m, W = 2 to 500 m, and the thickness of the oxide layer is in the range of 0.02 to 0.1 m.

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall MOSFETs MOS - metal oxide semicondutor structure (original devices had metal gates, now they are silicon) NMOS - n-channel MOSFET PMOS - p-channel MOSFET CMOS - complementary MOS, both n-channel and p-channel devices used in conjuction with each other (most popular in IC’s) MESFET - metal semiconductor structure, used in high-speed GaAs devices JFET - junction FET, early type of FET

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Cross section of a CMOS integrated circuit. Note that the PMOS transistor is formed in a separate n-type region, known as an n well. CMOS

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall If V GS > V T (threshold voltage), an induced, conducting n-channel forms between the drain and source. The channel conductance is proportional to v GS - V t.

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Symbols and conventions n-channel several slightly different symbols (source is often connected to the substrate which is usually grounded) + V DS - + V GS - drain source gate

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Symbols and conventions p-channel several slightly different symbols (source is often connected to V DD ) + V DS - + V GS - drain source gate

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall An n-channel MOSFET with V GS and V DS applied and with the normal directions of current flow indicated. Output characteristics (n-channel) (linear) + V DS -

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Input characteristics (n-channel) + V DS - I D = K(V GS -V T ) 2

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Summary of MOSFET behavior (n-channel) V GS > V T (threshold voltage) for the device to be on V DS > V GS - V T for device to be in saturation region I D = K(V GS -V T ) 2 Enhancement mode device, V T > 0 Depletion mode device, V T < 0 (conducts with V GS = 0)

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Comparison of BJT and FET FET voltage controlled V GS > V T for device to be on operates in saturation region (amplifier); V DS > V GS - V T I D = K(V GS -V T ) 2 BJT current controlled V BE 0.7 V for device to be on operates in linear region (amplifier); BE junction forward biased, BC junction reversed biased I C = I B

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall MOSFET aspect ratio I D = K(V GS -V T ) 2 K = transconductance parameter K = 1/2 K' (W/L) K' = n C ox, where n is the mobility of electrons, and C ox is the capacitance of the oxide W/L is the aspect ratio, W is the width of the gate, L is the length of the gate. I D W/L

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Prob 5.41(a) Given: V T = 2V, K = (1/2).5 mA/V 2 (a) Find V 1 Use, I D = K(V GS -V T ) 2 10uA = (1/2).5 (V GS - 2) 2 Solve for V GS V GS = 2.2V V 1 = - 2.2V V1V1 V GS - + IDID I G = 0 n channel

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Prob 5.41(b) Given: V T = 2V, K = (1/2).5 mA/V 2 (b) Find V 2 Use, I D = K(V GS -V T ) 2 10uA = (1/2).5 (V GS - 2) 2 Solve for V GS V GS = 2.2V V 2 = V GS = 2.2V V2V2 V GS - + IDID I G = 0 n channel

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Prob 5.41(f) Given: V T = 2V, K = (1/2).5 mA/V 2 (f) Find V GS Equate current in load and transistor Current in transistor: I D = K(V GS -V T ) 2 Current in resistor: I = (5 - V GS ) /100K Equate currents (5 - V GS ) /100K = (1/2).5 (V GS - 2) 2 Solve for V GS V GS = 2.33V V GS - + IDID I G = 0 n channel

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall MOSFETS at DC DC problem Find I D, and V GS, and V DS V GS = 5V V GS > VT, so device is on Assume device is in saturation I D = K(V GS -V T ) 2 I D = (0.05 mA/V 2 )(5-1) 2 I D = 0.8 mA V DS = V DD - I D R D V DS = 10 - (0.8)6 V DS = 5.2V V T = 1V K = 0.05 mA/V 2 (typical values) + V GS - + V DS - IDID IDID I G = 0

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall General DC problem + V GS - + V DS - IDID I G = 0 DC problem Find I D, and V GS Assume device is in saturation I D = K(V GS -V T ) 2 I D = K(5 - I D R S -V T ) 2 18I D I D + 8 = 0 Solve for I D, use quadratic formula I D = 0.89mA, 0.5mA, which is correct? For I D = 0.89mA, V GS = 5 - (0.89)6 = V For I D = 0.5mA, V GS = 5 - (05)6 = 2V Only for I D = 0.5mA, is transistor on! V T = 1V, K = 0.5 mA/V 2

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall DC problem: two FETs in series Find V If devices are identical IDID I G = 0 n channel I G = 0V VDD = 5V Ground device V =V DD /2 = 2.5V

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall MOSFET as an amplifier g d s n channel ac model d s g v gs + - R o = 1/slope of the output characteristics. d s g v gs + - SPICE model Ro

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Transconductance Useful relation: g m = 2 K I D Transconductance = g m = dI D /dV GS = d [K(V GS -V T ) 2 ]/dV GS = 2 K(V GS -V T )

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Prob (a) Find the resistance of an enhancement load R in R in = resistance of current source || R o resistance of current source = voltage across current source / current in current source resistance of current source = v gs / g m v gs = 1/g m Replace current source by a resistor of resistance 1/g m ac model +V-+V- I g d s

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Prob (a) Find the resistance of an enhancement load Often, Ro >> 1/gm

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Prob (b) To raise the resistance of the transistor by a factor of 3, what must be done? R 1/g m = 1 / 2 K I D = [1/2 ] [1/ K] [ 1/ I D ] = [1/2 ] [1/ 1/2 K W/L ] [ 1/ I D ] Decrease I D by a factor of 9 Decrease W by a factor of 9 Increase L by a factor of 9

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Integrated Circuit MOSFET amplifiers Resistors take up too much space on an integrated ciruit (IC) Use transistors as loads Typical amplifier DC analysis Equate current in Q1 and load I in Q1 = I in load K(V GS -V T ) 2 = I in load IDID IDID

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall ac analysis of MOSFET amplifiers IDID IDID ac circuit R in = R out = R load || R o d s g v gs + - RinRout

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall ac analysis of MOSFET amplifiers A i = i out / i in = A v = v out /v in = - g m v gs (R o || R load ) / v gs = - g m (R o || R load ) d s g v gs + - v out + - i in = 0 - g m v gs

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Transistor loads: depletion load I V + - Depletion load V GS = 0 R = R o || resistance of current source with 0 magnitude = R o || = R o R o = |V A | / I Resistance is current dependent

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall CMOS amp I ref I Q2 and Q3 form a p-channel current mirror load for Q1 Q4 and Q3 establish I ref I = I ref due to current mirror Given: |V T | = 1V, |V A | = 50V p-channel p C ox = 20 A/V 2 n-channel n C ox = 40 A/V 2 W Q1 = W p = 100 m W Q4 = 50 m L = 10 m

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall CMOS amp: power I ref I Given: |V T | = 1V, |V A | = 50V p-channel p C ox = 20 A/V 2 n-channel n C ox = 40 A/V 2 W Q1 = W p = 100 m W Q4 = 50 m L = 10 m Find Total power consumed Power consumed = 2I ref V DD Equate currents in Q3 and Q4 to find I ref I Q3 = I Q4 = K 3 (V GS -V T ) 2 = K 4 (V GS -V T ) 2 Note that K’s are the same: K3 = (1/2)(20)(100/10) = K4 = (1/2)(40)(50/10) Therefore, Q3 and Q4 behave the same, so V GS3 = V GS4 = 2.5V Iref = K 4 (V GS -V T ) 2 = (1/2)(40)(50/10) ( ) 2 = 225 A Power consumed = (2) 5V 225 A = 2.25mW

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall CMOS amp: DC analysis Given: |V T | = 1V, |V A | = 50V p-channel p C ox = 20 A/V 2 n-channel n C ox = 40 A/V 2 W Q1 = W p = 100 m W Q4 = 50 m L = 10 m Find V out Consider current in Q1 or Q2 Using Q1, I Q1 = K 1 (V GS -V T ) 2 where V GS = V out 225 A = (1/2)(40)(100/10) (V GS - 1) 2 Solve for V GS, V GS = V out = 1.75V I ref + V out -

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall CMOS amp: ac analysis Given: |V T | = 1V, |V A | = 50V p-channel p C ox = 20 A/V 2 n-channel n C ox = 40 A/V 2 W Q1 = W p = 100 m W Q4 = 50 m L = 10 m I ref + V out - Find A v A v = - g m1 (Ro 1 || Ro 2 ) Ro 1 = Ro 2 = 50/ 225 A = 222K g m = 2 K I D = (2) [(1/2)(40)(100/10)] 1/2 225 A = 300 A/V A v = - g m1 (Ro 1 || Ro 2 ) = -300(.222/2) -33

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall CMOS multistage amp: ac analysis DC circuit ac circuit (neglects resistances of current sources)

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall CMOS multistage amp: ac analysis Av of stage 1: V out1 /V gs1 = -g m1 V gs1 R o1 /V gs1 = -g m1 r o1 Av of stage 2: V out2 /V gs2 = -g m2 V gs2 R o2 /V gs2 = -g m2 r o2 Overall Av = (-g m1 r o1 ) ( -g m2 r o2 ) = g m1 g m2 r o1 r o2

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Multistage CMOS amp: DC analysis Q3 and Q6 form a PMOS current mirror load for Q4 Q1 and Q5 form an NMOS current mirror load for Q2 Q5 and Q6 establish the current in Q1,Q2,Q3 and Q4 The width of Q5 is adjusted to give a particular I ref I ref

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ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall Multistage CMOS amp: DC analysis Equate currents in Q5 and Q6 I Q5 = I Q6 = K 5 (V GS5 -V T ) 2 = K 6 ((V GS5 - V DD )-V T ) 2 Solve for V GS5, Use V GS5 to find I ref Other current s are multiples of I ref K 3 /K 6 = I Q3 /I ref K 1 /K 5 = I Q1 /I ref Find V D4, and V D1 = V out from currents in those transistors I ref Given K P = 80 A/V 2, K N = 100 A/V 2, |V T | = 1V, V DD = 9V 100(V GS5 - 1) 2 = 80((V GS5 - 9) - (- 1)) 2, V GS5 = 5.14V, 48.9V Find I ref, 100( ) 2 = 1.7mA I Q3 = I Q4 = I Q2 = I Q1 because all K N ’s and K P ’s are equal

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