Advanced Analog Building Blocks. Prof. Dr. Peter Fischer, Dr. Wei Shen, Dr. Albert Comerma, Dr. Johannes Schemmel, etc

Transcription

1 Advanced Analog Building Blocks Prof. Dr. Peter Fischer, Dr. Wei Shen, Dr. Albert Comerma, Dr. Johannes Schemmel, etc 1

2 Topics 1. S domain and Laplace Transform Zeros and Poles 2. Basic and Advanced current mirrors 3. Single ended stages 4. Stability 5. Differential stages and 2 stage amplifiers 6. Compensation schemes 7. Low noise preamplifier and shaping 8. Voltage and current references and bandgaps 9. Layout of components, parasitic, corners and MC simulation 10. Standard cells and transistor level digital 2

3 Advanced Analog Building Blocks Laplace Transformation & Transfer Function Wei SHEN (KIP) 3

4 Kirchhoff s law and circuit every circuit node I = every circuit loop U = 0 List variables : # variables = # equations Using Kirchhoff s 1 & 2 Law 4

5 Components Resistor Capacitor Inductor I = U/R I = C du/dt U = L di/dt Transistor Transformer bipolar cmos 5

6 Example... 1 loop & 2 nodes + 3 variables (U1, U2 and U3) U = U 1 + U 2 + U 3 U = U 1 + U 2 I R = IC = I L U 1 R = C du 2 dt = 0 t U 3 L dt 6

7 Fourier Transformation F ω = f t e -j ωt dt used in Frequency analysis (covered later) Concise expressions in derivative and integral Concise expression for convolution 7

8 Important Properties of FT Shift in both domains symmetric convolution Most important for L & C, if initial value is zero, For non-zeros, covered later 8

9 In Angular Frequency domain... capacitor inductor C du 2 dt = j ω C U 2 0 t U 2 L dt = U 2 jωl U = U 1 + U 2 U = U 1 + U 2 U 1 R = C du 2 dt = 0 t U 3 L dt U 1 R = j ω C U 2 = U 3 jωl 9

10 Solve the equations U = U 1 + U 2 U 1 = jωcr ω 2 LC+jωCR+1 U U 1 R = j ω C U 2 = U 3 jωl U 2 = U 3 = ω 2 LC ω 2 LC+jωCR+1 U 1 ω 2 LC+jωCR+1 U Mathematical tool is recommended Later on, we will only focus on U3 10

11 If we assume RC > 4L/R and Inverse Transform f t = 1 2π F ω e j ωt dω the switch turns on and off for a rather short time (delta function) U 3 = 1 1 LC R2 4L 2 exp Rt 2L sin (t 1 LC R2 4L 2) 11

12 Laplace Transformation X s = f t e -s t dt same as Fourier Transformation except jω replaced by s However, mathematically, Not every function in the time domain can be calculated by the Fourier Transformation e.g. exp(a*t), a > 0, but calculable in the s domain complex frequencies F ω = f t e σt e -j ωt dt 12

13 RoC Region of Convergence the power of decay in e -σt example: f(t) = exp(-at) u(t), where u(t) is the step function at t=0 L f t = exp at u t exp st dt = exp at st dt = 0 exp[ s+a t] s+a the transformation only converges when Re(s) > -a, i.e. σ>-a. 13

14 Important Properties of LT Keeps all nice features of FT since, Laplace is the mentor of Fourier derivative inegration convolution 14

15 Inverse Laplace Transformation Fourier Transformation with convergence factor F σ + jω = f t e σt Fourier Transformation with convergence factor f t e σt = 1 2π F(σ + jω) e -j ωt dt e j ωt dω Multiply e σt on both sides, and take s = σ+jω f t = 1 2πj σ+j σ j F(s) e st ds 15

16 Inverse Laplace Transformation However, direct calculation using the definitions is difficult (knowledge on complex analysis) Calculation based on the Residue Theorem (complex analysis) f t = 1 2πj σ+j σ j F(s) e st ds n = k=1 Res F s e st, sk, (t > 0) As f(t) = 0, t<0 for real signals 16

17 Residue Theorem for ILT Res(f,c) = lim s c f(s) for the first order pole s c 1 d Res(f,c) = lim n_ 1 s c n 1! ds n_ 1 [ s c n f s ] for higher orders Example : If we assume F(s) = 5s 1 (s+1)(s 2), calculate f(t). f(t) = 5s 1 s 2 est s= s 1 s+1 est s=2 = 2 e t + 3e 2t 17

18 Decomposition Different from the Residue Theorem, decomposition based on the poles First order : F(s) = P(s) Q(s) = and higher orders: F(s) = P(s) Q(s) = A 0 = P(a) Q 2 (a) P(s) = A + P 1 (s), s a Q 1 (s) s a Q 1 (s) P(s) = A 0 + s a m Q 2 (s) s a m d k P s ( ) ds k Q s 2 s = a, A k = 1 k! and complex poles : F(s) = P(s) C = 1 b Q(s) = P(s) [ s a 2 +b 2 ]Q 3 (s) P(a+jb) Im( Q 3 (a+jb) ), D= 1 P(a+jb) Re( b Q 3 (a+jb) ) A = P(s) Q 1 (s) s=a = P(a) Q 1 (a) A P 2 (s) s a m_ 1 Q 2 (s) = C s a +Db s a 2 +b 2 + P 3 (s) Q 3 (s) 18

19 Example F(s) = s 2 +2s+1 (s 2 2s+5)(s 3), calculate f(t) F(s) = 2 s 3 s 1 + s s Calculate each using Residue Theorem / Table f(t) = 2e 3t - e t cos2t - e t sin2t 19

20 Laplace Transformation Table 20

21 Laplace transformed equations U = U 1 + U 2 U = U 1 + U 2 U 1 R = j ω C U 2 = U 3 jωl U 1 R = sc U 2 = U 3 sl U 1 = U 2 = scr s 2 LC+sCR+1 U s 2 LC s 2 LC+sCR+1 U U 3 = 1 U s 2 LC+sCR+1 21

22 When to use what transformation? Laplace : circuit analysis, when oscillation is possible, no convergence required, Fourier : frequency analysis, check out poles, zeros, stability, compensation schemes, phase margin etc... Procedures: first use Laplace to get the expression, then replace s by jω to get the frequency plot 22

23 Components in the s domain time domain s domain Capacitor I = C du/dt sc U Inductor U = L di/dt sl I Capacitor and inductor can be treated same as resistor in the s domain, except with the impedance 1/sC and sl. The two Kirchhoff s Law can be applied in the s domain taking these models NO DIFFERENTIAL EQUATIONS NEEDED initial value often zero 23

24 Voltage Division U 1 = U 2 = scr s 2 LC+sCR+1 U s 2 LC s 2 LC+sCR+1 U U 3 = 1 s 2 LC+sCR+1 U U 3 = 1/SC R+sL+1/SC U = 1 s 2 LC+sCR+1 U 24

25 Transfer Function 25

26 Transfer function input system output Transfer function A output A input (A can be any electrical quantity) 26

27 Linear System x(t) input system output y(t) the system can be treated as an operator H, which converts x(t) to y(t) y(t) = H [x(t)] and follows the linear relation α y 1 (t) + β y 2 (t) = H [α x 1 (t) +β x 2 (t)] 27

28 Impulse Response H is difficult to calculate, it is easier to first calculate the response by sending an impulse at the input. The impulse is a delta function δ(t). the output is defined as the impulse response of the system in the time domain, h(t) h(t) = H [δ(t)] Other inputs are often used, if the input is a step funciton, the output is defined as the step response of the system. The most important is the impulse response. 28

29 Convolution For an arbitrary input waveform x t = x τ 0 δ( t τ ) d τ can be interpreted as a sum of different delta functions δ(t τ ) with amplitude x(τ ), the impulse response for δ(t τ ) is h(t τ ) by using the linear property of the system, x(τ)as the coefficient of δ(t τ ) y t = H x t = x τ = x τ 0 0 h( t τ ) d τ H[δ( t τ )] d τ For an arbitrary waveform input, the output is the convolution of the input signal and the impulse response of the system 29

30 Why we need transfer function signal transferred by using the convolution in the time domain, but multiplying in the s domain input system1 system2 output easier for cascading the processing systems H(s) = H 1 (s) H 2 (s) 30

31 Noise Transfer noise inside the circuit is a random process which has no clear definition for the polarity of the corresponding current and voltage, it is always expressed and treated in terms of noise power P = σ 2 v = s(ω)dω 0 Mean square of noise voltage equals power. s(ω) is called power spectral density, usually is predicted by different noise theories. e.g. s R (ω) = 4kTR for resistor The transfer function for the noise power is s o (ω) = H(ω) 2 s i (ω) H(ω) is the Laplace form of the transfer function replacing s by jω, and then taking the Norm (Absolut value) 31

32 verify the average power in period T P T = 1 T T/2 v 2 T T/2 dt If T, P T P, by using Paserval s Theorem 1 P = σ 2 v lim T T T/2 v 2 T T/2 1 dt = lim T T 1 By definition from last slide, s(ω) = lim V T T T(ω) 2. V T (ω) 2 dω 1 s o (ω) = = lim T T V o(ω) 2 1 = lim T T v i ω H(ω) 2 = H(ω) 2 si(ω) 32

33 Zeros and Poles 33

34 Poles and Zeros H(s) = b m sm +bm -1 s m-1 + +b 0 = b m s z 1 s z 2 (s zm) a n s n +an -1 s n-1 + +a 0 a n s p 1 s p 2 (s pn) Real coefficients, complex poles/zeros appear with conjugates 6 types of poles : t Null > X(s)/s x τ dτ 0 a positive real number > 1/(s+a) exp ( -a t ) a negative real number (diverge) imaginary conjugates (a=0, diverge) > b/[(s+a) 2 +b 2 ] exp(-a t) sin(bt) a complex conjugate with a positive real part (a>0, converge) a complex conjugate with a negative real part (a<0, diverge) In order for the waveform to converge in the time domain, a must be positive, poles have to be on the left half complex plane, otherwise oscillation in the circuit 34

35 Poles and Zeros sx(s) x(0) dx/dt Zeros are usually real numbers, they can be treated as adding an additional timing derivative to the original function in the time domain. Similarly, adding a pole with large time constant approximates adding an integration stage to the circuit 35

36 Frequency response of a signal/transfer function The frequency plot can be generated by the transfer function in the s domain. jω replaces s in the expression. Take the Norm of the expression and then express in db This type of frequency response is called Bode Plot 36

37 Bode Plot real zeros zeros with complex conjugates real poles poles with complex conjugates 37

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