analog bicmos design practices and pitfalls phần 2 doc
analog bicmos design practices and pitfalls phần 2 doc
... Equation 1 .24 into Equation 1 .27 C J = C J0 1+ V R Ψ o (1 .28 ) where C J0 = A qN D 2 o (1 .29 ) Equations 1 .29 and 1 .27 apply to the single-sided junction with uniform doping in the p-sides and n-sides. ... level E 1 , is occupied by N 1 electrons and set 2, at energy level E 2 , is occupied by N 2 electrons. The Boltzmann assumption is that N 2 N 1 = e − E 2 −E 1...
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analog bicmos design practices and pitfalls phần 4 docx
... I d (M2) (3.34) I d (M1) = W 1 L 1 KP 2 (V gs1 − V th ) 2 where V gs1 = V th +2 V (3.35) I d (M2) = W 2 L 2 KP 2 (V gs2 − V th ) 2 where V gs2 = V th +∆V (3.36) Substituting Equation 3.34 and Equation ... Circuit Design, Layout and Simu- lation, IEEE Press, New York, c. 1998. [2] Gray, Paul R., and Mayer, Robert G., Analysis and Design of Analog Integrated Circuits,...
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analog bicmos design practices and pitfalls phần 10 pdf
... R 1 and R 2 , their ratio is R 1 R 2 = R 1 +∆R 1 R 2 +∆R 2 = 1+ ∆R1 R 1 1+ ∆R 2 R 2 = 1 − ∆W 1 W 1 1 − ∆W 2 W 2 ≈ 1 − ∆W 1 − ∆W 2 W =1− 2 W W (9.5) There is a tradeoff between resistor size and ... to the bases of N1 and N2, so that I N2 = I P 2 , then I out will be zero. If the ratio of I P 1 /I P 2 is off by 20 %, I N1 /I N2 must also be off by 20 % to achieve I...
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analog bicmos design practices and pitfalls phần 1 potx
... n-type silicon. Practices and Pitfalls Analog BiCMOS DESIGN Contents 1Devices 1.1Introduction 1.2SiliconConductivity 1 .2. 1DriftCurrent 1 .2. 2EnergyBands 1 .2. 3SheetResistance 1 .2. 4DiffusionCurrent 1.3PnJunctions 1.3.1BreakdownVoltage 1.3.2JunctionCapacitance 1.3.3TheLawoftheJunction 1.3.4DiffusionCapacitance 1.4DiodeCurrent 1.5BipolarTransistors 1.5.1CollectorCurrent 1.5.2Base...
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analog bicmos design practices and pitfalls phần 3 pps
... Ibc2 (2. 15) Ibe1=IS exp V BE NF V T − 1 (2. 16) Ibe2=ISE exp V BE NE V T − 1 (2. 17) Ibc1=IS exp V bc NR V T − 1 (2. 18) Ibc2=ISC exp V bc NC V T − 1 (2. 19) Kqb = Kq1 1+(1+4Kq2) NK 2 (2. 20) where ... the following I E = −(I be1 − I bc1 ) − I pe β F (2. 8) I C = I be1 − I bc1 − I ce β R (2. 9) Equations2. 8and2 .9arerepresentedschematicallyinFigure2.3. Equations...
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analog bicmos design practices and pitfalls phần 5 potx
... P 2 , acts to hold I 1 = I 2 . I 1 = nI s e V be 1 V T = I 2 = I s e V be 2 V T V T ln[n]=V be 2 − V be 1 = R 2 I 1 solving for I = I 1 I = V T ln[n] R 2 (4 .2) The voltage at the base of N 2 is ... with respect to temperature and setting it equal to zero and rearranging terms 2 R 1 R 2 ln[n]= 0.0 02 8.62x10 −5 R 1 R 2 = 11.6 ln[n] If n =4, R 1 R 2 =8.4. Example...
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analog bicmos design practices and pitfalls phần 6 pps
... common-mode output signals v od = v o1 − v o2 and v oc = v o1 − v o2 2 We can identify v o1 and v o2 in terms of v od and v oc v o1 = v od 2 + v oc and v o2 = − v od 2 + v oc The differential-mode gain ... zero. That is G m = d(∆I D ) d(∆V I ) = 2I SS −2k∆V 2 I 2 I SS K −∆V 2 I = √ 2KI SS ∆V I =0 ∆V I =0 G m = 2 W L µC OX I SS 2 = 2 W L µC OX I D Figure 5 .20 Di...
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analog bicmos design practices and pitfalls phần 7 potx
... = 11.7V 5KΩ =2. 34mA Then I B (N3) = 2. 34mA 100 =23 .4µA We can then write I C (P 2) − I C (N2) = 23 .4µA I C (P 1) = 1.02I C (N2) I C (P 2) = 100 − I C (P 1) Rearranging, we have I C (P 2) − I C (P 1) 1. 02 =23 .4µA 100µA − I C (P ... V 1=1V , the collector currents in P1 and P2 are equal. Then I C (P 1)=50µA =1.02I C (N1), or I C (N1) = 49. 02 A Thus, I C (P 2) = 50µA and I c...
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analog bicmos design practices and pitfalls phần 8 ppsx
... −4.2V . References [1] Baker, R. Jacob, et al., CMOS Circuit Design, Layout and Simu- lation, IEEE Press, New York, c. 1998. [2] Gray, Paul R., and Mayer, Robert G., Analysis and Design of Analog ... V I becomes negative, Q2 reaches V BE (on) and conducts, eventually resulting in V I = −VCC and V o = −VCC− V BE . Note the deadband of 2V BE (on) centered at the origin. This dea...
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analog bicmos design practices and pitfalls phần 9 pdf
... is HIGH, P 1 is off, P 2 is on, and the output is HIGH. Hysteresis is achieved by the current mirror N 1 and N 2 . With P 2 on, N 1 and N 2 turn on. N 2 pulls the base of P 2 to one V be below the ... voltages as low as 5 .23 V for the 14 V process, and 12. 27 V for the 30 V process Figure 8 .23 A small current turns on the base-emitter diode of P 4 and clamps the base of...
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