Electronics in High Energy Physics Introduction to electronics in HEP

1. Electronics in High Energy PhysicsIntroduction to electronics in HEP Electrical Circuits (based on P.Farthoaut lecture at Cern)

2. Electrical Circuits • Generators • Thevenin / Norton representation • Power • Components • Sinusoidal signal • Laplace transform • Impedance • Transfer function • Bode diagram • RC-CR networks • Quadrupole

3. I r + v R - Sources • Voltage Generator • Current Generator

4. A A Rth Vth B B Thevenin theorem (1) • Any two-terminal network of resistors and sources is equivalent to a single resistor with a single voltage source • Vth = open-circuit voltage • Rth = Vth / Ishort

5. A A R1 + R1//R2 V R2 - Thevenin theorem (2) • Voltage divider

6. A A Rth Vth Rno Ino B B Norton representation • Any voltage source followed by an impedance can be represented by a current source with a resistor in parallel

7. I r + v R - Power transfer • Power in the load R • P is maximum for R = r

8. Sinusoidal regime

9. Complex notation • Signal : v1(t) = V cos( t + ) • v2(t) = V sin( t + ) • v(t) = v1 + j v2 = V e j( t + ) = V ej ej t = S ej t • Interest: • S = V ej contains only phase and amplitude • ej t contains time and frequency • Real signal = R [ S ej t ] • In case of several signals of same  only complex amplitude are significant and one can forget ej t • One can separate phase and time

10. Complex impedance • In a linear network with v(t) and i(t), the instantaneous ratio v/i is often meaningless as it changes during a period • To i(t) and v(t) one can associate J ej t and S ej t • S / J is now independent of the time and characterizes the linear network • Z = S / J is the complex impedance of the network • Z = R + j X = z ej • R is the resistance, X the reactance • z is the module,  is the phase • z, R and X are in Ohms • Examples of impedances: • Resistor Z = R • Capacitance (perfect) Z = -j / C; Phase = - /2 • 100 pF at 1MHz  1600 Ohms • 100 pF at 100 MHz  16 Ohms • Inductance (perfect) Z = jL; Phase = + /2 • 100 nH at 1 MHz  0.63 Ohms • 100 nH at 100 MHz  63 Ohms

11. Power in sinusoidal regime • i = IM cos  t in an impedance Z = R + j X = z ej • v = z IM cos( t + ) = R IM cos t - X IM sin t • p = v i = R IM2 cos2t – X IM2 cost sin t = R IM2 /2 (1+cos2t ) - X IM2 /2 sin2 t • p = P (1+cos2 t ) - Pqsin2 t = pa + pq • pa is the active power (Watts); pa = P (1+ cos2t) • Mean value > 0; R IM2 /2 • pq is the reactive power (volt-ampere); pq = Pq sin2t • Mean value = 0 • Pq = X IM2 /2 • In an inductance X = L ; Pq > 0 : the inductance absorbs some reactive energy • In a capacitance X = -1/C; Pq < 0 : the capacitance gives some reactive energy

12. Cp Cs Rs Rp Real capacitance • A perfect capacitance does not absorb any active power • it exchanges reactive power with the source Pq = - IM2 /2C • In reality it does absorb an active power P • Loss coefficient • tg  = |P/Pq| • Equivalent circuit • Resistor in series or in parallel • tg  = RsCs • tg  = 1/RpCp

13. Lp Ls Rs Rp Real inductance • Similarly a quality coefficient is defined • Q = Pq/P • Equivalent circuit • Resistor in series or in parallel • Q = Ls/Rs • Q = Rp/Lp

14. Laplace Transform (1) • v = f(i) integro-differential relations • In sinusoidal regime, one can use the complex notation and the complex impedance • V = Z I • Laplace transform allows to extend it to any kind of signals • Two important functions • Heaviside (t) • = 0 for t < 0 • = 1 for t  0 • Dirac impulsion (t) = ’(t) • = 0 for t  0

15. Laplace Transform (2) • Examples • Linearity • Derivation, Integration • Translation

16. Laplace Transform (3) • Change of time scale • Derivation, Integration of the Laplace transform • Initial and final value

17. I(p) i(t) Z(p) Z V(p) v(t) Impedances • Network v(t), I(t) • Generalisation • V(p) = Z(p) I(p)

18. I1 I2 TransferFunction V2 V1 Transfer Functions • Input V1, I1;Output V2, I2 • Voltage gain V2(p) / V1(p) • Current gain I1(p) / I2(p) • Transadmittance I2(p) / V1(p) • Transimpedance V2(p) / I1(p) • Transfer function Out(p) = F(p) In(p) • Convolution in time domain:

19. Bode diagram (1) • Replacing p with j in F(p), one obtains the imaginary form of the function transfer • F(j) = |F| ej() • Logarithmic unit: Decibel • In decibel the module |F| will be • The phase of each separate functions add • Functions to be studied

20. 6 dB per octave 20 dB per decade a |F|dB  [rad/s] 3 dB error Bode diagram (2) • F(p) = p + a ; |F1|db= 20 log | j  + a| • Bode diagram = asymptotic diagram •  < a, |F1| approximated with A = 20 log(a) •  > a, |F1| approximated with A = 20 log() • 6 dB per octave (20 log2) or 20 dB per decade (20 log10) • Maximum error when  = a • 20 log| j a + a| - 20 log(a) = 20 log (21/2) = 3 dB

21.  [rad/s] -6 dB per octave |F|dB a - 20 dB per decade 3 dB error Bode diagram (3) • |F2|db= - 20 log | j  + a| • Bode diagram = asymptotic diagram •  < a, |F2| approximated with A = - 20 log(a) •  > a, |F2| approximated with A = - 20 log() • - 6 dB per octave (20 log2) or - 20 dB per decade (20 log10) • Maximum error when  = a • 20 log| j a + a| - 20 log(a) = 20 log (21/2) = 3 dB

22.  [rad/s] |F|dB -20 dB per decade -40 dB per decade Bode diagram (4) • As before but: • Slope 6*n dB per octave (20*n dB per decade) • Error at =a is 3*n dB • Low pass filters

23. Bode diagram (5) • Phase of F1(j ) = (j  + a) • tg  = /a • Asymptotic diagram •  = 0 when  < a •  = /4 when  = a •  = /2 when  > a

24. Bode diagram (6) • Phase of F2(j ) = 1/(j  + a) • tg  =- /a • Asymptotic diagram •  = 0 when  < a •  = - /4 when  = a •  = - /2 when  > a

25. 40 dB per decade b Bode diagram (7) • |F3|dB = 20 log|b2 - 2 + 2aj| • Asymptotic diagram •  --> 0 A = 40 log b •  --> ∞ A’ = 20 log 2 = 40 log  • A = A’ for  = b • Error depends on a and b • p2 + 2a p + b2 = b2[(p/b)2 + 2(a/b)(p/b) + 1] • Z = a/b U = /b

26. b -40 dB per decade Bode diagram (8) • |F4|dB = - 20 log|b2 - 2 + 2aj| • Asymptotic diagram •  --> 0 A = - 40 log b •  --> ∞ A’ = - 20 log 2 = - 40 log  • A = A’ for  = b • Error depends on a and b • Z = a/b U = /b

27. Z = 1 Z = 0.1 Bode diagram (9) • Phase of F3(j) = (b2 - 2 + 2aj) and F4(j) = 1/(b2 - 2 + 2aj) • tg  = 2a/ (b2 - 2) • Asymptotic diagram •  = 0 when  < b •  = ± /2 when  = b •  = ±  when  > b

28. V R Dirac response Heaviside response V2 V1 C t/RC RC-CR networks (1) • Integrator; RC = time constant

29. R V2 V1 C RC-CR networks (2) • Low pass filter • c = 1/RC

30. V Dirac response Heaviside response RC = 1 C R V1 V2 t/RC RC-CR networks (3) • Derivator; RC = time constant

31. Dirac response Heaviside response RiCi = RC V R2 C1 C2 V1 R1 V2 t/RC RC-CR networks (4)

32. R2 C1 C2 V1 R1 V2 RC-CR networks (5) • Band pass filter

33. y(t) = x(t) * f(t) Y(f) = X(f) F(f) f(t) F(f) x(t) X(f) Time or frequency analysis (1) • A signal x(t) has a spectral representation |X(f)|; X(f) = Fourier transform of x(t) • The transfer function of a circuit has also a Fourier transform F(f) • The transformation of a signal when applied to this circuit can be looked at in time or frequency domain

34. Time or frequency analysis (2) • The 2 types of analysis are useful • Simple example: Pulse signal (100 ns width) • (1) What happens when going through a R-C network? • Time analysis • (2) How can we avoid to distort it? • Frequency analysis

35. R Y(t) X(t) C Time or frequency analysis (3) • Time analysis

36. Time or frequency analysis (4) • Contains all frequencies • Most of the signal within 10 MHz • To avoid huge distortion the minimum bandwidth is 10-20 MHz • Used to define the optimum filter to increase signal-to-noise ratio

37. I1 I2 V1 V2 Quadrupole • Passive • Network of R, C and L • Active • Internal linked sources • Parameters • V1, V2, I1, I2 • Matrix representation

38. Parameters • Impedances • Admittances • Hybrids

39. Input and output impedances • Input impedance: as seen when output loaded • Zin = Z11 - (Z12 Z21 / (Z22 + Zu)) • Zin = h11 - (h12 h21 / (h22 + 1/Zu)) • Output impedance: as seen from output when input loaded with the output impedance of the previous stage • Zout = Z22 - (Z12 Z21 / (Z11 + Zg)) • 1/Zout = h22 - (h12 h21 / (h11+ Zg))